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Posts Tagged ‘wave equation’

Plane wave solutions of Maxwell’s equation using Geometric Algebra

Posted by peeterjoot on September 3, 2012

[Click here for a PDF of this post with nicer formatting]

Motivation

Study of reflection and transmission of radiation in isotropic, charge and current free, linear matter utilizes the plane wave solutions to Maxwell’s equations. These have the structure of phasor equations, with some specific constraints on the components and the exponents.

These constraints are usually derived starting with the plain old vector form of Maxwell’s equations, and it is natural to wonder how this is done directly using Geometric Algebra. [1] provides one such derivation, using the covariant form of Maxwell’s equations. Here’s a slightly more pedestrian way of doing the same.

Maxwell’s equations in media

We start with Maxwell’s equations for linear matter as found in [2]

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 0\end{aligned} \hspace{\stretch{1}}(1.2.1a)$

$\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = -\frac{\partial {\mathbf{B}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(1.2.1b)$

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} = 0\end{aligned} \hspace{\stretch{1}}(1.2.1c)$

$\begin{aligned}\boldsymbol{\nabla} \times \mathbf{B} = \mu\epsilon \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.1d)$

We merge these using the geometric identity

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{a} + I \boldsymbol{\nabla} \times \mathbf{a} = \boldsymbol{\nabla} \mathbf{a},\end{aligned} \hspace{\stretch{1}}(1.2.2)$

where $I$ is the 3D pseudoscalar $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$ , to find

$\begin{aligned}\boldsymbol{\nabla} \mathbf{E} = -I \frac{\partial {\mathbf{B}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(1.2.3a)$

$\begin{aligned}\boldsymbol{\nabla} \mathbf{B} = I \mu\epsilon \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.3b)$

We want dimensions of $1/L$ for the derivative operator on the RHS of 1.2.3b, so we divide through by $\sqrt{\mu\epsilon} I$ for

$\begin{aligned}-I \frac{1}{{\sqrt{\mu\epsilon}}} \boldsymbol{\nabla} \mathbf{B} = \sqrt{\mu\epsilon} \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.4)$

This can now be added to 1.2.3a for

$\begin{aligned}\left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \left( \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \right)= 0.\end{aligned} \hspace{\stretch{1}}(1.2.5)$

This is Maxwell’s equation in linear isotropic charge and current free matter in Geometric Algebra form.

Phasor solutions

We write the electromagnetic field as

$\begin{aligned}F = \left( \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \right),\end{aligned} \hspace{\stretch{1}}(1.3.6)$

so that for vacuum where $1/\sqrt{\mu \epsilon} = c$ we have the usual $F = \mathbf{E} + I c \mathbf{B}$ . Assuming a phasor solution of

$\begin{aligned}\tilde{F} = F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}\end{aligned} \hspace{\stretch{1}}(1.3.7)$

where $F_0$ is allowed to be complex, and the actual field is obtained by taking the real part

$\begin{aligned}F = \text{Real} \tilde{F} = \text{Real}(F_0) \cos(\mathbf{k} \cdot \mathbf{x} - \omega t)-\text{Imag}(F_0) \sin(\mathbf{k} \cdot \mathbf{x} - \omega t).\end{aligned} \hspace{\stretch{1}}(1.3.8)$

Note carefully that we are using a scalar imaginary $i$ , as well as the multivector (pseudoscalar) $I$ , despite the fact that both have the square to scalar minus one property.

We now seek the constraints on $\mathbf{k}$ , $\omega$ , and $F_0$ that allow this to be a solution to 1.2.5

$\begin{aligned}0 = \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F}.\end{aligned} \hspace{\stretch{1}}(1.3.9)$

As usual in the non-geometric algebra treatment, we observe that any such solution $F$ to Maxwell’s equation is also a wave equation solution. In GA we can do so by right multiplying an operator that has a conjugate form,

$\begin{aligned}\begin{aligned}0 &= \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F} \\ &= \left(\boldsymbol{\nabla} - \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F} \\ &=\left( \boldsymbol{\nabla}^2 - \mu\epsilon \frac{\partial^2}{\partial t^2} \right) \tilde{F} \\ &=\left( \boldsymbol{\nabla}^2 - \frac{1}{{v^2}} \frac{\partial^2}{\partial t^2} \right) \tilde{F},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.10)$

where $v = 1/\sqrt{\mu\epsilon}$ is the speed of the wave described by this solution.

Inserting the exponential form of our assumed solution 1.3.7 we find

$\begin{aligned}0 = -(\mathbf{k}^2 - \omega^2/v^2) F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)},\end{aligned} \hspace{\stretch{1}}(1.3.11)$

which implies that the wave number vector $\mathbf{k}$ and the angular frequency $\omega$ are related by

$\begin{aligned}v^2 \mathbf{k}^2 = \omega^2.\end{aligned} \hspace{\stretch{1}}(1.3.12)$

Our assumed solution must also satisfy the first order system 1.3.9

$\begin{aligned}\begin{aligned}0 &= \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) F_0e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)} \\ &=i\left(\mathbf{e}_m k_m - \frac{\omega}{v}\right) F_0e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)} \\ &=i k ( \hat{\mathbf{k}} - 1 ) F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.13)$

The constraints on $F_0$ must then be given by

$\begin{aligned}0 = ( \hat{\mathbf{k}} - 1 ) F_0.\end{aligned} \hspace{\stretch{1}}(1.3.14)$

With

$\begin{aligned}F_0 = \mathbf{E}_0 + I v \mathbf{B}_0,\end{aligned} \hspace{\stretch{1}}(1.3.15)$

we must then have all grades of the multivector equation equal to zero

$\begin{aligned}0 = ( \hat{\mathbf{k}} - 1 ) \left(\mathbf{E}_0 + I v \mathbf{B}_0\right).\end{aligned} \hspace{\stretch{1}}(1.3.16)$

Writing out all the geometric products, noting that $I$ commutes with all of $\hat{\mathbf{k}}$ , $\mathbf{E}_0$ , and $\mathbf{B}_0$ and employing the identity $\mathbf{a} \mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \wedge \mathbf{b}$ we have

$\begin{aligned}\begin{array}{l l l l l}0 &= \hat{\mathbf{k}} \cdot \mathbf{E}_0 & - \mathbf{E}_0 & + \hat{\mathbf{k}} \wedge \mathbf{E}_0 & I v \hat{\mathbf{k}} \cdot \mathbf{B}_0 \\ & & + I v \hat{\mathbf{k}} \wedge \mathbf{B}_0 & + I v \mathbf{B}_0 &\end{array}\end{aligned} \hspace{\stretch{1}}(1.3.17)$

This is

$\begin{aligned}0 = \hat{\mathbf{k}} \cdot \mathbf{E}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18a)$

$\begin{aligned}\mathbf{E}_0 =- \hat{\mathbf{k}} \times v \mathbf{B}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18b)$

$\begin{aligned}v \mathbf{B}_0 = \hat{\mathbf{k}} \times \mathbf{E}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18c)$

$\begin{aligned}0 = \hat{\mathbf{k}} \cdot \mathbf{B}_0.\end{aligned} \hspace{\stretch{1}}(1.3.18d)$

This and 1.3.12 describe all the constraints on our phasor that are required for it to be a solution. Note that only one of the two cross product equations in are required because the two are not independent. This can be shown by crossing $\hat{\mathbf{k}}$ with 1.3.18b and using the identity

$\begin{aligned}\mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = - \mathbf{a}^2 \mathbf{b} + a (\mathbf{a} \cdot \mathbf{b}).\end{aligned} \hspace{\stretch{1}}(1.3.19)$

One can find easily that 1.3.18b and 1.3.18c provide the same relationship between the $\mathbf{E}_0$ and $\mathbf{B}_0$ components of $F_0$ . Writing out the complete expression for $F_0$ we have

$\begin{aligned}\begin{aligned}F_0 &= \mathbf{E}_0 + I v \mathbf{B}_0 \\ &=\mathbf{E}_0 + I \hat{\mathbf{k}} \times \mathbf{E}_0 \\ &=\mathbf{E}_0 + \hat{\mathbf{k}} \wedge \mathbf{E}_0.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.20)$

Since $\hat{\mathbf{k}} \cdot \mathbf{E}_0 = 0$ , this is

$\begin{aligned}F_0 = (1 + \hat{\mathbf{k}}) \mathbf{E}_0.\end{aligned} \hspace{\stretch{1}}(1.3.21)$

Had we been clever enough this could have been deduced directly from the 1.3.14 directly, since we require a product that is killed by left multiplication with $\hat{\mathbf{k}} - 1$ . Our complete plane wave solution to Maxwell’s equation is therefore given by

$\begin{aligned}\begin{aligned}F &= \text{Real}(\tilde{F}) = \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \\ \tilde{F} &= (1 \pm \hat{\mathbf{k}}) \mathbf{E}_0 e^{i (\mathbf{k} \cdot \mathbf{x} \mp \omega t)} \\ 0 &= \hat{\mathbf{k}} \cdot \mathbf{E}_0 \\ \mathbf{k}^2 &= \omega^2 \mu \epsilon.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.22)$

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

Posted in Math and Physics Learning. | Tagged: angular frequency, bivector, cross product, geometric algebra, geometric product, imaginary, Maxwell's equation, multivector, phasor, plane wave, pseudoscalar, real, scalar, vector, wave equation, wave number, wedge product | Leave a Comment »

Continuum mechanics elasticity review.

Posted by peeterjoot on April 23, 2012

[Click here for a PDF of this post with nicer formatting]

Motivation.

Review of key ideas and equations from the theory of elasticity portion of the class.

Strain Tensor

Identifying a point in a solid with coordinates $x_i$ and the coordinates of that portion of the solid after displacement, we formed the difference as a measure of the displacement

$\begin{aligned}u_i = x_i' - x_i.\end{aligned} \hspace{\stretch{1}}(2.1)$

With $du_i = {\partial {u_i}}/{\partial {x_j}} dx_j$ , we computed the difference in length (squared) for an element of the displaced solid and found

$\begin{aligned}dx_k' dx_k' - dx_k dx_k = \left( \frac{\partial {u_j}}{\partial {x_i}} + \frac{\partial {u_i}}{\partial {x_j}} + \frac{\partial {u_k}}{\partial {x_i}} \frac{\partial {u_k}}{\partial {x_j}} \right) dx_i dx_j,\end{aligned} \hspace{\stretch{1}}(2.2)$

or defining the \textit{strain tensor} $e_{ij}$ , we have

$\begin{aligned}(d\mathbf{x}')^2 - (d\mathbf{x})^2= 2 e_{ij} dx_i dx_j\end{aligned} \hspace{\stretch{1}}(2.3a)$

$\begin{aligned}e_{ij}=\frac{1}{{2}}\left( \frac{\partial {u_j}}{\partial {x_i}} + \frac{\partial {u_i}}{\partial {x_j}} + \frac{\partial {u_k}}{\partial {x_i}} \frac{\partial {u_k}}{\partial {x_j}} \right).\end{aligned} \hspace{\stretch{1}}(2.3b)$

In this course we use only the linear terms and write

$\begin{aligned}e_{ij}=\frac{1}{{2}}\left( \frac{\partial {u_j}}{\partial {x_i}} + \frac{\partial {u_i}}{\partial {x_j}} \right).\end{aligned} \hspace{\stretch{1}}(2.4)$

Unresolved: Relating displacement and position by strain

In [1] it is pointed out that this strain tensor simply relates the displacement vector coordinates $u_i$ to the coordinates at the point at which it is measured

$\begin{aligned}u_i = e_{ij} x_j.\end{aligned} \hspace{\stretch{1}}(2.5)$

When we get to fluid dynamics we perform a linear expansion of $du_i$ and find something similar

$\begin{aligned}dx_i' - dx_i = du_i = \frac{\partial {u_i}}{\partial {x_k}} dx_k = e_{ij} dx_k + \omega_{ij} dx_k\end{aligned} \hspace{\stretch{1}}(2.6)$

where

$\begin{aligned}\omega_{ij} = \frac{1}{{2}} \left( \frac{\partial {u_j}}{\partial {x_i}} +\frac{\partial {u_i}}{\partial {x_j}} \right).\end{aligned} \hspace{\stretch{1}}(2.7)$

Except for the antisymmetric term, note the structural similarity of 2.5 and 2.6. Why is it that we neglect the vorticity tensor in statics?

Diagonal strain representation.

In a basis for which the strain tensor is diagonal, it was pointed out that we can write our difference in squared displacement as (for $k = 1, 2, 3$ , no summation convention)

$\begin{aligned}(dx_k')^2 - (dx_k)^2 = 2 e_{kk} dx_k dx_k\end{aligned} \hspace{\stretch{1}}(2.8)$

from which we can rearrange, take roots, and apply a first order Taylor expansion to find (again no summation convention)

$\begin{aligned}dx_k' \approx (1 + e_{kk}) dx_k.\end{aligned} \hspace{\stretch{1}}(2.9)$

An approximation of the displaced volume was then found in terms of the strain tensor trace (summation convention back again)

$\begin{aligned}dV' \approx (1 + e_{kk}) dV,\end{aligned} \hspace{\stretch{1}}(2.10)$

allowing us to identify this trace as a relative difference in displaced volume

$\begin{aligned}e_{kk} \approx \frac{dV' - dV}{dV}.\end{aligned} \hspace{\stretch{1}}(2.11)$

Strain in cylindrical coordinates.

Useful in many practice problems are the cylindrical coordinate representation of the strain tensor

$\begin{aligned}2 e_{rr} &= \frac{\partial {u_r}}{\partial {r}} \\ 2 e_{\phi\phi} &= \frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\phi}} +\frac{1}{{r}} u_r \\ 2 e_{zz} &= \frac{\partial {u_z}}{\partial {z}} \\ 2 e_{zr} &= \frac{\partial {u_r}}{\partial {z}} + \frac{\partial {u_z}}{\partial {r}} \\ 2 e_{r\phi} &= \frac{\partial {u_\phi}}{\partial {r}} - \frac{1}{{r}} u_\phi + \frac{1}{{r}} \frac{\partial {u_r}}{\partial {\phi}} \\ 2 e_{\phi z} &= \frac{\partial {u_\phi}}{\partial {z}} +\frac{1}{{r}} \frac{\partial {u_z}}{\partial {\phi}}.\end{aligned} \hspace{\stretch{1}}(2.12)$

This can be found in [2]. It was not derived there or in class, but is not too hard, even using the second order methods we used for the Cartesian form of the tensor.

An easier way to do this derivation (and understand what the coordinates represent) follows from the relation found in section 6 of [3]

$\begin{aligned}2 \mathbf{e}_i e_{ij} n_j = 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u} + \hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u}),\end{aligned} \hspace{\stretch{1}}(2.18)$

where $\hat{\mathbf{n}}$ is the normal to the surface at which we are measuring a force applied to the solid (our Cauchy tetrahedron).

The cylindrical tensor coordinates of 2.12 follow from
2.18 nicely taking $\hat{\mathbf{n}} = \hat{\mathbf{r}}, \hat{\boldsymbol{\phi}}, \hat{\mathbf{z}}$ in turn.

Compatibility condition.

For a 2D strain tensor we found an interrelationship between the components of the strain tensor

$\begin{aligned}2 \frac{\partial^2 e_{12}}{\partial x_1 \partial x_2} =\frac{\partial^2 {{e_{22}}}}{\partial {{x_1}}^2} +\frac{\partial^2 {{e_{11}}}}{\partial {{x_2}}^2},\end{aligned} \hspace{\stretch{1}}(2.19)$

and called this the compatibility condition. It was claimed, but not demonstrated that this is what is required to ensure a deformation maintained a coherent solid geometry.

I wasn’t able to find any references to this compatibility condition in any of the texts I have, but found [4], [5], and [6]. It’s not terribly surprising to see Christoffel symbol and differential forms references on those pages, since one can imagine that we’d wish to look at the mappings of all the points in the object as it undergoes the transformation from the original to the deformed state.

Even with just three points in a plane, say $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ , the general deformation of an object doesn’t seem like it’s the easiest thing to describe. We can imagine that these have trajectories in the deformation process $\mathbf{a} = \mathbf{a}(\alpha$ , $\mathbf{b} = \mathbf{b}(\beta)$ , $\mathbf{c} = \mathbf{c}(\gamma)$ , with $\mathbf{a}', \mathbf{b}', \mathbf{c}'$ at the end points of the trajectories. We’d want to look at displacement vectors $\mathbf{u}_a, \mathbf{u}_b, \mathbf{u}_c$ along each of these trajectories, and then see how they must be related. Doing that carefully must result in this compatibility condition.

Stress tensor.

By sought and found a representation of the force per unit area acting on a body by expressing the components of that force as a set of divergence relations

$\begin{aligned}f_i = \partial_k \sigma_{i k},\end{aligned} \hspace{\stretch{1}}(3.20)$

and call the associated tensor $\sigma_{ij}$ the \textit{stress}.

Unlike the strain, we don’t have any expectation that this tensor is symmetric, and identify the diagonal components (no sum) $\sigma_{i i}$ as quantifying the amount of compressive or contractive force per unit area, whereas the cross terms of the stress tensor introduce shearing deformations in the solid.

With force balance arguments (the Cauchy tetrahedron) we found that the force per unit area on the solid, for a surface with unit normal pointing into the solid, was

$\begin{aligned}\mathbf{t} = \mathbf{e}_i t_i = \mathbf{e}_i \sigma_{ij} n_j.\end{aligned} \hspace{\stretch{1}}(3.21)$

Constitutive relation.

In the scope of this course we considered only Newtonian materials, those for which the stress and strain tensors are linearly related

$\begin{aligned}\sigma_{ij} = c_{ijkl} e_{kl},\end{aligned} \hspace{\stretch{1}}(3.22)$

and further restricted our attention to isotropic materials, which can be shown to have the form

$\begin{aligned}\sigma_{ij} = \lambda e_{kk} \delta_{ij} + 2 \mu e_{ij},\end{aligned} \hspace{\stretch{1}}(3.23)$

where $\lambda$ and $\mu$ are the Lame parameters and $\mu$ is called the shear modulus (and viscosity in the context of fluids).

By computing the trace of the stress $\sigma_{ii}$ we can invert this to find

$\begin{aligned}2 \mu e_{ij} = \sigma_{ij} - \frac{\lambda}{3 \lambda + 2 \mu} \sigma_{kk} \delta_{ij}.\end{aligned} \hspace{\stretch{1}}(3.24)$

Uniform hydrostatic compression.

With only normal components of the stress (no shear), and the stress having the same value in all directions, we find

$\begin{aligned}\sigma_{ij} = ( 3 \lambda + 2 \mu ) e_{ij},\end{aligned} \hspace{\stretch{1}}(3.25)$

and identify this combination $-3 \lambda - 2 \mu$ as the pressure, linearly relating the stress and strain tensors

$\begin{aligned}\sigma_{ij} = -p e_{ij}.\end{aligned} \hspace{\stretch{1}}(3.26)$

With $e_{ii} = (dV' - dV)/dV = \Delta V/V$ , we formed the Bulk modulus $K$ with the value

$\begin{aligned}K = \left( \lambda + \frac{2 \mu}{3} \right) = -\frac{p V}{\Delta V}.\end{aligned} \hspace{\stretch{1}}(3.27)$

Uniaxial stress. Young’s modulus. Poisson’s ratio.

For the special case with only one non-zero stress component (we used $\sigma_{11}$ ) we were able to compute Young’s modulus $E$ , the ratio between stress and strain in that direction

$\begin{aligned}E = \frac{\sigma_{11}}{e_{11}} = \frac{\mu(3 \lambda + 2 \mu)}{\lambda + \mu } = \frac{3 K \mu}{K + \mu/3}.\end{aligned} \hspace{\stretch{1}}(3.28)$

Just because only one component of the stress is non-zero, does not mean that we have no deformation in any other directions. Introducing Poisson’s ratio $\nu$ in terms of the ratio of the strains relative to the strain in the direction of the force we write and then subsequently found

$\begin{aligned}\nu = -\frac{e_{22}}{e_{11}} = -\frac{e_{33}}{e_{11}} = \frac{\lambda}{2(\lambda + \mu)}.\end{aligned} \hspace{\stretch{1}}(3.29)$

We were also able to find

We can also relate the Poisson’s ratio $\nu$ to the shear modulus $\mu$

$\begin{aligned}\mu = \frac{E}{2(1 + \nu)}\end{aligned} \hspace{\stretch{1}}(3.30)$

$\begin{aligned}\lambda = \frac{E \nu}{(1 - 2 \nu)(1 + \nu)}\end{aligned} \hspace{\stretch{1}}(3.31)$

$\begin{aligned}e_{11} &= \frac{1}{{E}}\left( \sigma_{11} - \nu(\sigma_{22} + \sigma_{33}) \right) \\ e_{22} &= \frac{1}{{E}}\left( \sigma_{22} - \nu(\sigma_{11} + \sigma_{33}) \right) \\ e_{33} &= \frac{1}{{E}}\left( \sigma_{33} - \nu(\sigma_{11} + \sigma_{22}) \right)\end{aligned} \hspace{\stretch{1}}(3.32)$

Displacement propagation

It was argued that the equation relating the time evolution of a one of the vector displacement coordinates was given by

$\begin{aligned}\rho \frac{\partial^2 {{u_i}}}{\partial {{t}}^2} = \frac{\partial {\sigma_{ij}}}{\partial {x_j}} + f_i,\end{aligned} \hspace{\stretch{1}}(4.35)$

where the divergence term ${\partial {\sigma_{ij}}}/{\partial {x_j}}$ is the internal force per unit volume on the object and $f_i$ is the external force. Employing the constitutive relation we showed that this can be expanded as

$\begin{aligned}\rho \frac{\partial^2 {{u_i}}}{\partial {{t}}^2} = (\lambda + \mu) \frac{\partial^2 u_k}{\partial x_i \partial x_k}+ \mu\frac{\partial^2 u_i}{\partial x_j^2},\end{aligned} \hspace{\stretch{1}}(4.36)$

or in vector form

$\begin{aligned}\rho \frac{\partial^2 {\mathbf{u}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \mu \boldsymbol{\nabla}^2 \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(4.37)$

P-waves

Operating on 4.37 with the divergence operator, and writing $\Theta = \boldsymbol{\nabla} \cdot \mathbf{u}$ , a quantity that was our relative change in volume in the diagonal strain basis, we were able to find this divergence obeys a wave equation

$\begin{aligned}\frac{\partial^2 {{\Theta}}}{\partial {{t}}^2} = \frac{\lambda + 2 \mu}{\rho} \boldsymbol{\nabla}^2 \Theta.\end{aligned} \hspace{\stretch{1}}(4.38)$

We called these P-waves.

S-waves

Similarly, operating on 4.37 with the curl operator, and writing $\boldsymbol{\omega} = \boldsymbol{\nabla} \times \mathbf{u}$ , we were able to find this curl also obeys a wave equation

$\begin{aligned}\rho \frac{\partial^2 {{\boldsymbol{\omega}}}}{\partial {{t}}^2} = \mu \boldsymbol{\nabla}^2 \boldsymbol{\omega}.\end{aligned} \hspace{\stretch{1}}(4.39)$

These we called S-waves. We also noted that the (transverse) compression waves (P-waves) with speed $C_T = \sqrt{\mu/\rho}$ , traveled faster than the (longitudinal) vorticity (S) waves with speed $C_L = \sqrt{(\lambda + 2 \mu)/\rho}$ since $\lambda > 0$ and $\mu > 0$ , and

$\begin{aligned}\frac{C_L}{C_T} = \sqrt{\frac{ \lambda + 2 \mu}{\mu}} = \sqrt{ \frac{\lambda}{\mu} + 2}.\end{aligned} \hspace{\stretch{1}}(4.40)$

Scalar and vector potential representation.

Assuming a vector displacement representation with gradient and curl components

$\begin{aligned}\mathbf{u} = \boldsymbol{\nabla} \phi + \boldsymbol{\nabla} \times \mathbf{H},\end{aligned} \hspace{\stretch{1}}(4.41)$

We found that the displacement time evolution equation split nicely into curl free and divergence free terms

$\begin{aligned}\boldsymbol{\nabla}\left(\rho \frac{\partial^2 {{\phi}}}{\partial {{t}}^2} - (\lambda + 2\mu) \boldsymbol{\nabla}^2 \phi\right)+\boldsymbol{\nabla} \times\left(\rho \frac{\partial^2 {\mathbf{H}}}{\partial {{t}}^2} - \mu \boldsymbol{\nabla}^2 \mathbf{H}\right)= 0.\end{aligned} \hspace{\stretch{1}}(4.42)$

When neglecting boundary value effects this could be written as a pair of independent equations

$\begin{aligned}\rho \frac{\partial^2 {{\phi}}}{\partial {{t}}^2} - (\lambda + 2\mu) \boldsymbol{\nabla}^2 \phi = 0\end{aligned} \hspace{\stretch{1}}(4.43a)$

$\begin{aligned}\rho \frac{\partial^2 {\mathbf{H}}}{\partial {{t}}^2} - \mu \boldsymbol{\nabla}^2 \mathbf{H}= 0.\end{aligned} \hspace{\stretch{1}}(4.43b)$

This are the irrotational (curl free) P-wave and solenoidal (divergence free) S-wave equations respectively.

Phasor description.

It was mentioned that we could assume a phasor representation for our potentials, writing

$\begin{aligned}\phi = A \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right) \end{aligned} \hspace{\stretch{1}}(4.44a)$

$\begin{aligned}\mathbf{H} = \mathbf{B} \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right)\end{aligned} \hspace{\stretch{1}}(4.44b)$

finding

$\begin{aligned}\mathbf{u} = i \mathbf{k} \phi + i \mathbf{k} \times \mathbf{H}.\end{aligned} \hspace{\stretch{1}}(4.45)$

We did nothing with neither the potential nor the phasor theory for solid displacement time evolution, and presumably won’t on the exam either.

Some wave types

Some time was spent on non-qualitative descriptions and review of descriptions for solutions to the time evolution equations we did not attempt

P-waves [7]. Irrotational, non volume preserving body wave.
S-waves [8]. Divergence free body wave. Shearing forces are present and volume is preserved (slower than S-waves)
Rayleigh wave [9]. A surface wave that propagates near the surface of a body without penetrating into it.
Love wave [10]. A polarized shear surface wave with the shear displacements moving perpendicular to the direction of propagation.

For reasons that aren’t clear both the midterm and last years final ask us to spew this sort of stuff (instead of actually trying to do something analytic associated with them).

References

[1] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics.[Lectures on physics], chapter Elastic Materials. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963.

[2] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of Elasticity: Vol. 7 of Course of Theoretical Physics. 1960.

[3] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.

[4] Wikipedia. Compatibility (mechanics) — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 23-April-2012]. http://en.wikipedia.org/w/index.php?title=Compatibility_(mechanics)&oldid=463812965.

[5] Wikipedia. Infinitesimal strain theory — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 23-April-2012]. http://en.wikipedia.org/w/index.php?title=Infinitesimal_strain_theory&oldid=478640283.

[6] Wikipedia. Saint-venant’s compatibility condition — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 23-April-2012]. http://en.wikipedia.org/w/index.php?title=Saint-Venant\%27s_compatibility_condition&oldid=436103127.

[7] Wikipedia. P-wave — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 1-February-2012]. http://en.wikipedia.org/w/index.php?title=P-wave&oldid=474119033.

[8] Wikipedia. S-wave — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 1-February-2012]. http://en.wikipedia.org/w/index.php?title=S-wave&oldid=468110825.

[9] Wikipedia. Rayleigh wave — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 4-February-2012]. http://en.wikipedia.org/w/index.php?title=Rayleigh_wave&oldid=473693354.

[10] Wikipedia. Love wave — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 4-February-2012]. http://en.wikipedia.org/w/index.php?title=Love_wave&oldid=474355253.

Posted in Math and Physics Learning. | Tagged: bulk modulus, Cauchy tetrahedron, Compatibility condition, constitutive relation, displacement potentials, displacement vector, Lame parameters, Love wave, P-waves, phasor, PHY454H1, PHY454H1S, Poisson's ratio, Rayleigh wave, S-waves, shear modulus, strain, stress, uniaxial stress, Uniform hydrostatic compression, wave equation, Young's modulus | Leave a Comment »

PHY454H1S Continuum Mechanics. Lecture 8: Phasor description of elastic waves. Fluid dynamics. Taught by Prof. K. Das.

Posted by peeterjoot on February 6, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Review. Elastic wave equation

Starting with

$\begin{aligned}\rho \frac{\partial^2 {\mathbf{e}}}{\partial {{t}}^2} = (\lambda + \mu) \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{e}) + \mu \boldsymbol{\nabla}^2 \mathbf{e}\end{aligned} \hspace{\stretch{1}}(2.1)$

and applying a divergence operation we find

$\begin{aligned}\rho \frac{\partial^2 {{\theta}}}{\partial {{t}}^2} &= C_L^2 \boldsymbol{\nabla}^2 \theta \\ \theta &= \boldsymbol{\nabla} \cdot \mathbf{e} \\ C_L^2 &= \frac{\lambda + 2\mu}{\rho}.\end{aligned} \hspace{\stretch{1}}(2.2)$

This is the P-wave equation. Applying a curl operation we find

$\begin{aligned}\rho \frac{\partial^2 {{\boldsymbol{\omega}}}}{\partial {{t}}^2} &= C_T^2 \boldsymbol{\nabla}^2 \boldsymbol{\omega} \\ \boldsymbol{\omega} &= \boldsymbol{\nabla} \times \mathbf{e} \\ C_T^2 &= \frac{\lambda + 2\mu}{\rho}.\end{aligned} \hspace{\stretch{1}}(2.5)$

This is the S-wave equation. We also found that

$\begin{aligned}\frac{C_L}{C_T} > 1,\end{aligned} \hspace{\stretch{1}}(2.8)$

and concluded that P waves are faster than S waves. What we haven’t shown is that the P waves are longitudinal, and that the S waves are transverse.

Assuming a gradient and curl description of our displacement

$\begin{aligned}\mathbf{e} = \boldsymbol{\nabla} \phi + \boldsymbol{\nabla} \times \mathbf{H} = \mathbf{P} + \mathbf{S},\end{aligned} \hspace{\stretch{1}}(2.9)$

we found

$\begin{aligned}(\lambda + 2 \mu) \boldsymbol{\nabla}^2 \phi - \rho \frac{\partial^2 {{\phi}}}{\partial {{t}}^2} &= 0 \\ \mu \boldsymbol{\nabla}^2 \mathbf{H} - \rho \frac{\partial^2 {\mathbf{H}}}{\partial {{t}}^2} &= 0,\end{aligned} \hspace{\stretch{1}}(2.10)$

allowing us to separately solve for the P and the S wave solutions respectively. Now, let’s introduce a phasor representation (again following section 22 of the text [1])

$\begin{aligned}\phi &= A \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right) \\ \mathbf{H} &= \mathbf{B} \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right)\end{aligned} \hspace{\stretch{1}}(2.12)$

Operating with the gradient we find

$\begin{aligned}\mathbf{P}&= \boldsymbol{\nabla} \phi \\ &= \mathbf{e}_k \partial_k A \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right) \\ &= \mathbf{e}_k \partial_k A \exp\left( i ( k_m x_m - \omega t) \right) \\ &= \mathbf{e}_k i k_k A \exp\left( i ( k_m x_m - \omega t) \right) \\ &= i \mathbf{k} A \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right) \\ &= i \mathbf{k} \phi\end{aligned}$

We can also write

$\begin{aligned}\mathbf{P} = \mathbf{k} \phi'\end{aligned} \hspace{\stretch{1}}(2.14)$

where $\phi'$ is the derivative of $\phi$ “with respect to its argument”. Here argument must mean the entire phase $\mathbf{k} \cdot \mathbf{x} - \omega t$ .

$\begin{aligned}\phi' = \frac{ d\phi( \mathbf{k} \cdot \mathbf{x} - \omega t )}{ d(\mathbf{k} \cdot \mathbf{x} - \omega t) } = i \phi\end{aligned} \hspace{\stretch{1}}(2.15)$

Actually, argument is a good label here, since we can use the word in the complex number sense.

For the curl term we find

$\begin{aligned}\mathbf{S}&= \boldsymbol{\nabla} \times \mathbf{H} \\ &= \mathbf{e}_a \partial_b H_c \epsilon_{a b c} \\ &= \mathbf{e}_a \partial_b \epsilon_{a b c} B_c \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right) \\ &= \mathbf{e}_a \partial_b \epsilon_{a b c} B_c \exp\left( i ( k_m x_m - \omega t) \right) \\ &= \mathbf{e}_a i k_b \epsilon_{a b c} B_c \exp\left( i ( \mathbf{k} \cdot \mathbf{x} - \omega t) \right) \\ &= i \mathbf{k} \times \mathbf{H}\end{aligned}$

Again writing

$\begin{aligned}\mathbf{H}' = \frac{ d\mathbf{H}( \mathbf{k} \cdot \mathbf{x} - \omega t )}{ d(\mathbf{k} \cdot \mathbf{x} - \omega t) } = i \mathbf{H}\end{aligned} \hspace{\stretch{1}}(2.16)$

we can write the S wave as

$\begin{aligned}\mathbf{S} = \mathbf{k} \times \mathbf{H}'\end{aligned} \hspace{\stretch{1}}(2.17)$

Some waves illustrated.

The following wave types were noted, but not defined:

\begin{itemize}
\item Rayleigh wave. This is discussed in section 24 of the text (a wave that propagates near the surface of a body without penetrating into it). Wikipedia has an illustration of one possible mode of propagation [2].
\item Love wave. These aren’t discussed in the text, but wikipedia [3] describes them as polarized shear waves (where the figure indicates that the shear displacements are perpendicular to the direction of propagation).
\end{itemize}

Some illustrations from the class notes were also shown. Hopefully we’ll have some homework assignments where we do some problems to get a feel for how to apply the formalism.

Fluid dynamics.

In fluid dynamics we look at displacements with respect to time as illustrated in figure (\ref{fig:continuumL8:continuumL8fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL8fig1}
\caption{Differential displacement.}
\end{figure}

$\begin{aligned}d\mathbf{x}' = d\mathbf{x} + d\mathbf{u} \delta t\end{aligned} \hspace{\stretch{1}}(3.18)$

In index notation

$\begin{aligned}dx_i'&= dx_i + du_i \delta t \\ &= dx_i + \frac{\partial {u_i}}{\partial {x_j}} dx_j \delta t\end{aligned}$

We define

$\begin{aligned}e_{ij} = \frac{1}{{2}} \left(\frac{\partial {u_i}}{\partial {x_j}} +\frac{\partial {u_j}}{\partial {x_i}} \right)\end{aligned} \hspace{\stretch{1}}(3.19)$

a symmetric tensor. We also define

$\begin{aligned}\omega_{ij} = \frac{1}{{2}} \left(\frac{\partial {u_i}}{\partial {x_j}}-\frac{\partial {u_j}}{\partial {x_i}} \right)\end{aligned} \hspace{\stretch{1}}(3.20)$

Effect of $e_{ij}$ when diagonalized

$\begin{aligned}e_{ij} ==\begin{bmatrix}e_{11} & 0 & 0 \\ 0 & e_{22} & 0 \\ 0 & 0 & e_{33}\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.21)$

so that in this frame of reference we have

$\begin{aligned}dx_1' &= ( 1 + e_{11} \delta t) dx_1 \\ dx_2' &= ( 1 + e_{22} \delta t) dx_2 \\ dx_3' &= ( 1 + e_{33} \delta t) dx_3\end{aligned} \hspace{\stretch{1}}(3.22)$

Let’s find the matrix form of the antisymmetric tensor. We find

$\begin{aligned}\omega_{11} = \omega_{22} = \omega_{33} = 0\end{aligned} \hspace{\stretch{1}}(3.25)$

Introducing a vorticity vector

$\begin{aligned}\boldsymbol{\omega} = \boldsymbol{\nabla} \times \mathbf{u}\end{aligned} \hspace{\stretch{1}}(3.26)$

we find

$\begin{aligned}\omega_{12} &= \frac{1}{{2}}\left( \frac{\partial {u_1}}{\partial {x_2}} -\frac{\partial {u_2}}{\partial {x_1}} \right) = - \frac{1}{{2}} (\boldsymbol{\nabla} \times \mathbf{u})_3 \\ \omega_{23} &= \frac{1}{{2}}\left( \frac{\partial {u_2}}{\partial {x_3}} -\frac{\partial {u_3}}{\partial {x_2}} \right) = - \frac{1}{{2}} (\boldsymbol{\nabla} \times \mathbf{u})_1 \\ \omega_{31} &= \frac{1}{{2}}\left( \frac{\partial {u_3}}{\partial {x_1}} -\frac{\partial {u_1}}{\partial {x_3}} \right) = - \frac{1}{{2}} (\boldsymbol{\nabla} \times \mathbf{u})_2\end{aligned} \hspace{\stretch{1}}(3.27)$

Writing

$\begin{aligned}\Omega_i = \frac{1}{{2}} \omega_i\end{aligned} \hspace{\stretch{1}}(3.30)$

we find the matrix form of this antisymmetric tensor

$\begin{aligned}\omega_{ij}=\begin{bmatrix}0 & -\Omega_3 & \Omega_2 \\ \Omega_3 & 0 & -\Omega_1 \\ -\Omega_2 & \Omega_1 & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.31)$

$\begin{aligned}dx_1'&= dx_1 + \left( \not{{\omega_{11}}} dx_1 + \omega_{12} dx_2 + \omega_{13} dx_3 \right) \delta t \\ &= dx_1 + \left( \omega_{12} dx_2 + \omega_{13} dx_3 \right) \delta t \\ &= dx_1 + \left( \Omega_2 dx_3 - \Omega_3 dx_2 \right) \delta t\end{aligned}$

Doing this for all components we find

$\begin{aligned}d\mathbf{x}' = d\mathbf{x} + (\boldsymbol{\Omega} \times d\mathbf{x}) \delta t.\end{aligned} \hspace{\stretch{1}}(3.32)$

The tensor $\omega_{ij}$ implies rotation of a control volume with an angular velocity $\boldsymbol{\Omega} = \boldsymbol{\omega}/2$ (half the vorticity vector).

In general we have

$\begin{aligned}dx_i' = dx_i + e_{ij} dx_j \delta t + \omega_{ij} dx_j \delta t\end{aligned} \hspace{\stretch{1}}(3.33)$

Making sense of things.

After this first fluid dynamics lecture I was left troubled. We’d just been barraged with a set of equations pulled out of a magic hat, with no notion of where they came from. Unlike the contiuum strain tensor, which was derived by considering differences in squared displacements, we have an antisymmetric term now. Why did we have no such term considering solids?

After a bit of thought I think I see where things are coming from. We have essentially looked at a first order decomposition of the displacement (per unit time) of a point in terms of symmetric and antisymmetric terms. This is really just a gradient evaluation, split into coordinates

$\begin{aligned}x_i' &= x_i + (\boldsymbol{\nabla} u_i) \cdot d\mathbf{x} \delta t \\ &= x_i + \frac{\partial {u_i}}{\partial {x_j}} dx_j \delta t \\ &= x_i + \frac{1}{{2}}\left(\frac{\partial {u_i}}{\partial {x_j}} +\frac{\partial {u_i}}{\partial {x_j}} \right)dx_j \delta t +\frac{1}{{2}}\left(\frac{\partial {u_i}}{\partial {x_j}} -\frac{\partial {u_i}}{\partial {x_j}} \right)dx_j \delta t \\ &=x_i + e_{ij} dx_j \delta t + \omega_{ij} dx_j \delta t\end{aligned}$

Here, as in the solids case, we have

$\begin{aligned}\mathbf{u} = \mathbf{x}' - \mathbf{x}\end{aligned} \hspace{\stretch{1}}(3.34)$

References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

[2] Wikipedia. Rayleigh wave — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 4-February-2012]. http://en.wikipedia.org/w/index.php?title=Rayleigh_wave&oldid=473693354.

[3] Wikipedia. Love wave — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 4-February-2012]. http://en.wikipedia.org/w/index.php?title=Love_wave&oldid=474355253.

Posted in Math and Physics Learning. | Tagged: elastic wave, elastic wave equation, Love wave, p-wave, phasor, PHY454, Rayleigh wave, s-wave, vorticity vector, wave equation | Leave a Comment »

PHY450H1S. Relativistic Electrodynamics Lecture 18 (Taught by Prof. Erich Poppitz). Green’s function solution to Maxwell’s equation.

Posted by peeterjoot on March 12, 2011

[Click here for a PDF of this post with nicer formatting]

Reading.

Covering chapter 8 material from the text [1].

Covering lecture notes pp. 136-146: continued reminder of electrostatic Greens function (136); the retarded Greens function of the d’Alembert operator: derivation and properties (137-140); the solution of the d’Alembert equation with a source: retarded potentials (141-142)

Solving the forced wave equation.

See the notes for a complex variables and Fourier transform method of deriving the Green’s function. In class, we’ll just pull it out of a magic hat. We wish to solve

$\begin{aligned}\square A^k = \partial_i \partial^i A^k = \frac{4 \pi}{c} j^k\end{aligned} \hspace{\stretch{1}}(2.1)$

(with a $\partial_i A^i = 0$ gauge choice).

Our Green’s method utilizes

$\begin{aligned}\square_{(\mathbf{x}, t)} G(\mathbf{x} - \mathbf{x}', t - t') = \delta^3( \mathbf{x} - \mathbf{x}') \delta( t - t')\end{aligned} \hspace{\stretch{1}}(2.2)$

If we know such a function, our solution is simple to obtain

$\begin{aligned}A^k(\mathbf{x}, t)= \int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t') G(\mathbf{x} - \mathbf{x}', t - t')\end{aligned} \hspace{\stretch{1}}(2.3)$

Proof:

$\begin{aligned}\square_{(\mathbf{x}, t)} A^k(\mathbf{x}, t)&=\int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t')\square_{(\mathbf{x}, t)}G(\mathbf{x} - \mathbf{x}', t - t') \\ &=\int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t')\delta^3( \mathbf{x} - \mathbf{x}') \delta( t - t') \\ &=\frac{4 \pi}{c} j^k(\mathbf{x}, t)\end{aligned}$

Claim:

$\begin{aligned}G(\mathbf{x}, t) = \frac{\delta(t - {\left\lvert{\mathbf{x}}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }\end{aligned} \hspace{\stretch{1}}(2.4)$

This is the retarded Green’s function of the operator $\square$ , where

$\begin{aligned}\square G(\mathbf{x}, t) = \delta^3(\mathbf{x}) \delta(t)\end{aligned} \hspace{\stretch{1}}(2.5)$

Proof of the d’Alembertian Green’s function

Our Prof is excellent at motivating any results that he pulls out of magic hats. He’s said that he’s included a derivation using Fourier transforms and tricky contour integration arguments in the class notes for anybody who is interested (and for those who also know how to do contour integration). For those who don’t know contour integration yet (some people are taking it concurrently), one can actually prove this by simply applying the wave equation operator to this function. This treats the delta function as a normal function that one can take the derivatives of, something that can be well defined in the context of generalized functions. Chugging ahead with this approach we have

$\begin{aligned}\square G(\mathbf{x}, t)=\left(\frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta\right)\frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\frac{\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi c^2 {\left\lvert{\mathbf{x}}\right\rvert} }- \Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }.\end{aligned} \hspace{\stretch{1}}(2.6)$

This starts things off and now things get a bit hairy. It’s helpful to consider a chain rule expansion of the Laplacian

$\begin{aligned}\Delta (u v)&=\partial_{\alpha\alpha} (u v) \\ &=\partial_{\alpha} (v \partial_\alpha u+ u\partial_\alpha v) \\ &=(\partial_\alpha v) (\partial_\alpha u ) + v \partial_{\alpha\alpha} u+(\partial_\alpha u) (\partial_\alpha v ) + u \partial_{\alpha\alpha} v).\end{aligned}$

In vector form this is

$\begin{aligned}\Delta (u v) = u \Delta v + 2 (\boldsymbol{\nabla} u) \cdot (\boldsymbol{\nabla} v) + v \Delta u.\end{aligned} \hspace{\stretch{1}}(2.7)$

Applying this to the Laplacian portion of 2.6 we have

$\begin{aligned}\Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)\Delta\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}+\left(\boldsymbol{\nabla} \frac{1}{{2 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\right)\cdot\left(\boldsymbol{\nabla}\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \right)+\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\Delta\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right).\end{aligned} \hspace{\stretch{1}}(2.8)$

Here we make the identification

$\begin{aligned}\Delta \frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }} = - \delta^3(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.9)$

This could be considered a given from our knowledge of electrostatics, but it’s not too much work to just do so.

An aside. Proving the Laplacian Green’s function.

If $-1/{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }$ is a Green’s function for the Laplacian, then the Laplacian of the convolution of this with a test function should recover that test function

$\begin{aligned}\Delta \int d^3 \mathbf{x}' \left(-\frac{1}{{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} \right) f(\mathbf{x}') = f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.10)$

We can directly evaluate the LHS of this equation, following the approach in [2]. First note that the Laplacian can be pulled into the integral and operates only on the presumed Green’s function. For that operation we have

$\begin{aligned}\Delta \left(-\frac{1}{{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} \right)=-\frac{1}{{4 \pi}} \boldsymbol{\nabla} \cdot \boldsymbol{\nabla} {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(2.11)$

It will be helpful to compute the gradient of various powers of ${\left\lvert{\mathbf{x}}\right\rvert}$

$\begin{aligned}\boldsymbol{\nabla} {\left\lvert{\mathbf{x}}\right\rvert}^a&=e_\alpha \partial_\alpha (x^\beta x^\beta)^{a/2} \\ &=e_\alpha \left(\frac{a}{2}\right) 2 x^\beta {\delta_\beta}^\alpha {\left\lvert{\mathbf{x}}\right\rvert}^{a - 2}.\end{aligned}$

In particular we have, when $\mathbf{x} \ne 0$ , this gives us

$\begin{aligned}\boldsymbol{\nabla} {\left\lvert{\mathbf{x}}\right\rvert} &= \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} &= -\frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}^3}} &= -3 \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^5}.\end{aligned} \hspace{\stretch{1}}(2.12)$

For the Laplacian of $1/{\left\lvert{\mathbf{x}}\right\rvert}$ , at the points $\mathbf{e} \ne 0$ where this is well defined we have

$\begin{aligned}\Delta \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} &=\boldsymbol{\nabla} \cdot \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} \\ &= -\partial_\alpha \frac{x^\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} - x^\alpha \partial_\alpha \frac{1}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} - \mathbf{x} \cdot \boldsymbol{\nabla} \frac{1}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} + 3 \frac{\mathbf{x}^2}{{\left\lvert{\mathbf{x}}\right\rvert}^5}\end{aligned}$

So we have a zero. This means that the Laplacian operation

$\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \Delta \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}},\end{aligned} \hspace{\stretch{1}}(2.15)$

can only have a value in a neighborhood of point $\mathbf{x}$ . Writing $\Delta = \boldsymbol{\nabla} \cdot \boldsymbol{\nabla}$ we have

$\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \boldsymbol{\nabla} \cdot -\frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(2.16)$

Observing that $\boldsymbol{\nabla} \cdot f(\mathbf{x} -\mathbf{x}') = -\boldsymbol{\nabla}' f(\mathbf{x} - \mathbf{x}')$ we can put this in a form that allows for use of Stokes theorem so that we can convert this to a surface integral

$\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') &=\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \boldsymbol{\nabla}' \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &=\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^2 \mathbf{x}' \mathbf{n} \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &= \int_{\phi=0}^{2\pi} \int_{\theta = 0}^\pi \epsilon^2 \sin\theta d\theta d\phi \frac{\mathbf{x}' - \mathbf{x}}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &= -\int_{\phi=0}^{2\pi} \int_{\theta = 0}^\pi \epsilon^2 \sin\theta d\theta d\phi \frac{\epsilon^2}{\epsilon^4}\end{aligned}$

where we use $(\mathbf{x}' - \mathbf{x})/{\left\lvert{\mathbf{x}' - \mathbf{x}}\right\rvert}$ as the outwards normal for a sphere centered at $\mathbf{x}$ of radius $\epsilon$ . This integral is just $-4 \pi$ , so we have

$\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{-4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.17)$

The convolution of $f(\mathbf{x})$ with $-\Delta/4 \pi {\left\lvert{\mathbf{x}}\right\rvert}$ produces $f(\mathbf{x})$ , allowing an identification of this function with a delta function, since the two have the same operational effect

$\begin{aligned}\int d^3 \mathbf{x}' \delta(\mathbf{x} - \mathbf{x}') f(\mathbf{x}') =f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.18)$

Returning to the d’Alembertian Green’s function.

We need two additional computations to finish the job. The first is the gradient of the delta function

$\begin{aligned}\boldsymbol{\nabla} \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= ? \\ \Delta \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= ?\end{aligned}$

Consider $\boldsymbol{\nabla} f(g(\mathbf{x}))$ . This is

$\begin{aligned}\boldsymbol{\nabla} f(g(\mathbf{x}))&=e_\alpha \frac{\partial {f(g(\mathbf{x}))}}{\partial {x^\alpha}} \\ &=e_\alpha \frac{\partial {f}}{\partial {g}} \frac{\partial {g}}{\partial {x^\alpha}},\end{aligned}$

so we have

$\begin{aligned}\boldsymbol{\nabla} f(g(\mathbf{x}))=\frac{\partial {f}}{\partial {g}} \boldsymbol{\nabla} g.\end{aligned} \hspace{\stretch{1}}(2.19)$

The Laplacian is similar

$\begin{aligned}\Delta f(g)&= \boldsymbol{\nabla} \cdot \left(\frac{\partial {f}}{\partial {g}} \boldsymbol{\nabla} g \right) \\ &= \partial_\alpha \left(\frac{\partial {f}}{\partial {g}} \partial_\alpha g \right) \\ &= \left( \partial_\alpha \frac{\partial {f}}{\partial {g}} \right) \partial_\alpha g +\frac{\partial {f}}{\partial {g}} \partial_{\alpha\alpha} g \\ &= \frac{\partial^2 {{f}}}{\partial {{g}}^2} \left( \partial_\alpha g \right) (\partial_\alpha g)+\frac{\partial {f}}{\partial {g}} \Delta g,\end{aligned}$

so we have

$\begin{aligned}\Delta f(g)= \frac{\partial^2 {{f}}}{\partial {{g}}^2} (\boldsymbol{\nabla} g)^2 +\frac{\partial {f}}{\partial {g}} \Delta g\end{aligned} \hspace{\stretch{1}}(2.20)$

With $g(\mathbf{x}) = {\left\lvert{\mathbf{x}}\right\rvert}$ , we’ll need the Laplacian of this vector magnitude

$\begin{aligned}\Delta {\left\lvert{\mathbf{x}}\right\rvert}&=\partial_\alpha \frac{x_\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ &=\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}} + x_\alpha \partial_\alpha (x^\beta x^\beta)^{-1/2} \\ &=\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}} - \frac{x_\alpha x_\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \end{aligned}$

So that we have

$\begin{aligned}\boldsymbol{\nabla} \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ \Delta \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &=\frac{1}{{c^2}} \delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \end{aligned} \hspace{\stretch{1}}(2.21)$

Now we have all the bits and pieces of 2.8 ready to assemble

$\begin{aligned}\Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }&=-\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) \\ &\quad +\frac{1}{{2\pi}} \left( - \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \right)\cdot-\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ &\quad +\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\left(\frac{1}{{c^2}} \delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \right) \\ &=-\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) +\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} c^2 }}\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \end{aligned}$

Since we also have

$\begin{aligned}\frac{1}{{c^2}} \partial_{tt}\frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\frac{\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} c^2}\end{aligned} \hspace{\stretch{1}}(2.23)$

The $\delta''$ terms cancel out in the d’Alembertian, leaving just

$\begin{aligned}\square \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) \end{aligned} \hspace{\stretch{1}}(2.24)$

Noting that the spatial delta function is non-zero only when $\mathbf{x} = 0$ , which means $\delta(t - {\left\lvert{\mathbf{x}}\right\rvert}/c) = \delta(t)$ in this product, and we finally have

$\begin{aligned}\square \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta(t) \delta^3(\mathbf{x}) \end{aligned} \hspace{\stretch{1}}(2.25)$

We write

$\begin{aligned}G(\mathbf{x}, t) = \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} },\end{aligned} \hspace{\stretch{1}}(2.26)$

Elaborating on the wave equation Green’s function

The Green’s function 2.26 is a distribution that is non-zero only on the future lightcone. Observe that for $t < 0$ we have

$\begin{aligned}\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)&=\delta\left(-{\left\lvert{t}\right\rvert} - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \\ &= 0.\end{aligned}$

We say that $G$ is supported only on the future light cone. At $\mathbf{x} = 0$ , only the contributions for $t > 0$ matter. Note that in the “old days”, Green’s functions used to be called influence functions, a name that works particularly well in this case. We have other Green’s functions for the d’Alembertian. The one above is called the retarded Green’s functions and we also have an advanced Green’s function. Writing $+$ for advanced and $-$ for retarded these are

$\begin{aligned}G_{\pm} = \frac{\delta\left(t \pm \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.27)$

There are also causal and non-causal variations that won’t be of interest for this course.

This arms us now to solve any problem in the Lorentz gauge

$\begin{aligned}A^k(\mathbf{x}, t) = \frac{1}{{c}} \int d^3 \mathbf{x}' dt' \frac{\delta\left(t - t' - \frac{{\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}j^k(\mathbf{x}', t')+\text{An arbitrary collection of EM waves.}\end{aligned} \hspace{\stretch{1}}(3.28)$

The additional EM waves are the possible contributions from the homogeneous equation.

Since $\delta(t - t' - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c)$ is non-zero only when $t' = t - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c)$ , the non-homogeneous parts of 3.28 reduce to

$\begin{aligned}A^k(\mathbf{x}, t) = \frac{1}{{c}} \int d^3 \mathbf{x}' \frac{j^k(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(3.29)$

Our potentials at time $t$ and spatial position $\mathbf{x}$ are completely specified in terms of the sums of the currents acting at the retarded time $t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c$ . The field can only depend on the charge and current distribution in the past. Specifically, it can only depend on the charge and current distribution on the past light cone of the spacetime point at which we measure the field.

Example of the Green’s function. Consider a charged particle moving on a worldline

$\begin{aligned}(c t, \mathbf{x}_c(t))\end{aligned} \hspace{\stretch{1}}(4.30)$

( $c$ for classical)

For this particle

$\begin{aligned}\rho(\mathbf{x}, t) &= e \delta^3(\mathbf{x} - \mathbf{x}_c(t)) \\ \mathbf{j}(\mathbf{x}, t) &= e \dot{\mathbf{x}}_c(t) \delta^3(\mathbf{x} - \mathbf{x}_c(t))\end{aligned} \hspace{\stretch{1}}(4.31)$

$\begin{aligned}\begin{bmatrix}A^0(\mathbf{x}, t)\mathbf{A}(\mathbf{x}, t)\end{bmatrix}&=\frac{1}{{c}}\int d^3 \mathbf{x}' dt'\frac{ \delta( t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\begin{bmatrix}c e \\ e \dot{\mathbf{x}}_c(t)\end{bmatrix}\delta^3(\mathbf{x} - \mathbf{x}_c(t)) \\ &=\int_{-\infty}^\infty\frac{ \delta( t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}_c(t')}\right\rvert}/c }{{\left\lvert{\mathbf{x}_c(t') - \mathbf{x}}\right\rvert}}\begin{bmatrix}e \\ e \frac{\dot{\mathbf{x}}_c(t)}{c}\end{bmatrix}\end{aligned}$

PICTURE: light cones, and curved worldline. Pick an arbitrary point $(\mathbf{x}_0, t_0)$ , and draw the past light cone, looking at where this intersects with the trajectory

For the arbitrary point $(\mathbf{x}_0, t_0)$ we see that this point and the retarded time $(\mathbf{x}_c(t_r), t_r)$ obey the relation

$\begin{aligned}c (t_0 - t_r) = {\left\lvert{\mathbf{x}_0 - \mathbf{x}_c(t_r)}\right\rvert}\end{aligned} \hspace{\stretch{1}}(4.33)$

This retarded time is unique. There is only one such intersection.

Our job is to calculate

$\begin{aligned}\int_{-\infty}^\infty \delta(f(x)) g(x) = \frac{g(x_{*})}{f'(x_{*})}\end{aligned} \hspace{\stretch{1}}(4.34)$

where $f(x_{*}) = 0$ .

$\begin{aligned}f(t') = t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}_c(t')}\right\rvert}/c\end{aligned} \hspace{\stretch{1}}(4.35)$

$\begin{aligned}\frac{\partial {f}}{\partial {t'}}&= -1 - \frac{1}{{c}} \frac{\partial {}}{\partial {t'}} \sqrt{ (\mathbf{x} - \mathbf{x}_c(t')) \cdot (\mathbf{x} - \mathbf{x}_c(t')) } \\ &= -1 + \frac{1}{{c}} \frac{\partial {}}{\partial {t'}} \frac{(\mathbf{x} - \mathbf{x}_c(t')) \cdot \mathbf{v}_c(t_r)}{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert}}\end{aligned}$

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

[2] M. Schwartz. Principles of Electrodynamics. Dover Publications, 1987.

Posted in Math and Physics Learning. | Tagged: d'Alembertian, delta function, four current, four potential, Green's function, laplacian, retarded time, stokes theorem, wave equation | Leave a Comment »

PHY450H1S. Relativistic Electrodynamics Lecture 15 (Taught by Prof. Erich Poppitz). Fourier solution of Maxwell’s vacuum wave equation in the Coulomb gauge.

Posted by peeterjoot on March 2, 2011

[Click here for a PDF of this post with nicer formatting]

Reading.

Covering chapter 6 material from the text [1].

Covering lecture notes pp. 115-127: reminder on wave equations (115); reminder on Fourier series and integral (115-117); Fourier expansion of the EM potential in Coulomb gauge and equation of motion for the spatial Fourier components (118-119); the general solution of Maxwell’s equations in vacuum (120-121) [Tuesday, Mar. 1]

Review of wave equation results obtained.

Maxwell’s equations in vacuum lead to Coulomb gauge and the Lorentz gauge.

\paragraph{Coulomb gauge}

$\begin{aligned}A^0 &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{A} &= 0 \\ \left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) \mathbf{A} &= 0\end{aligned} \hspace{\stretch{1}}(2.1)$

\paragraph{Lorentz gauge}

$\begin{aligned}\partial_i A^i &= 0 \\ \left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) A^i &= 0\end{aligned} \hspace{\stretch{1}}(2.4)$

Note that $\partial_i A^i = 0$ is invariant under gauge transformations

$\begin{aligned}A^i \rightarrow A^i + \partial^i \chi\end{aligned} \hspace{\stretch{1}}(2.6)$

where

$\begin{aligned}\partial_i \partial^i \chi = 0,\end{aligned} \hspace{\stretch{1}}(2.7)$

So if one uses the Lorentz gauge, this has to be fixed.

However, in both cases we have

$\begin{aligned}\left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) f = 0\end{aligned} \hspace{\stretch{1}}(2.8)$

where

$\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \end{aligned} \hspace{\stretch{1}}(2.9)$

is the wave operator.

Consider

$\begin{aligned}\Delta = \frac{\partial^2 {{}}}{\partial {{x}}^2}\end{aligned} \hspace{\stretch{1}}(2.10)$

where we are looking for a solution that is independent of $y, z$ . Recall that the general solution for this equation has the form

$\begin{aligned}f(t, x) = F_1 \left(t - \frac{x}{c}\right)+F_2 \left(t + \frac{x}{c}\right)\end{aligned} \hspace{\stretch{1}}(2.11)$

PICTURE: superposition of two waves with $F_1$ moving along the x-axis in the positive direction, and $F_2$ in the negative x direction.

It is notable that the text derives 2.11 in a particularly slick way. It’s still black magic, since one has to know the solution to find it, but very very cool.

Review of Fourier methods.

It is often convienent to impose periodic boundary conditions

$\begin{aligned}\mathbf{A}(\mathbf{x} + \mathbf{e}_i L) = \mathbf{A}(\mathbf{x}), i = 1,2,3\end{aligned} \hspace{\stretch{1}}(3.12)$

In one dimension

$\begin{aligned}f(x + L) = f(x)\end{aligned} \hspace{\stretch{1}}(3.13)$

$\begin{aligned}f(x) = \sum_{n=-\infty}^\infty e^{i \frac{2 \pi n}{L} x} \tilde{f}_n\end{aligned} \hspace{\stretch{1}}(3.14)$

When $f(x)$ is real we also have

$\begin{aligned}f^{*}(x) = \sum_{n = -\infty}^\infty e^{-i \frac{2 \pi n}{L} x} (\tilde{f}_n)^{*}\end{aligned} \hspace{\stretch{1}}(3.15)$

which implies

$\begin{aligned}{\tilde{f}^{*}}_{n} = \tilde{f}_{-n}.\end{aligned} \hspace{\stretch{1}}(3.16)$

We introduce a wave number

$\begin{aligned}k_n = \frac{2 \pi n}{L},\end{aligned} \hspace{\stretch{1}}(3.17)$

allowing a slightly simpler expression of the Fourier decomposition

$\begin{aligned}f(x) = \sum_{n=-\infty}^\infty e^{i k_n x} \tilde{f}_{k_n}.\end{aligned} \hspace{\stretch{1}}(3.18)$

The inverse transform is obtained by integration over some length $L$ interval

$\begin{aligned}\tilde{f}_{k_n} = \frac{1}{{L}} \int_{-L/2}^{L/2} dx e^{-i k_n x} f(x)\end{aligned} \hspace{\stretch{1}}(3.19)$

\paragraph{Verify:}

We should be able to recover the Fourier coefficient by utilizing the above

$\begin{aligned}\frac{1}{{L}} \int_{-L/2}^{L/2} dx e^{-i k_n x} \sum_{m=-\infty}^\infty e^{i k_m x} \tilde{f}_{k_m} \\ &= \sum_{m = -\infty}^\infty \tilde{f}_{k_m} \delta_{mn} = \tilde{f}_{k_n},\end{aligned}$

where we use the easily verifyable fact that

$\begin{aligned}\frac{1}{{L}} \int_{-L/2}^{L/2} dx e^{i (k_m - k_n) x} = \begin{array}{l l}0 & \quad \mbox{if$ latex m \ne n$} \\ 1 & \quad \mbox{if $m = n$ } \\ \end{array}.\end{aligned} \hspace{\stretch{1}}(3.20)$

It is conventional to absorb $\tilde{f}_{k_n} = \tilde{f}(k_n)$ for

$\begin{aligned}f(x) &= \frac{1}{{L}} \sum_n \tilde{f}(k_n) e^{i k_n x} \\ \tilde{f}(k_n) &= \int_{-L/2}^{L/2} dx f(x) e^{-i k_n x}\end{aligned} \hspace{\stretch{1}}(3.21)$

To take $L \rightarrow \infty$ notice

$\begin{aligned}k_n = \frac{2 \pi}{L} n\end{aligned} \hspace{\stretch{1}}(3.23)$

when $n$ changes by $\Delta n = 1$ , $k_n$ changes by $\Delta k_n = \frac{2 \pi}{L} \Delta n$

Using this

$\begin{aligned}f(x) = \frac{1}{{2\pi}} \sum_n \left( \frac{2\pi}{L} \Delta n \right) \tilde{f}(k_n) e^{i k_n x}\end{aligned} \hspace{\stretch{1}}(3.24)$

With $L \rightarrow \infty$ , and $\Delta k_n \rightarrow 0$

$\begin{aligned}f(x) &= \int_{-\infty}^\infty \frac{dk}{2\pi} \tilde{f}(k) e^{i k x} \\ \tilde{f}(k) &= \int_{-\infty}^\infty dx f(x) e^{-i k x}\end{aligned} \hspace{\stretch{1}}(3.25)$

\paragraph{Verify:}

A loose verification of the inversion relationship (the most important bit) is possible by substitution

$\begin{aligned}\int \frac{dk}{2\pi} e^{i k x} \tilde{f}(k) &= \iint \frac{dk}{2\pi} e^{i k x} dx' f(x') e^{-i k x'} \\ &= \int dx' f(x') \frac{1}{{2\pi}} \int dk e^{i k (x - x')}\end{aligned}$

Now we employ the old physics ploy where we identify

$\begin{aligned}\frac{1}{{2\pi}} \int dk e^{i k (x - x')} = \delta(x - x').\end{aligned} \hspace{\stretch{1}}(3.27)$

With that we see that we recover the function $f(x)$ above as desired.

In three dimensions

$\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= \int \frac{d^3 \mathbf{k}}{(2\pi)^3} \tilde{\mathbf{A}}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x}} \\ \tilde{\mathbf{A}}(\mathbf{x}, t) &= \int d^3 \mathbf{x} \mathbf{A}(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x}}\end{aligned} \hspace{\stretch{1}}(3.28)$

Application to the wave equation

$\begin{aligned}0 &= \left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) \mathbf{A}(\mathbf{x}, t) \\ &=\left( \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta \right) \int \frac{d^3 \mathbf{k}}{(2\pi)^3} \tilde{\mathbf{A}}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=\int \frac{d^3 \mathbf{k}}{(2\pi)^3} \left( \frac{1}{{c^2}} \partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) + \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right)e^{i \mathbf{k} \cdot \mathbf{x}} \end{aligned}$

Now operate with $\int d^3 \mathbf{x} e^{-i \mathbf{p} \cdot \mathbf{x} }$

$\begin{aligned}0 &=\int d^3 \mathbf{x} e^{-i \mathbf{p} \cdot \mathbf{x} }\int \frac{d^3 \mathbf{k}}{(2\pi)^3} \left( \frac{1}{{c^2}} \partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) + \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right)e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=\int d^3 \mathbf{k}\delta^3(\mathbf{p} -\mathbf{k}) \left( \frac{1}{{c^2}} \partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) + \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right)\end{aligned}$

Since this is true for all $\mathbf{p}$ we have

$\begin{aligned}\partial_{tt} \tilde{\mathbf{A}}(\mathbf{p}, t) = -c^2 \mathbf{p}^2 \tilde{\mathbf{A}}(\mathbf{p}, t) \end{aligned} \hspace{\stretch{1}}(3.30)$

For every value of momentum we have a harmonic osciallator!

$\begin{aligned}\dot{d}{x} = -\omega^2 x\end{aligned} \hspace{\stretch{1}}(3.31)$

Fourier modes of EM potential in vacuum obey

$\begin{aligned}\partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) = -c^2 \mathbf{k}^2 \tilde{\mathbf{A}}(\mathbf{k}, t)\end{aligned} \hspace{\stretch{1}}(3.32)$

Because we are operating in the Coulomb gauge we must also have zero divergence. Let’s see how that translates to our Fourier representation

implies

$\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \mathbf{A}(\mathbf{x}, t) \\ &= \int \frac{d^3 \mathbf{k} }{(2 \pi)^3} \boldsymbol{\nabla} \cdot \left( e^{i \mathbf{k} \cdot \mathbf{x}} \cdot \tilde{\mathbf{A}}(\mathbf{k}, t) \right)\end{aligned}$

The chain rule for the divergence in this case takes the form

$\begin{aligned}\boldsymbol{\nabla} \cdot (\phi \mathbf{B}) = (\boldsymbol{\nabla} \phi) \cdot \mathbf{B} + \phi \boldsymbol{\nabla} \cdot \mathbf{B}.\end{aligned} \hspace{\stretch{1}}(3.33)$

But since our vector function $\tilde{\mathbf{A}}$ is not a function of spatial coordinates we have

$\begin{aligned}0 = \int \frac{d^3 \mathbf{k} }{(2 \pi)^3} e^{i \mathbf{k} \cdot \mathbf{x}} (i \mathbf{k} \cdot \tilde{\mathbf{A}}(\mathbf{k}, t)).\end{aligned} \hspace{\stretch{1}}(3.34)$

This has two immediate consequences. The first is that our momentum space potential is perpendicular to the wave number vector at all points in momentum space, and the second gives us a conjugate relation (substitute $\mathbf{k} \rightarrow -\mathbf{k}'$ after taking conjugates for that one)

$\begin{aligned}\mathbf{k} \cdot \tilde{\mathbf{A}}(\mathbf{k}, t) &= 0 \\ \tilde{\mathbf{A}}(-\mathbf{k}, t) &= \tilde{\mathbf{A}}^{*}(\mathbf{k}, t).\end{aligned} \hspace{\stretch{1}}(3.35)$

$\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \int \frac{d^3 \mathbf{k}}{(2\pi)^3} e^{i \mathbf{k} \cdot \mathbf{x}} \left( \frac{1}{{2}} \tilde{\mathbf{A}}(\mathbf{k}, t) + \frac{1}{{2}} \tilde{\mathbf{A}}^{*}(- \mathbf{k}, t) \right)\end{aligned} \hspace{\stretch{1}}(3.37)$

Since out system is essentially a harmonic oscillator at each point in momentum space

$\begin{aligned}\partial_{tt} \tilde{\mathbf{A}}(\mathbf{k}, t) &= - \omega_k^2 \tilde{\mathbf{A}}(\mathbf{k}, t) \\ \omega_k^2 &= c^2 \mathbf{k}^2\end{aligned} \hspace{\stretch{1}}(3.38)$

our general solution is of the form

$\begin{aligned}\tilde{\mathbf{A}}(\mathbf{k}, t) &= e^{i \omega_k t} \mathbf{a}_{+}(\mathbf{k}) +e^{-i \omega_k t} \mathbf{a}_{-}(\mathbf{k}) \\ \tilde{\mathbf{A}}^{*}(\mathbf{k}, t) &= e^{-i \omega_k t} \mathbf{a}_{+}^{*}(\mathbf{k}) +e^{i \omega_k t} \mathbf{a}_{-}^{*}(\mathbf{k})\end{aligned} \hspace{\stretch{1}}(3.40)$

$\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \int \frac{d^3 \mathbf{k}}{(2 pi)^3} e^{i \mathbf{k} \cdot \mathbf{x}} \frac{1}{{2}} \left( e^{i \omega_k t} (\mathbf{a}_{+}(\mathbf{k}) + \mathbf{a}_{-}^{*}(-\mathbf{k})) +e^{-i \omega_k t} (\mathbf{a}_{-}(\mathbf{k}) + \mathbf{a}_{+}^{*}(-\mathbf{k})) \right)\end{aligned} \hspace{\stretch{1}}(3.42)$

Define

$\begin{aligned}\boldsymbol{\beta}(\mathbf{k}) \equiv \frac{1}{{2}} (\mathbf{a}_{-}(\mathbf{k}) + \mathbf{a}_{+}^{*}(-\mathbf{k}) )\end{aligned} \hspace{\stretch{1}}(3.43)$

so that

$\begin{aligned}\boldsymbol{\beta}(-\mathbf{k}) = \frac{1}{{2}} (\mathbf{a}_{+}^{*}(\mathbf{k}) + \mathbf{a}_{-}(-\mathbf{k}))\end{aligned} \hspace{\stretch{1}}(3.44)$

Our solution now takes the form

$\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \int \frac{d^3\mathbf{k}}{(2 \pi)^3} \left( e^{i (\mathbf{k} \cdot \mathbf{x} + \omega_k t)} \boldsymbol{\beta}^{*}(-\mathbf{k})+e^{i (\mathbf{k} \cdot \mathbf{x} - \omega_k t)} \boldsymbol{\beta}(\mathbf{k})\right)\end{aligned} \hspace{\stretch{1}}(3.45)$

\paragraph{Claim:}

This is now manifestly real. To see this, consider the first term with $\mathbf{k} = -\mathbf{k}'$ , noting that $\int_{-\infty}^\infty dk = \int_{\infty}^\infty -dk' = \int_{-\infty}^\infty dk'$ with $dk = -dk'$

$\begin{aligned}\int \frac{d^3\mathbf{k}'}{(2 \pi)^3} e^{i (-\mathbf{k}' \cdot \mathbf{x} + \omega_k t)} \boldsymbol{\beta}^{*}(\mathbf{k}')\end{aligned} \hspace{\stretch{1}}(3.46)$

Dropping primes this is the conjugate of the second term.

\paragraph{Claim:}

We have $\mathbf{k} \cdot \boldsymbol{\beta}(\mathbf{k}) = 0$ .

Since we have $\mathbf{k} \cdot \tilde{\mathbf{A}}(\mathbf{k}, t) = 0$ , 3.40 implies that we have $\mathbf{k} \cdot \mathbf{a}_{\pm}(\mathbf{k}) = 0$ . With each of these vector integration constants being perpendicular to $\mathbf{k}$ at that point in momentum space, so must be the linear combination of these constants $\boldsymbol{\beta}(\mathbf{k})$ .

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

Posted in Math and Physics Learning. | Tagged: Coulomb gauge, fourier series, fourier transform, gauge transformation, Lorentz Gauge, momentum space, PHY450, wave equation | Leave a Comment »

PHY450H1S. Relativistic Electrodynamics Lecture 13 (Taught by Prof. Erich Poppitz). Variational principle for the field.

Posted by peeterjoot on February 22, 2011

[Click here for a PDF of this post with nicer formatting]

Reading.

Covering chapter 4 material from the text [1].

Covering lecture notes pp.103-113: variational principle for the electromagnetic field and the relevant boundary conditions (103-105); the second set of Maxwell’s equations from the variational principle (106-108); Maxwell’s equations in vacuum and the wave equation in the non-relativistic Coulomb gauge (109-111)

Review. Our action.

$\begin{aligned}S&= S_{\text{particles}} + S_{\text{interaction}} + S_{\text{EM field}}&= \sum_A \int_{x_A^i(\tau)} ds ( -m_A c )- \sum_A\frac{e_A}{c}\int dx_A^i A_i(x_A)- \frac{1}{{16 \pi c}} \int d^4 x F^{ij } F_{ij}.\end{aligned}$

Our dynamics variables are

$\begin{aligned}\left\{\begin{array}{l l}x_A^i(\tau) & \quad \mbox{$ A = 1, \cdots, N$} \\ A^i(x) & \quad \mbox{$A = 1, \cdots, N$}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.1)$

We saw that the interaction term could also be written in terms of a delta function current, with

$\begin{aligned}S_{\text{interaction}}= -\frac{1}{{c^2}} \int d^4x j^i(x) A_i(x),\end{aligned} \hspace{\stretch{1}}(2.2)$

and

$\begin{aligned}j^i(x) = \sum_A c e_A \int dx_A^i \delta^4( x - x_A(\tau)).\end{aligned} \hspace{\stretch{1}}(2.3)$

Variation with respect to $x_A^i(\tau)$ gave us

$\begin{aligned}m c \frac{d{{u^i_A}}}{ds} = \frac{e}{c} u_A^j F_{ij}.\end{aligned} \hspace{\stretch{1}}(2.4)$

Note that it’s easy to get the sign mixed up here. With our $(+,-,-,-)$ metric tensor, if the second index is the summation index, we have a positive sign.

Only the $S_{\text{particles}}$ and $S_{\text{interaction}}$ depend on $x_A^i(\tau)$ .

The field action variation.

\paragraph{Today:} We’ll find the EOM for $A^i(x)$ . The dynamical degrees of freedom are $A^i(\mathbf{x},t)$

$\begin{aligned}S[A^i(\mathbf{x}, t)] = -\frac{1}{{16 \pi c}} \int d^4x F_{ij}F^{ij} - \frac{1}{{c^2}} \int d^4 x A^i j_i.\end{aligned} \hspace{\stretch{1}}(3.5)$

Here $j^i$ are treated as “sources”.

We demand that

$\begin{aligned}\delta S = S[ A^i(\mathbf{x}, t) + \delta A^i(\mathbf{x}, t)] - S[ A^i(\mathbf{x}, t) ] = 0 + O(\delta A)^2.\end{aligned} \hspace{\stretch{1}}(3.6)$

We need to impose two conditions.
\begin{itemize}
\item At spatial $\infty$ , i.e. at ${\left\lvert{\mathbf{x}}\right\rvert} \rightarrow \infty, \forall t$ , we’ll impose the condition

$\begin{aligned}{\left.{{A^i(\mathbf{x}, t)}}\right\vert}_{{{\left\lvert{\mathbf{x}}\right\rvert} \rightarrow \infty}} \rightarrow 0.\end{aligned} \hspace{\stretch{1}}(3.7)$

This is sensible, because fields are created by charges, and charges are assumed to be localized in a bounded region. The field outside charges will $\rightarrow 0$ at ${\left\lvert{\mathbf{x}}\right\rvert} \rightarrow \infty$ . Later we will treat the integration range as finite, and bounded, then later allow the boundary to go to infinity.

\item at $t = -T$ and $t = T$ we’ll imagine that the values of $A^i(\mathbf{x}, \pm T)$ are fixed.

This is analogous to $x(t_i) = x_1$ and $x(t_f) = x_2$ in particle mechanics.

Since $A^i(\mathbf{x}, \pm T)$ is given, and equivalent to the initial and final field configurations, our extremes at the boundary is zero

$\begin{aligned}\delta A^i(\mathbf{x}, \pm T) = 0.\end{aligned} \hspace{\stretch{1}}(3.8)$

\end{itemize}

PICTURE: a cylinder in spacetime, with an attempt to depict the boundary.

Computing the variation.

$\begin{aligned}\delta S[A^i(\mathbf{x}, t)]= -\frac{1}{{16 \pi c}} \int d^4 x \delta (F_{ij}F^{ij}) - \frac{1}{{c^2}} \int d^4 x \delta(A^i) j_i.\end{aligned} \hspace{\stretch{1}}(4.9)$

Looking first at the variation of just the $F^2$ bit we have

$\begin{aligned}\delta (F_{ij}F^{ij})&=\delta(F_{ij}) F^{ij} + F_{ij} \delta(F^{ij}) \\ &=2 \delta(F^{ij}) F_{ij} \\ &=2 \delta(\partial^i A^j - \partial^j A^i) F_{ij} \\ &=2 \delta(\partial^i A^j) F_{ij} - 2 \delta(\partial^j A^i) F_{ij} \\ &=2 \delta(\partial^i A^j) F_{ij} - 2 \delta(\partial^i A^j) F_{ji} \\ &=4 \delta(\partial^i A^j) F_{ij} \\ &=4 F_{ij} \partial^i \delta(A^j).\end{aligned}$

Our variation is now reduced to

$\begin{aligned}\delta S[A^i(\mathbf{x}, t)]&= -\frac{1}{{4 \pi c}} \int d^4 x F_{ij} \partial^i \delta(A^j) - \frac{1}{{c^2}} \int d^4 x j^i \delta(A_i) \\ &= -\frac{1}{{4 \pi c}} \int d^4 x F^{ij} \frac{\partial {}}{\partial {x^i}} \delta(A_j) - \frac{1}{{c^2}} \int d^4 x j^i \delta(A_i).\end{aligned}$

We can integrate this first term by parts

$\begin{aligned}\int d^4 x F^{ij} \frac{\partial {}}{\partial {x^i}} \delta(A_j)&=\int d^4 x \frac{\partial {}}{\partial {x^i}} \left( F^{ij} \delta(A_j) \right)-\int d^4 x \left( \frac{\partial {}}{\partial {x^i}} F^{ij} \right) \delta(A_j) \end{aligned}$

The first term is a four dimensional divergence, with the contraction of the four gradient $\partial_i$ with a four vector $B^i = F^{ij} \delta(A_j)$ .

Prof. Poppitz chose $dx^0 d^3 \mathbf{x}$ split of $d^4 x$ to illustrate that this can be viewed as regular old spatial three vector divergences. It is probably more rigorous to mandate that the four volume element is oriented $d^4 x = (1/4!)\epsilon_{ijkl} dx^i dx^j dx^k dx^l$ , and then utilize the 4D version of the divergence theorem (or its Stokes Theorem equivalent). The completely antisymmetric tensor should do most of the work required to express the oriented boundary volume.

Because we have specified that $A^i$ is zero on the boundary, so is $F^{ij}$ , so these boundary terms are killed off. We are left with

$\begin{aligned}\delta S[A^i(\mathbf{x}, t)]&= -\frac{1}{{4 \pi c}} \int d^4 x \delta (A_j) \partial_i F^{ij} - \frac{1}{{c^2}} \int d^4 x j^i \delta(A_i) \\ &=\int d^4 x \delta A_j(x)\left(-\frac{1}{{4 \pi c}} \partial_i F^{ij}(x) - \frac{1}{{c^2}} j^i\right) \\ &= 0.\end{aligned}$

This gives us

$\begin{aligned}\boxed{\partial_i F^{ij} = \frac{4 \pi}{c} j^j}\end{aligned} \hspace{\stretch{1}}(4.10)$

Unpacking these.

Recall that the Bianchi identity

$\begin{aligned}\epsilon^{ijkl} \partial_j F_{kl} = 0,\end{aligned} \hspace{\stretch{1}}(5.11)$

gave us

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{E} &= -\frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(5.12)$

How about the EOM that we have found by varying the action? One of those equations is

$\begin{aligned}\partial_\alpha F^{\alpha 0} = \frac{4 \pi}{c} j^0 = 4 \pi \rho,\end{aligned} \hspace{\stretch{1}}(5.14)$

since $j^0 = c \rho$ .

Because

$\begin{aligned}F^{\alpha 0} = (\mathbf{E})^\alpha,\end{aligned} \hspace{\stretch{1}}(5.15)$

we have

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 4 \pi \rho.\end{aligned} \hspace{\stretch{1}}(5.16)$

The messier one to deal with is

$\begin{aligned}\partial_i F^{i\alpha} = \frac{4 \pi}{c} j^\alpha.\end{aligned} \hspace{\stretch{1}}(5.17)$

Splitting out the spatial and time indexes for the four gradient we have

$\begin{aligned}\partial_i F^{i\alpha}&= \partial_\beta F^{\beta \alpha} + \partial_0 F^{0 \alpha} \\ &= \partial_\beta F^{\beta \alpha} - \frac{1}{{c}} \frac{\partial {(\mathbf{E})^\alpha}}{\partial {t}} \\ \end{aligned}$

The spatial index tensor element is

$\begin{aligned}F^{\beta \alpha} &= \partial^\beta A^\alpha - \partial^\alpha A^\beta \\ &= - \frac{\partial {A^\alpha}}{\partial {x^\beta}} + \frac{\partial {A^\beta}}{\partial {x^\alpha}} \\ &= \epsilon^{\alpha\beta\gamma} B^\gamma,\end{aligned}$

so the sum becomes

$\begin{aligned}\partial_i F^{i\alpha}&= \partial_\beta ( \epsilon^{\alpha\beta\gamma} B^\gamma) - \frac{1}{{c}} \frac{\partial {(\mathbf{E})^\alpha}}{\partial {t}} \\ &= \epsilon^{\beta\gamma\alpha} \partial_\beta B^\gamma - \frac{1}{{c}} \frac{\partial {(\mathbf{E})^\alpha}}{\partial {t}} \\ &= (\boldsymbol{\nabla} \times \mathbf{B})^\alpha - \frac{1}{{c}} \frac{\partial {(\mathbf{E})^\alpha}}{\partial {t}}.\end{aligned}$

This gives us

$\begin{aligned}\frac{4 \pi}{c} j^\alpha= (\boldsymbol{\nabla} \times \mathbf{B})^\alpha - \frac{1}{{c}} \frac{\partial {(\mathbf{E})^\alpha}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(5.18)$

or in vector form

$\begin{aligned}\boldsymbol{\nabla} \times \mathbf{B} - \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}} = \frac{4 \pi}{c} \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(5.19)$

Summarizing what we know so far, we have

$\begin{aligned}\boxed{\begin{aligned}\partial_i F^{ij} &= \frac{4 \pi}{c} j^j \\ \epsilon^{ijkl} \partial_j F_{kl} &= 0\end{aligned}}\end{aligned} \hspace{\stretch{1}}(5.20)$

or in vector form

$\begin{aligned}\boxed{\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} &= 4 \pi \rho \\ \boldsymbol{\nabla} \times \mathbf{B} -\frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}} &= \frac{4 \pi}{c} \mathbf{j} \\ \boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{E} +\frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} &= 0\end{aligned}}\end{aligned} \hspace{\stretch{1}}(5.21)$

Speed of light

\paragraph{Claim}: “ $c$ ” is the speed of EM waves in vacuum.

Study equations in vacuum (no sources, so $j^i = 0$ ) for $A^i = (\phi, \mathbf{A})$ .

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{B} &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(6.22)$

where

$\begin{aligned}\mathbf{E} &= - \boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A}\end{aligned} \hspace{\stretch{1}}(6.24)$

In terms of potentials

$\begin{aligned}0 &= \boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{A}) \\ &= \boldsymbol{\nabla} \times \mathbf{B} \\ &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}} \\ &= \frac{1}{{c}} \frac{\partial {}}{\partial {t}} \left( - \boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \right) \\ &= -\frac{1}{{c}} \frac{\partial {}}{\partial {t}} \boldsymbol{\nabla} \phi - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}}{\partial t^2} \end{aligned}$

Since we also have

$\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{A}) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) - \boldsymbol{\nabla}^2 \mathbf{A},\end{aligned} \hspace{\stretch{1}}(6.26)$

some rearrangement gives

$\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) = \boldsymbol{\nabla}^2 \mathbf{A} -\frac{1}{{c}} \frac{\partial {}}{\partial {t}} \boldsymbol{\nabla} \phi - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}}{\partial t^2}.\end{aligned} \hspace{\stretch{1}}(6.27)$

The remaining equation $\boldsymbol{\nabla} \cdot \mathbf{E} = 0$ , in terms of potentials is

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = - \boldsymbol{\nabla}^2 \phi - \frac{1}{{c}} \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}}}{\partial {t}} \end{aligned} \hspace{\stretch{1}}(6.28)$

We can make a gauge transformation that completely eliminates 6.28, and reduces 6.27 to a wave equation.

$\begin{aligned}(\phi, \mathbf{A}) \rightarrow (\phi', \mathbf{A}')\end{aligned} \hspace{\stretch{1}}(6.29)$

with

$\begin{aligned}\phi &= \phi' - \frac{1}{{c}} \frac{\partial {\chi}}{\partial {t}} \\ \mathbf{A} &= \mathbf{A}' + \boldsymbol{\nabla} \chi\end{aligned} \hspace{\stretch{1}}(6.30)$

Can choose $\chi(\mathbf{x}, t)$ to make $\phi' = 0$ ( $\forall \phi \exists \chi, \phi' = 0$ )

$\begin{aligned}\frac{1}{{c}} \frac{\partial {}}{\partial {t}} \chi(\mathbf{x}, t) = \phi(\mathbf{x}, t)\end{aligned} \hspace{\stretch{1}}(6.32)$

$\begin{aligned}\chi(\mathbf{x}, t) = c \int_{-\infty}^t dt' \phi(\mathbf{x}, t')\end{aligned} \hspace{\stretch{1}}(6.33)$

Can also find a transformation that also allows $\boldsymbol{\nabla} \cdot \mathbf{A} = 0$

\paragraph{Q:} What would that second transformation be explicitly?
\paragraph{A:} To be revisited next lecture, when this is covered in full detail.

This is the Coulomb gauge

$\begin{aligned}\phi &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{A} &= 0\end{aligned} \hspace{\stretch{1}}(6.34)$

From 6.27, we then have

$\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}'}{\partial t^2} -\boldsymbol{\nabla}^2 \mathbf{A}' = 0\end{aligned} \hspace{\stretch{1}}(6.36)$

which is the wave equation for the propagation of the vector potential $\mathbf{A}'(\mathbf{x}, t)$ through space at velocity $c$ , confirming that $c$ is the speed of electromagnetic propagation (the speed of light).

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

Posted in Math and Physics Learning. | Tagged: Coulomb gauge, field Lagrangian, gauge transformation, lagrangian density, Maxwell action, Maxwell field equation, Maxwell field Lagrangian, PHY450, speed of light, variational calculus, wave equation | Leave a Comment »

PHY450H1S. Relativistic Electrodynamics Lecture 14 (Taught by Simon Freedman). Wave equation in Coulomb and Lorentz gauges.

Posted by peeterjoot on February 17, 2011

[Click here for a PDF of this post with nicer formatting]

Reading.

Covering chapter 4 material from the text [1].

Covering lecture notes pp.103-114: the wave equation in the relativistic Lorentz gauge (114-114) [Tuesday, Feb. 15; Wednesday, Feb.16]…

Covering lecture notes pp. 114-127: reminder on wave equations (114); reminder on Fourier series and integral (115-117); Fourier expansion of the EM potential in Coulomb gauge and equation of motion for the spatial Fourier components (118-119); the general solution of Maxwell’s equations in vacuum (120-121) [Tuesday, Mar. 1]; properties of monochromatic plane EM waves (122-124); energy and energy flux of the EM field and energy conservation from the equations of motion (125-127) [Wednesday, Mar. 2]

Trying to understand “c”

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{B} &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(2.1)$

Maxwell’s equations in a vacuum were

$\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) &= \boldsymbol{\nabla}^2 \mathbf{A} -\frac{1}{{c}} \frac{\partial {}}{\partial {t}} \boldsymbol{\nabla} \phi - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}}{\partial t^2} \\ \boldsymbol{\nabla} \cdot \mathbf{E} &= - \boldsymbol{\nabla}^2 \phi - \frac{1}{{c}} \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}}}{\partial {t}} \end{aligned} \hspace{\stretch{1}}(2.3)$

There’s a redundancy here since we can change $\phi$ and $\mathbf{A}$ without changing the EOM

$\begin{aligned}(\phi, \mathbf{A}) \rightarrow (\phi', \mathbf{A}')\end{aligned} \hspace{\stretch{1}}(2.5)$

with

$\begin{aligned}\phi &= \phi' + \frac{1}{{c}} \frac{\partial {\chi}}{\partial {t}} \\ \mathbf{A} &= \mathbf{A}' - \boldsymbol{\nabla} \chi\end{aligned} \hspace{\stretch{1}}(2.6)$

$\begin{aligned}\chi(\mathbf{x}, t) = c \int dt \phi(\mathbf{x}, t)\end{aligned} \hspace{\stretch{1}}(2.8)$

which gives

$\begin{aligned}\phi' = 0\end{aligned} \hspace{\stretch{1}}(2.9)$

$\begin{aligned}(\phi, \mathbf{A}) \sim (\phi = 0, \mathbf{A}')\end{aligned} \hspace{\stretch{1}}(2.10)$

Maxwell’s equations are now

$\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}') &= \boldsymbol{\nabla}^2 \mathbf{A}' - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}'}{\partial t^2} \\ \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}'}}{\partial {t}} &= 0\end{aligned}$

Can we make $\boldsymbol{\nabla} \cdot \mathbf{A}'' = 0$ , while $\phi'' = 0$ .

$\begin{aligned}\underbrace{\phi}_{=0} &= \underbrace{\phi'}_{=0} + \frac{1}{{c}} \frac{\partial {\chi'}}{\partial {t}} \\ \end{aligned} \hspace{\stretch{1}}(2.11)$

We need

$\begin{aligned}\frac{\partial {\chi'}}{\partial {t}} = 0\end{aligned} \hspace{\stretch{1}}(2.13)$

How about $\mathbf{A}'$

$\begin{aligned}\mathbf{A}' = \mathbf{A}'' - \boldsymbol{\nabla} \chi'\end{aligned} \hspace{\stretch{1}}(2.14)$

We want the divergence of $\mathbf{A}'$ to be zero, which means

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A}' = \underbrace{\boldsymbol{\nabla} \cdot \mathbf{A}''}_{=0} - \boldsymbol{\nabla}^2 \chi'\end{aligned} \hspace{\stretch{1}}(2.15)$

So we want

$\begin{aligned}\boldsymbol{\nabla}^2 \chi' = \boldsymbol{\nabla} \cdot \mathbf{A}'\end{aligned} \hspace{\stretch{1}}(2.16)$

Can we solve this?

Recall that in electrostatics we have

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(2.17)$

and

$\begin{aligned}\mathbf{E} = -\boldsymbol{\nabla} \phi\end{aligned} \hspace{\stretch{1}}(2.18)$

which meant that we had

$\begin{aligned}\boldsymbol{\nabla}^2 \phi = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(2.19)$

This has the identical form (with $\phi \sim \chi$ , and $4 \pi \rho \sim \boldsymbol{\nabla} \cdot \mathbf{A}'$ ).

While we aren’t trying to actually solve this (just show that it can be solved). One way to look at this problem is that it is just a Laplace equation, and we could utilize a Green’s function solution if desired.

On the Green’s function.

Recall that the Green’s function for the Laplacian was

$\begin{aligned}G(\mathbf{x}, \mathbf{x}') = \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}}\end{aligned} \hspace{\stretch{1}}(2.20)$

with the property

$\begin{aligned}\boldsymbol{\nabla}^2 G(\mathbf{x}, \mathbf{x}') = \delta(\mathbf{x} - \mathbf{x}')\end{aligned} \hspace{\stretch{1}}(2.21)$

Our LDE to solve by Green’s method is

$\begin{aligned}\boldsymbol{\nabla}^2 \phi = 4 \pi \rho,\end{aligned} \hspace{\stretch{1}}(2.22)$

We let this equation (after switching to primed coordinates) operate on the Green’s function

$\begin{aligned}\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}') =\int d^3 \mathbf{x}' 4 \pi \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}').\end{aligned} \hspace{\stretch{1}}(2.23)$

Assuming that the left action of the Green’s function on the test function $\phi(\mathbf{x}')$ is the same as the right action (i.e. $\phi(\mathbf{x}')$ and $G(\mathbf{x}, \mathbf{x}')$ commute), we have for the LHS

$\begin{aligned}\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}') &=\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 G(\mathbf{x}, \mathbf{x}') \phi(\mathbf{x}') \\ &=\int d^3 \mathbf{x}' \delta(\mathbf{x} - \mathbf{x}') \phi(\mathbf{x}') \\ &=\phi(\mathbf{x}).\end{aligned}$

Substitution of $G(\mathbf{x}, \mathbf{x}') = 1/{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}$ on the RHS then gives us the general solution

$\begin{aligned}\phi(\mathbf{x}) = 4 \pi \int d^3 \mathbf{x}' \frac{\rho(\mathbf{x}') }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(2.24)$

Back to Maxwell’s equations in vacuum.

What are the Maxwell’s vacuum equations now?

With the second gauge substitution we have

$\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}'') &= \boldsymbol{\nabla}^2 \mathbf{A}'' - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}''}{\partial t^2} \\ \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}''}}{\partial {t}} &= 0\end{aligned}$

but we can utilize

to reduce Maxwell’s equations (after dropping primes) to just

$\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}''}{\partial t^2} - \Delta \mathbf{A} = 0\end{aligned} \hspace{\stretch{1}}(2.26)$

where

$\begin{aligned}\Delta = \boldsymbol{\nabla}^2 = \boldsymbol{\nabla} \cdot \boldsymbol{\nabla} = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial y^2}\end{aligned} \hspace{\stretch{1}}(2.27)$

Note that for this to be correct we have to also explicitly include the gauge condition used. This particular gauge is called the \underline{Coulomb gauge}.

$\begin{aligned}\phi &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{A}'' &= 0 \end{aligned} \hspace{\stretch{1}}(2.28)$

Claim: EM waves propagate with speed $c$ and are transverse.

\paragraph{Note:} Is the Coulomb gauge Lorentz invariant?
\paragraph{No.} We can boost which will introduce a non-zero $\phi$ .

The gauge that is Lorentz Invariant is the “Lorentz gauge”. This one uses

$\begin{aligned}\partial_i A^i = 0\end{aligned} \hspace{\stretch{1}}(3.30)$

Recall that Maxwell’s equations are

$\begin{aligned}\partial_i F^{ij} = j^j = 0\end{aligned} \hspace{\stretch{1}}(3.31)$

where

$\begin{aligned}\partial_i &= \frac{\partial {}}{\partial {x^i}} \\ \partial^i &= \frac{\partial {}}{\partial {x_i}}\end{aligned} \hspace{\stretch{1}}(3.32)$

Writing out the equations in terms of potentials we have

$\begin{aligned}0 &= \partial_i (\partial^i A^j - \partial^j A^i) \\ &= \partial_i \partial^i A^j - \partial_i \partial^j A^i \\ &= \partial_i \partial^i A^j - \partial^j \partial_i A^i \\ \end{aligned}$

So, if we pick the gauge condition $\partial_i A^i = 0$ , we are left with just

$\begin{aligned}0 = \partial_i \partial^i A^j\end{aligned} \hspace{\stretch{1}}(3.34)$

Can we choose ${A'}^i$ such that $\partial_i A^i = 0$ ?

Our gauge condition is

$\begin{aligned}A^i = {A'}^i + \partial^i \chi\end{aligned} \hspace{\stretch{1}}(3.35)$

Hit it with a derivative for

$\begin{aligned}\partial_i A^i = \partial_i {A'}^i + \partial_i \partial^i \chi\end{aligned} \hspace{\stretch{1}}(3.36)$

If we want $\partial_i A^i = 0$ , then we have

$\begin{aligned}-\partial_i {A'}^i = \partial_i \partial^i \chi = \left( \frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} - \Delta \right) \chi\end{aligned} \hspace{\stretch{1}}(3.37)$

This is the physicist proof. Yes, it can be solved. To really solve this, we’d want to use Green’s functions. I seem to recall the Green’s function is a retarded time version of the Laplacian Green’s function, and we can figure that exact form out by switching to a Fourier frequency domain representation.

Anyways. Returning to Maxwell’s equations we have

$\begin{aligned}0 &= \partial_i \partial^i A^j \\ 0 &= \partial_i A^i ,\end{aligned} \hspace{\stretch{1}}(3.38)$

where the first is Maxwell’s equation, and the second is our gauge condition.

Observe that the gauge condition is now a Lorentz scalar.

$\begin{aligned}\partial^i A_i \rightarrow \partial^j {O_j}^i {O_i}^k A_k\end{aligned} \hspace{\stretch{1}}(3.40)$

But the Lorentz transform matrices multiply out to identity, in the same way that they do for the transformation of a plain old four vector dot product $x^i y_i$ .

What happens with a Massive vector field?

$\begin{aligned}S = \int d^4 x \left( \frac{1}{{4}} F^{ij} F_{ij} + \frac{m^2}{2} A^i A_i \right)\end{aligned} \hspace{\stretch{1}}(4.41)$

An aside on units

“Note that this action is expressed in dimensions where $\hbar = c = 1$ , making the action is unit-less (energy and time are inverse units of each other). The $d^4x$ has units of $m^{-4}$ (since $[x] = \hbar/mc$ ), so $F$ has units of $m^2$ , and then $A$ has units of mass. Therefore $d^4x A A$ has units of $m^{-2}$ and therefore you need something that has units of $m^2$ to make the action unit-less. When you don’t take $c=1$ , then you’ve got to worry about those factors, but I think you’ll see it works out fine.”

For what it’s worth, I can adjust the units of this action to those that we’ve used in class with,

$\begin{aligned}S = \int d^4 x \left( -\frac{1}{{16 \pi c}} F^{ij} F_{ij} - \frac{m^2 c^2}{8 \hbar^2} A^i A_i \right)\end{aligned} \hspace{\stretch{1}}(4.42)$

Back to the problem.

The variation of the field invariant is

$\begin{aligned}\delta (F_{ij} F^{ij})&=2 (\delta F_{ij}) F^{ij}) \\ &=2 (\delta(\partial_i A_j -\partial_j A_i)) F^{ij}) \\ &=2 (\partial_i \delta(A_j) -\partial_j \delta(A_i)) F^{ij}) \\ &=4 F^{ij} \partial_i \delta(A_j) \\ &=4 \partial_i (F^{ij} \delta(A_j)) - 4 (\partial_i F^{ij}) \delta(A_j).\end{aligned}$

Variation of the $A^2$ term gives us

$\begin{aligned}\delta (A^j A_j) = 2 A^j \delta(A_j),\end{aligned} \hspace{\stretch{1}}(4.43)$

so we have

$\begin{aligned}0 &= \delta S \\ &= \int d^4 x \delta(A_j) \left( -\partial_i F^{ij} + m^2 A^j \right)+ \int d^4 x \partial_i (F^{ij} \delta(A_j))\end{aligned}$

The last integral vanishes on the boundary with the assumption that $\delta(A_j) = 0$ on that boundary.

Since this must be true for all variations, this leaves us with

$\begin{aligned}\partial_i F^{ij} = m^2 A^j\end{aligned} \hspace{\stretch{1}}(4.44)$

The RHS can be expanded into wave equation and divergence parts

$\begin{aligned}\partial_i F^{ij}&=\partial_i (\partial^i A^j - \partial^j A^i) \\ &=(\partial_i \partial^i) A^j - \partial^j (\partial_i A^i) \\ \end{aligned}$

With $\square$ for the wave equation operator

$\begin{aligned}\square = \partial_i \partial^i = \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta,\end{aligned} \hspace{\stretch{1}}(4.45)$

we can manipulate the EOM to pull out an $A_i$ factor

$\begin{aligned}0 &= \left( \square -m^2 \right) A^j - \partial^j (\partial_i A^i) \\ &= \left( \square -m^2 \right) g^{ij} A_i - \partial^j (\partial^i A_i) \\ &= \left( \left( \square -m^2 \right) g^{ij} - \partial^j \partial^i \right) A_i.\end{aligned}$

If we hit this with a derivative we get

$\begin{aligned}0 &= \partial_j \left( \left( \square -m^2 \right) g^{ij} - \partial^j \partial^i \right) A_i \\ &= \left( \left( \square -m^2 \right) \partial^i - \partial_j \partial^j \partial^i \right) A_i \\ &= \left( \left( \square -m^2 \right) \partial^i - \square \partial^i \right) A_i \\ &= \left( \square -m^2 - \square \right) \partial^i A_i \\ &= -m^2 \partial^i A_i \\ \end{aligned}$

Since $m$ is presumed to be non-zero here, this means that the Lorentz gauge is already chosen for us by the equations of motion.

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

Posted in Math and Physics Learning. | Tagged: Coulomb gauge, gauge transformation, Lorentz Gauge, maxwells equations, PHY450, speed of light, wave equation | Leave a Comment »

Electrodynamic field energy for vacuum (reworked)

Posted by peeterjoot on December 21, 2009

[Click here for a PDF of this post with nicer formatting]

Previous version.

This is a reworked version of a previous post ([also in PDF]

Reducing the products in the Dirac basis makes life more complicated then it needs to be (became obvious when attempting to derive an expression for the Poynting integral).

Motivation.

From Energy and momentum for Complex electric and magnetic field phasors [PDF] how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism is now understood. To recap, for the field $F = \mathbf{E} + I c \mathbf{B}$ , where $\mathbf{E}$ and $\mathbf{B}$ may be complex vectors we have for Maxwell’s equation

$\begin{aligned}\nabla F = J/\epsilon_0 c.\end{aligned} \quad\quad\quad(1)$

This is a doubly complex representation, with the four vector pseudoscalar $I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$ acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution $F$ of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary $i$ , as in $\mathbf{E} = \text{Real}(\mathbf{E}) + i \text{Imag}(\mathbf{E})$ , is a commuting imaginary, commuting with all the multivector elements in the algebra.

The real valued, four vector, energy momentum tensor $T(a)$ was found to be

$\begin{aligned}T(a) = \frac{\epsilon_0}{4} \Bigl( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} \Bigr) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \quad\quad\quad(2)$

To supply some context that gives meaning to this tensor the associated conservation relationship was found to be

$\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(3)$

and in particular for $a = \gamma^0$ , this four vector divergence takes the form

$\begin{aligned}\frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ \boldsymbol{\nabla} \cdot \frac{1}{{\mu_0}} \text{Real} (\mathbf{E} \times {\mathbf{B}}^{*} )+ \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) = 0,\end{aligned} \quad\quad\quad(4)$

relating the energy term $T^{00} = T(\gamma^0) \cdot \gamma^0$ and the Poynting spatial vector $T(\gamma^0) \wedge \gamma^0$ with the current density and electric field product that constitutes the energy portion of the Lorentz force density.

Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [2]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.

Setup

Let’s assume a Fourier representation for the four vector potential $A$ for the field $F = \nabla \wedge A$ . That is

$\begin{aligned}A = \sum_{\mathbf{k}} A_\mathbf{k}(t) e^{i \mathbf{k} \cdot \mathbf{x}},\end{aligned} \quad\quad\quad(5)$

where summation is over all angular wave number triplets $\mathbf{k} = 2 \pi (k_1/\lambda_1, k_2/\lambda_2, k_3/\lambda_3)$ . The Fourier coefficients $A_\mathbf{k} = {A_\mathbf{k}}^\mu \gamma_\mu$ are allowed to be complex valued, as is the resulting four vector $A$ , and the associated bivector field $F$ .

Fourier inversion, with $V = \lambda_1 \lambda_2 \lambda_3$ , follows from

$\begin{aligned}\delta_{\mathbf{k}', \mathbf{k}} =\frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} e^{ i \mathbf{k}' \cdot \mathbf{x}} e^{-i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(6)$

but only this orthogonality relationship and not the Fourier coefficients themselves

$\begin{aligned}A_\mathbf{k} = \frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} A(\mathbf{x}, t) e^{- i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(7)$

will be of interest here. Evaluating the curl for this potential yields

$\begin{aligned}F = \nabla \wedge A= \sum_{\mathbf{k}} \left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^m \wedge A_\mathbf{k} \frac{2 \pi i k_m}{\lambda_m} \right) e^{i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(8)$

Since the four vector potential has been expressed using an explicit split into time and space components it will be natural to re express the bivector field in terms of scalar and (spatial) vector potentials, with the Fourier coefficients. Writing $\sigma_m = \gamma_m \gamma_0$ for the spatial basis vectors, ${A_\mathbf{k}}^0 = \phi_\mathbf{k}$ , and $\mathbf{A} = A^k \sigma_k$ , this is

$\begin{aligned}A_\mathbf{k} = (\phi_\mathbf{k} + \mathbf{A}_\mathbf{k}) \gamma_0.\end{aligned} \quad\quad\quad(9)$

The Faraday bivector field $F$ is then

$\begin{aligned}F = \sum_\mathbf{k} \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) e^{i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(10)$

This is now enough to express the energy momentum tensor $T(\gamma^\mu)$

$\begin{aligned}T(\gamma^\mu) &= -\frac{\epsilon_0}{2} \sum_{\mathbf{k},\mathbf{k}'}\text{Real} \left(\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}'})}}^{*} + i \mathbf{k}' {{\phi_{\mathbf{k}'}}}^{*} - i \mathbf{k}' \wedge {{\mathbf{A}_{\mathbf{k}'}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x}}\right).\end{aligned} \quad\quad\quad(11)$

It will be more convenient to work with a scalar plus bivector (spatial vector) form of this tensor, and right multiplication by $\gamma_0$ produces such a split

$\begin{aligned}T(\gamma^\mu) \gamma_0 = \left\langle{{T(\gamma^\mu) \gamma_0}}\right\rangle + \sigma_a \left\langle{{ \sigma_a T(\gamma^\mu) \gamma_0 }}\right\rangle\end{aligned} \quad\quad\quad(12)$

The primary object of this treatment will be consideration of the $\mu = 0$ components of the tensor, which provide a split into energy density $T(\gamma^0) \cdot \gamma_0$ , and Poynting vector (momentum density) $T(\gamma^0) \wedge \gamma_0$ .

Our first step is to integrate (12) over the volume $V$ . This integration and the orthogonality relationship (6), removes the exponentials, leaving

$\begin{aligned}\int T(\gamma^\mu) \cdot \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) \gamma_0 }}\right\rangle \\ \int T(\gamma^\mu) \wedge \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) \gamma_0}}\right\rangle \end{aligned} \quad\quad\quad(13)$

Because $\gamma_0$ commutes with the spatial bivectors, and anticommutes with the spatial vectors, the remainder of the Dirac basis vectors in these expressions can be eliminated

$\begin{aligned}\int T(\gamma^0) \cdot \gamma_0&= -\frac{\epsilon_0 V }{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(15)$

$\begin{aligned}\int T(\gamma^0) \wedge \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(16)$

$\begin{aligned}\int T(\gamma^m) \cdot \gamma_0&= \frac{\epsilon_0 V }{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \sigma_m\left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(17)$

$\begin{aligned}\int T(\gamma^m) \wedge \gamma_0&= \frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \sigma_m\left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle.\end{aligned} \quad\quad\quad(18)$

Expanding the energy momentum tensor components.

Energy

In (15) only the bivector-bivector and vector-vector products produce any scalar grades. Except for the bivector product this can be done by inspection. For that part we utilize the identity

$\begin{aligned}\left\langle{{ (\mathbf{k} \wedge \mathbf{a}) (\mathbf{k} \wedge \mathbf{b}) }}\right\rangle= (\mathbf{a} \cdot \mathbf{k}) (\mathbf{b} \cdot \mathbf{k}) - \mathbf{k}^2 (\mathbf{a} \cdot \mathbf{b}).\end{aligned} \quad\quad\quad(19)$

This leaves for the energy $H = \int T(\gamma^0) \cdot \gamma_0$ in the volume

$\begin{aligned}H = \frac{\epsilon_0 V}{2} \sum_\mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2 +\mathbf{k}^2 \left( {\left\lvert{\phi_\mathbf{k}}\right\rvert}^2 + {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right) - {\left\lvert{\mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2+ \frac{2}{c} \text{Real} \left( i {{\phi_\mathbf{k}}}^{*} \cdot \dot{\mathbf{A}}_\mathbf{k} \right)\right)\end{aligned} \quad\quad\quad(20)$

We are left with a completely real expression, and one without any explicit Geometric Algebra. This does not look like the Harmonic oscillator Hamiltonian that was expected. A gauge transformation to eliminate $\phi_\mathbf{k}$ and an observation about when $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum.

Momentum

Now move on to (16). For the factors other than $\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form

$\begin{aligned}\sigma_a \left\langle{{ \sigma_a \mathbf{a} (\mathbf{k} \wedge \mathbf{c}) }}\right\rangle&=\sigma_a (\mathbf{c} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{k}) - \sigma_a (\mathbf{k} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{c}) \\ &=\mathbf{c} (\mathbf{a} \cdot \mathbf{k}) - \mathbf{k} (\mathbf{a} \cdot \mathbf{c}),\end{aligned}$

and the other

$\begin{aligned}\sigma_a \left\langle{{ \sigma_a (\mathbf{k} \wedge \mathbf{c}) \mathbf{a} }}\right\rangle&=\sigma_a (\mathbf{k} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{c}) - \sigma_a (\mathbf{c} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{k}) \\ &=\mathbf{k} (\mathbf{a} \cdot \mathbf{c}) - \mathbf{c} (\mathbf{a} \cdot \mathbf{k}).\end{aligned}$

The momentum $\mathbf{P} = \int T(\gamma^0) \wedge \gamma_0$ in this volume follows by computation of

$\begin{aligned}&\sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \\ &= i \mathbf{A}_\mathbf{k} \left( \left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} \right) \cdot \mathbf{k} \right) - i \mathbf{k} \left( \left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} \right) \cdot \mathbf{A}_\mathbf{k} \right) \\ &- i \mathbf{k} \left( \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} \right) \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right) + i {{\mathbf{A}_{\mathbf{k}}}}^{*} \left( \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} \right) \cdot \mathbf{k} \right)\end{aligned}$

All the products are paired in nice conjugates, taking real parts, and premultiplication with $-\epsilon_0 V/2$ gives the desired result. Observe that two of these terms cancel, and another two have no real part. Those last are

$\begin{aligned}-\frac{\epsilon_0 V \mathbf{k}}{2 c} \text{Real} \left( i {{(\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{A}_\mathbf{k}+\dot{\mathbf{A}}_\mathbf{k} \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right)&=-\frac{\epsilon_0 V \mathbf{k}}{2 c} \text{Real} \left( i \frac{d}{dt} \mathbf{A}_\mathbf{k} \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right)\end{aligned}$

Taking the real part of this pure imaginary $i {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2$ is zero, leaving just

$\begin{aligned}\mathbf{P} &= \epsilon_0 V \sum_{\mathbf{k}}\text{Real} \left(i \mathbf{A}_\mathbf{k} \left( \frac{1}{{c}} {{\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{k} \right)+ \mathbf{k}^2 \phi_\mathbf{k} {{ \mathbf{A}_\mathbf{k} }}^{*}- \mathbf{k} {{\phi_\mathbf{k}}}^{*} (\mathbf{k} \cdot \mathbf{A}_\mathbf{k})\right)\end{aligned} \quad\quad\quad(21)$

I am not sure why exactly, but I actually expected a term with ${\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2$ , quadratic in the vector potential. Is there a mistake above?

Gauge transformation to simplify the Hamiltonian.

In (20) something that looked like the Harmonic oscillator was expected. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.

If we are to change our four vector potential $A \rightarrow A + \nabla \psi$ , then Maxwell’s equation takes the form

$\begin{aligned}J/\epsilon_0 c = \nabla (\nabla \wedge (A + \nabla \psi) = \nabla (\nabla \wedge A) + \nabla (\underbrace{\nabla \wedge \nabla \psi}_{=0}),\end{aligned} \quad\quad\quad(22)$

which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form

$\begin{aligned}A^\mu \rightarrow A^\mu + \partial_\mu \psi,\end{aligned} \quad\quad\quad(23)$

and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a $\psi$ such that $A^0 + \partial_0 \psi = 0$ . That is

$\begin{aligned}\sum_\mathbf{k} \phi_\mathbf{k}(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \frac{1}{{c}} \partial_t \psi = 0.\end{aligned} \quad\quad\quad(24)$

Or,

$\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x},0) -\frac{1}{{c}} \sum_\mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x}} \int_{\tau=0}^t \phi_\mathbf{k}(\tau).\end{aligned} \quad\quad\quad(25)$

With such a transformation, the $\phi_\mathbf{k}$ and $\dot{\mathbf{A}}_\mathbf{k}$ cross term in the Hamiltonian (20) vanishes, as does the $\phi_\mathbf{k}$ term in the four vector square of the last term, leaving just

$\begin{aligned}H = \frac{\epsilon_0}{c^2} V \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} \Bigl((c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 + {\left\lvert{ ( c \mathbf{k}) \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2+ {\left\lvert{ c \mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(26)$

Additionally, wedging (5) with $\gamma_0$ now does not loose any information so our potential Fourier series is reduced to just

$\begin{aligned}\mathbf{A} &= \sum_{\mathbf{k}} \mathbf{A}_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \\ \mathbf{A}_\mathbf{k} &= \frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} \mathbf{A}(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3.\end{aligned} \quad\quad\quad(27)$

The desired harmonic oscillator form would be had in (26) if it were not for the $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While $\mathbf{A} = A \wedge \gamma_0$ , the lack of an $A^0$ component means that this can be inverted as

$\begin{aligned}A = \mathbf{A} \gamma_0 = -\gamma_0 \mathbf{A}.\end{aligned} \quad\quad\quad(29)$

The gradient can also be factored scalar and spatial vector components

$\begin{aligned}\nabla = \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) = ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0.\end{aligned} \quad\quad\quad(30)$

So, with this $A^0 = 0$ gauge choice the bivector field $F$ is

$\begin{aligned}F = \nabla \wedge A = \frac{1}{{2}} \left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \end{aligned} \quad\quad\quad(31)$

From the left the gradient action on $A$ is

$\begin{aligned}\stackrel{ \rightarrow }{\nabla} A &= ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0 (-\gamma_0 \mathbf{A}) \\ &= ( -\partial_0 + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} ) \mathbf{A},\end{aligned}$

and from the right

$\begin{aligned}A \stackrel{ \leftarrow }{\nabla}&= \mathbf{A} \gamma_0 \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \mathbf{A} ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \partial_0 \mathbf{A} + \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \end{aligned}$

Taking the difference we have

$\begin{aligned}F &= \frac{1}{{2}} \Bigl( -\partial_0 \mathbf{A} + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \partial_0 \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \Bigr).\end{aligned}$

Which is just

$\begin{aligned}F = -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}.\end{aligned} \quad\quad\quad(32)$

For this vacuum case, premultiplication of Maxwell’s equation by $\gamma_0$ gives

$\begin{aligned}0 &= \gamma_0 \nabla ( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= (\partial_0 + \boldsymbol{\nabla})( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} - \partial_0 \boldsymbol{\nabla} \cdot \mathbf{A} - \partial_0 \boldsymbol{\nabla} \wedge \mathbf{A} + \partial_0 ( \boldsymbol{\nabla} \wedge \mathbf{A} ) + \underbrace{\boldsymbol{\nabla} \cdot ( \boldsymbol{\nabla} \wedge \mathbf{A} ) }_{\boldsymbol{\nabla}^2 \mathbf{A} - \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A})}+ \underbrace{\boldsymbol{\nabla} \wedge ( \boldsymbol{\nabla} \wedge \mathbf{A} )}_{=0} \\ \end{aligned}$

The spatial bivector and trivector grades are all zero. Equating the remaining scalar and vector components to zero separately yields a pair of equations in $\mathbf{A}$

$\begin{aligned}0 &= \partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) \\ 0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) \end{aligned} \quad\quad\quad(33)$

If the divergence of the vector potential is constant we have just a wave equation. Let’s see what that divergence is with the assumed Fourier representation

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A} &=\sum_{\mathbf{k} \ne (0,0,0)} {\mathbf{A}_\mathbf{k}}^m 2 \pi i \frac{k_m}{\lambda_m} e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=i \sum_{\mathbf{k} \ne (0,0,0)} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=i \sum_\mathbf{k} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} \end{aligned}$

Since $\mathbf{A}_\mathbf{k} = \mathbf{A}_\mathbf{k}(t)$ , there are two ways for $\partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) = 0$ . For each $\mathbf{k}$ there must be a requirement for either $\mathbf{A}_\mathbf{k} \cdot \mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k} = \text{constant}$ . The constant $\mathbf{A}_\mathbf{k}$ solution to the first equation appears to represent a standing spatial wave with no time dependence. Is that of any interest?

The more interesting seeming case is where we have some non-static time varying state. In this case, if $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$ , the second of these Maxwell’s equations is just the vector potential wave equation, since the divergence is zero. That is

$\begin{aligned}0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} \end{aligned} \quad\quad\quad(35)$

Solving this isn’t really what is of interest, since the objective was just to determine if the divergence could be assumed to be zero. This shows then, that if the transverse solution to Maxwell’s equation is picked, the Hamiltonian for this field, with this gauge choice, becomes

$\begin{aligned}H = \frac{\epsilon_0}{c^2} V \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} (c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right).\end{aligned} \quad\quad\quad(36)$

How does the gauge choice alter the Poynting vector? From (21), all the $\phi_\mathbf{k}$ dependence in that integrated momentum density is lost

The $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$ solutions to Maxwell’s equation are seen to result in zero momentum for this infinite periodic field. My expectation was something of the form $c \mathbf{P} = H \hat{\mathbf{k}}$ , so intuition is either failing me, or my math is failing me, or this contrived periodic field solution leads to trouble.

Conclusions and followup.

The objective was met, a reproduction of Bohm’s Harmonic oscillator result using a complex exponential Fourier series instead of separate sine and cosines.

The reason for Bohm’s choice to fix zero divergence as the gauge choice upfront is now clear. That automatically cuts complexity from the results. Figuring out how to work this problem with complex valued potentials and also using the Geometric Algebra formulation probably also made the work a bit more difficult since blundering through both simultaneously was required instead of just one at a time.

This was an interesting exercise though, since doing it this way I am able to understand all the intermediate steps. Bohm employed some subtler argumentation to eliminate the scalar potential $\phi$ upfront, and I have to admit I did not follow his logic, whereas blindly following where the math leads me all makes sense.

As a bit of followup, I’d like to consider the constant $\mathbf{A}_\mathbf{k}$ case in more detail, and any implications of the freedom to pick $\mathbf{A}_0$ .

The general calculation of $T^{\mu\nu}$ for the assumed Fourier solution should be possible too, but was not attempted. Doing that general calculation with a four dimensional Fourier series is likely tidier than working with scalar and spatial variables as done here.

Now that the math is out of the way (except possibly for the momentum which doesn’t seem right), some discussion of implications and applications is also in order. My preference is to let the math sink-in a bit first and mull over the momentum issues at leisure.

References

[2] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

Posted in Math and Physics Learning. | Tagged: bivector, clifford algebra, divergence theorem, energy-momentum, fourier series, geometric algebra, hamiltonian, harmonic oscillator, maxwells equations, spacetime gradient, wave equation | 2 Comments »

Electrodynamic field energy for vacuum.

Posted by peeterjoot on December 19, 2009

[Click here for a PDF of this post with nicer formatting]

Motivation.

We now know how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism. To recap, for the field $F = \mathbf{E} + I c \mathbf{B}$ , where $\mathbf{E}$ and $\mathbf{B}$ may be complex vectors we have for Maxwell’s equation

$\begin{aligned}\nabla F = J/\epsilon_0 c.\end{aligned} \quad\quad\quad(1)$

The real valued, four vector, energy momentum tensor $T(a)$ was found to be

To supply some context that gives meaning to this tensor the associated conservation relationship was found to be

$\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(3)$

and in particular for $a = \gamma^0$ , this four vector divergence takes the form

Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [1]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.

Setup

Let’s assume a Fourier representation for the four vector potential $A$ for the field $F = \nabla \wedge A$ . That is

$\begin{aligned}A = \sum_{\mathbf{k}} A_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}},\end{aligned} \quad\quad\quad(5)$

where summation is over all wave number triplets $\mathbf{k} = (p/\lambda_1,q/\lambda_2,r/\lambda_3)$ . The Fourier coefficients $A_\mathbf{k} = {A_\mathbf{k}}^\mu \gamma_\mu$ are allowed to be complex valued, as is the resulting four vector $A$ , and the associated bivector field $F$ .

Fourier inversion follows from

$\begin{aligned}\delta_{\mathbf{k}', \mathbf{k}} =\frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} e^{2 \pi i \mathbf{k}' \cdot \mathbf{x}} e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(6)$

but only this orthogonality relationship and not the Fourier coefficients themselves

$\begin{aligned}A_\mathbf{k} = \frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} A(\mathbf{x}, t) e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(7)$

will be of interest here. Evaluating the curl for this potential yields

$\begin{aligned}F = \nabla \wedge A= \sum_{\mathbf{k}} \left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \sum_{m=1}^3 \gamma^m \wedge A_\mathbf{k} \frac{2 \pi i k_m}{\lambda_m} \right) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(8)$

We can now form the energy density

$\begin{aligned}U = T(\gamma^0) \cdot \gamma^0=-\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} \gamma^0 F \gamma^0 \Bigr).\end{aligned} \quad\quad\quad(9)$

With implied summation over all repeated integer indexes (even without matching uppers and lowers), this is

$\begin{aligned}U =-\frac{\epsilon_0}{2} \sum_{\mathbf{k}', \mathbf{k}} \text{Real} \left\langle{{\left( \frac{1}{{c}} \gamma^0 \wedge {{\dot{A}_{\mathbf{k}'}}}^{*} - \gamma^m \wedge {{A_{\mathbf{k}'}}}^{*} \frac{2 \pi i k_m'}{\lambda_m} \right) e^{-2 \pi i \mathbf{k}' \cdot \mathbf{x}}\gamma^0\left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^n \wedge A_\mathbf{k} \frac{2 \pi i k_n}{\lambda_n} \right) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}\gamma^0}}\right\rangle.\end{aligned} \quad\quad\quad(10)$

The grade selection used here doesn’t change the result since we already have a scalar, but will just make it convenient to filter out any higher order products that will cancel anyways. Integrating over the volume element and taking advantage of the orthogonality relationship (6), the exponentials are removed, leaving the energy contained in the volume

$\begin{aligned}H = -\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2}\sum_{\mathbf{k}} \text{Real} \left\langle{{\left( \frac{1}{{c}} \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} - \gamma^m \wedge {{A_{\mathbf{k}}}}^{*} \frac{2 \pi i k_m}{\lambda_m} \right) \gamma^0\left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^n \wedge A_\mathbf{k} \frac{2 \pi i k_n}{\lambda_n} \right) \gamma^0}}\right\rangle.\end{aligned} \quad\quad\quad(11)$

First reduction of the Hamiltonian.

Let’s take the products involved in sequence one at a time, and evaluate, later adding and taking real parts if required all of

$\begin{aligned}\frac{1}{{c^2}}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) \gamma^0 (\gamma^0 \wedge \dot{A}_\mathbf{k}) \gamma^0 }}\right\rangle &=-\frac{1}{{c^2}}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) (\gamma^0 \wedge \dot{A}_\mathbf{k}) }}\right\rangle \end{aligned} \quad\quad\quad(12)$

$\begin{aligned}- \frac{2 \pi i k_m}{c \lambda_m} \left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) \gamma^0 ( \gamma^0 \wedge \dot{A}_\mathbf{k} ) \gamma^0}}\right\rangle &=\frac{2 \pi i k_m}{c \lambda_m} \left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) ( \gamma^0 \wedge \dot{A}_\mathbf{k} ) }}\right\rangle \end{aligned} \quad\quad\quad(13)$

$\begin{aligned}\frac{2 \pi i k_n}{c \lambda_n} \left\langle{{ ( \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) \gamma^0 ( \gamma^n \wedge A_\mathbf{k} ) \gamma^0}}\right\rangle &=-\frac{2 \pi i k_n}{c \lambda_n} \left\langle{{ ( \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) ( \gamma^n \wedge A_\mathbf{k} ) }}\right\rangle \end{aligned} \quad\quad\quad(14)$

$\begin{aligned}-\frac{4 \pi^2 k_m k_n}{\lambda_m \lambda_n}\left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) \gamma^0(\gamma^n \wedge A_\mathbf{k} ) \gamma^0}}\right\rangle. &\end{aligned} \quad\quad\quad(15)$

The expectation is to obtain a Hamiltonian for the field that has the structure of harmonic oscillators, where the middle two products would have to be zero or sum to zero or have real parts that sum to zero. The first is expected to contain only products of ${\left\lvert{{\dot{A}_\mathbf{k}}^m}\right\rvert}^2$ , and the last only products of ${\left\lvert{{A_\mathbf{k}}^m}\right\rvert}^2$ .

While initially guessing that (13) and (14) may cancel, this isn’t so obviously the case. The use of cyclic permutation of multivectors within the scalar grade selection operator $\left\langle{{A B}}\right\rangle = \left\langle{{B A}}\right\rangle$ plus a change of dummy summation indexes in one of the two shows that this sum is of the form $Z + {{Z}}^{*}$ . This sum is intrinsically real, so we can neglect one of the two doubling the other, but we will still be required to show that the real part of either is zero.

Lets reduce these one at a time starting with (12), and write $\dot{A}_\mathbf{k} = \kappa$ temporarily

$\begin{aligned}\left\langle{{ (\gamma^0 \wedge {{\kappa}}^{*} ) (\gamma^0 \wedge \kappa }}\right\rangle &={\kappa^m}^{{*}} \kappa^{m'}\left\langle{{ \gamma^0 \gamma_m \gamma^0 \gamma_{m'} }}\right\rangle \\ &=-{\kappa^m}^{{*}} \kappa^{m'}\left\langle{{ \gamma_m \gamma_{m'} }}\right\rangle \\ &={\kappa^m}^{{*}} \kappa^{m'}\delta_{m m'}.\end{aligned}$

So the first of our Hamiltonian terms is

$\begin{aligned}\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_\mathbf{k}}}^{*} ) (\gamma^0 \wedge \dot{A}_\mathbf{k} }}\right\rangle &=\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2}{\left\lvert{{{\dot{A}}_{\mathbf{k}}}^m}\right\rvert}^2.\end{aligned} \quad\quad\quad(16)$

Note that summation over $m$ is still implied here, so we’d be better off with a spatial vector representation of the Fourier coefficients $\mathbf{A}_\mathbf{k} = A_\mathbf{k} \wedge \gamma_0$ . With such a notation, this contribution to the Hamiltonian is

$\begin{aligned}\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2} \dot{\mathbf{A}}_\mathbf{k} \cdot {{\dot{\mathbf{A}}_\mathbf{k}}}^{*}.\end{aligned} \quad\quad\quad(17)$

To reduce (13) and (13), this time writing $\kappa = A_\mathbf{k}$ , we can start with just the scalar selection

$\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) ( \gamma^0 \wedge \dot{\kappa} ) }}\right\rangle &=\Bigl( \gamma^m {{(\kappa^0)}}^{*} - {{\kappa}}^{*} \underbrace{(\gamma^m \cdot \gamma^0)}_{=0} \Bigr) \cdot \dot{\kappa} \\ &={{(\kappa^0)}}^{*} \dot{\kappa}^m\end{aligned}$

Thus the contribution to the Hamiltonian from (13) and (13) is

$\begin{aligned}\frac{2 \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \pi k_m}{c \lambda_m} \text{Real} \Bigl( i {{(A_\mathbf{k}^0)}}^{*} \dot{A_\mathbf{k}}^m \Bigl)=\frac{2 \pi \epsilon_0 \lambda_1 \lambda_2 \lambda_3}{c} \text{Real} \Bigl( i {{(A_\mathbf{k}^0)}}^{*} \mathbf{k} \cdot \dot{\mathbf{A}}_\mathbf{k} \Bigl).\end{aligned} \quad\quad\quad(18)$

Most definitively not zero in general. Our final expansion (15) is the messiest. Again with $A_\mathbf{k} = \kappa$ for short, the grade selection of this term in coordinates is

$\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) \gamma^0 (\gamma^n \wedge \kappa ) \gamma^0 }}\right\rangle&=- {{\kappa_\mu}}^{*} \kappa^\nu \left\langle{{ (\gamma^m \wedge \gamma^\mu) \gamma^0 (\gamma_n \wedge \gamma_\nu) \gamma^0 }}\right\rangle\end{aligned} \quad\quad\quad(19)$

Expanding this out yields

$\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) \gamma^0 (\gamma^n \wedge \kappa ) \gamma^0 }}\right\rangle&=- ( {\left\lvert{\kappa_0}\right\rvert}^2 - {\left\lvert{A^a}\right\rvert}^2 ) \delta_{m n} + {{A^n}}^{*} A^m.\end{aligned} \quad\quad\quad(20)$

The contribution to the Hamiltonian from this, with $\phi_\mathbf{k} = A^0_\mathbf{k}$ , is then

$\begin{aligned}2 \pi^2 \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \Bigl(-\mathbf{k}^2 {{\phi_\mathbf{k}}}^{*} \phi_\mathbf{k} + \mathbf{k}^2 ({{\mathbf{A}_\mathbf{k}}}^{*} \cdot \mathbf{A}_\mathbf{k})+ (\mathbf{k} \cdot {{\mathbf{A}_k}}^{*}) (\mathbf{k} \cdot \mathbf{A}_k)\Bigr).\end{aligned} \quad\quad\quad(21)$

A final reassembly of the Hamiltonian from the parts (17) and (18) and (21) is then

$\begin{aligned}H = \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2 c^2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{2 \pi}{c} \text{Real} \Bigl( i {{ \phi_\mathbf{k} }}^{*} (\mathbf{k} \cdot \dot{\mathbf{A}}_\mathbf{k}) \Bigl)+2 \pi^2 \Bigl(\mathbf{k}^2 ( -{\left\lvert{\phi_\mathbf{k}}\right\rvert}^2 + {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 ) + {\left\lvert{\mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(22)$

This is finally reduced to a completely real expression, and one without any explicit Geometric Algebra. All the four vector Fourier vector potentials written out explicitly in terms of the spacetime split $A_\mathbf{k} = (\phi_\mathbf{k}, \mathbf{A}_\mathbf{k})$ , which is natural since an explicit time and space split was the starting point.

Gauge transformation to simplify the Hamiltonian.

While (22) has considerably simpler form than (11), what was expected, was something that looked like the Harmonic oscillator. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.

If we are to change our four vector potential $A \rightarrow A + \nabla \psi$ , then Maxwell’s equation takes the form

$\begin{aligned}J/\epsilon_0 c = \nabla (\nabla \wedge (A + \nabla \psi) = \nabla (\nabla \wedge A) + \nabla (\underbrace{\nabla \wedge \nabla \psi}_{=0}),\end{aligned} \quad\quad\quad(23)$

which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form

$\begin{aligned}A^\mu \rightarrow A^\mu + \partial_\mu \psi,\end{aligned} \quad\quad\quad(24)$

and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a $\psi$ such that $A^0 + \partial_0 \psi = 0$ . That is

$\begin{aligned}\sum_\mathbf{k} \phi_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} + \frac{1}{{c}} \partial_t \psi = 0.\end{aligned} \quad\quad\quad(25)$

Or,

$\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x},0) -\frac{1}{{c}} \sum_\mathbf{k} e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \int_{\tau=0}^t \phi_\mathbf{k}(\tau).\end{aligned} \quad\quad\quad(26)$

With such a transformation, the $\phi_\mathbf{k}$ and $\dot{\mathbf{A}}_\mathbf{k}$ cross term in the Hamiltonian (22) vanishes, as does the $\phi_\mathbf{k}$ term in the four vector square of the last term, leaving just

$\begin{aligned}H = \frac{\epsilon_0}{c^2} \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} \Bigl((2 \pi c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 + {\left\lvert{ ( 2 \pi c \mathbf{k}) \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(27)$

Additionally, wedging (5) with $\gamma_0$ now does not loose any information so our potential Fourier series is reduced to just

$\begin{aligned}\mathbf{A} &= \sum_{\mathbf{k}} \mathbf{A}_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \\ \mathbf{A}_\mathbf{k} &= \frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} \mathbf{A}(\mathbf{x}, t) e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3.\end{aligned} \quad\quad\quad(28)$

The desired harmonic oscillator form would be had in (27) if it were not for the $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While $\mathbf{A} = A \wedge \gamma_0$ , the lack of an $A^0$ component means that this can be inverted as

$\begin{aligned}A = \mathbf{A} \gamma_0 = -\gamma_0 \mathbf{A}.\end{aligned} \quad\quad\quad(30)$

The gradient can also be factored scalar and spatial vector components

$\begin{aligned}\nabla = \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) = ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0.\end{aligned} \quad\quad\quad(31)$

So, with this $A^0 = 0$ gauge choice the bivector field $F$ is

$\begin{aligned}F = \nabla \wedge A = \frac{1}{{2}} \left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \end{aligned} \quad\quad\quad(32)$

From the left the gradient action on $A$ is

and from the right

Taking the difference we have

Which is just

$\begin{aligned}F = -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}.\end{aligned} \quad\quad\quad(33)$

For this vacuum case, premultiplication of Maxwell’s equation by $\gamma_0$ gives

The spatial bivector and trivector grades are all zero. Equating the remaining scalar and vector components to zero separately yields a pair of equations in $\mathbf{A}$

If the divergence of the vector potential is constant we have just a wave equation. Let’s see what that divergence is with the assumed Fourier representation

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A} &=\sum_{k \ne (0,0,0)} {\mathbf{A}_\mathbf{k}}^m 2 \pi i \frac{k_m}{\lambda_m} e^{2\pi i \mathbf{k} \cdot \mathbf{x}} \\ &=2 \pi i \sum_{k \ne (0,0,0)} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{2\pi i \mathbf{k} \cdot \mathbf{x}} \\ \end{aligned}$

Since $\mathbf{A}_\mathbf{k} = \mathbf{A}_\mathbf{k}(t)$ , there are two ways for $\partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) = 0$ . For each $\mathbf{k} \ne 0$ there must be a requirement for either $\mathbf{A}_\mathbf{k} \cdot \mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k} = \text{constant}$ . The constant $\mathbf{A}_\mathbf{k}$ solution to the first equation appears to represent a standing spatial wave with no time dependence. Is that of any interest?

The more interesting seeming case is where we have some non-static time varying state. In this case, if $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$ for all $\mathbf{k} \ne 0$ the second of these Maxwell’s equations is just the vector potential wave equation, since the divergence is zero. That is

$\begin{aligned}0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} \end{aligned} \quad\quad\quad(36)$

$\begin{aligned}H = \frac{\epsilon_0}{c^2} \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} (2 \pi c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right).\end{aligned} \quad\quad\quad(37)$

Conclusions and followup.

The objective was met, a reproduction of Bohm’s Harmonic oscillator result using a complex exponential Fourier series instead of separate sine and cosines.

As a bit of followup, I’d like to consider the constant $\mathbf{A}_\mathbf{k}$ case, and any implications of the freedom to pick $\mathbf{A}_0$ . I’d also like to construct the Poynting vector $T(\gamma^0) \wedge \gamma_0$ , and see what the structure of that is with this Fourier representation.

A general calculation of $T^{\mu\nu}$ for an assumed Fourier solution should be possible too, but working in spatial quantities for the general case is probably torture. A four dimensional Fourier series is likely a superior option for the general case.

References

[1] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

Reader notes for Jackson 12.11, Retarded time solution to the wave equation.

Posted by peeterjoot on September 20, 2009

[Click here for a PDF of this sequence of posts with nicer formatting]

Motivation

In [1] I blundered my way towards the retarded time Green’s function solution to the 3D wave equation. Jackson’s [2] (section 12.11) covers this in a much more coherent fashion. It is however somewhat terse, and some details that were not immediately obvious to me were omitted.

Here are my notes for this section in case I want to refer to it again later.

Guts

The starting point is the electrodynamic wave equation

$\begin{aligned}\partial_\alpha F^{\alpha\beta} = \frac{4 \pi}{c} J^\beta\end{aligned} \quad\quad\quad(1)$

A substitution of $F^{\alpha \beta} = \partial^\alpha A^\beta - \partial^\beta A^\alpha$ gives us

$\begin{aligned}\partial_\alpha F^{\alpha\beta} = \partial_\alpha \partial^\alpha A^\beta - \partial_\alpha \partial^\beta A^\alpha= \square A^\beta - \partial^\beta (\partial_\alpha A^\alpha)\end{aligned} \quad\quad\quad(2)$

Thus with the Lorentz condition $\partial_\alpha A^\alpha = 0$ we have

$\begin{aligned}\square A^\beta = \frac{4 \pi}{c} J^\beta\end{aligned} \quad\quad\quad(3)$

A set of four non-homogeneous wave equations to solve. It is assumed that a Green’s function of the form

$\begin{aligned}\square_x D(x - x') = \delta^4(x - x')\end{aligned} \quad\quad\quad(4)$

can be found. Jackson states that this is possible in the absense of boundary surfaces, which seems to imply that the more general case would require $\square_x D(x, x') = \delta^4(x - x')$ , where $D$ is not neccessarily a function of the four vector difference $x - x'$ .

What is really meant by this Green’s function? It only takes meaning in the context of the convolution integral. Namely

$\begin{aligned}A^\beta = \int d^4 x' D(x, x') \frac{4 \pi}{c} J^\beta(x') \end{aligned} \quad\quad\quad(5)$

So that

$\begin{aligned}\square_x A^\beta &= \int d^4 x' \square_x D(x, x') \frac{4 \pi}{c} J^\beta(x') \\ &= \frac{4 \pi}{c} \int d^4 x' \delta^4(x - x') J^\beta(x') \\ &= \frac{4 \pi}{c} J^\beta(x) \\ \end{aligned}$

So if a function with this delta filtering property under the Delambertian can be found we can find the non-homogeneous solutions directly by four-volume convolution.

It is implied in the text (probably stated explicitly somewhere earlier) that the asymmetric convention for the Fourier transform pairs is being used

$\begin{aligned}\tilde{f}(k) &= \int d^4 z f(z) e^{i k \cdot z} \\ f(z) &= \frac{1}{{(2\pi)^4}} \int d^4 k \tilde{f}(k) e^{-i k \cdot z} \end{aligned} \quad\quad\quad(6)$

where $d^4 k = dk_0 dk_1 dk_2 dk_3$ , and $d^4 z = dz^0 dz^1 dz^2 dz^3$ , and $k \cdot z = k_\mu z^\mu = k^\mu z_\mu$ .

Assuming the validity of this transform pair, even for the delta distribution, we can find an integral representation of the delta using the transform pairs. For the Fourier transform of delta we have

$\begin{aligned}\tilde{\delta^4}(k) &= \int d^4 z \delta^4(z) e^{i k \cdot z} \\ &= e^{i k \cdot 0} \\ &= 1\end{aligned}$

Performing the inverse transformation provides the delta function exponential integral representation

$\begin{aligned}\delta^4(z) &= \frac{1}{{(2\pi)^4}} \int d^4 k \tilde{\delta^4}(k) e^{-i k \cdot z} \\ &= \frac{1}{{(2\pi)^4}} \int d^4 k e^{-i k \cdot z} \\ \end{aligned}$

Just as a Fourier representation of the delta can be found, we can integrate by parts to find an integral representation of the Green’s function that we seek. Taking Fourier transforms

$\begin{aligned}\mathcal{F}(\square_x D(z))(k) &= \int d^4 z \partial_\alpha \partial^\alpha D(z) e^{i k \cdot z} \\ &= -\int d^4 z \partial^\alpha D(z) \partial_\alpha e^{i k_\beta z^\beta} \\ &= -\int d^4 z \partial^\alpha D(z) i k_\alpha e^{i k_\beta z^\beta} \\ &= \int d^4 z D(z) i k_\alpha \partial^\alpha e^{i k^\beta z_\beta} \\ &= -\int d^4 z D(z) k_\alpha k^\alpha e^{i k \cdot z } \\ &= - k^2 \tilde{D}(k)\end{aligned}$

Usign the assumed delta function property of this Green’s function we also have

$\begin{aligned}\mathcal{F}(\square_x D(z))(k) &= \int d^4 z \delta^4(z) e^{i k \cdot z} \\ &= 1\end{aligned}$

This completely specifies the Fourier transform of the Green’s function

$\begin{aligned}\tilde{D}(k) &= - \frac{1}{{k^2}}\end{aligned} \quad\quad\quad(8)$

and we can inverse transform to complete the task of finding an initial representation of the Green’s function itself. That is

$\begin{aligned}D(z) = -\frac{1}{{(2\pi)^4}} \int d^4 k \frac{1}{{k^2}} e^{-i k \cdot z} \end{aligned} \quad\quad\quad(9)$

With an explicit spacetime split we have our integral prepped for the contour integration

$\begin{aligned}D(z) = -\frac{1}{{(2\pi)^4}} \int d^3 k e^{i \mathbf{k} \cdot \mathbf{z}} \int_{-\infty}^\infty dk_0 \frac{1}{{k_0^2 - \mathbf{k}^2}} e^{-i k_0 z_0} \end{aligned} \quad\quad\quad(10)$

Here $\kappa = {\left\lvert{\mathbf{k}}\right\rvert}$ is used as in the text. If we let $k_0 = R e^{i\theta}$ take on complex values, integrating over a semicircular arc, we have for the exponential

$\begin{aligned}{\left\lvert{e^{-i k_0 z_0}}\right\rvert}&= {\left\lvert{e^{-i R (\cos\theta + i \sin\theta) z_0} }\right\rvert} \\ &= {\left\lvert{e^{ z_0 R \sin\theta} e^{ -i z_0 R \cos\theta } }\right\rvert} \\ &= {\left\lvert{e^{ z_0 R \sin\theta} }\right\rvert}\end{aligned}$

In the upper half plane $\theta \in [0,\pi]$ , so $\sin\theta$ is never negative, and the integral on an upper half plane semi-circular contour can only vanish as desired for $z_0 0$ for a lower half plane contour. This is mentioned in the text but I felt it more clear just writing out the exponential as above.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{retardedContourBoth}
\caption{Contours strictly above the $k_0 = 0$ axis}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{retardedContourAroundPole}
\caption{Contour around pole}
\end{figure}

Having established the value on the loop at infinity we can now integrate over the contour $r_1$ as depicted in figure (\ref{fig:jacksonRet:retardedContourBoth}). The problem is mainly reduced to an integral of the form figure (\ref{fig:jacksonRet:retardedContourAroundPole}) around the simple poles at $\alpha = \pm \kappa$

$\begin{aligned}I_\alpha = \oint \frac{f(z)}{z - \alpha} dz\end{aligned} \quad\quad\quad(11)$

With $z = \alpha + R e^{i\theta}$ , and $\theta \in [\pi/2, 5\pi/2]$ , we have

$\begin{aligned}I_\alpha = \int \frac{f(z)}{R e^{i\theta}} R i e^{i\theta} d\theta\end{aligned} \quad\quad\quad(12)$

with $R \rightarrow 0$ , we are left with

$\begin{aligned}I_\alpha = 2 \pi i f(\alpha)\end{aligned} \quad\quad\quad(13)$

There are six arcs on the contour of interest. For the first two around the poles lets lable the integral contributions $I_\kappa$ and $I_{-\kappa}$ . Along the infinite semicircular contour the integral vanishes with the right sign choice for $z_0$ . For the remainder lets write the integral contributions $I$ .

Summing over the complete contour, specially chosen to enclose no poles, we have

$\begin{aligned}I + I_\kappa + I_{-\kappa} + 0 = 0 \end{aligned} \quad\quad\quad(14)$

For this $z_0 > 0$ integral we are left with the residue sum

$\begin{aligned}\int_{-\infty}^\infty dk_0 \frac{1}{{k_0^2 - \mathbf{k}^2}} e^{-i k_0 z_0} &= - 2 \pi i \left( {\left. \frac{1}{{k_0 - \kappa}} e^{-i k_0 z_0} \right\vert}_{k_0 = -\kappa}+{\left. \frac{1}{{k_0 + \kappa}} e^{-i k_0 z_0} \right\vert}_{k_0 = \kappa}\right) \\ &= \frac{2 \pi i^2}{\kappa} \sin(\kappa z_0)\end{aligned}$

Since I can never remember the signs and integral orientations for the residue formula so I’ve always done it “manually” as above picking a zero valued contour.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{retardedContourOnAxis}
\caption{Contour exactly on the $k_0 = 0$ axis?}
\end{figure}

Now, the issue of where to place the contour wasn’t really discussed mathematically. Physically this makes the difference between causal and acausal behaviour, but why put the contour strictly above or below the axis and not right on it. If we put the contour exactly on the $k_0 = 0$ axis as in (\ref{fig:jacksonRet:retardedContourOnAxis}), then our integrals around the two half circular poles gives us a result off by a factor of two? There is also an (implied) limiting procedure required to place the contour strictly above the axis, and the details of this aren’t mentioned (and I also haven’t thought them through). Some of these would be worth thinking through in more detail, but for now lets ignore these. We are left with

$\begin{aligned}D(z) = \frac{\theta(z_0)}{(2\pi)^3} \int d^3 k e^{i \mathbf{k} \cdot \mathbf{z}} \frac{1}{{\kappa}} \sin(\kappa z_0)\end{aligned} \quad\quad\quad(15)$

How to reduce this to the single variable integral in $\kappa$ was not immediately clear to me. Aligning $\mathbf{z}$ with the $\mathbf{e}_3$ axis, and using a spherical polar representation for $\mathbf{k}$ we can write $\mathbf{z} \cdot \mathbf{k} = R \kappa \cos\theta$ . With this and the volume element $d^3 k = \kappa^2 \sin\theta d\theta d\phi d\kappa$ , we have

$\begin{aligned}D(z) = \frac{\theta(z_0)}{(2\pi)^3} \int_0^\infty d\kappa \sin(\kappa z_0) \int_0^{2\pi} d\phi \int_0^\pi d\theta \kappa e^{i R \kappa \cos\theta} \sin\theta\end{aligned} \quad\quad\quad(16)$

This now happily submits to a nice variable substitution, unlike an integral like $\int e^{i \mu \cos\theta} d\theta = J_0({\left\lvert{\mu}\right\rvert})$ which can be evaluated, but only in terms of Bessel functions or messy series expansion. Writing $\tau = \kappa \cos\theta$ , and $-d\tau = \kappa \sin\theta d\theta$ we have

$\begin{aligned}\int_0^\pi d\theta \kappa e^{i R \kappa \cos\theta} \sin\theta&=-\int_{\kappa}^{-\kappa} d\tau e^{i R \tau} \\ &=\frac{e^{i R \kappa}}{i R} -\frac{e^{-i R \kappa}}{i R} \\ &=2 \frac{1}{{R}} \sin(R \kappa)\end{aligned}$

Our Green’s function is now reduced to

$\begin{aligned}D(z) = \frac{\theta(z_0)}{2 \pi^2 R} \int_0^\infty d\kappa \sin(\kappa z_0) \sin(\kappa R)\end{aligned} \quad\quad\quad(17)$

Expanding out these sines in terms of exponentials we have

$\begin{aligned}D(z) &= -\frac{\theta(z_0)}{8 \pi^2 R} \int_0^\infty d\kappa ( e^{i\kappa(z_0+R)} + e^{-i\kappa(z_0+R)} -e^{i\kappa(R-z_0)} - e^{i\kappa(z_0-R)} ) \\ &= -\frac{\theta(z_0)}{8 \pi^2 R} \left(\int_0^\infty d\kappa \left( e^{i\kappa(z_0+R)} -e^{i\kappa(R-z_0)} \right) +\int_0^{-\infty} -d\kappa \left( e^{i\kappa(z_0+R)} - e^{-i\kappa(z_0-R)} \right) \right)\\ &= \frac{\theta(z_0)}{8 \pi^2 R} \int_{-\infty}^\infty d\kappa \left( e^{i\kappa(R-z_0)} -e^{i\kappa(z_0+R)} \right) \\ \end{aligned}$

the sign in this first exponential differs from what Jackson obtained but it won’t change the end result. Did I make a mistake or did he? Wonder what the third edition shows? Using $\delta(x) = \int e^{-ikx} dk/2\pi$ we have

$\begin{aligned}D(z) = \frac{\theta(z_0)}{4 \pi R} \left( \delta(z_0 -R) - \delta(-(z_0 + R)) \right)\end{aligned} \quad\quad\quad(18)$

With $R = {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} \ge 0$ , and $z_0 = c(t - t') > 0$ , this second delta cannot contribute, and we are left with the retarded Green’s function

$\begin{aligned}D_r(z) = \frac{\theta(z_0)}{4 \pi R} \delta(c(t -t') - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}) \end{aligned} \quad\quad\quad(19)$

Very slick. I like the procedure, despite a few magic steps (like the choice to offset the contour).

References

[1] Peeter Joot. {Poisson and retarded Potential Green’s functions from Fourier kernels} [online]. http://sites.google.com/site/peeterjoot/math2009/poisson.pdf.

[2] JD Jackson. Classical Electrodynamics Wiley. 2nd edition, 1975.

Posted in Math and Physics Learning. | Tagged: contour integral, Green's function, jackson, retarded time potential, wave equation | 1 Comment »

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