# Peeter Joot's (OLD) Blog.

• ## Archives

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 293,778

## 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.