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# Archive for February, 2011

## PHY450H1S. Relativistic Electrodynamics Lecture 11 (Taught by Prof. Erich Poppitz). Unpacking Lorentz force equation. Lorentz transformations of the strength tensor, Lorentz field invariants, Bianchi identity, and first half of Maxwell’s.

Posted by peeterjoot on February 24, 2011

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

Covering chapter 3 material from the text [1].

Covering lecture notes pp. 74-83: Lorentz transformation of the strength tensor (82) [Tuesday, Feb. 8] [extra reading for the mathematically minded: gauge field, strength tensor, and gauge transformations in differential form language, not to be covered in class (83)]

Covering lecture notes pp. 84-102: Lorentz invariants of the electromagnetic field (84-86); Bianchi identity and the first half of Maxwell’s equations (87-90)

# Chewing on the four vector form of the Lorentz force equation.

After much effort, we arrived at

\begin{aligned}\frac{d{{(m c u_l) }}}{ds} = \frac{e}{c} \left( \partial_l A_i - \partial_i A_l \right) u^i\end{aligned} \hspace{\stretch{1}}(2.1)

or

\begin{aligned}\frac{d{{ p_l }}}{ds} = \frac{e}{c} F_{l i} u^i\end{aligned} \hspace{\stretch{1}}(2.2)

## Elements of the strength tensor

\paragraph{Claim}: there are only 6 independent elements of this matrix (tensor)

\begin{aligned}\begin{bmatrix}0 & . & . & . \\ & 0 & . & . \\ & & 0 & . \\ & & & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)

This is a no-brainer, for we just have to mechanically plug in the elements of the field strength tensor

Recall

\begin{aligned}A^i &= (\phi, \mathbf{A}) \\ A_i &= (\phi, -\mathbf{A})\end{aligned} \hspace{\stretch{1}}(2.4)

\begin{aligned}F_{0\alpha} &= \partial_0 A_\alpha - \partial_\alpha A_0 \\ &= -\partial_0 (\mathbf{A})_\alpha - \partial_\alpha \phi \\ \end{aligned}

\begin{aligned}F_{0\alpha} = E_\alpha\end{aligned} \hspace{\stretch{1}}(2.6)

For the purely spatial index combinations we have

\begin{aligned}F_{\alpha\beta} &= \partial_\alpha A_\beta - \partial_\beta A_\alpha \\ &= -\partial_\alpha (\mathbf{A})_\beta + \partial_\beta (\mathbf{A})_\alpha \\ \end{aligned}

Written out explicitly, these are

\begin{aligned}F_{1 2} &= \partial_2 (\mathbf{A})_1 -\partial_1 (\mathbf{A})_2 \\ F_{2 3} &= \partial_3 (\mathbf{A})_2 -\partial_2 (\mathbf{A})_3 \\ F_{3 1} &= \partial_1 (\mathbf{A})_3 -\partial_3 (\mathbf{A})_1 .\end{aligned} \hspace{\stretch{1}}(2.7)

We can compare this to the elements of $\mathbf{B}$

\begin{aligned}\mathbf{B} = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \partial_1 & \partial_2 & \partial_3 \\ A_x & A_y & A_z\end{vmatrix}\end{aligned} \hspace{\stretch{1}}(2.10)

We see that

\begin{aligned}(\mathbf{B})_z &= \partial_1 A_y - \partial_2 A_x \\ (\mathbf{B})_x &= \partial_2 A_z - \partial_3 A_y \\ (\mathbf{B})_y &= \partial_3 A_x - \partial_1 A_z\end{aligned} \hspace{\stretch{1}}(2.11)

So we have

\begin{aligned}F_{1 2} &= - (\mathbf{B})_3 \\ F_{2 3} &= - (\mathbf{B})_1 \\ F_{3 1} &= - (\mathbf{B})_2.\end{aligned} \hspace{\stretch{1}}(2.14)

These can be summarized as simply

\begin{aligned}F_{\alpha\beta} = - \epsilon_{\alpha\beta\gamma} B_\gamma.\end{aligned} \hspace{\stretch{1}}(2.17)

This provides all the info needed to fill in the matrix above

\begin{aligned}{\left\lVert{ F_{i j} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.18)

## Index raising of rank 2 tensor

To raise indexes we compute

\begin{aligned}F^{i j} = g^{i l} g^{j k} F_{l k}.\end{aligned} \hspace{\stretch{1}}(2.19)

### Justifying the raising operation.

To justify this consider raising one index at a time by applying the metric tensor to our definition of $F_{l k}$. That is

\begin{aligned}g^{a l} F_{l k} &=g^{a l} (\partial_l A_k - \partial_k A_l) \\ &=\partial^a A_k - \partial_k A^a.\end{aligned}

Now apply the metric tensor once more

\begin{aligned}g^{b k} g^{a l} F_{l k} &=g^{b k} (\partial^a A_k - \partial_k A^a) \\ &=\partial^a A^b - \partial^b A^a.\end{aligned}

This is, by definition $F^{a b}$. Since a rank 2 tensor has been defined as an object that transforms like the product of two pairs of coordinates, it makes sense that this particular tensor raises in the same fashion as would a product of two vector coordinates (in this case, it happens to be an antisymmetric product of two vectors, and one of which is an operator, but we have the same idea).

### Consider the components of the raised $F_{i j}$ tensor.

\begin{aligned}F^{0\alpha} &= -F_{0\alpha} \\ F^{\alpha\beta} &= F_{\alpha\beta}.\end{aligned} \hspace{\stretch{1}}(2.20)

\begin{aligned}{\left\lVert{ F^{i j} }\right\rVert} = \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.22)

## Back to chewing on the Lorentz force equation.

\begin{aligned}m c \frac{d{{ u_i }}}{ds} = \frac{e}{c} F_{i j} u^j\end{aligned} \hspace{\stretch{1}}(2.23)

\begin{aligned}u^i &= \gamma \left( 1, \frac{\mathbf{v}}{c} \right) \\ u_i &= \gamma \left( 1, -\frac{\mathbf{v}}{c} \right)\end{aligned} \hspace{\stretch{1}}(2.24)

For the spatial components of the Lorentz force equation we have

\begin{aligned}m c \frac{d{{ u_\alpha }}}{ds} &= \frac{e}{c} F_{\alpha j} u^j \\ &= \frac{e}{c} F_{\alpha 0} u^0+ \frac{e}{c} F_{\alpha \beta} u^\beta \\ &= \frac{e}{c} (-E_{\alpha}) \gamma+ \frac{e}{c} (- \epsilon_{\alpha\beta\gamma} B_\gamma ) \frac{v^\beta}{c} \gamma \end{aligned}

But

\begin{aligned}m c \frac{d{{ u_\alpha }}}{ds} &= -m \frac{d{{(\gamma \mathbf{v}_\alpha)}}}{ds} \\ &= -m \frac{d(\gamma \mathbf{v}_\alpha)}{c \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} dt} \\ &= -\gamma \frac{d(m \gamma \mathbf{v}_\alpha)}{c dt}.\end{aligned}

Canceling the common $-\gamma/c$ terms, and switching to vector notation, we are left with

\begin{aligned}\frac{d( m \gamma \mathbf{v}_\alpha)}{dt} = e \left( E_\alpha + \frac{1}{{c}} (\mathbf{v} \times \mathbf{B})_\alpha \right).\end{aligned} \hspace{\stretch{1}}(2.26)

Now for the energy term. We have

\begin{aligned}m c \frac{d{{u_0}}}{ds} &= \frac{e}{c} F_{0\alpha} u^\alpha \\ &= \frac{e}{c} E_{\alpha} \gamma \frac{v^\alpha}{c} \\ \frac{d{{ m c \gamma }}}{ds} &=\end{aligned}

Putting the final two lines into vector form we have

\begin{aligned}\frac{d{{ (m c^2 \gamma)}}}{dt} = e \mathbf{E} \cdot \mathbf{v},\end{aligned} \hspace{\stretch{1}}(2.27)

or

\begin{aligned}\frac{d{{ \mathcal{E} }}}{dt} = e \mathbf{E} \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.28)

# Transformation of rank two tensors in matrix and index form.

## Transformation of the metric tensor, and some identities.

With

\begin{aligned}\hat{G} = {\left\lVert{ g_{i j} }\right\rVert} = {\left\lVert{ g^{i j} }\right\rVert}\end{aligned} \hspace{\stretch{1}}(3.29)

\paragraph{We claim:}
The rank two tensor $\hat{G}$ transforms in the following sort of sandwich operation, and this leaves it invariant

\begin{aligned}\hat{G} \rightarrow \hat{O} \hat{G} \hat{O}^\text{T} = \hat{G}.\end{aligned} \hspace{\stretch{1}}(3.30)

To demonstrate this let’s consider a transformed vector in coordinate form as follows

\begin{aligned}{x'}^i &= O^{i j} x_j = {O^i}_j x^j \\ {x'}_i &= O_{i j} x^j = {O_i}^j x_j.\end{aligned} \hspace{\stretch{1}}(3.31)

We can thus write the equation in matrix form with

\begin{aligned}X &= {\left\lVert{x^i}\right\rVert} \\ X' &= {\left\lVert{{x'}^i}\right\rVert} \\ \hat{O} &= {\left\lVert{{O^i}_j}\right\rVert} \\ X' &= \hat{O} X\end{aligned} \hspace{\stretch{1}}(3.33)

Our invariant for the vector square, which is required to remain unchanged is

\begin{aligned}{x'}^i {x'}_i &= (O^{i j} x_j)(O_{i k} x^k) \\ &= x^k (O^{i j} O_{i k}) x_j.\end{aligned}

This shows that we have a delta function relationship for the Lorentz transform matrix, when we sum over the first index

\begin{aligned}O^{a i} O_{a j} = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(3.37)

It appears we can put 3.37 into matrix form as

\begin{aligned}\hat{G} \hat{O}^\text{T} \hat{G} \hat{O} = I\end{aligned} \hspace{\stretch{1}}(3.38)

Now, if one considers that the transpose of a rotation is an inverse rotation, and the transpose of a boost leaves it unchanged, the transpose of a general Lorentz transformation, a composition of an arbitrary sequence of boosts and rotations, must also be a Lorentz transformation, and must then also leave the norm unchanged. For the transpose of our Lorentz transformation $\hat{O}$ lets write

\begin{aligned}\hat{P} = \hat{O}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.39)

For the action of this on our position vector let’s write

\begin{aligned}{x''}^i &= P^{i j} x_j = O^{j i} x_j \\ {x''}_i &= P_{i j} x^j = O_{j i} x^j\end{aligned} \hspace{\stretch{1}}(3.40)

so that our norm is

\begin{aligned}{x''}^a {x''}_a &= (O_{k a} x^k)(O^{j a} x_j) \\ &= x^k (O_{k a} O^{j a} ) x_j \\ &= x^j x_j \\ \end{aligned}

We must then also have an identity when summing over the second index

\begin{aligned}{\delta_{k}}^j = O_{k a} O^{j a} \end{aligned} \hspace{\stretch{1}}(3.42)

Armed with these facts on the products of $O_{i j}$ and $O^{i j}$ we can now consider the transformation of the metric tensor.

The rule (definition) supplied to us for the transformation of an arbitrary rank two tensor, is that this transforms as its indexes transform individually. Very much as if it was the product of two coordinate vectors and we transform those coordinates separately. Doing so for the metric tensor we have

\begin{aligned}g^{i j} &\rightarrow {O^i}_k g^{k m} {O^j}_m \\ &= ({O^i}_k g^{k m}) {O^j}_m \\ &= O^{i m} {O^j}_m \\ &= O^{i m} (O_{a m} g^{a j}) \\ &= (O^{i m} O_{a m}) g^{a j}\end{aligned}

However, by 3.42, we have $O_{a m} O^{i m} = {\delta_a}^i$, and we prove that

\begin{aligned}g^{i j} \rightarrow g^{i j}.\end{aligned} \hspace{\stretch{1}}(3.43)

Finally, we wish to put the above transformation in matrix form, look more carefully at the very first line

\begin{aligned}g^{i j}&\rightarrow {O^i}_k g^{k m} {O^j}_m \\ \end{aligned}

which is

\begin{aligned}\hat{G} \rightarrow \hat{O} \hat{G} \hat{O}^\text{T} = \hat{G}\end{aligned} \hspace{\stretch{1}}(3.44)

We see that this particular form of transformation, a sandwich between $\hat{O}$ and $\hat{O}^\text{T}$, leaves the metric tensor invariant.

## Lorentz transformation of the electrodynamic tensor

Having identified a composition of Lorentz transformation matrices, when acting on the metric tensor, leaves it invariant, it is a reasonable question to ask how this form of transformation acts on our electrodynamic tensor $F^{i j}$?

\paragraph{Claim:} A transformation of the following form is required to maintain the norm of the Lorentz force equation

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T} ,\end{aligned} \hspace{\stretch{1}}(3.45)

where $\hat{F} = {\left\lVert{F^{i j}}\right\rVert}$. Observe that our Lorentz force equation can be written exclusively in upper index quantities as

\begin{aligned}m c \frac{d{{u^i}}}{ds} = \frac{e}{c} F^{i j} g_{j l} u^l\end{aligned} \hspace{\stretch{1}}(3.46)

Because we have a vector on one side of the equation, and it transforms by multiplication with by a Lorentz matrix in SO(1,3)

\begin{aligned}\frac{du^i}{ds} \rightarrow \hat{O} \frac{du^i}{ds} \end{aligned} \hspace{\stretch{1}}(3.47)

The LHS of the Lorentz force equation provides us with one invariant

\begin{aligned}(m c)^2 \frac{d{{u^i}}}{ds} \frac{d{{u_i}}}{ds}\end{aligned} \hspace{\stretch{1}}(3.48)

so the RHS must also provide one

\begin{aligned}\frac{e^2}{c^2} F^{i j} g_{j l} u^lF_{i k} g^{k m} u_m=\frac{e^2}{c^2} F^{i j} u_jF_{i k} u^k.\end{aligned} \hspace{\stretch{1}}(3.49)

Let’s look at the RHS in matrix form. Writing

\begin{aligned}U = {\left\lVert{u^i}\right\rVert},\end{aligned} \hspace{\stretch{1}}(3.50)

we can rewrite the Lorentz force equation as

\begin{aligned}m c \dot{U} = \frac{e}{c} \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.51)

In this matrix formalism our invariant 3.49 is

\begin{aligned}\frac{e^2}{c^2} (\hat{F} \hat{G} U)^\text{T} G \hat{F} \hat{G} U=\frac{e^2}{c^2} U^\text{T} \hat{G} \hat{F}^\text{T} G \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.52)

If we compare this to the transformed Lorentz force equation we have

\begin{aligned}m c \hat{O} \dot{U} = \frac{e}{c} \hat{F'} \hat{G} \hat{O} U.\end{aligned} \hspace{\stretch{1}}(3.53)

Our invariant for the transformed equation is

\begin{aligned}\frac{e^2}{c^2} (\hat{F'} \hat{G} \hat{O} U)^\text{T} G \hat{F'} \hat{G} \hat{O} U&=\frac{e^2}{c^2} U^\text{T} \hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} G \hat{F'} \hat{G} \hat{O} U \\ \end{aligned}

Thus the transformed electrodynamic tensor $\hat{F}'$ must satisfy the identity

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} G \hat{F'} \hat{G} \hat{O} = \hat{G} \hat{F}^\text{T} G \hat{F} \hat{G} \end{aligned} \hspace{\stretch{1}}(3.54)

With the substitution $\hat{F}' = \hat{O} \hat{F} \hat{O}^\text{T}$ the LHS is

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} \hat{G} \hat{F'} \hat{G} \hat{O} &= \hat{O}^\text{T} \hat{G} ( \hat{O} \hat{F} \hat{O}^\text{T})^\T \hat{G} (\hat{O} \hat{F} \hat{O}^\text{T}) \hat{G} \hat{O} \\ &= (\hat{O}^\text{T} \hat{G} \hat{O}) \hat{F}^\text{T} (\hat{O}^\text{T} \hat{G} \hat{O}) \hat{F} (\hat{O}^\text{T} \hat{G} \hat{O}) \\ \end{aligned}

We’ve argued that $\hat{P} = \hat{O}^\text{T}$ is also a Lorentz transformation, thus

\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{O}&=\hat{P} \hat{G} \hat{O}^\text{T} \\ &=\hat{G}\end{aligned}

This is enough to make both sides of 3.54 match, verifying that this transformation does provide the invariant properties desired.

## Direct computation of the Lorentz transformation of the electrodynamic tensor.

We can construct the transformed field tensor more directly, by simply transforming the coordinates of the four gradient and the four potential directly. That is

\begin{aligned}F^{i j} = \partial^i A^j - \partial^j A^i&\rightarrow {O^i}_a {O^j}_b \left( \partial^a A^b - \partial^b A^a \right) \\ &={O^i}_a F^{a b} {O^j}_b \end{aligned}

By inspection we can see that this can be represented in matrix form as

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.55)

# Four vector invariants

For three vectors $\mathbf{A}$ and $\mathbf{B}$ invariants are

\begin{aligned}\mathbf{A} \cdot \mathbf{B} = A^\alpha B_\alpha\end{aligned} \hspace{\stretch{1}}(4.56)

For four vectors $A^i$ and $B^i$ invariants are

\begin{aligned}A^i B_i = A^i g_{i j} B^j \end{aligned} \hspace{\stretch{1}}(4.57)

For $F_{i j}$ what are the invariants? One invariant is

\begin{aligned}g^{i j} F_{i j} = 0,\end{aligned} \hspace{\stretch{1}}(4.58)

but this isn’t interesting since it is uniformly zero (product of symmetric and antisymmetric).

The two invariants are

\begin{aligned}F_{i j}F^{i j}\end{aligned} \hspace{\stretch{1}}(4.59)

and

\begin{aligned}\epsilon^{i j k l} F_{i j}F_{k l}\end{aligned} \hspace{\stretch{1}}(4.60)

where

\begin{aligned}\epsilon^{i j k l} =\left\{\begin{array}{l l}0 & \quad \mbox{if any two indexes coincide} \\ 1 & \quad \mbox{for even permutations ofi j k l=0123$} \\ -1 & \quad \mbox{for odd permutations of$i j k l=0123} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(4.61)

We can show (homework) that

\begin{aligned}F_{i j}F^{i j} \propto \mathbf{E}^2 - \mathbf{B}^2\end{aligned} \hspace{\stretch{1}}(4.62)

\begin{aligned}\epsilon^{i j k l} F_{i j}F_{k l} \propto \mathbf{E} \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(4.63)

This first invariant serves as the action density for the Maxwell field equations.

There’s some useful properties of these invariants. One is that if the fields are perpendicular in one frame, then will be in any other.

From the first, note that if ${\left\lvert{\mathbf{E}}\right\rvert} > {\left\lvert{\mathbf{B}}\right\rvert}$, the invariant is positive, and must be positive in all frames, or if ${\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}$ in one frame, we can transform to a frame with only $\mathbf{E}'$ component, solve that, and then transform back. Similarly if ${\left\lvert{\mathbf{E}}\right\rvert} < {\left\lvert{\mathbf{B}}\right\rvert}$ in one frame, we can transform to a frame with only $\mathbf{B}'$ component, solve that, and then transform back.

# The first half of Maxwell’s equations.

\paragraph{Claim: } The source free portions of Maxwell’s equations are a consequence of the definition of the field tensor alone.

Given

\begin{aligned}F_{i j} = \partial_i A_j - \partial_j A_i,\end{aligned} \hspace{\stretch{1}}(5.64)

where

\begin{aligned}\partial_i = \frac{\partial {}}{\partial {x^i}}\end{aligned} \hspace{\stretch{1}}(5.65)

This alone implies half of Maxwell’s equations. To show this we consider

\begin{aligned}e^{m k i j} \partial_k F_{i j} = 0.\end{aligned} \hspace{\stretch{1}}(5.66)

This is the Bianchi identity. To demonstrate this identity, we’ll have to swap indexes, employ derivative commutation, and then swap indexes once more

\begin{aligned}e^{m k i j} \partial_k F_{i j} &= e^{m k i j} \partial_k (\partial_i A_j - \partial_j A_i) \\ &= 2 e^{m k i j} \partial_k \partial_i A_j \\ &= 2 e^{m k i j} \frac{1}{{2}} \left( \partial_k \partial_i A_j + \partial_i \partial_k A_j \right) \\ &= e^{m k i j} \partial_k \partial_i A_j e^{m i k j} \partial_k \partial_i A_j \\ &= (e^{m k i j} - e^{m k i j}) \partial_k \partial_i A_j \\ &= 0 \qquad \square\end{aligned}

This is the 4D analogue of

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} f) = 0\end{aligned} \hspace{\stretch{1}}(5.67)

i.e.

\begin{aligned}e^{\alpha\beta\gamma} \partial_\beta \partial_\gamma f = 0\end{aligned} \hspace{\stretch{1}}(5.68)

Let’s do this explicitly, starting with

\begin{aligned}{\left\lVert{ F_{i j} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.69)

For the $m= 0$ case we have

\begin{aligned}\epsilon^{0 k i j} \partial_k F_{i j}&=\epsilon^{\alpha \beta \gamma} \partial_\alpha F_{\beta \gamma} \\ &= \epsilon^{\alpha \beta \gamma} \partial_\alpha (-\epsilon_{\beta \gamma \delta} B_\delta) \\ &= -\epsilon^{\alpha \beta \gamma} \epsilon_{\delta \beta \gamma }\partial_\alpha B_\delta \\ &= - 2 {\delta^\alpha}_\delta \partial_\alpha B_\delta \\ &= - 2 \partial_\alpha B_\alpha \end{aligned}

We must then have

\begin{aligned}\partial_\alpha B_\alpha = 0.\end{aligned} \hspace{\stretch{1}}(5.70)

This is just Gauss’s law for magnetism

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} = 0.\end{aligned} \hspace{\stretch{1}}(5.71)

Let’s do the spatial portion, for which we have three equations, one for each $\alpha$ of

\begin{aligned}e^{\alpha j k l} \partial_j F_{k l}&=e^{\alpha 0 \beta \gamma} \partial_0 F_{\beta \gamma}+e^{\alpha 0 \gamma \beta} \partial_0 F_{\gamma \beta}+e^{\alpha \beta 0 \gamma} \partial_\beta F_{0 \gamma}+e^{\alpha \beta \gamma 0} \partial_\beta F_{\gamma 0}+e^{\alpha \gamma 0 \beta} \partial_\gamma F_{0 \beta}+e^{\alpha \gamma \beta 0} \partial_\gamma F_{\beta 0} \\ &=2 \left( e^{\alpha 0 \beta \gamma} \partial_0 F_{\beta \gamma}+e^{\alpha \beta 0 \gamma} \partial_\beta F_{0 \gamma}+e^{\alpha \gamma 0 \beta} \partial_\gamma F_{0 \beta}\right) \\ &=2 e^{0 \alpha \beta \gamma} \left(-\partial_0 F_{\beta \gamma}+\partial_\beta F_{0 \gamma}- \partial_\gamma F_{0 \beta}\right)\end{aligned}

This implies

\begin{aligned}0 =-\partial_0 F_{\beta \gamma}+\partial_\beta F_{0 \gamma}- \partial_\gamma F_{0 \beta}\end{aligned} \hspace{\stretch{1}}(5.72)

Referring back to the previous expansions of 2.6 and 2.17, we have

\begin{aligned}0 =\partial_0 \epsilon_{\beta\gamma\mu} B_\mu+\partial_\beta E_\gamma- \partial_\gamma E_{\beta},\end{aligned} \hspace{\stretch{1}}(5.73)

or

\begin{aligned}\frac{1}{{c}} \frac{\partial {B_\alpha}}{\partial {t}} + (\boldsymbol{\nabla} \times \mathbf{E})_\alpha = 0.\end{aligned} \hspace{\stretch{1}}(5.74)

These are just the components of the Maxwell-Faraday equation

\begin{aligned}0 = \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} + \boldsymbol{\nabla} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(5.75)

# Appendix. Some additional index gymnastics.

## Transposition of mixed index tensor.

Is the transpose of a mixed index object just a substitution of the free indexes? This wasn’t obvious to me that it would be the case, especially since I’d made an error in some index gymnastics that had me temporarily convinced differently. However, working some examples clears the fog. For example let’s take the transpose of 3.37.

\begin{aligned}{\left\lVert{ {\delta^i}_j }\right\rVert}^\text{T} &= {\left\lVert{ O^{a i} O_{a j} }\right\rVert}^\text{T} \\ &= \left( {\left\lVert{ O^{j i} }\right\rVert} {\left\lVert{ O_{i j} }\right\rVert} \right)^\text{T} \\ &={\left\lVert{ O_{i j} }\right\rVert}^\text{T}{\left\lVert{ O^{j i} }\right\rVert}^\text{T} \\ &={\left\lVert{ O_{j i} }\right\rVert}{\left\lVert{ O^{i j} }\right\rVert} \\ &={\left\lVert{ O_{a i} O^{a j} }\right\rVert} \\ \end{aligned}

If the transpose of a mixed index tensor just swapped the indexes we would have

\begin{aligned}{\left\lVert{ {\delta^i}_j }\right\rVert}^\text{T} = {\left\lVert{ O_{a i} O^{a j} }\right\rVert} \end{aligned} \hspace{\stretch{1}}(6.76)

From this it does appear that all we have to do is switch the indexes and we will write

\begin{aligned}{\delta^j}_i = O_{a i} O^{a j} \end{aligned} \hspace{\stretch{1}}(6.77)

We can consider a more general operation

\begin{aligned}{\left\lVert{{A^i}_j}\right\rVert}^\text{T}&={\left\lVert{ A^{i m} g_{m j} }\right\rVert}^\text{T} \\ &={\left\lVert{ g_{i j} }\right\rVert}^\text{T}{\left\lVert{ A^{i j} }\right\rVert}^\text{T} \\ &={\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ A^{j i} }\right\rVert} \\ &={\left\lVert{ g_{i m} A^{j m} }\right\rVert} \\ &={\left\lVert{ {A^{j}}_i }\right\rVert}\end{aligned}

So we see that we do just have to swap indexes.

## Transposition of lower index tensor.

We’ve saw above that we had

\begin{aligned}{\left\lVert{ {A^{i}}_j }\right\rVert}^\text{T} &= {\left\lVert{ {A_{j}}^i }\right\rVert} \\ {\left\lVert{ {A_{i}}^j }\right\rVert}^\text{T} &= {\left\lVert{ {A^{j}}_i }\right\rVert} \end{aligned} \hspace{\stretch{1}}(6.78)

which followed by careful treatment of the transposition in terms of $A^{i j}$ for which we defined a transpose operation. We assumed as well that

\begin{aligned}{\left\lVert{ A_{i j} }\right\rVert}^\text{T} = {\left\lVert{ A_{j i} }\right\rVert}.\end{aligned} \hspace{\stretch{1}}(6.80)

However, this does not have to be assumed, provided that $g^{i j} = g_{i j}$, and $(AB)^\text{T} = B^\text{T} A^\text{T}$. We see this by expanding this transposition in products of $A^{i j}$ and $\hat{G}$

\begin{aligned}{\left\lVert{ A_{i j} }\right\rVert}^\text{T}&= \left( {\left\lVert{g_{i j}}\right\rVert} {\left\lVert{ A^{i j} }\right\rVert} {\left\lVert{g_{i j}}\right\rVert} \right)^\text{T} \\ &= \left( {\left\lVert{g^{i j}}\right\rVert} {\left\lVert{ A^{i j} }\right\rVert} {\left\lVert{g^{i j}}\right\rVert} \right)^\text{T} \\ &= {\left\lVert{g^{i j}}\right\rVert}^\text{T} {\left\lVert{ A^{i j}}\right\rVert}^\text{T} {\left\lVert{g^{i j}}\right\rVert}^\text{T} \\ &= {\left\lVert{g^{i j}}\right\rVert} {\left\lVert{ A^{j i}}\right\rVert} {\left\lVert{g^{i j}}\right\rVert} \\ &= {\left\lVert{g_{i j}}\right\rVert} {\left\lVert{ A^{i j}}\right\rVert} {\left\lVert{g_{i j}}\right\rVert} \\ &= {\left\lVert{ A_{j i}}\right\rVert} \end{aligned}

It would be worthwhile to go through all of this index manipulation stuff and lay it out in a structured axiomatic form. What is the minimal set of assumptions, and how does all of this generalize to non-diagonal metric tensors (even in Euclidean spaces).

## Translating the index expression of identity from Lorentz products to matrix form

A verification that the matrix expression 3.38, matches the index expression 3.37 as claimed is worthwhile. It would be easy to guess something similar like $\hat{O}^\text{T} \hat{G} \hat{O} \hat{G}$ is instead the matrix representation. That was in fact my first erroneous attempt to form the matrix equivalent, but is the transpose of 3.38. Either way you get an identity, but the indexes didn’t match.

Since we have $g^{i j} = g_{i j}$ which do we pick to do this verification? This appears to be dictated by requirements to match lower and upper indexes on the summed over index. This is probably clearest by example, so let’s expand the products on the LHS explicitly

\begin{aligned}{\left\lVert{ g^{i j} }\right\rVert} {\left\lVert{ {O^{i}}_j }\right\rVert} ^\text{T}{\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ {O^{i}}_j }\right\rVert} &=\left( {\left\lVert{ {O^{i}}_j }\right\rVert} {\left\lVert{ g^{i j} }\right\rVert} \right) ^\text{T}{\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ {O^{i}}_j }\right\rVert} \\ &=\left( {\left\lVert{ {O^{i}}_k g^{k j} }\right\rVert} \right) ^\text{T}{\left\lVert{ g_{i m} {O^{m}}_j }\right\rVert} \\ &={\left\lVert{ O^{i j} }\right\rVert} ^\text{T}{\left\lVert{ O_{i j} }\right\rVert} \\ &={\left\lVert{ O^{j i} }\right\rVert} {\left\lVert{ O_{i j} }\right\rVert} \\ &={\left\lVert{ O^{k i} O_{k j} }\right\rVert} \\ \end{aligned}

This matches the ${\left\lVert{{\delta^i}_j}\right\rVert}$ that we have on the RHS, and all is well.

# References

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

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

Posted by peeterjoot on February 22, 2011

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.

## Exploring Stokes Theorem in tensor form.

Posted by peeterjoot on February 22, 2011

# Obsolete with potential errors.

This post may be in error.  I wrote this before understanding that the gradient used in Stokes Theorem must be projected onto the tangent space of the parameterized surface, as detailed in Alan MacDonald’s Vector and Geometric Calculus.

See the post ‘stokes theorem in geometric algebra‘ [PDF], where this topic has been revisited with this in mind.

# Original Post:

## Motivation.

I’ve worked through Stokes theorem concepts a couple times on my own now. One of the first times, I was trying to formulate this in a Geometric Algebra context. I had to resort to a tensor decomposition, and pictures, before ending back in the Geometric Algebra description. Later I figured out how to do it entirely with a Geometric Algebra description, and was able to eliminate reliance on the pictures that made the path to generalization to higher dimensional spaces unclear.

It’s my expectation that if one started with a tensor description, the proof entirely in tensor form would not be difficult. This is what I’d like to try this time. To start off, I’ll temporarily use the Geometric Algebra curl expression so I know what my tensor equation starting point will be, but once that starting point is found, we can work entirely in coordinate representation. For somebody who already knows that this is the starting point, all of this initial motivation can be skipped.

# Translating the exterior derivative to a coordinate representation.

Our starting point is a curl, dotted with a volume element of the same grade, so that the result is a scalar

\begin{aligned}\int d^n x \cdot (\nabla \wedge A).\end{aligned} \hspace{\stretch{1}}(2.1)

Here $A$ is a blade of grade $n-1$, and we wedge this with the gradient for the space

\begin{aligned}\nabla \equiv e^i \partial_i = e_i \partial^i,\end{aligned} \hspace{\stretch{1}}(2.2)

where we with with a basis (not necessarily orthonormal) $\{e_i\}$, and the reciprocal frame for that basis $\{e^i\}$ defined by the relation

\begin{aligned}e^i \cdot e_j = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(2.3)

Our coordinates in these basis sets are

\begin{aligned}x \cdot e^i & \equiv x^i \\ x \cdot e_i & \equiv x_i\end{aligned} \hspace{\stretch{1}}(2.4)

so that

\begin{aligned}x = x^i e_i = x_i e^i.\end{aligned} \hspace{\stretch{1}}(2.6)

The operator coordinates of the gradient are defined in the usual fashion

\begin{aligned}\partial_i & \equiv \frac{\partial }{\partial {x^i}} \\ \partial^i & \equiv \frac{\partial}{\partial {x_i}}\end{aligned} \hspace{\stretch{1}}(2.7)

The volume element for the subspace that we are integrating over we will define in terms of an arbitrary parametrization

\begin{aligned}x = x(\alpha_1, \alpha_2, \cdots, \alpha_n)\end{aligned} \hspace{\stretch{1}}(2.9)

The subspace can be considered spanned by the differential elements in each of the respective curves where all but the $i$th parameter are held constant.

\begin{aligned}dx_{\alpha_i}= d\alpha_i \frac{\partial x}{\partial {\alpha_i}}= d\alpha_i \frac{\partial {x^j}}{\partial {\alpha_i}} e_j.\end{aligned} \hspace{\stretch{1}}(2.10)

We assume that the integral is being performed in a subspace for which none of these differential elements in that region are linearly dependent (i.e. our Jacobean determinant must be non-zero).

The magnitude of the wedge product of all such differential elements provides the volume of the parallelogram, or parallelepiped (or higher dimensional analogue), and is

\begin{aligned}d^n x=d\alpha_1 d\alpha_2\cdots d\alpha_n\frac{\partial x}{\partial {\alpha_n}} \wedge\cdots \wedge\frac{\partial x}{\partial {\alpha_2}}\wedge\frac{\partial x}{\partial {\alpha_1}}.\end{aligned} \hspace{\stretch{1}}(2.11)

The volume element is a oriented quantity, and may be adjusted with an arbitrary sign (or equivalently an arbitrary permutation of the differential elements in the wedge product), and we’ll see that it is convenient for the translation to tensor form, to express these in reversed order.

Let’s write

\begin{aligned}d^n \alpha = d\alpha_1 d\alpha_2 \cdots d\alpha_n,\end{aligned} \hspace{\stretch{1}}(2.12)

so that our volume element in coordinate form is

\begin{aligned}d^n x = d^n \alpha\frac{\partial {x^i}}{\partial {\alpha_1}}\frac{\partial {x^j}}{\partial {\alpha_2}}\cdots \frac{\partial {x^k}}{\partial {\alpha_{n-1}}}\frac{\partial {x^l}}{\partial {\alpha_n}}( e_l \wedge e_k \wedge \cdots \wedge e_j \wedge e_i ).\end{aligned} \hspace{\stretch{1}}(2.13)

Our curl will also also be a grade $n$ blade. We write for the $n-1$ grade blade

\begin{aligned}A = A_{b c \cdots d} (e^b \wedge e^c \wedge \cdots e^d),\end{aligned} \hspace{\stretch{1}}(2.14)

where $A_{b c \cdots d}$ is antisymmetric (i.e. $A = a_1 \wedge a_2 \wedge \cdots a_{n-1}$ for a some set of vectors $a_i, i \in 1 .. n-1$).

With our gradient in coordinate form

\begin{aligned}\nabla = e^a \partial_a,\end{aligned} \hspace{\stretch{1}}(2.15)

the curl is then

\begin{aligned}\nabla \wedge A = \partial_a A_{b c \cdots d} (e^a \wedge e^b \wedge e^c \wedge \cdots e^d).\end{aligned} \hspace{\stretch{1}}(2.16)

The differential form for our integral can now be computed by expanding out the dot product. We want

\begin{aligned}( e_l \wedge e_k \wedge \cdots \wedge e_j \wedge e_i )\cdot(e^a \wedge e^b \wedge e^c \wedge \cdots e^d)=((((( e_l \wedge e_k \wedge \cdots \wedge e_j \wedge e_i ) \cdot e^a ) \cdot e^b ) \cdot e^c ) \cdot \cdots ) \cdot e^d.\end{aligned} \hspace{\stretch{1}}(2.17)

Evaluation of the interior dot products introduces the intrinsic antisymmetry required for Stokes theorem. For example, with

\begin{aligned}( e_n \wedge e_{n-1} \wedge \cdots \wedge e_2 \wedge e_1 ) \cdot e^a a & =( e_n \wedge e_{n-1} \wedge \cdots \wedge e_3 \wedge e_2 ) (e_1 \cdot e^a) \\ & -( e_n \wedge e_{n-1} \wedge \cdots \wedge e_3 \wedge e_1 ) (e_2 \cdot e^a) \\ & +( e_n \wedge e_{n-1} \wedge \cdots \wedge e_2 \wedge e_1 ) (e_3 \cdot e^a) \\ & \cdots \\ & (-1)^{n-1}( e_{n-1} \wedge e_{n-2} \wedge \cdots \wedge e_2 \wedge e_1 ) (e_n \cdot e^a)\end{aligned}

Since $e_i \cdot e^a = {\delta_i}^a$ our end result is a completely antisymmetric set of permutations of all the deltas

\begin{aligned}( e_l \wedge e_k \wedge \cdots \wedge e_j \wedge e_i )\cdot(e^a \wedge e^b \wedge e^c \wedge \cdots e^d)={\delta^{[a}}_i{\delta^b}_j\cdots {\delta^{d]}}_l,\end{aligned} \hspace{\stretch{1}}(2.18)

and the curl integral takes it’s coordinate form

\begin{aligned}\int d^n x \cdot ( \nabla \wedge A ) =\int d^n \alpha\frac{\partial {x^i}}{\partial {\alpha_1}}\frac{\partial {x^j}}{\partial {\alpha_2}}\cdots \frac{\partial {x^k}}{\partial {\alpha_{n-1}}}\frac{\partial {x^l}}{\partial {\alpha_n}}\partial_a A_{b c \cdots d}{\delta^{[a}}_i{\delta^b}_j\cdots {\delta^{d]}}_l.\end{aligned} \hspace{\stretch{1}}(2.19)

One final contraction of the paired indexes gives us our Stokes integral in its coordinate representation

\begin{aligned}\boxed{\int d^n x \cdot ( \nabla \wedge A ) =\int d^n \alpha\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^b}}{\partial {\alpha_2}}\cdots \frac{\partial {x^c}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{d]}}}{\partial {\alpha_n}}\partial_a A_{b c \cdots d}}\end{aligned} \hspace{\stretch{1}}(2.20)

We now have a starting point that is free of any of the abstraction of Geometric Algebra or differential forms. We can identify the products of partials here as components of a scalar hypervolume element (possibly signed depending on the orientation of the parametrization)

\begin{aligned}d\alpha_1 d\alpha_2\cdots d\alpha_n\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^b}}{\partial {\alpha_2}}\cdots \frac{\partial {x^c}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{d]}}}{\partial {\alpha_n}}\end{aligned} \hspace{\stretch{1}}(2.21)

This is also a specific computation recipe for these hypervolume components, something that may not be obvious when we allow for general metrics for the space. We are also allowing for non-orthonormal coordinate representations, and arbitrary parametrization of the subspace that we are integrating over (our integral need not have the same dimension as the underlying vector space).

Observe that when the number of parameters equals the dimension of the space, we can write out the antisymmetric term utilizing the determinant of the Jacobian matrix

\begin{aligned}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^b}}{\partial {\alpha_2}}\cdots \frac{\partial {x^c}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{d]}}}{\partial {\alpha_n}}= \epsilon^{a b \cdots d} {\left\lvert{ \frac{\partial(x^1, x^2, \cdots x^n)}{\partial(\alpha_1, \alpha_2, \cdots \alpha_n)} }\right\rvert}\end{aligned} \hspace{\stretch{1}}(2.22)

When the dimension of the space $n$ is greater than the number of parameters for the integration hypervolume in question, the antisymmetric sum of partials is still the determinant of a Jacobian matrix

\begin{aligned}\frac{\partial {x^{[a_1}}}{\partial {\alpha_1}}\frac{\partial {x^{a_2}}}{\partial {\alpha_2}}\cdots \frac{\partial {x^{a_{n-1}}}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{a_n]}}}{\partial {\alpha_n}}= {\left\lvert{ \frac{\partial(x^{a_1}, x^{a_2}, \cdots x^{a_n})}{\partial(\alpha_1, \alpha_2, \cdots \alpha_n)} }\right\rvert},\end{aligned} \hspace{\stretch{1}}(2.23)

however, we will have one such Jacobian for each unique choice of indexes.

# The Stokes work starts here.

The task is to relate our integral to the boundary of this volume, coming up with an explicit recipe for the description of that bounding surface, and determining the exact form of the reduced rank integral. This job is essentially to reduce the ranks of the tensors that are being contracted in our Stokes integral. With the derivative applied to our rank $n-1$ antisymmetric tensor $A_{b c \cdots d}$, we can apply the chain rule and examine the permutations so that this can be rewritten as a contraction of $A$ itself with a set of rank $n-1$ surface area elements.

\begin{aligned}\int d^n \alpha\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^b}}{\partial {\alpha_2}}\cdots \frac{\partial {x^c}}{\partial {\alpha_{n-1}}}\frac{\partial {x^{d]}}}{\partial {\alpha_n}}\partial_a A_{b c \cdots d} = ?\end{aligned} \hspace{\stretch{1}}(3.24)

Now, while the setup here has been completely general, this task is motivated by study of special relativity, where there is a requirement to work in a four dimensional space. Because of that explicit goal, I’m not going to attempt to formulate this in a completely abstract fashion. That task is really one of introducing sufficiently general notation. Instead, I’m going to proceed with a simpleton approach, and do this explicitly, and repeatedly for each of the rank 1, rank 2, and rank 3 tensor cases. It will be clear how this all generalizes by doing so, should one wish to work in still higher dimensional spaces.

## The rank 1 tensor case.

The equation we are working with for this vector case is

\begin{aligned}\int d^2 x \cdot (\nabla \wedge A) =\int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\partial_a A_{b}(\alpha_1, \alpha_2)\end{aligned} \hspace{\stretch{1}}(3.25)

Expanding out the antisymmetric partials we have

\begin{aligned}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}} & =\frac{\partial {x^{a}}}{\partial {\alpha_1}}\frac{\partial {x^{b}}}{\partial {\alpha_2}}-\frac{\partial {x^{b}}}{\partial {\alpha_1}}\frac{\partial {x^{a}}}{\partial {\alpha_2}},\end{aligned}

with which we can reduce the integral to

\begin{aligned}\int d^2 x \cdot (\nabla \wedge A) & =\int \left( d{\alpha_1}\frac{\partial {x^{a}}}{\partial {\alpha_1}}\frac{\partial {A_{b}}}{\partial {x^a}} \right)\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-\left( d{\alpha_2}\frac{\partial {x^{a}}}{\partial {\alpha_2}}\frac{\partial {A_{b}}}{\partial {x^a}} \right)\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1} \\ & =\int \left( d\alpha_1 \frac{\partial {A_b}}{\partial {\alpha_1}} \right)\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-\left( d\alpha_2 \frac{\partial {A_b}}{\partial {\alpha_2}} \right)\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1} \\ \end{aligned}

Now, if it happens that

\begin{aligned}\frac{\partial}{\partial {\alpha_1}}\frac{\partial {x^{a}}}{\partial {\alpha_2}} = \frac{\partial}{\partial {\alpha_2}}\frac{\partial {x^{a}}}{\partial {\alpha_1}} = 0\end{aligned} \hspace{\stretch{1}}(3.26)

then each of the individual integrals in $d\alpha_1$ and $d\alpha_2$ can be carried out. In that case, without any real loss of generality we can designate the integration bounds over the unit parametrization space square $\alpha_i \in [0,1]$, allowing this integral to be expressed as

\begin{aligned}\begin{aligned} & \int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\partial_a A_{b}(\alpha_1, \alpha_2) \\ & =\int \left( A_b(1, \alpha_2) - A_b(0, \alpha_2) \right)\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-\left( A_b(\alpha_1, 1) - A_b(\alpha_1, 0) \right)\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.27)

It’s also fairly common to see ${\left.{{A}}\right\vert}_{{\partial \alpha_i}}$ used to designate evaluation of this first integral on the boundary, and using this we write

\begin{aligned}\int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\partial_a A_{b}(\alpha_1, \alpha_2)=\int {\left.{{A_b}}\right\vert}_{{\partial \alpha_1}}\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-{\left.{{A_b}}\right\vert}_{{\partial \alpha_2}}\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1}.\end{aligned} \hspace{\stretch{1}}(3.28)

Also note that since we are summing over all $a,b$, and have

\begin{aligned}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}=-\frac{\partial {x^{[b}}}{\partial {\alpha_1}}\frac{\partial {x^{a]}}}{\partial {\alpha_2}},\end{aligned} \hspace{\stretch{1}}(3.29)

we can write this summing over all unique pairs of $a,b$ instead, which eliminates a small bit of redundancy (especially once the dimension of the vector space gets higher)

\begin{aligned}\boxed{\sum_{a < b}\int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\left( \partial_a A_{b}-\partial_b A_{a} \right)=\int {\left.{{A_b}}\right\vert}_{{\partial \alpha_1}}\frac{\partial {x^{b}}}{\partial {\alpha_2}} d{\alpha_2}-{\left.{{A_b}}\right\vert}_{{\partial \alpha_2}}\frac{\partial {x^{b}}}{\partial {\alpha_1}} d{\alpha_1}.}\end{aligned} \hspace{\stretch{1}}(3.30)

In this form we have recovered the original geometric structure, with components of the curl multiplied by the component of the area element that shares the orientation and direction of that portion of the curl bivector.

This form of the result with evaluation at the boundaries in this form, assumed that ${\partial {x^a}}/{\partial {\alpha_1}}$ was not a function of $\alpha_2$ and ${\partial {x^a}}/{\partial {\alpha_2}}$ was not a function of $\alpha_1$. When that is not the case, we appear to have a less pretty result

\begin{aligned}\boxed{\sum_{a < b}\int d{\alpha_1} d{\alpha_2}\frac{\partial {x^{[a}}}{\partial {\alpha_1}}\frac{\partial {x^{b]}}}{\partial {\alpha_2}}\left( \partial_a A_{b}-\partial_b A_{a} \right)=\int d\alpha_2\int d\alpha_1\frac{\partial {A_b}}{\partial {\alpha_1}}\frac{\partial {x^{b}}}{\partial {\alpha_2}}-\int d\alpha_2\int d\alpha_1\frac{\partial {A_b}}{\partial {\alpha_2}}\frac{\partial {x^{b}}}{\partial {\alpha_1}}}\end{aligned} \hspace{\stretch{1}}(3.31)

Can this be reduced any further in the general case? Having seen the statements of Stokes theorem in it’s differential forms formulation, I initially expected the answer was yes, and only when I got to evaluating my $\mathbb{R}^{4}$ spacetime example below did I realize that the differentials displacements for the parallelogram that constituted the area element were functions of both parameters. Perhaps this detail is there in the differential forms version of the general Stokes theorem too, but is just hidden in a tricky fashion by the compact notation.

### Sanity check: $\mathbb{R}^{2}$ case in rectangular coordinates.

For $x^1 = x, x^2 = y$, and $\alpha_1 = x, \alpha_2 = y$, we have for the LHS

\begin{aligned} & \int_{x=x_0}^{x_1}\int_{y=y_0}^{y_1}dx dy\left(\frac{\partial {x^{1}}}{\partial {\alpha_1}}\frac{\partial {x^{2}}}{\partial {\alpha_2}}-\frac{\partial {x^{2}}}{\partial {\alpha_1}}\frac{\partial {x^{1}}}{\partial {\alpha_2}}\right)\partial_1 A_{2}+\left(\frac{\partial {x^{2}}}{\partial {\alpha_1}}\frac{\partial {x^{1}}}{\partial {\alpha_2}}-\frac{\partial {x^{1}}}{\partial {\alpha_1}}\frac{\partial {x^{2}}}{\partial {\alpha_2}}\right)\partial_2 A_{1} \\ & =\int_{x=x_0}^{x_1}\int_{y=y_0}^{y_1}dx dy\left( \frac{\partial {A_y}}{\partial x} - \frac{\partial {A_x}}{\partial y} \right)\end{aligned}

Our RHS expands to

\begin{aligned} & \int_{y=y_0}^{y_1} dy\left(\left( A_1(x_1, y) - A_1(x_0, y) \right)\frac{\partial {x^{1}}}{\partial y}+\left( A_2(x_1, y) - A_2(x_0, y) \right)\frac{\partial {x^{2}}}{\partial y}\right) \\ & \qquad-\int_{x=x_0}^{x_1} dx\left(\left( A_1(x, y_1) - A_1(x, y_0) \right)\frac{\partial {x^{1}}}{\partial x}+\left( A_2(x, y_1) - A_2(x, y_0) \right)\frac{\partial {x^{2}}}{\partial x}\right) \\ & =\int_{y=y_0}^{y_1} dy\left( A_y(x_1, y) - A_y(x_0, y) \right)-\int_{x=x_0}^{x_1} dx\left( A_x(x, y_1) - A_x(x, y_0) \right)\end{aligned}

We have

\begin{aligned}\begin{aligned} & \int_{x=x_0}^{x_1}\int_{y=y_0}^{y_1}dx dy\left( \frac{\partial {A_y}}{\partial x} - \frac{\partial {A_x}}{\partial y} \right) \\ & =\int_{y=y_0}^{y_1} dy\left( A_y(x_1, y) - A_y(x_0, y) \right)-\int_{x=x_0}^{x_1} dx\left( A_x(x, y_1) - A_x(x, y_0) \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.32)

The RHS is just a positively oriented line integral around the rectangle of integration

\begin{aligned}\int A_x(x, y_0) \hat{\mathbf{x}} \cdot ( \hat{\mathbf{x}} dx )+ A_y(x_1, y) \hat{\mathbf{y}} \cdot ( \hat{\mathbf{y}} dy )+ A_x(x, y_1) \hat{\mathbf{x}} \cdot ( -\hat{\mathbf{x}} dx )+ A_y(x_0, y) \hat{\mathbf{y}} \cdot ( -\hat{\mathbf{y}} dy )= \oint \mathbf{A} \cdot d\mathbf{r}.\end{aligned} \hspace{\stretch{1}}(3.33)

This special case is also recognizable as Green’s theorem, evident with the substitution $A_x = P$, $A_y = Q$, which gives us

\begin{aligned}\int_A dx dy \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)=\oint_C P dx + Q dy.\end{aligned} \hspace{\stretch{1}}(3.34)

Strictly speaking, Green’s theorem is more general, since it applies to integration regions more general than rectangles, but that generalization can be arrived at easily enough, once the region is broken down into adjoining elementary regions.

### Sanity check: $\mathbb{R}^{3}$ case in rectangular coordinates.

It is expected that we can recover the classical Kelvin-Stokes theorem if we use rectangular coordinates in $\mathbb{R}^{3}$. However, we see that we have to consider three different parametrizations. If one picks rectangular parametrizations $(\alpha_1, \alpha_2) = \{ (x,y), (y,z), (z,x) \}$ in sequence, in each case holding the value of the additional coordinate fixed, we get three different independent Green’s function like relations

\begin{aligned}\int_A dx dy \left( \frac{\partial {A_y}}{\partial x} - \frac{\partial {A_x}}{\partial y} \right) & = \oint_C A_x dx + A_y dy \\ \int_A dy dz \left( \frac{\partial {A_z}}{\partial y} - \frac{\partial {A_y}}{\partial z} \right) & = \oint_C A_y dy + A_z dz \\ \int_A dz dx \left( \frac{\partial {A_x}}{\partial z} - \frac{\partial {A_z}}{\partial x} \right) & = \oint_C A_z dz + A_x dx.\end{aligned} \hspace{\stretch{1}}(3.35)

Note that we cannot just add these to form a complete integral $\oint \mathbf{A} \cdot d\mathbf{r}$ since the curves are all have different orientations. To recover the $\mathbb{R}^{3}$ Stokes theorem in rectangular coordinates, it appears that we’d have to consider a Riemann sum of triangular surface elements, and relate that to the loops over each of the surface elements. In that limiting argument, only the boundary of the complete surface would contribute to the RHS of the relation.

All that said, we shouldn’t actually have to go to all this work. Instead we can stick to a two variable parametrization of the surface, and use 3.30 directly.

### An illustration for a $\mathbb{R}^{4}$ spacetime surface.

Suppose we have a particle trajectory defined by an active Lorentz transformation from an initial spacetime point

\begin{aligned}x^i = O^{ij} x_j(0) = O^{ij} g_{jk} x^k = {O^{i}}_k x^k(0)\end{aligned} \hspace{\stretch{1}}(3.38)

Let the Lorentz transformation be formed by a composition of boost and rotation

\begin{aligned}{O^i}_j & = {L^i}_k {R^k}_j \\ {L^i}_j & =\begin{bmatrix}\cosh_\alpha & -\sinh\alpha & 0 & 0 \\ -\sinh_\alpha & \cosh\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ {R^i}_j & =\begin{bmatrix}1 & 0 & 0 & 0 \\ \cos_\alpha & \sin\alpha & 0 & 0 \\ -\sin_\alpha & \cos\alpha & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.39)

Different rates of evolution of $\alpha$ and $\theta$ define different trajectories, and taken together we have a surface described by the two parameters

\begin{aligned}x^i(\alpha, \theta) = {L^i}_k {R^k}_j x^j(0, 0).\end{aligned} \hspace{\stretch{1}}(3.42)

We can compute displacements along the trajectories formed by keeping either $\alpha$ or $\theta$ fixed and varying the other. Those are

\begin{aligned}\frac{\partial {x^i}}{\partial {\alpha}} d\alpha & = \frac{d{L^i}_k}{d\alpha} {R^k}_j x^j(0, 0) \\ \frac{\partial {x^i}}{\partial {\theta}} d\theta & = {L^i}_k \frac{d{R^k}_j}{d\theta} x^j(0, 0) .\end{aligned} \hspace{\stretch{1}}(3.43)

Writing $y^i = x^i(0,0)$ the computation of the partials above yields

\begin{aligned}\frac{\partial {x^i}}{\partial {\alpha}} & =\begin{bmatrix}\sinh\alpha y^0 -\cosh\alpha (\cos\theta y^1 + \sin\theta y^2) \\ -\cosh\alpha y^0 +\sinh\alpha (\cos\theta y^1 + \sin\theta y^2) \\ 0 \\ 0\end{bmatrix} \\ \frac{\partial {x^i}}{\partial {\theta}} & =\begin{bmatrix}-\sinh\alpha (-\sin\theta y^1 + \cos\theta y^2 ) \\ \cosh\alpha (-\sin\theta y^1 + \cos\theta y^2 ) \\ -(\cos\theta y^1 + \sin\theta y^2 ) \\ 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.45)

Different choices of the initial point $y^i$ yield different surfaces, but we can get the idea by picking a simple starting point $y^i = (0, 1, 0, 0)$ leaving

\begin{aligned}\frac{\partial {x^i}}{\partial {\alpha}} & =\begin{bmatrix}-\cosh\alpha \cos\theta \\ \sinh\alpha \cos\theta \\ 0 \\ 0\end{bmatrix} \\ \frac{\partial {x^i}}{\partial {\theta}} & =\begin{bmatrix}\sinh\alpha \sin\theta \\ -\cosh\alpha \sin\theta \\ -\cos\theta \\ 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.47)

We can now compute our Jacobian determinants

\begin{aligned}\frac{\partial {x^{[a}}}{\partial {\alpha}} \frac{\partial {x^{b]}}}{\partial {\theta}}={\left\lvert{\frac{\partial(x^a, x^b)}{\partial(\alpha, \theta)}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(3.49)

Those are

\begin{aligned}{\left\lvert{\frac{\partial(x^0, x^1)}{\partial(\alpha, \theta)}}\right\rvert} & = \cos\theta \sin\theta \\ {\left\lvert{\frac{\partial(x^0, x^2)}{\partial(\alpha, \theta)}}\right\rvert} & = \cosh\alpha \cos^2\theta \\ {\left\lvert{\frac{\partial(x^0, x^3)}{\partial(\alpha, \theta)}}\right\rvert} & = 0 \\ {\left\lvert{\frac{\partial(x^1, x^2)}{\partial(\alpha, \theta)}}\right\rvert} & = -\sinh\alpha \cos^2\theta \\ {\left\lvert{\frac{\partial(x^1, x^3)}{\partial(\alpha, \theta)}}\right\rvert} & = 0 \\ {\left\lvert{\frac{\partial(x^2, x^3)}{\partial(\alpha, \theta)}}\right\rvert} & = 0\end{aligned} \hspace{\stretch{1}}(3.50)

Using this, let’s see a specific 4D example in spacetime for the integral of the curl of some four vector $A^i$, enumerating all the non-zero components of 3.31 for this particular spacetime surface

\begin{aligned}\sum_{a < b}\int d{\alpha} d{\theta}{\left\lvert{\frac{\partial(x^a, x^b)}{\partial(\alpha, \theta)}}\right\rvert}\left( \partial_a A_{b}-\partial_b A_{a} \right)=\int d\theta\int d\alpha\frac{\partial {A_b}}{\partial {\alpha}}\frac{\partial {x^{b}}}{\partial {\theta}}-\int d\theta\int d\alpha\frac{\partial {A_b}}{\partial {\theta}}\frac{\partial {x^{b}}}{\partial {\alpha}}\end{aligned} \hspace{\stretch{1}}(3.56)

The LHS is thus found to be

\begin{aligned} & \int d{\alpha} d{\theta}\left({\left\lvert{\frac{\partial(x^0, x^1)}{\partial(\alpha, \theta)}}\right\rvert} \left( \partial_0 A_{1} -\partial_1 A_{0} \right)+{\left\lvert{\frac{\partial(x^0, x^2)}{\partial(\alpha, \theta)}}\right\rvert} \left( \partial_0 A_{2} -\partial_2 A_{0} \right)+{\left\lvert{\frac{\partial(x^1, x^2)}{\partial(\alpha, \theta)}}\right\rvert} \left( \partial_1 A_{2} -\partial_2 A_{1} \right)\right) \\ & =\int d{\alpha} d{\theta}\left(\cos\theta \sin\theta \left( \partial_0 A_{1} -\partial_1 A_{0} \right)+\cosh\alpha \cos^2\theta \left( \partial_0 A_{2} -\partial_2 A_{0} \right)-\sinh\alpha \cos^2\theta \left( \partial_1 A_{2} -\partial_2 A_{1} \right)\right)\end{aligned}

On the RHS we have

\begin{aligned}\int d\theta\int d\alpha & \frac{\partial {A_b}}{\partial {\alpha}}\frac{\partial {x^{b}}}{\partial {\theta}}-\int d\theta\int d\alpha\frac{\partial {A_b}}{\partial {\theta}}\frac{\partial {x^{b}}}{\partial {\alpha}} \\ & =\int d\theta\int d\alpha\begin{bmatrix}\sinh\alpha \sin\theta & -\cosh\alpha \sin\theta & -\cos\theta & 0\end{bmatrix}\frac{\partial}{\partial {\alpha}}\begin{bmatrix}A_0 \\ A_1 \\ A_2 \\ A_3 \\ \end{bmatrix} \\ & -\int d\theta\int d\alpha\begin{bmatrix}-\cosh\alpha \cos\theta & \sinh\alpha \cos\theta & 0 & 0\end{bmatrix}\frac{\partial}{\partial {\theta}}\begin{bmatrix}A_0 \\ A_1 \\ A_2 \\ A_3 \\ \end{bmatrix} \\ \end{aligned}

\begin{aligned}\begin{aligned} & \int d{\alpha} d{\theta}\cos\theta \sin\theta \left( \partial_0 A_{1} -\partial_1 A_{0} \right) \\ & \qquad+\int d{\alpha} d{\theta}\cosh\alpha \cos^2\theta \left( \partial_0 A_{2} -\partial_2 A_{0} \right) \\ & \qquad-\int d{\alpha} d{\theta}\sinh\alpha \cos^2\theta \left( \partial_1 A_{2} -\partial_2 A_{1} \right) \\ & =\int d\theta \sin\theta \int d\alpha \left( \sinh\alpha \frac{\partial {A_0}}{\partial {\alpha}} - \cosh\alpha \frac{\partial {A_1}}{\partial {\alpha}} \right) \\ & \qquad-\int d\theta \cos\theta \int d\alpha \frac{\partial {A_2}}{\partial {\alpha}} \\ & \qquad+\int d\alpha \cosh\alpha \int d\theta \cos\theta \frac{\partial {A_0}}{\partial {\theta}} \\ & \qquad-\int d\alpha \sinh\alpha \int d\theta \cos\theta \frac{\partial {A_1}}{\partial {\theta}}\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.57)

Because of the complexity of the surface, only the second term on the RHS has the “evaluate on the boundary” characteristic that may have been expected from a Green’s theorem like line integral.

It is also worthwhile to point out that we have had to be very careful with upper and lower indexes all along (and have done so with the expectation that our application would include the special relativity case where our metric determinant is minus one.) Because we worked with upper indexes for the area element, we had to work with lower indexes for the four vector and the components of the gradient that we included in our curl evaluation.

## The rank 2 tensor case.

Let’s consider briefly the terms in the contraction sum

\begin{aligned}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{bc}\end{aligned} \hspace{\stretch{1}}(3.58)

For any choice of a set of three distinct indexes $(a, b, c) \in (0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)$), we have $6 = 3!$ ways of permuting those indexes in this sum

\begin{aligned}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{bc} & =\sum_{a < b < c} {\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{bc} + {\left\lvert{ \frac{\partial(x^a, x^c, x^b)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{cb} + {\left\lvert{ \frac{\partial(x^b, x^c, x^a)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_b A_{ca} \\ & \qquad + {\left\lvert{ \frac{\partial(x^b, x^a, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_b A_{ac} + {\left\lvert{ \frac{\partial(x^c, x^a, x^b)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_c A_{ab} + {\left\lvert{ \frac{\partial(x^c, x^b, x^a)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_c A_{ba} \\ & =2!\sum_{a < b < c}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert}\left( \partial_a A_{bc} + \partial_b A_{c a} + \partial_c A_{a b} \right)\end{aligned}

Observe that we have no sign alternation like we had in the vector (rank 1 tensor) case. That sign alternation in this summation expansion appears to occur only for odd grade tensors.

Returning to the problem, we wish to expand the determinant in order to apply a chain rule contraction as done in the rank-1 case. This can be done along any of rows or columns of the determinant, and we can write any of

\begin{aligned}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} & =\frac{\partial {x^a}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {x^a}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {x^a}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ & =\frac{\partial {x^b}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^c, x^a)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {x^b}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^c, x^a)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {x^b}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^c, x^a)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ & =\frac{\partial {x^c}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^a, x^b)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {x^c}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^a, x^b)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {x^c}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^a, x^b)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ \end{aligned}

This allows the contraction of the index $a$, eliminating it from the result

\begin{aligned}{\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \partial_a A_{bc} & =\left( \frac{\partial {x^a}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {x^a}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {x^a}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \right) \frac{\partial {A_{bc}}}{\partial {x^a}} \\ & =\frac{\partial {A_{bc}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {A_{bc}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {A_{bc}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ & =2!\sum_{b < c}\frac{\partial {A_{bc}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert}-\frac{\partial {A_{bc}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert}+\frac{\partial {A_{bc}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert} \\ \end{aligned}

Dividing out the common $2!$ terms, we can summarize this result as

\begin{aligned}\boxed{\begin{aligned}\sum_{a < b < c} & \int d\alpha_1 d\alpha_2 d\alpha_3 {\left\lvert{ \frac{\partial(x^a, x^b, x^c)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert}\left( \partial_a A_{bc} + \partial_b A_{c a} + \partial_c A_{a b} \right) \\ & =\sum_{b < c}\int d\alpha_2 d\alpha_3 \int d\alpha_1\frac{\partial {A_{bc}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_2, \alpha_3)} }\right\rvert} \\ & -\sum_{b < c}\int d\alpha_1 d\alpha_3 \int d\alpha_2\frac{\partial {A_{bc}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_3)} }\right\rvert} \\ & +\sum_{b < c}\int d\alpha_1 d\alpha_2 \int d\alpha_3\frac{\partial {A_{bc}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c)}{\partial(\alpha_1, \alpha_2)} }\right\rvert}\end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.59)

In general, as observed in the spacetime surface example above, the two index Jacobians can be functions of the integration variable first being eliminated. In the special cases where this is not the case (such as the $\mathbb{R}^{3}$ case with rectangular coordinates), then we are left with just the evaluation of the tensor element $A_{bc}$ on the boundaries of the respective integrals.

## The rank 3 tensor case.

The key step is once again just a determinant expansion

\begin{aligned} {\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \\ & =\frac{\partial {x^a}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_2, \alpha_3, \alpha_4)} }\right\rvert}-\frac{\partial {x^a}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_3, \alpha_4)} }\right\rvert}+\frac{\partial {x^a}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_4)} }\right\rvert}+\frac{\partial {x^a}}{\partial {\alpha_4}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert}\\ \end{aligned}

so that the sum can be reduced from a four index contraction to a 3 index contraction

\begin{aligned} {\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a A_{bcd} \\ & =\frac{\partial {A_{bcd}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_2, \alpha_3, \alpha_4)} }\right\rvert}-\frac{\partial {A_{bcd}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_3, \alpha_4)} }\right\rvert}+\frac{\partial {A_{bcd}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_4)} }\right\rvert}+\frac{\partial {A_{bcd}}}{\partial {\alpha_4}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert}\end{aligned}

That’s the essence of the theorem, but we can play the same combinatorial reduction games to reduce the built in redundancy in the result

\begin{aligned}\boxed{\begin{aligned}\frac{1}{{3!}} & \int d^4 \alpha {\left\lvert{ \frac{\partial(x^a, x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a A_{bcd} \\ & =\sum_{a < b < c < d}\int d^4 \alpha {\left\lvert{ \frac{\partial(x^a, x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \left( \partial_a A_{bcd} -\partial_b A_{cda} +\partial_c A_{dab} -\partial_d A_{abc} \right) \\ & =\qquad \sum_{b < c < d}\int d\alpha_2 d\alpha_3 d\alpha_4 \int d\alpha_1\frac{\partial {A_{bcd}}}{\partial {\alpha_1}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \\ & \qquad -\sum_{b < c < d}\int d\alpha_1 d\alpha_3 d\alpha_4 \int d\alpha_2\frac{\partial {A_{bcd}}}{\partial {\alpha_2}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_3, \alpha_4)} }\right\rvert} \\ & \qquad +\sum_{b < c < d}\int d\alpha_1 d\alpha_2 d\alpha_4 \int d\alpha_3\frac{\partial {A_{bcd}}}{\partial {\alpha_3}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_4)} }\right\rvert} \\ & \qquad +\sum_{b < c < d}\int d\alpha_1 d\alpha_2 d\alpha_3 \int d\alpha_4\frac{\partial {A_{bcd}}}{\partial {\alpha_4}} {\left\lvert{ \frac{\partial(x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3)} }\right\rvert} \\ \end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.60)

## A note on Four diverence.

Our four divergence integral has the following form

\begin{aligned}\int d^4 \alpha {\left\lvert{ \frac{\partial(x^1, x^2, x^2, x^4)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a A^a\end{aligned} \hspace{\stretch{1}}(3.61)

We can relate this to the rank 3 Stokes theorem with a duality transformation, multiplying with a pseudoscalar

\begin{aligned}A^a = \epsilon^{abcd} T_{bcd},\end{aligned} \hspace{\stretch{1}}(3.62)

where $T_{bcd}$ can also be related back to the vector by the same sort of duality transformation

\begin{aligned}A^a \epsilon_{a b c d} = \epsilon^{abcd} \epsilon_{a b c d} T_{bcd} = 4! T_{bcd}.\end{aligned} \hspace{\stretch{1}}(3.63)

The divergence integral in terms of the rank 3 tensor is

\begin{aligned}\int d^4 \alpha {\left\lvert{ \frac{\partial(x^1, x^2, x^2, x^4)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a \epsilon^{abcd} T_{bcd}=\int d^4 \alpha {\left\lvert{ \frac{\partial(x^a, x^b, x^c, x^d)}{\partial(\alpha_1, \alpha_2, \alpha_3, \alpha_4)} }\right\rvert} \partial_a T_{bcd},\end{aligned} \hspace{\stretch{1}}(3.64)

and we are free to perform the same Stokes reduction of the integral. Of course, this is particularly simple in rectangular coordinates. I still have to think though one sublty that I feel may be important. We could have started off with an integral of the following form

\begin{aligned}\int dx^1 dx^2 dx^3 dx^4 \partial_a A^a,\end{aligned} \hspace{\stretch{1}}(3.65)

and I think this differs from our starting point slightly because this has none of the antisymmetric structure of the signed 4 volume element that we have used. We do not take the absolute value of our Jacobians anywhere.

## Magnetic monopole discovery at the local dollar store was a fraud.

Posted by peeterjoot on February 20, 2011

The dollar store has available little plaster moulding kits for the kids to make little dinosaurs.  According to the instructions, these are supposed to have included one sided magnets.  Sadly, they are perfectly regular two sided magnets, and the search for Dirac’s monopoles must still continue

“Remember that the magnets only work on one side so before placing into the mixture make sure that the magnetic side is facing upwards.”

Posted in Incoherent ramblings | 2 Comments »

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

Posted by peeterjoot on February 17, 2011

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)

\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

\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}}(2.25)

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.

## PHY450H1S. Relativistic Electrodynamics Lecture 11 (Taught by Prof. Erich Poppitz). Action for the field.

Posted by peeterjoot on February 10, 2011

Covering chapter 3 material from the text [1].

Covering lecture notes pp. 84-102: relativity, gauge invariance, and superposition principles and the action for the electromagnetic field coupled to charged particles (91-95); the 4-current and its physical interpretation (96-102), including a needed mathematical interlude on delta-functions of functions (98-100) [Wednesday, Feb. 8; Thursday, Feb. 10]

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 nonrelativistic Coulomb gauge (109-111); the wave equation in the relativistic Lorentz gauge (112-113) [Tuesday, Feb. 15; Wednesday, Feb.16]…

# Where we are.

\begin{aligned}F_{ij} = \partial_i A_j - \partial_j A_i\end{aligned} \hspace{\stretch{1}}(2.1)

We learned that one half of Maxwell’s equations comes from the Bianchi identity

\begin{aligned}\epsilon^{ijkl} \partial_j F_{kl} = 0\end{aligned} \hspace{\stretch{1}}(2.2)

the other half (for vacuum) is

\begin{aligned}\partial_j F_{ji} = 0\end{aligned} \hspace{\stretch{1}}(2.3)

To get here we have to consider the action for the field.

# Generalizing the action to multiple particles.

We’ve learned that the action for a single particle is

\begin{aligned}S &= S_{\text{matter}} + S_{\text{interaction}} \\ &= -m c \int ds - \frac{e}{c} \int ds^i A_i\end{aligned}

This generalizes to more particles

\begin{aligned}S_{\text{particles in field''}}= -\sum_A m_A c \int_{x^A(\tau)} ds - \sum_A \frac{e_A}{c} \int dx^i_A A_i(x_A(\tau))\end{aligned} \hspace{\stretch{1}}(3.4)

$A$ lables the particles, and $x^A(\tau)$, $\{x^A(\tau), A= 1 \cdots N\}$ is the worldline of particle $A$.

# Action for the field.

However, $\mathbf{E}$ and $\mathbf{B}$ are created by charged particles and can “move” or “propagate” on their own. EM field is its own dynamical system. The variables are $A^i(\mathbf{x}, t)$. These are the “$q_a(t)$”.

The values of $\{A^i(\mathbf{x}, t), \forall \mathbf{x}\}$ is the dynamical degrees of freedom. This is a system with a continum of dynamical degrees of freedom.

We need to write an action for this continuous field system $A^i(\mathbf{x},t)$, and need some principles to guide the construction of this action.

When we have an action with many degrees of freedom, we sum over all the particles. The action for the electromagnetic field

\begin{aligned}S_{\text{EM field}} = \int dt \int d^3\mathbf{x} \mathcal{L}(A^i(\mathbf{x}, t))\end{aligned} \hspace{\stretch{1}}(4.5)

The quantity

\begin{aligned}\mathcal{L}(A^i(\mathbf{x}, t))\end{aligned} \hspace{\stretch{1}}(4.6)

is called the Lagrangian density, since the quantity

\begin{aligned}\int d^3\mathbf{x} \mathcal{L}(A^i(\mathbf{x}, t))\end{aligned} \hspace{\stretch{1}}(4.7)

is actually the Lagrangian.

While this may seem non-relativistic, with both $t$ and $\mathbf{x}$ in the integration range, because we have both, it is actually relativistic. We are integrating over all of spacetime, or the region where the EM fields are non-zero.

We write

\begin{aligned}\int d^4 x = c \int dt \int d^3 \mathbf{x},\end{aligned} \hspace{\stretch{1}}(4.8)

which is a Lorentz scalar.

We write our action as

\begin{aligned}S_{\text{EM field}} = \int d^4 x \mathcal{L}(A^i(\mathbf{x}, t))\end{aligned} \hspace{\stretch{1}}(4.9)

and demand that the Lagrangian density $\mathcal{L}$ must also be an invariant (Lorentz) scalar in $SO(1,3)$.

\paragraph{Analogy}: 3D rotations

\begin{aligned}\int d^3 \mathbf{x} \phi(\mathbf{x})\end{aligned} \hspace{\stretch{1}}(4.10)

Here $\phi$ is a 3-scalar, invariant under rotations.

\paragraph{Principles for the action}

\begin{enumerate}
\item Relativity.
\item Gauge invariance. Whatever $\mathcal{L}$ we write, it must be gauge invariant, impliying that it be a function of $F_{ij}$ only. Recall that we can adjust $A^i$ by a four-gradient of any scalar, but the quantities $\mathbf{E}$ and $\mathbf{B}$ were gauge invariant, and so $F^{ij}$ must also be.

If we don’t impose gauge invariance, then the resulting dynamical system will contain more than just $\mathbf{E}$ and $\mathbf{B}$. i.e. It will not be electromagnetism.

\item Superposition principle. The sum of two solutions is a solution. This implies linearity of the equations for $A^i$.

\item Locality. Could write

\begin{aligned}\int d^4 x \mathcal{L}_1(A) \int d^4 y \mathcal{L}_2(A)\end{aligned} \hspace{\stretch{1}}(4.11)

This would allow for fields that have aspects that effect the result from disjoint positions or times. This would probably result in non-causal results as well as the possibility of non-local results.
\end{enumerate}

Principle 1 means we must have

\begin{aligned}\mathcal{L}(A(\mathbf{x}, t))\end{aligned} \hspace{\stretch{1}}(4.12)

and principle 2

\begin{aligned}\mathcal{L}(F^{ij}(\mathbf{x}, t))\end{aligned} \hspace{\stretch{1}}(4.13)

and principle 1, means we must have a four scalar.

Without principle 3, we could have products of these, but we rule this out due to violation of non-linearity.

\paragraph{Example. Lagrangian for the Harmonic oscillator}

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \dot{q}^2 - \frac{1}{{2}} m \omega^2 q^2 \end{aligned} \hspace{\stretch{1}}(4.14)

This gives

\begin{aligned}\dot{d}{q} \propto q\end{aligned} \hspace{\stretch{1}}(4.15)

However, if we have

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \dot{q}^2 - \frac{1}{{2}} m \omega^2 q^2 - \lambda q^3\end{aligned} \hspace{\stretch{1}}(4.16)

we get

\begin{aligned}\dot{d}{q} \propto q + q^3\end{aligned} \hspace{\stretch{1}}(4.17)

In HW3, you’ll show that

\begin{aligned}\int dt dx \mathbf{E} \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(4.18)

only depends on $A^i$ at $\infty$ (the boundary). Because this depends only on $A^i$ spatial or time infinities, it can not affect the variational principle.

This is very much like in classical mechanics where we can add any total derivative to the Lagrangian. This does not change the Euler-Lagrange equation evaluation in any way. The $\mathbf{E} \cdot \mathbf{B}$ invariant has the same effect.

The invariants possible are $\mathbf{E}^2 - \mathbf{B}^2$, $(\mathbf{E} \cdot \mathbf{B})^2$, …, but we are now done, and know what is required. Our action must depend on $F$ squared.

Written in full with the constants in the right places we have

\begin{aligned}S_{\text{particles in field''}}= \sum_A \left( -m_A c \int_{x^A(\tau)} ds - \frac{e_A}{c} \int dx^i_A A_i(x_A(\tau))\right)- \frac{1}{{16 \pi c}} \int d^4 x F_{ij} F^{ij}\end{aligned} \hspace{\stretch{1}}(4.19)

To get the equation of motion for $A^i(\mathbf{x}, t)$ we need to vary $S_{\text{int}} + S_{\text{EM field}}$.

# Current density distribution.

Before we do the variation, we want to show that

\begin{aligned}S_{\text{int}} &= -\sum_A \frac{e_A}{c} \int_{x_A(\tau)} dx^i_A A_i(x_A(\tau) \\ &= -\frac{1}{{c^2}} \int d^4 x A_i(x) j^i(x) \end{aligned}

where

\begin{aligned}j^i(x) = \sum_A c e_A \int ds u^i_A \int_{x(\tau)}\delta(x^0 - x^0_A(\tau))\delta(x^1 - x^1_A(\tau))\delta(x^2 - x^2_A(\tau))\delta(x^3 - x^3_A(\tau)).\end{aligned} \hspace{\stretch{1}}(5.20)

We substitute in the integral

\begin{aligned}&\sum_A \int d^4 x A_i(x) j^i(x) \\ &= c e_A \sum_A \int d^4 x A_i(x) \int_{x(\tau)}ds u^i_A \delta(x^0 - x^0_A(\tau))\delta(x^1 - x^1_A(\tau))\delta(x^2 - x^2_A(\tau))\delta(x^3 - x^3_A(\tau)) \\ &= c e_A \sum_A \int d^4 x \int_{x(\tau)}dx^i_A A_i(x) \delta(x^0 - x^0_A(\tau))\delta(x^1 - x^1_A(\tau))\delta(x^2 - x^2_A(\tau))\delta(x^3 - x^3_A(\tau)) \\ &=c e_A \sum_A \int_{x_A(\tau)}dx^i_A A_i(x_A(\tau)) \end{aligned}

From this we see that we have

\begin{aligned}S_{\text{int}} = -\frac{1}{{c^2}} \int d^4 x A_i(x) j^i(x) \end{aligned} \hspace{\stretch{1}}(5.21)

Physical meaning of $j^i$

Minkowski diagram at angle $\arctan(v/c)$, with $x^0$ axis up and $x^1$ axis on horiztontal.

\begin{aligned}x^0(\tau) &= c \tau \\ x^1(\tau) &= v \tau \\ x^2(\tau) &= 0 \\ x^3(\tau) &= 0\end{aligned} \hspace{\stretch{1}}(5.22)

\begin{aligned}j^i(x) = e c \int dx^i(\tau) \delta^4 (x - x(\tau))\end{aligned} \hspace{\stretch{1}}(5.26)

\begin{aligned}j^0(x) &= e c^2 \int_{-\infty}^\infty d\tau \delta(x^0 - c \tau) \delta(x^1 - v \tau) \delta(x^2) \delta(x^3) \\ j^1(x) &= e c v \int_{-\infty}^\infty d\tau \delta(x^0 - c \tau) \delta(x^1 - v \tau) \delta(x^2) \delta(x^3) \\ j^2(x) &= 0 \\ j^3(x) &= 0\end{aligned} \hspace{\stretch{1}}(5.27)

To evaluate the $j^0$ integral, we have only the contribution from $\tau = x^0/c$. Recall that

\begin{aligned}\int dx \delta( cx - a) = \frac{1}{{{\left\lvert{c}\right\rvert}}} f\left( \frac{a}{c} \right)\end{aligned} \hspace{\stretch{1}}(5.31)

This $-c\tau$ scaling of the delta function, kills a factor of $c$ above, and leaves us with

\begin{aligned}j^0(x) &= e c \delta(x^1 - v x^0/c) \delta(x^2) \delta(x^3) \\ j^1(x) &= e v \delta(x^1 - v x^0/c) \delta(x^2) \delta(x^3) \\ j^2(x) &= 0 \\ j^3(x) &= 0 \end{aligned} \hspace{\stretch{1}}(5.32)

The current is non-zero only on the worldline of the particle. We identify

\begin{aligned}\rho(ct, x^1, x^2, x^3) = e \delta(x^1 - v x^0/c) \delta(x^2) \delta(x^3) \end{aligned} \hspace{\stretch{1}}(5.36)

so that our current can be interpretted as the charge and current density

\begin{aligned}j^0 &= c \rho(x) \\ j^\alpha(x) &= (\mathbf{v})^\alpha \rho(x)\end{aligned} \hspace{\stretch{1}}(5.37)

Except for the delta functions these are just the quantities that we are familiar with from the RHS of Maxwell’s equations.

# References

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

## PHY450H1S. Relativistic Electrodynamics Lecture 10 (Taught by Prof. Erich Poppitz). Lorentz force equation energy term, and four vector formulation of the Lorentz force equation.

Posted by peeterjoot on February 8, 2011

Covering chapter 3 material from the text [1].

Covering lecture notes pp. 74-83: gauge transformations in 3-vector language (74); energy of a relativistic particle in EM field (75); variational principle and equation of motion in 4-vector form (76-77); the field strength tensor (78-80); the fourth equation of motion (81)

# What is the significance to the gauge invariance of the action?

We had argued that under a gauge transformation

\begin{aligned}A_i \rightarrow A_i + \frac{\partial {\chi}}{\partial {x^i}},\end{aligned} \hspace{\stretch{1}}(2.1)

the action for a particle changes by a boundary term

\begin{aligned}- \frac{e}{c} ( \chi(x_b) - \chi(x_a) ).\end{aligned} \hspace{\stretch{1}}(2.2)

Because $S$ changes by a boundary term only, variation problem is not affected. The extremal trajectories are then the same, hence the EOM are the same.

## A less high brow demonstration.

With our four potential split into space and time components

\begin{aligned}A^i = (\phi, \mathbf{A}),\end{aligned} \hspace{\stretch{1}}(2.3)

the lower index representation of the same vector is

\begin{aligned}A_i = (\phi, -\mathbf{A}).\end{aligned} \hspace{\stretch{1}}(2.4)

Our gauge transformation is then

\begin{aligned}A_0 &\rightarrow A_0 + \frac{\partial {\chi}}{\partial {x^0}} \\ -\mathbf{A} &\rightarrow -\mathbf{A} + \frac{\partial {\chi}}{\partial {\mathbf{x}}}\end{aligned} \hspace{\stretch{1}}(2.5)

or

\begin{aligned}\phi &\rightarrow \phi + \frac{1}{{c}}\frac{\partial {\chi}}{\partial {t}} \\ \mathbf{A} &\rightarrow \mathbf{A} - \boldsymbol{\nabla} \chi.\end{aligned} \hspace{\stretch{1}}(2.7)

Now observe how the electric and magnetic fields are transformed

\begin{aligned}\mathbf{E} &= - \boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \\ &\rightarrow - \boldsymbol{\nabla} \left( \phi + \frac{1}{{c}}\frac{\partial {\chi}}{\partial {t}} \right) - \frac{1}{{c}}\frac{\partial {}}{\partial {t}} \left( \mathbf{A} - \boldsymbol{\nabla} \chi \right) \\ \end{aligned}

Sufficient continuity of $\chi$ is assumed, allowing commutation of the space and time derivatives, and we are left with just $\mathbf{E}$

For the magnetic field we have

\begin{aligned}\mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A} \\ &\rightarrow \boldsymbol{\nabla} \times (\mathbf{A} - \boldsymbol{\nabla} \chi) \\ \end{aligned}

Again with continuity assumptions, $\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \chi) = 0$, and we are left with just $\mathbf{B}$. The electromagnetic fields (as opposed to potentials) do not change under gauge transformations.

We conclude that the $\{A_i\}$ description is hugely redundant, but despite that, local $\mathcal{L}$ and $H$ can only be written in terms of the potentials $A_i$.

## Energy term of the Lorentz force. Three vector approach.

With the Lagrangian for the particle given by

\begin{aligned}\mathcal{L} = - mc^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + \frac{e}{c} \mathbf{A} \cdot \mathbf{v} - e \phi,\end{aligned} \hspace{\stretch{1}}(2.9)

we define the energy as

\begin{aligned}\mathcal{E} = \mathbf{v} \cdot \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} - \mathcal{L}\end{aligned} \hspace{\stretch{1}}(2.10)

This is not necessarily a conserved quantity, but we define it as the energy anyways (we don’t really have a Hamiltonian when the fields are time dependent). Associated with this quantity is the general relationship

\begin{aligned}\frac{d{{\mathcal{E}}}}{dt} = -\frac{\partial {\mathcal{L}}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(2.11)

and when the Lagrangian is invariant with respect to time translation the energy $\mathcal{E}$ will be a conserved quantity (and also the Hamiltonian).

Our canonical momentum is

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \gamma m \mathbf{v} + \frac{e}{c} \mathbf{A}\end{aligned} \hspace{\stretch{1}}(2.12)

So our energy is

\begin{aligned}\mathcal{E} = \gamma m \mathbf{v}^2 + \frac{e}{c} \mathbf{A} \cdot \mathbf{v} - \left( - mc^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + \frac{e}{c} \mathbf{A} \cdot \mathbf{v} - e \phi \right).\end{aligned}

Or

\begin{aligned}\mathcal{E} = \underbrace{\frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}}_{({*})} + e \phi.\end{aligned} \hspace{\stretch{1}}(2.13)

The contribution of $({*})$ to the energy $\mathcal{E}$ comes from the free (kinetic) particle portion of the Lagrangian $\mathcal{L} = -m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}}$, and we identify the remainder as a potential energy

\begin{aligned}\mathcal{E} = \frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} + \underbrace{e \phi}_{\text{"potential"}}.\end{aligned} \hspace{\stretch{1}}(2.14)

For the kinetic portion we can also show that we have

\begin{aligned}\frac{d}{dt} \mathcal{E}_{\text{kinetic}} =\frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} = e \mathbf{E} \cdot \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(2.15)

To show this observe that we have

\begin{aligned}\frac{d}{dt} \mathcal{E}_{\text{kinetic}} &= m c^2 \frac{d\gamma}{dt} \\ &= m c^2 \frac{d}{dt} \frac{1}{{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}} \\ &= m c^2 \frac{\frac{\mathbf{v}}{c^2} \cdot \frac{d\mathbf{v}}{dt}}{\left(1 - \frac{\mathbf{v}^2}{c^2}\right)^{3/2}} \\ &= \frac{m \gamma \mathbf{v} \cdot \frac{d\mathbf{v}}{dt}}{1 - \frac{\mathbf{v}^2}{c^2}}\end{aligned}

We also have

\begin{aligned}\mathbf{v} \cdot \frac{d{\mathbf{p}}}{dt} &= \mathbf{v} \cdot \frac{d{{}}}{dt} \frac{m \mathbf{v}}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} \\ &= m\mathbf{v}^2 \frac{d{{\gamma}}}{dt} + m \gamma \mathbf{v} \cdot \frac{d{\mathbf{v}}}{dt} \\ &= m\mathbf{v}^2 \frac{d{{\gamma}}}{dt} + m c^2 \frac{d{{\gamma}}}{dt} \left( 1 - \frac{\mathbf{v}^2}{c^2} \right) \\ &= m c^2 \frac{d{{\gamma}}}{dt}.\end{aligned}

Utilizing the Lorentz force equation, we have

\begin{aligned}\mathbf{v} \cdot \frac{d{\mathbf{p}}}{dt} = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right) \cdot \mathbf{v} = e \mathbf{E} \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.16)

and are able to assemble the above, and find that we have

\begin{aligned}\frac{d{{(m c^2 \gamma)}}}{dt} = e \mathbf{E} \cdot \mathbf{v} \end{aligned} \hspace{\stretch{1}}(2.17)

# Four vector Lorentz force

Using $ds = \sqrt{ dx^i dx_i }$ our action can be rewritten

\begin{aligned}S &= \int \left( -m c ds - \frac{e}{c} u^i A_i ds \right) \\ &= \int \left( -m c ds - \frac{e}{c} dx^i A_i \right) \\ &= \int \left( -m c \sqrt{ dx^i dx_i} - \frac{e}{c} dx^i A_i \right) \\ \end{aligned}

$x^i(\tau)$ is a worldline $x^i(0) = a^i$, $x^i(1) = b^i$,

We want $\delta S = S[ x + \delta x ] - S[ x ] = 0$ (to linear order in $\delta x$)

The variation of our proper length is

\begin{aligned}\delta ds &=\delta \sqrt{ dx^i dx_i } \\ &= \frac{1}{{ 2 \sqrt{ dx^i dx_i }}} \delta (dx^j dx_j)\end{aligned}

Observe that for the numerator we have

\begin{aligned}\delta (dx^j dx_j) &= \delta ( dx^j g_{jk} dx^k ) \\ &= \delta ( dx^j ) g_{jk} dx^k + dx^j g_{jk} \delta ( dx^k ) \\ &= \delta ( dx^j ) g_{jk} dx^k + dx^k g_{kj} \delta ( dx^j ) \\ &= 2 \delta ( dx^j ) g_{jk} dx^k \\ &= 2 \delta ( dx^j ) dx_j \end{aligned}

\paragraph{TIP:} If this goes too quick, or there is any disbelief, write these all out explicitly as $dx^j dx_j = dx^0 dx_0 + dx^1 dx_1 + dx^2 dx_2 + dx^3 dx_3$ and compute it that way.

For the four vector potential our variation is

\begin{aligned}\delta A_i = A_i(x + \delta x) - A_i = \frac{\partial {A_i}}{\partial {x^j}} \delta x^j = \partial_j A_i \delta x^j\end{aligned} \hspace{\stretch{1}}(3.18)

(i.e. By chain rule)

Completing the proper length variations above we have

\begin{aligned}\delta \sqrt{ dx^i dx_i } &= \frac{1}{{ \sqrt{ dx^i dx_i }}} \delta (dx^j) dx_j \\ &= \delta (dx^j) \frac{d{{x_j}}}{ds} \\ &= \delta (dx^j) u_j \\ &= d \delta x^j u_j\end{aligned}

We are now ready to assemble results and do the integration by parts

\begin{aligned}\delta S &= \int \left( -m c d (\delta x^j) u_j- \frac{e}{c} d (\delta x^i) A_i - \frac{e}{c} dx^i \partial_j A_i \delta x^j\right) \\ &= {\left. \left( -m c (\delta x^j) u_j - \frac{e}{c} (\delta x^i) A_i \right)\right\vert}_a^b+\int \left( m c \delta x^j d u_j+ \frac{e}{c} (\delta x^i) d A_i - \frac{e}{c} dx^i \partial_j A_i \delta x^j\right) \\ \end{aligned}

Our variation at the endpoints is zero ${\left.{{\delta x^i}}\right\vert}_{{a}} = {\left.{{\delta x^i}}\right\vert}_{{b}} = 0$, killing the non-integral terms

\begin{aligned}\delta S &= \int \delta x^j\left( m c d u_j+ \frac{e}{c} d A_j - \frac{e}{c} dx^i \partial_j A_i \right).\end{aligned}

Observe that our differential can also be expanded by chain rule

\begin{aligned}d A_j = \frac{\partial {A_j}}{\partial {x^i}} dx^i = \partial_i A_j dx^i,\end{aligned} \hspace{\stretch{1}}(3.19)

which simplifies the variation further

\begin{aligned}\delta S &= \int \delta x^j\left( m c d u_j+ \frac{e}{c} dx^i ( \partial_i A_j - \partial_j A_i )\right) \\ &= \int \delta x^j ds\left( m c \frac{d u_j}{ds}+ \frac{e}{c} u^i ( \partial_i A_j - \partial_j A_i )\right) \\ \end{aligned}

Since this is true for all variations $\delta x^j$, which is arbitrary, the interior part is zero everywhere in the trajectory. The antisymmetric portion, a rank 2 4-tensor is called the electromagnetic field strength tensor, and written

\begin{aligned}\boxed{F_{ij} = \partial_i A_j - \partial_j A_i.}\end{aligned} \hspace{\stretch{1}}(3.20)

In matrix form this is

\begin{aligned}{\left\lVert{ F_{ij} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.21)

In terms of the field strength tensor our Lorentz force equation takes the form

\begin{aligned}\boxed{\frac{d{{(m c u_i)}}}{ds} = \frac{e}{c} F_{ij} u^j.}\end{aligned} \hspace{\stretch{1}}(3.22)

# References

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

## Energy term of the Lorentz force equation.

Posted by peeterjoot on February 8, 2011

# Motivation.

In class this week, the Lorentz force was derived from an action (the simplest Lorentz invariant, gauge invariant, action that could be constructed)

\begin{aligned}S = - m c \int ds - \frac{e}{c} \int ds A^i u_i.\end{aligned} \hspace{\stretch{1}}(1.1)

We end up with the familiar equation, with the exception that the momentum includes the relativistically required gamma factor

\begin{aligned}\frac{d (\gamma m \mathbf{v})}{dt} = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(1.2)

I asked what the energy term of this equation would be and was answered that we would get to it, and it could be obtained by a four vector minimization of the action which produces the Lorentz force equation of the following form

\begin{aligned}\frac{du^i}{d\tau} \propto e F^{ij} u_j.\end{aligned} \hspace{\stretch{1}}(1.3)

Let’s see if we can work this out without the four-vector approach, using the action expressed with an explicit space time split, then also work it out in the four vector form and compare as a consistency check.

# Three vector approach.

## The Lorentz force derivation.

For completeness, let’s work out the Lorentz force equation from the action 1.1. Parameterizing by time we have

\begin{aligned}S &= -m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} \gamma \left( 1, \frac{1}{{c}} \mathbf{v}\right) \cdot (\phi, \mathbf{A}) \\ &= -m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \int dt \left( \phi - \frac{1}{{c}} \mathbf{A} \cdot \mathbf{v} \right)\end{aligned}

Our Lagrangian is therefore

\begin{aligned}\mathcal{L}(\mathbf{x}, \mathbf{v}, t) = -m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \phi(\mathbf{x}, t) + \frac{e}{c} \mathbf{A}(\mathbf{x}, t) \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.4)

We can calculate our conjugate momentum easily enough

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \gamma m \mathbf{v} + \frac{e}{c} \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.5)

and for the gradient portion of the Euler-Lagrange equations we have

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{x}}} = -e \boldsymbol{\nabla} \phi + e \boldsymbol{\nabla} \left( \frac{\mathbf{v}}{c} \cdot \mathbf{A} \right).\end{aligned} \hspace{\stretch{1}}(2.6)

Utilizing the convective derivative (i.e. chain rule in fancy clothes)

\begin{aligned}\frac{d}{dt} = \mathbf{v} \cdot \boldsymbol{\nabla} + \frac{\partial {}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(2.7)

This gives us

\begin{aligned}-e \boldsymbol{\nabla} \phi + e \boldsymbol{\nabla} \left( \frac{\mathbf{v}}{c} \cdot \mathbf{A} \right) = \frac{d(\gamma m \mathbf{v})}{dt} + \frac{e}{c} (\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}+ \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(2.8)

and a final bit of rearranging gives us

\begin{aligned}\frac{d(\gamma m \mathbf{v})}{dt} =e \left( -\boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}}\right)+ \frac{e}{c} \left( \boldsymbol{\nabla} \left( \mathbf{v} \cdot \mathbf{A} \right) - (\mathbf{v} \cdot \boldsymbol{\nabla}) \mathbf{A}\right).\end{aligned} \hspace{\stretch{1}}(2.9)

The first set of derivatives we identify with the electric field $\mathbf{E}$. For the second, utilizing the vector triple product identity [1]

\begin{aligned}\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b} (\mathbf{a} \cdot \mathbf{c}) - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c},\end{aligned} \hspace{\stretch{1}}(2.10)

we recognize as related to the magnetic field $\mathbf{v} \times \mathbf{B} = \mathbf{v} \times (\boldsymbol{\nabla} \times \mathbf{A})$.

## The power (energy) term.

When we start with an action explicitly constructed with Lorentz invariance as a requirement, it is somewhat odd to end up with a result that has only the spatial vector portion of what should logically be a four vector result. We have an equation for the particle momentum, but not one for the energy. In tutorial Simon provided the hint of how to approach this, and asked if we had calculated the Hamiltonian for the Lorentz force. We had only calculated the Hamiltonian for the free particle.

Considering this, we can only actually calculate a Hamiltonian for the case where $\phi(\mathbf{x}, t) = \phi(\mathbf{x})$ and $\mathbf{A}(\mathbf{x}, t) = \mathbf{A}(\mathbf{x})$, because when the potentials have any sort of time dependence we do not have a Lagrangian that is invariant under time translation. Returning to the derivation of the Hamiltonian conservation equation, we see that we must modify the argument slightly when there is a time dependence and get instead

\begin{aligned}\frac{d}{dt} \left( \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} \cdot \mathcal{L} - \mathcal{L} \right) + \frac{\partial {\mathcal{L}}}{\partial {t}} = 0.\end{aligned} \hspace{\stretch{1}}(2.11)

Only when there is no time dependence in the Lagrangian, do we have our conserved quantity, what we label as energy, or Hamiltonian.

From 2.5, we have

\begin{aligned}0 &= \frac{d}{dt} \left( \left( \gamma m \mathbf{v} + \frac{e}{c} \mathbf{A} \right) \cdot \mathbf{v} +m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + e \phi - \frac{e}{c} \mathbf{A} \cdot \mathbf{v}\right) - e \frac{\partial {\phi}}{\partial {t}} + \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ \end{aligned}

Our $\mathbf{A} \cdot \mathbf{v}$ terms cancel, and we can combine the $\gamma$ and $\gamma^{-1}$ terms, then apply the convective derivative again

\begin{aligned}\frac{d}{dt} \left( \gamma m c^2 \right) &= - e \left( \mathbf{v} \cdot \boldsymbol{\nabla} + \frac{\partial {}}{\partial {t}} \right) \phi + e \frac{\partial {\phi}}{\partial {t}} - \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ &= - e \mathbf{v} \cdot \boldsymbol{\nabla} \phi - \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} \cdot \mathbf{v} \\ &= + e \mathbf{v} \cdot \left( - \boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \right).\end{aligned}

This is just

\begin{aligned}\frac{d}{dt} \left( \gamma m c^2 \right) = e \mathbf{v} \cdot \mathbf{E},\end{aligned} \hspace{\stretch{1}}(2.12)

and we find the rate of change of energy term of our four momentum equation

\begin{aligned}\frac{d}{dt}\left( \frac{E}{c}, \mathbf{p}\right) = e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.13)

Specified explicilty, this is

\begin{aligned}\frac{d}{dt}\left( \gamma m \left( c, \mathbf{v} \right) \right)= e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.14)

While this was the result I was looking for, once written it now stands out as incomplete relativistically. We have an equation that specifies the time derivative of a four vector. What about the spatial derivatives? We really ought to have a rank two tensor result, and not a four vector result relating the fields and the energy and momentum of the particle. The Lorentz force equation, even when expanded to four vector form, does not seem complete relativistically.

With $u^i = dx^i/ds$, we can rewrite 2.14 as

\begin{aligned}\partial_0 (\gamma m u^i) = e \left( \frac{\mathbf{v}}{c} \cdot \mathbf{E}, \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(2.15)

If we were to vary the action with respect to a spatial coordinate instead of time, we should end up with a similar equation of the form $\partial_i (\gamma m u^i) = ?$. Having been pointed at the explicitly invariant result, I wonder if those equations are independent. Let’s defer exploring this, until at least after calculating the result using a four vector form of the action.

# Four vector approach.

## The Lorentz force derivation from invariant action.

We can rewrite our action, parameterizing with proper time. This is

\begin{aligned}S = -m c^2 \int d\tau \sqrt{ \frac{dx^i}{d\tau} \frac{dx_i}{d\tau} }- \frac{e}{c} \int d\tau A_i \frac{dx^i}{d\tau}\end{aligned} \hspace{\stretch{1}}(3.16)

Writing $\dot{x}^i = dx^i/d\tau$, our Lagrangian is then

\begin{aligned}\mathcal{L}(x^i, \dot{x^i}, \tau)= -m c^2 \sqrt{ \dot{x}^i \dot{x}_i }- \frac{e}{c} A_i \dot{x}^i\end{aligned} \hspace{\stretch{1}}(3.17)

The Euler-Lagrange equations take the form

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x^i}} = \frac{d}{d\tau} \frac{\partial {\mathcal{L}}}{\partial {\dot{x}^i}} .\end{aligned} \hspace{\stretch{1}}(3.18)

Our gradient and conjugate momentum are

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x^i}} &= - \frac{e}{c} \frac{\partial {A_j}}{\partial {x^i}} \dot{x}^j \\ \frac{\partial {\mathcal{L}}}{\partial {\dot{x}^i}} &= -m \dot{x}_i - \frac{e}{c} A_i.\end{aligned} \hspace{\stretch{1}}(3.19)

With our convective derivative taking the form

\begin{aligned}\frac{d}{d\tau} = \dot{x}^i \frac{\partial {}}{\partial {x^i}},\end{aligned} \hspace{\stretch{1}}(3.21)

we have

\begin{aligned}m \frac{d^2 x_i}{d\tau^2} &= \frac{e}{c} \frac{\partial {A_j}}{\partial {x^i}} \dot{x}^j- \frac{e}{c} \dot{x}^j \frac{\partial {A_i}}{\partial {x^j}} \\ &=\frac{e}{c} \dot{x}^j \left( \frac{\partial {A_j}}{\partial {x^i}} -\frac{\partial {A_i}}{\partial {x^j}} \right) \\ &=\frac{e}{c} \dot{x}^j \left( \partial_i A_j - \partial_j A_i\right) \\ &=\frac{e}{c} \dot{x}^j F_{ij}\end{aligned}

Our Prof wrote this with indexes raised and lowered respectively

\begin{aligned}m \frac{d^2 x^i}{d\tau^2} = \frac{e}{c} F^{ij} \dot{x}_j .\end{aligned} \hspace{\stretch{1}}(3.22)

Following the text [2] he also writes $u^i = dx^i/ds = (1/c) dx^i/d\tau$, and in that form we have

\begin{aligned}\frac{d (m c u^i)}{ds} = \frac{e}{c} F^{ij} u_j.\end{aligned} \hspace{\stretch{1}}(3.23)

## Expressed explicitly in terms of the three vector fields.

### The power term.

From 3.23, lets extract the $i=0$ term, relating the rate of change of energy to the field and particle velocity. With

\begin{aligned}\frac{d{{}}}{d\tau} = \frac{dt}{d\tau} \frac{d}{dt} = \gamma \frac{d{{}}}{dt},\end{aligned} \hspace{\stretch{1}}(3.24)

we have

\begin{aligned}\frac{d{{(m \gamma \frac{dx^i}{dt})}}}{dt} = \frac{e}{c} F^{ij} \frac{d{{x_j}}}{dt}.\end{aligned} \hspace{\stretch{1}}(3.25)

For $i=0$ we have

\begin{aligned}F^{0j} \frac{d{{x_j}}}{dt} = -F^{0\alpha} \frac{d{{x^\alpha}}}{dt} \end{aligned} \hspace{\stretch{1}}(3.26)

That component of the field is

\begin{aligned}F^{\alpha 0} &=\partial^\alpha A^0 - \partial^0 A^\alpha \\ &=-\frac{\partial {\phi}}{\partial {x^\alpha}} - \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} \\ &= \left( -\boldsymbol{\nabla} \phi - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \right)^\alpha.\end{aligned}

This verifies the result obtained with considerably more difficulty, using the Hamiltonian like conservation relation obtained for a time translation of a time dependent Lagrangian

\begin{aligned}\frac{d{{(m \gamma c^2 )}}}{dt} = e \mathbf{E} \cdot \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(3.27)

### The Lorentz force terms.

Let’s also verify the signs for the $i > 0$ terms. For those we have

\begin{aligned}\frac{d{{(m \gamma \frac{dx^\alpha}{dt})}}}{dt} &= \frac{e}{c} F^{\alpha j} \frac{d{{x_j}}}{dt} \\ &= \frac{e}{c} F^{\alpha 0} \frac{d{{x_0}}}{dt}+\frac{e}{c} F^{\alpha \beta} \frac{d{{x_\beta}}}{dt} \\ &= e E^\alpha- \sum_{\alpha \beta} \frac{e}{c} \left( \partial^\alpha A^\beta - \partial^\beta A^\alpha\right)v^\beta \\ \end{aligned}

Since we have only spatial indexes left, lets be sloppy and imply summation over all repeated indexes, even if unmatched upper and lower. This leaves us with

\begin{aligned}-\left( \partial^\alpha A^\beta - \partial^\beta A^\alpha \right) v^\beta &=\left( \partial_\alpha A^\beta - \partial_\beta A^\alpha \right) v^\beta \\ &=\epsilon_{\alpha \beta \gamma} B^\gamma\end{aligned}

With the $v^\beta$ contraction we have

\begin{aligned}\epsilon_{\alpha \beta \gamma} B^\gamma v^\beta = (\mathbf{v} \times \mathbf{B})^\alpha,\end{aligned} \hspace{\stretch{1}}(3.28)

leaving our first result obtained by the time parameterization of the Lagrangian

\begin{aligned}\frac{d{{(m \gamma \mathbf{v})}}}{dt} = e \left(\mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(3.29)

This now has a nice symmetrical form. It’s slightly disappointing not to have a rank two tensor on the LHS like we have with the symmetric stress tensor with Poynting Vector and energy and other similar terms that relates field energy and momentum with $\mathbf{E} \cdot \mathbf{J}$ and the charge density equivalents of the Lorentz force equation. Is there such a symmetric relationship for particles too?

# References

[1] Wikipedia. Triple product — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 7-February-2011]. http://en.wikipedia.org/w/index.php?title=Triple_product&oldid=407455209.

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

## PHY450H1S. Relativistic Electrodynamics Tutorial 3 (TA: Simon Freedman). Relativistic motion in constant uniform electric or magnetic fields.

Posted by peeterjoot on February 4, 2011

# Motion in an constant uniform Electric field.

Given

\begin{aligned}\mathbf{E} = E \hat{\mathbf{x}},\end{aligned} \hspace{\stretch{1}}(1.1)

We want to solve the problem

\begin{aligned}\mathbf{F} = \frac{d\mathbf{p}}{dt} =e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right) = e \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.2)

Unlike second year classical physics, we will use relativistic momentum, so for only a constant electric field, our Lorentz force equation to solve becomes

\begin{aligned}\frac{d\mathbf{p}}{dt} = \frac{d (m \gamma \mathbf{v})}{dt} = e \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.3)

In components this is

\begin{aligned}\dot{p}_x &= e E \\ \dot{p}_y &= \text{constant}\end{aligned} \hspace{\stretch{1}}(1.4)

Integrating the $x$ component we have

\begin{aligned}e E t + p_x(0)=\frac{m \dot{x}}{\sqrt{1 - (\dot{x}^2 + \dot{y}^2)/c^2}} \end{aligned} \hspace{\stretch{1}}(1.6)

If we let $p_x(0) = 0$, square and rearrange a bit we have

\begin{aligned}\frac{m^2}{(e E t)^2} \dot{x}^2 = 1 - \frac{\dot{x}^2 + \dot{y}^2}{c^2}\end{aligned} \hspace{\stretch{1}}(1.7)

For

\begin{aligned}\dot{x}^2 = \frac{c^2 - \dot{y}^2}{1 + (\frac{mc}{eEt})^2}.\end{aligned} \hspace{\stretch{1}}(1.8)

Now for the $y$ components, with $p_y(0) = p_0$, our equation to solve is

\begin{aligned}\frac{m \dot{y}}{\sqrt{1 - (\dot{x}^2 + \dot{y}^2)/c^2}} = p_0.\end{aligned} \hspace{\stretch{1}}(1.9)

Squaring this one we have

\begin{aligned}\frac{c^2 m^2}{p_0^2} \dot{y}^2 = c^2 - \dot{x}^2 - \dot{y}^2 ,\end{aligned} \hspace{\stretch{1}}(1.10)

and

\begin{aligned}\dot{y}^2 = \frac{ c^2 - \dot{x}^2}{1 + \frac{m^2 c^2}{p_0^2}}\end{aligned} \hspace{\stretch{1}}(1.11)

Observe that our energy is

\begin{aligned}\mathcal{E}^2 = p^2 c^2 + m^2 c^4,\end{aligned} \hspace{\stretch{1}}(1.12)

and for $t=0$

\begin{aligned}\mathcal{E}_0^2 = p_0^2 c^2 + m^2 c^4.\end{aligned} \hspace{\stretch{1}}(1.13)

We can then write

\begin{aligned}\dot{y}^2 = \frac{ c^2 p_0^2 (c^2 - \dot{x}^2)}{ \mathcal{E}_0^2 }.\end{aligned} \hspace{\stretch{1}}(1.14)

Some messy substitution, using 1.8, yields

\begin{aligned}\boxed{\begin{aligned}\dot{x} &= \frac{c^2 e E t}{\sqrt{ \mathcal{E}_0^2 + (e c E t)^2 }} \\ \dot{y} &= \frac{c^2 p_0 }{\sqrt{ \mathcal{E}_0^2 + (e c E t)^2 }}\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.15)

Solving for $x$ we have

\begin{aligned}x(t) = c^2 e E \int \frac{dt' t'}{\sqrt{ \mathcal{E}_0^2 + (e c E t')^2 }}\end{aligned} \hspace{\stretch{1}}(1.16)

Can solve with hyperbolic substitution or

\begin{aligned}x(t) = c^2 e E \int \frac{dt' t'}{\sqrt{ \mathcal{E}_0^2 + (e c E t')^2 }}\end{aligned} \hspace{\stretch{1}}(1.17)

\begin{aligned}d(u^2) = 2 u du \implies u du = \frac{1}{{2}} d(u^2)\end{aligned} \hspace{\stretch{1}}(1.18)

\begin{aligned}x(t) = \frac{c^2 e E}{2 \mathcal{E}_0} \int \frac{d (u^2)}{\sqrt{ 1 + \left(\frac{e c E}{\mathcal{E}_0}\right)^2 u^2 }}\end{aligned} \hspace{\stretch{1}}(1.19)

Now we have something of the form

\begin{aligned}\int \frac{d v}{\sqrt{1 + a v}} = \frac{2}{a} \sqrt{1 + a v},\end{aligned} \hspace{\stretch{1}}(1.20)

so our final solution for $x(t)$ is

\begin{aligned}x(t) = \frac{1}{{e E}} \sqrt{ \mathcal{E}_0^2 + (e c E t)^2 }\end{aligned} \hspace{\stretch{1}}(1.21)

or

\begin{aligned}x^2 - c^2 t^2 = \frac{\mathcal{E}_0^2}{ e^2 E^2 } = a^{-2}.\end{aligned} \hspace{\stretch{1}}(1.22)

Now for $y(t)$ we have

\begin{aligned}y(t) = c^2 p_0 \int \frac{dt}{ \sqrt{\mathcal{E}_0^2 + (e c E t)^2 }}\end{aligned} \hspace{\stretch{1}}(1.23)

\begin{aligned}t = \frac{\mathcal{E}_0}{ e c E} \sinh(u) \end{aligned} \hspace{\stretch{1}}(1.24)

\begin{aligned}dt = \frac{\mathcal{E}_0}{ e c E} \cosh(u) du\end{aligned} \hspace{\stretch{1}}(1.25)

\begin{aligned}y(t) &= \frac{c^2 p_0}{\mathcal{E}_0} \int \frac{dt}{\sqrt{1 + (\frac{e c E}{\mathcal{E}_0})^2 t^2 }} \\ &= \frac{c^2 p_0}{\mathcal{E}_0} \frac{\mathcal{E}_0}{ e c E} \int \frac{ du \cosh u }{\sqrt{1 + \sinh^2 u }} \\ &= \frac{c p_0}{ e E} u \end{aligned}

A final bit of substitution, including a sort of odd seeming parametrization of $x$ in terms of $y$ in terms of $t$, we have

\begin{aligned}\boxed{\begin{aligned}y(t) &= \frac{c p_0}{ e E} \sinh^{-1} \left( \frac{e c E t}{\mathcal{E}_0} \right) \\ x(y) &= \frac{\mathcal{E}_0}{c E \cosh \left( \frac{y e E }{ c p_0} \right) }\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.26)

## Checks

FIXME: check the checks.

\begin{aligned}v \rightarrow c, t \rightarrow \infty\end{aligned} \hspace{\stretch{1}}(1.27)

\begin{aligned}v << c, t \rightarrow 0\end{aligned} \hspace{\stretch{1}}(1.28)

\begin{aligned}m v_x &= e E t + ... \\ x &\sim t^2 \end{aligned}

\begin{aligned}m v_y = p_0 \rightarrow y \sim t\end{aligned} \hspace{\stretch{1}}(1.29)

\begin{aligned}x(y) \sim y^2\end{aligned} \hspace{\stretch{1}}(1.30)

(a parabola)

## An alternate way.

There’s also a tricky way (as in the text), with

\begin{aligned}\mathbf{p} &= m \gamma \mathbf{v} \\ \mathcal{E} &= \gamma m c^2 \end{aligned} \hspace{\stretch{1}}(1.31)

We can solve this for $\mathbf{p}$

\begin{aligned}m \gamma &= \frac{\mathbf{p} \cdot \mathbf{v}}{\mathbf{v}^2} = \frac{\mathcal{E}}{c^2} \\ \mathbf{p} \times \mathbf{v} &= 0\end{aligned}

With the cross product zero, $\mathbf{p}$ has only a component in the direction of $\mathbf{v}$, and we can invert to yield

\begin{aligned}\mathbf{p} = \frac{\mathcal{E} \mathbf{v}}{c^2}.\end{aligned} \hspace{\stretch{1}}(1.33)

This implies

\begin{aligned}\dot{x} = \frac{c^2 p_x}{\mathcal{E}},\end{aligned} \hspace{\stretch{1}}(1.34)

and one can work from there as well.

# Motion in an constant uniform Magnetic field.

## Work by the magnetic field

Note that the magnetic field does no work

\begin{aligned}\mathbf{F} = \frac{e}{c} \mathbf{v} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(2.35)

\begin{aligned}dW &= \mathbf{F} \cdot d\mathbf{l} \\ &=\frac{e}{c} (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} \\ &=\frac{e}{c} (\mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} dt \\ &= 0\end{aligned}

Because $\mathbf{v}$ and $\mathbf{v} \times \mathbf{B}$ are necessarily perpendicular we are reminded that the magnetic field does no work (even in this relativistic sense).

## Initial energy of the particle

Because no work is done, the particle’s energy is only the initial time value

\begin{aligned}\mathcal{E} = .... + e A^0\end{aligned} \hspace{\stretch{1}}(2.36)

Simon asked if we’d calculated this (i.e. the Hamiltonian in class). We’d calculated the conservation for time invariance, the Hamiltonian (and called it $E$). We’d also calculated the Hamiltonian for the free particle

\begin{aligned}\mathcal{E} = \mathbf{p}^2 c^2 + (m c^2)^2.\end{aligned} \hspace{\stretch{1}}(2.37)

We had not done this calculation for the Lorentz force Lagrangian, so lets do it now. Recall that this Lagrangian was

\begin{aligned}\mathcal{L} = - m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \phi + \frac{e}{c} \mathbf{v} \cdot \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.38)

with generalized momentum of

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \frac{m \mathbf{v}}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} + \frac{e}{c} \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(2.39)

Our Hamiltonian is thus

\begin{aligned}\mathcal{E} = \frac{m \mathbf{v}^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} + \frac{e}{c} \mathbf{A} \cdot \mathbf{v}+ m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + e \phi - \frac{e}{c} \mathbf{v} \cdot \mathbf{A},\end{aligned}

which gives us

\begin{aligned}\mathcal{E} = e \phi + \frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} \end{aligned}

So we see that our “energy”, defined as a quantity that is conserved, as a result of the symmetry of time translation invariance, has a component due to the electric field (but not the vector potential field $\mathbf{A}$), plus the free particle “energy”.

Is this right? With $\mathbf{A}$ and $\phi$ being functions of space and time, perhaps we need to be more careful with this argument. Perhaps this actually only applies to a statics case where $\mathbf{A}$ and $\phi$ are constant.

Since it was hinted to us that the energy component of the Lorentz force equation was proportional to $F^{0j} u_j$, and we can peek ahead to find that $F^{ij} = \partial^i A^j - \partial^j A^i$, let’s compare to that

\begin{aligned}e F^{0 j} u_j&=e (\partial^0 A^j - \partial^j A^0) u_j \\ &=e (\partial^0 A^\alpha - \partial^\alpha A^0) u_\alpha \\ &=e \left( \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} + \partial_\alpha A^0 \right) \frac{1}{{c}} \frac{dx_\alpha}{d\tau} \\ &=-e \left( \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} + \frac{\partial {\phi}}{\partial {x^\alpha}} \right) \frac{1}{{c}} \frac{dx^\alpha}{dt} \gamma,\end{aligned}

which is

\begin{aligned}e F^{0 j} u_j = e \left(\mathbf{E} \cdot \frac{\mathbf{v}}{c}\right) \gamma.\end{aligned} \hspace{\stretch{1}}(2.40)

So if we have

\begin{aligned}\frac{d\mathbf{p}}{dt} = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right)\end{aligned} \hspace{\stretch{1}}(2.41)

I’d guess that we have

\begin{aligned}\frac{d(\mathcal{E}/c)}{d\tau} \propto e F^{0 j} u_j,\end{aligned} \hspace{\stretch{1}}(2.42)

which is, using 2.40

\begin{aligned}\frac{d(\mathcal{E}/c)}{dt} \propto e \left(\mathbf{E} \cdot \frac{\mathbf{v}}{c}\right) \end{aligned} \hspace{\stretch{1}}(2.43)

Can the left hand side be integrated to yield $e \phi$? Yes, but only in the statics case when ${\partial {\mathbf{A}}}/{\partial {t}} = 0$, and $\phi(\mathbf{x},t) = \phi(\mathbf{x})$ for which we have

\begin{aligned}\mathcal{E} &\propto e \int \mathbf{E} \cdot \mathbf{v} dt \\ &= -e \int (\boldsymbol{\nabla} \phi) \cdot \frac{\mathbf{x}}{dt} dt \\ &= -e \int \frac{\partial {\phi}}{\partial {x^\alpha}} \frac{\partial {x^\alpha}}{\partial {t}} dt&= -e \phi\end{aligned}

FIXME: My suspicion is that the result 2.43, is generally true, but that we have dropped terms from the Hamiltonian calculation that need to be retained when $\phi$ and $\mathbf{A}$ are functions of time.

## Expressing the field and the force equation.

We will align our field with the $z$ axis, and write

\begin{aligned}\mathbf{B} = H \hat{\mathbf{z}},\end{aligned} \hspace{\stretch{1}}(2.44)

or, in components

\begin{aligned}\delta_{\alpha 3} H = H_\alpha.\end{aligned} \hspace{\stretch{1}}(2.45)

Because the energy is only due to the initial value, we write

\begin{aligned}\mathcal{E}(t) = \mathcal{E}_0\end{aligned} \hspace{\stretch{1}}(2.46)

\begin{aligned}\mathbf{p} = \mathcal{E} \frac{\mathbf{v}}{c^2} = \mathcal{E}_0 \frac{\mathbf{v}}{c^2}\end{aligned} \hspace{\stretch{1}}(2.47)

implies

\begin{aligned}\mathbf{v} = \mathbf{p} \frac{c^2}{\mathcal{E}_0}\end{aligned} \hspace{\stretch{1}}(2.48)

\begin{aligned}\dot{\mathbf{v}} = \dot{\mathbf{p}} \frac{c^2}{\mathcal{E}_0}\end{aligned} \hspace{\stretch{1}}(2.49)

\begin{aligned}\dot{v}_\alpha = \frac{e c}{\mathcal{E}_0} \epsilon_{\alpha \beta \gamma} v_\beta H_\gamma\end{aligned} \hspace{\stretch{1}}(2.50)

write

\begin{aligned}\omega = \frac{e c H}{\mathcal{E}_0}\end{aligned} \hspace{\stretch{1}}(2.51)

Evaluating the delta

\begin{aligned}\dot{v}_\alpha = \omega \epsilon_{\alpha \beta 3} v_\beta \end{aligned} \hspace{\stretch{1}}(2.52)

\begin{aligned}\dot{v}_1 &= \omega \epsilon_{1 \beta 3} v_\beta = \omega v_2 \\ \dot{v}_2 &= \omega \epsilon_{2 \beta 3} v_\beta = - \omega v_1 \\ \dot{v}_3 &= \omega \epsilon_{3 \beta 3} v_\beta = 0\end{aligned} \hspace{\stretch{1}}(2.53)

Looks like circular motion, so it’s natural to use complex variables. With

\begin{aligned}z = v_1 + i v_2 \end{aligned} \hspace{\stretch{1}}(2.56)

Using this we have

\begin{aligned}\frac{d}{dt} ( v_1 + i v_2 ) &= \omega v_2 - i \omega v_1 \\ &= -i \omega ( v_1 + i v_2 ).\end{aligned}

which comes out nicely

\begin{aligned}\frac{dz}{dt} = -i \omega z\end{aligned} \hspace{\stretch{1}}(2.57)

for

\begin{aligned}z = V_0 e^{-i \omega z t + i \alpha}\end{aligned} \hspace{\stretch{1}}(2.58)

Real and imaginary parts

\begin{aligned}v_1(t) &= V_0 \cos( \omega z t + \alpha) \\ v_2(t) &= -V_0 \sin( \omega z t + \alpha)\end{aligned} \hspace{\stretch{1}}(2.59)

Integrating

\begin{aligned}x_1(t) &= x_1(0) + V_0 \sin( \omega z t + \alpha) \\ x_2(t) &= x_2(0) + V_0 \cos( \omega z t + \alpha)\end{aligned} \hspace{\stretch{1}}(2.61)

Which is a helix.
FIXME: PICTURE.

## PHY450H1S. Relativistic Electrodynamics Lecture 9 (Taught by Prof. Erich Poppitz). Dynamics in a vector field.

Posted by peeterjoot on February 3, 2011

Covering chapter 2 material from the text [1].

Covering lecture notes pp. 56.1-72: comments on mass, energy, momentum, and massless particles (56.1-58); particles in external fields: Lorentz scalar field (59-62); reminder of a vector field under spatial rotations (63) and a Lorentz vector field (64-65) [Tuesday, Feb. 1]; the action for a relativistic particle in an external 4-vector field (65-66); the equation of motion of a relativistic particle in an external electromagnetic (4-vector) field (67,68,73) [Wednesday, Feb. 2]; mathematical interlude: (69-72): on 3×3 antisymmetric matrices, 3-vectors, and totally antisymmetric 3-index tensor – please read by yourselves, preferably by Wed., Feb. 2 class! (this is important, well also soon need the 4-dimensional generalization)

# More on the action.

Action for a relativistic particle in an external 4-scalar field

\begin{aligned}S = -m c \int ds - g \int ds \phi(x)\end{aligned} \hspace{\stretch{1}}(2.1)

Unfortunately we have no 4-vector scalar fields (at least for particles that are long lived and stable).

PICTURE: 3-vector field, some arrows in various directions.

PICTURE: A vector $\mathbf{A}$ in an $x,y$ frame, and a rotated (counterclockwise by angle $\alpha$) $x', y'$ frame with the components in each shown pictorially.

We have

\begin{aligned}A_x'(x', y') &= \cos\alpha A_x(x,y) + \sin\alpha A_y(x,y) \\ A_y'(x', y') &= -\sin\alpha A_x(x,y) + \cos\alpha A_y(x,y) \end{aligned} \hspace{\stretch{1}}(2.2)

\begin{aligned}\begin{bmatrix}A_x'(x', y') \\ A_y'(x', y')\end{bmatrix}=\begin{bmatrix}\cos\alpha A_x(x,y) & \sin\alpha A_y(x,y) \\ -\sin\alpha A_x(x,y) & \cos\alpha A_y(x,y) \end{bmatrix}\begin{bmatrix}A_x(x, y) \\ A_y(x, y)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.4)

More generally we have

\begin{aligned}\begin{bmatrix}A_x'(x', y', z') \\ A_y'(x', y', z') \\ A_z'(x', y', z')\end{bmatrix}=\hat{O}\begin{bmatrix}A_x(x, y, z) \\ A_y(x, y, z) \\ A_z(x, y, z)\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.5)

Here $\hat{O}$ is an $SO(3)$ matrix rotating $x \rightarrow x'$

\begin{aligned}\mathbf{A}(\mathbf{x}) \cdot \mathbf{y} = \mathbf{A}'(\mathbf{x}') \cdot \mathbf{y}'\end{aligned} \hspace{\stretch{1}}(2.6)

\begin{aligned}\mathbf{A} \cdot \mathbf{B} = \text{invariant}\end{aligned} \hspace{\stretch{1}}(2.7)

A four vector field is $A^i(x)$, with $x = x^i, i = 0,1,2,3$ and we’d write

\begin{aligned}\begin{bmatrix}(x^0)' \\ (x^1)' \\ (x^2)' \\ (x^3)'\end{bmatrix}=\hat{O}\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.8)

Now $\hat{O}$ is an $SO(1,3)$ matrix. Our four vector field is then

\begin{aligned}\begin{bmatrix}(A^0)' \\ (A^1)' \\ (A^2)' \\ (A^3)'\end{bmatrix}=\hat{O}\begin{bmatrix}A^0 \\ A^1 \\ A^2 \\ A^3\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.9)

We have

\begin{aligned}A^i g_{ij} x^i = \text{invariant} = {A'}^i g_{ij} {x'}^i \end{aligned} \hspace{\stretch{1}}(2.10)

From electrodynamics we know that we have a scalar field, the electrostatic potential, and a vector field

What’s a plausible action?

\begin{aligned}\int ds x^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.11)

This isn’t translation invariant.

\begin{aligned}\int ds x^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.12)

Next simplest is

\begin{aligned}\int ds u^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.13)

Could also do

\begin{aligned}\int ds A^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.14)

but it turns out that this isn’t gauge invariant (to be defined and discussed in detail).

Note that the convention for this course is to write

\begin{aligned}u^i = \left( \gamma, \gamma \frac{\mathbf{v}}{c} \right) = \frac{dx^i}{ds}\end{aligned} \hspace{\stretch{1}}(2.15)

Where $u^i$ is dimensionless ($u^i u_i = 1$). Some authors use

\begin{aligned}u^i = \left( \gamma c, \gamma \mathbf{v} \right) = \frac{dx^i}{d\tau}\end{aligned} \hspace{\stretch{1}}(2.16)

The simplest action for a four vector field $A^i$ is then

\begin{aligned}S = - m c \int ds - \frac{e}{c} \int ds u^i A_i\end{aligned} \hspace{\stretch{1}}(2.17)

(Recall that $u^i A_i = u^i g_{ij} A^j$).

In this action $e$ is nothing but a Lorentz scalar, a property of the particle that describes how it “couples” (or “feels”) the electrodynamics field.

Similarily $mc$ is a Lorentz scalar which is a property of the particle (inertia).

It turns out that all the electric charges in nature are quantized, and there are some deep reasons (in magnetic monopoles exist) for this.

Another reason for charge quantitization apparently has to do with gauge invariance and associated compact groups. Poppitz is amusing himself a bit here, hinting at some stuff that we can eventually learn.

Returning to our discussion, we have

\begin{aligned}S = - m c \int ds - \frac{e}{c} \int ds u^i g_{ij} A^j\end{aligned} \hspace{\stretch{1}}(2.18)

with the electrodynamics four vector potential

\begin{aligned}A^i &= (\phi, \mathbf{A}) \\ u^i &= \left(\gamma, \gamma \frac{\mathbf{v}}{c} \right) \\ u^i g_{ij} A^j &= \gamma \phi - \gamma \frac{\mathbf{v} \cdot \mathbf{A}}{c}\end{aligned} \hspace{\stretch{1}}(2.19)

\begin{aligned}S &= - m c^2 \int dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - \frac{e}{c} \int c dt \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} \left( \gamma \phi - \gamma \frac{\mathbf{v}}{c} \cdot \mathbf{A} \right) \\ &= \int dt \left(- m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \phi(\mathbf{x}, t) + \frac{e}{c} \mathbf{v} \cdot \mathbf{A}(\mathbf{x}, t)\right) \\ \end{aligned}

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} \frac{\mathbf{v}}{c^2} + \frac{e}{c} \mathbf{A}(\mathbf{x}, t)\end{aligned} \hspace{\stretch{1}}(2.22)

\begin{aligned}\frac{d}{dt} \frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = m \frac{d}{dt} (\gamma \mathbf{v}) + \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {t}} + \frac{e}{c} \frac{\partial {\mathbf{A}}}{\partial {x^\alpha}} v^\alpha\end{aligned} \hspace{\stretch{1}}(2.23)

Here $\alpha,\beta = 1,2,3$ and are summed over.

For the other half of the Euler-Lagrange equations we have

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {x^\alpha}} = - e \frac{\partial {\phi}}{\partial {x^\alpha}} + \frac{e}{c} v^\beta \frac{\partial {A^\beta}}{\partial {x^\alpha}}\end{aligned} \hspace{\stretch{1}}(2.24)

Equating these, and switching to coordinates for 2.23, we have

\begin{aligned}m \frac{d}{dt} (\gamma v^\alpha) + \frac{e}{c} \frac{\partial {A^\alpha}}{\partial {t}} + \frac{e}{c} \frac{\partial {A^\alpha}}{\partial {x^\beta}} v^\beta= - e \frac{\partial {\phi}}{\partial {x^\alpha}} + \frac{e}{c} v^\beta \frac{\partial {A^\beta}}{\partial {x^\alpha}}\end{aligned} \hspace{\stretch{1}}(2.25)

A final rearrangement yields

\begin{aligned}\frac{d}{dt} m \gamma v^\alpha = e \underbrace{\left( - \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} - \frac{\partial {\phi}}{\partial {x^\alpha}} \right)}_{E^\alpha} + \frac{e}{c} v^\beta \left( \frac{\partial {A^\beta}}{\partial {x^\alpha}} - \frac{\partial {A^\alpha}}{\partial {x^\beta}} \right)\end{aligned} \hspace{\stretch{1}}(2.26)

We can identity the second term with the magnetic field but first have to introduce antisymmetric matrices.

# antisymmetric matrixes

\begin{aligned}M_{\mu\nu} &= \frac{\partial {A^\nu}}{\partial {x^\mu}} - \frac{\partial {A^\mu}}{\partial {x^\nu}} \\ &= \epsilon_{\mu\nu\lambda} B_\lambda,\end{aligned}

where

\begin{aligned}\epsilon_{\mu\nu\lambda} =\begin{array}{l l}0 & \quad \mbox{if any two indexes coincide} \\ 1 & \quad \mbox{for even permutations oflatex \mu\nu\lambda} \\ -1 & \quad \mbox{for odd permutations of $\mu\nu\lambda$}\end{array}\end{aligned} \hspace{\stretch{1}}(3.27)

Example:

\begin{aligned}\epsilon_{123} &= 1 \\ \epsilon_{213} &= -1 \\ \epsilon_{231} &= 1.\end{aligned}

We can show that

\begin{aligned}B_\lambda = \frac{1}{{2}} \epsilon_{\lambda\mu\nu} M_{\mu\nu}\end{aligned} \hspace{\stretch{1}}(3.28)

\begin{aligned}B_1 &= \frac{1}{{2}} ( \epsilon_{123} M_{23} + \epsilon_{132} M_{32}) \\ &= \frac{1}{{2}} ( M_{23} - M_{32}) \\ &= \partial_2 A_3 - \partial_3 A_2.\end{aligned}

Using

\begin{aligned}\epsilon_{\mu\nu\alpha} \epsilon_{\sigma\kappa\alpha} = \delta_{\mu\sigma} \delta_{\nu\kappa} - \delta_{\nu\sigma} \delta_{\mu\kappa},\end{aligned} \hspace{\stretch{1}}(3.29)

we can verify the identity 3.28 by expanding

\begin{aligned}\epsilon_{\mu\nu\lambda} B_\lambda&=\frac{1}{{2}} \epsilon_{\mu\nu\lambda} \epsilon_{\lambda\alpha\beta} M_{\alpha\beta} \\ &=\frac{1}{{2}} (\delta_{\mu\alpha} \delta_{\nu\beta} - \delta_{\nu\alpha} \delta_{\mu\beta})M_{\alpha\beta} \\ &=\frac{1}{{2}} (M_{\mu\nu} - M_{\nu\mu}) \\ &=M_{\mu\nu}\end{aligned}

Returning to the action evaluation we have

\begin{aligned}\frac{d}{dt} ( m \gamma v^\alpha ) = e E^\alpha + \frac{e}{c} \epsilon_{\alpha\beta\gamma} v^\beta B_\gamma,\end{aligned} \hspace{\stretch{1}}(3.30)

but

\begin{aligned}\epsilon_{\alpha\beta\gamma} B_\gamma = (\mathbf{v} \times \mathbf{B})_\alpha.\end{aligned} \hspace{\stretch{1}}(3.31)

So

\begin{aligned}\frac{d}{dt} ( m \gamma \mathbf{v} ) = e \mathbf{E} + \frac{e}{c} \mathbf{v} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.32)

or

\begin{aligned}\frac{d}{dt} ( \mathbf{p} ) = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right).\end{aligned} \hspace{\stretch{1}}(3.33)

\paragraph{What is the energy component of the Lorentz force equation}

I asked this, not because I don’t know (I could answer this myself from $dp/d\tau = F \cdot v/c$, in the geometric algebra formalism, but I was curious if he had a way of determining this from what we’ve derived so far (intuitively I’d expect this to be possible). Answer was:

Observe that this is almost a relativisitic equation, but we aren’t going to get to the full equation yet. The energy component can be obtained from

\begin{aligned}\frac{du^0}{ds} = e F^{0j} u_j\end{aligned} \hspace{\stretch{1}}(3.34)

Since the full equation is

\begin{aligned}\frac{du^i}{ds} = e F^{ij} u_j\end{aligned} \hspace{\stretch{1}}(3.35)

“take with a grain of salt, may be off by sign, or factors of $c$”.

Also curious is that he claimed the energy component of this equation was not very important. Why would that be?

# Gauge transformations.

Claim

\begin{aligned}S_{\text{interaction}} = - \frac{e}{c} \int ds u^i A_i\end{aligned} \hspace{\stretch{1}}(4.36)

changes by boundary terms only under

“gauge transformation” :

\begin{aligned}A_i = A_i' + \frac{\partial {\chi}}{\partial {x^i}}\end{aligned} \hspace{\stretch{1}}(4.37)

where $\chi$ is a Lorentz scalar. This ${\partial {}}/{\partial {x^i}}$ is the four gradient. Let’s see this

Therefore the equations of motion are the same in an external $A^i$ and ${A'}^i$.

Recall that the $\mathbf{E}$ and $\mathbf{B}$ fields do not change under such transformations. Let’s see how the action transforms

\begin{aligned}S &= - \frac{e}{c} \int ds u^i A_i \\ &= - \frac{e}{c} \int ds u^i \left( {A'}_i + \frac{\partial {\chi}}{\partial {x^i}} \right) \\ &= - \frac{e}{c} \int ds u^i {A'}_i - \frac{e}{c} \int ds \frac{dx^i}{ds} \frac{\partial {\chi}}{\partial {x^i}} \\ \end{aligned}

Observe that this last bit is just a chain rule expansion

\begin{aligned}\frac{d}{ds} \chi(x^0, x^1, x^2, x^3) &= \frac{\partial {\chi}}{\partial {x^0}}\frac{dx^0}{ds} + \frac{\partial {\chi}}{\partial {x^1}}\frac{dx^1}{ds} + \frac{\partial {\chi}}{\partial {x^2}}\frac{dx^2}{ds} + \frac{\partial {\chi}}{\partial {x^3}}\frac{dx^3}{ds} \\ &= \frac{\partial {\chi}}{\partial {x^i}} \frac{dx^i}{ds},\end{aligned}

so we have

\begin{aligned}S = - \frac{e}{c} \int ds u^i {A'}_i - \frac{e}{c} \int ds \frac{d \chi}{ds}.\end{aligned} \hspace{\stretch{1}}(4.38)

This allows the line integral to be evaluated, and we find that it only depends on the end points of the interval

\begin{aligned}S = - \frac{e}{c} \int ds u^i {A'}_i - \frac{e}{c} ( \chi(x_b) - \chi(x_a) ),\end{aligned} \hspace{\stretch{1}}(4.39)

which completes the proof of the claim that this gauge transformation results in an action difference that only depends on the end points of the interval.

\paragraph{What is the significance to this claim?}

# References

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