Energy-Momentum Tensor for the electromagnetic field

[spaghetti3451]

Homework Statement

Maxwell's Lagrangian for the electromagnetic field is $\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ and $A_{\mu}$ is the $4$-vector potential. Show that $\mathcal{L}$ is invariant under gauge transformations $A_{\mu} \rightarrow A_{\mu} + \partial_{\mu}\xi$ where $\xi=\xi(x)$ is a scalar field with arbitrary (differentiable) dependence on $x$.

Use Noether's theorem, and the spacetime translational invariance of the action, to construct the energy-momentum tensor $T^{\mu\nu}$ for the electromagnetic field. Show that the resulting object is neither symmetric nor gauge invariant.

Consider a new tensor given by $\Theta^{\mu\nu}=T^{\mu\nu}-F^{\rho\mu}\partial_{\rho}A^{\nu}$. Show that this object also defines four conserved currents. Moreover, show that it is symmetric, gauge invariant and traceless.

Comment: $T^{\mu\nu}$ and $\Theta^{\mu\nu}$ are both equally good definitions of the energy-momentum tensor. However $\Theta^{\mu\nu}$ clearly has the nicer properties. Moreover, if you couple Maxwell's Lagrangian to general relativity then it is $\Theta^{\mu\nu}$ which appears in Einstein's equations.

Homework Equations

The Attempt at a Solution

Under gauge transformations $A_{\mu} \rightarrow A_{\mu} + \partial_{\mu}\xi$ where $\xi=\xi(x)$ is a scalar field,

$\delta\mathcal{L} = -\frac{1}{4}\delta(F_{\mu\nu}F^{\mu\nu})$

$\implies \delta\mathcal{L}=-\frac{1}{4}[(\delta F_{\mu\nu})(F^{\mu\nu})+(F_{\mu\nu})(\delta F^{\mu\nu})]$

$\implies \delta\mathcal{L}=-\frac{1}{2}(F_{\mu\nu})(\delta F^{\mu\nu})$

$\implies \delta\mathcal{L}=-\frac{1}{2}(F_{\mu\nu})[\delta(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})]$

$\implies \delta\mathcal{L}=\frac{1}{2}(F_{\mu\nu})[\partial^{\mu}(\delta A^{\nu})-\partial^{\nu}(\delta A^{\mu})]$

$\implies \delta\mathcal{L}=\frac{1}{2}(F_{\mu\nu})[\partial^{\mu}\partial^{\nu}\xi-\partial^{\nu}\partial^{\mu}\xi)]$

$\implies \delta\mathcal{L}=\frac{1}{2}(F_{\mu\nu})[\partial^{\mu}\partial^{\nu}\xi-\partial^{\mu}\partial^{\nu}\xi)]$

$\implies \delta\mathcal{L}=0$.

Therefore, $\mathcal{L}$ is invariant.

P.S.: The problem mentions that $\xi=\xi(x)$ has arbitrary (differentiable) dependence on $x$. The differentiability of $\xi=\xi(x)$ is used in the lines $\delta\mathcal{L}=\frac{1}{2}(F_{\mu\nu})[\partial^{\mu}\partial^{\nu}\xi-\partial^{\nu}\partial^{\mu}\xi)]$ and $\delta\mathcal{L}=\frac{1}{2}(F_{\mu\nu})[\partial^{\mu}\partial^{\nu}\xi-\partial^{\mu}\partial^{\nu}\xi)]$.
.
Am I correct so far?

 

[TSny]

Am I correct so far?

I think it looks good.

 

[spaghetti3451]

Thanks!

Let me try to solve the next part of the problem.

Use Noether's theorem, and the spacetime translational invariance of the action, to construct the energy-momentum tensor $T^{\mu\nu}$ for the electromagnetic field. Show that the resulting object is neither symmetric nor gauge invariant.

$\delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial A_{\nu}}\delta A_{\nu}+\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\partial_{\rho}(\delta A_{\nu})$

$\implies \delta\mathcal{L}=\Big[\frac{\partial\mathcal{L}}{\partial A_{\nu}}-\partial_{\rho}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\Big)\Big]\delta A_{\nu}+\partial_{\rho}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\delta A_{\nu}\Big)$

$\implies \delta\mathcal{L}=\partial_{\rho}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\delta A_{\nu}\Big)$

Under the infinitesimal translation $x^{\sigma} \rightarrow x^{\sigma}-\epsilon^{\sigma}$,

$A_{\nu}(x) \rightarrow A_{\nu}(x) + \epsilon^{\sigma}\partial_{\sigma}A_{\nu}(x)$ and $\mathcal{L}(x) \rightarrow \mathcal{L}(x) + \epsilon^{\sigma}\partial_{\sigma}\mathcal{L}(x)$,

where the vector field configuration $A_{\nu}(x)$, and by extension, the Lagrangian $\mathcal{L}(x)$ transform actively.

Now, $\delta\mathcal{L}=\partial_{\rho}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\delta A_{\nu}\Big)$

$\implies \epsilon^{\sigma}\partial_{\sigma}\mathcal{L}=\partial_{\rho}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\epsilon^{\sigma}\partial_{\sigma}A_{\nu}\Big)$

$\implies \partial_{\rho}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\epsilon^{\sigma}\partial_{\sigma}A_{\nu}-\epsilon^{\sigma}\delta^{\rho}_{\sigma}\mathcal{L}\Big)=0$

$\implies \partial_{\rho}j^{\rho}=0$,

where the conserved Noether current $j^{\rho}=\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\epsilon^{\sigma}\partial_{\sigma}A_{\nu}-\epsilon^{\sigma}\delta^{\rho}_{\sigma}\mathcal{L}$.

Furthermore,

$j^{\rho}=\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\epsilon^{\sigma}\partial_{\sigma}A_{\nu}-\epsilon^{\sigma}\delta^{\rho}_{\sigma}\mathcal{L}=\epsilon^{\sigma}\Big(\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\partial_{\sigma}A_{\nu}-\delta^{\rho}_{\sigma}\mathcal{L}\Big) = \epsilon^{\sigma}{j^{\rho}}_{\sigma}= \epsilon^{\sigma}{T^{\rho}}_{\sigma}$,

where $\partial_{\rho}{j^{\rho}}_{\sigma}= \partial_{\rho}{T^{\rho}}_{\sigma}=0$ so that each of the four conserved Noether charges ${j^{\rho}}_{\sigma}= {T^{\rho}}_{\sigma}$ arise due to the translational invariance of the action in each of the four spacetime coordinates.

The conserved Noether currents ${j^{\rho\sigma}}= {T^{\rho\sigma}}=\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\partial^{\sigma}A_{\nu}-\eta^{\rho\sigma}\mathcal{L}$ form the energy-momentum tensor $T^{\rho\sigma}$.

For the electromagnetic field,

${T^{\rho\sigma}}=\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\partial^{\sigma}A_{\nu}-\eta^{\rho\sigma}\mathcal{L}$

$=\frac{\partial}{\partial (\partial_{\rho}A_{\nu})}\Big(-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}\Big)\partial^{\sigma}A_{\nu}-\eta^{\rho\sigma}\Big(-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}\Big)$

$=-\frac{1}{4}\frac{\partial}{\partial (\partial_{\rho}A_{\nu})}[(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha})(\partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha})]\partial^{\sigma}A_{\nu}-\eta^{\rho\sigma}\Big(-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}\Big)$

$=-\frac{1}{4}[({\eta^{\rho}}_{\alpha}{\eta^{\nu}}_{\beta}-{\eta^{\rho}}_{\beta}{\eta^{\nu}}_{\alpha})(\partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha})+(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha})(\eta^{\rho\alpha}\eta^{\nu\beta}-\eta^{\rho\beta}\eta^{\nu\alpha})]\partial^{\sigma}A_{\nu}+\frac{1}{4}\eta^{\rho\sigma}F_{\alpha\beta}F^{\alpha\beta}$

$=-\frac{1}{4}[\partial^{\rho}A^{\nu}-\partial^{\nu}A^{\rho}-\partial^{\nu}A^{\rho}+\partial^{\rho}A^{\nu}+\partial^{\rho}A^{\nu}-\partial^{\nu}A^{\rho}-\partial^{\nu}A^{\rho}+\partial^{\rho}A^{\nu}]\partial^{\sigma}A_{\nu}+\frac{1}{4}\eta^{\rho\sigma}F_{\alpha\beta}F^{\alpha\beta}$

$=-(\partial^{\rho}A^{\nu}-\partial^{\nu}A^{\rho})\partial^{\sigma}A_{\nu}+\frac{1}{4}\eta^{\rho\sigma}F_{\alpha\beta}F^{\alpha\beta}$

$=-F^{\rho\nu}\partial^{\sigma}A_{\nu}+\frac{1}{4}\eta^{\rho\sigma}F_{\alpha\beta}F^{\alpha\beta}$

So, the energy-momentum tensor $T^{\mu\nu}=-F^{\mu\rho}\partial^{\nu}A_{\rho}+\frac{1}{4}\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}$.

What do you think?

 

[TSny]

OK, it looks good to me. You effectively re-derived Noether's theorem for this case. Nothing wrong with that!

If the problem just wants you to use Noether's theorem, then you can shorten the amount of work by applying Noether's theorem using your result:

Under the infinitesimal translation $x^{\sigma} \rightarrow x^{\sigma}-\epsilon^{\sigma}$,

$A_{\nu}(x) \rightarrow A_{\nu}(x) + \epsilon^{\sigma}\partial_{\sigma}A_{\nu}(x)$ and $\mathcal{L}(x) \rightarrow \mathcal{L}(x) + \epsilon^{\sigma}\partial_{\sigma}\mathcal{L}(x)$

From the above, you can use Noether's theorem to quickly write the energy-momentum tensor as ${T^{\rho\sigma}}=\frac{\partial\mathcal{L}}{\partial (\partial_{\rho}A_{\nu})}\partial^{\sigma}A_{\nu}-\eta^{\rho\sigma}\mathcal{L}$

Anyway, I agree with your final result $T^{\mu\nu}=-F^{\mu\rho}\partial^{\nu}A_{\rho}+\frac{1}{4}\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}$.

 

[spaghetti3451]

Ok, now let me try and show that the energy-momentum tensor $T^{\mu\nu}=-F^{\mu\rho}\partial^{\nu}A_{\rho}+\frac{1}{4}\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}$ is neither symmetric nor gauge invariant.

Firstly,

$T^{\mu\nu}=-F^{\mu\rho}\partial^{\nu}A_{\rho}+\frac{1}{4}\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}$

$\implies T^{\nu\mu}=-F^{\nu\rho}\partial^{\mu}A_{\rho}+\frac{1}{4}\eta^{\nu\mu}F_{\rho\sigma}F^{\rho\sigma}$

$\implies T^{\nu\mu}=-F^{\nu\rho}\partial^{\mu}A_{\rho}+\frac{1}{4}\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}$

The second term is clearly symmetric under the interchange of indices $\mu$ and $\nu$. How do I proceed to show that the first term is not symmetric under the interchange of indices $\mu$ and $\nu$?

 

[TSny]

$T^{\nu\mu}=-F^{\nu\rho}\partial^{\mu}A_{\rho}+\frac{1}{4}\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}$

The second term is clearly symmetric under the interchange of indices $\mu$ and $\nu$. How do I proceed to show that the first term is not symmetric under the interchange of indices $\mu$ and $\nu$?

Can you construct a simple example of an E field and/or B field (with corresponding expressions for A^μ) where you can show explicitly that the first term is not symmetric? My first couple of attempts failed (the first term kept coming out symmetric!). But I think I now have an example. I would like to see what you come up with.

 

[spaghetti3451]

Can you construct a simple example of an E field and/or B field (with corresponding expressions for A^μ) where you can show explicitly that the first term is not symmetric? My first couple of attempts failed (the first term kept coming out symmetric!). But I think I now have an example. I would like to see what you come up with.

Let me try to understand your approach to the problem.

$F^{\nu\rho}$ is always anti-symmetric, so if we can show that $\partial^{\mu}A_{\rho}$ is symmetric for at least one choice of $A_{\mu}$, then we will have shown that the resulting object $F^{\nu\rho}\partial^{\mu}A_{\rho}$ is not always symmetric.

If $\partial^{\mu}A_{\rho}$ is symmetric, then $\partial^{\mu}A_{\rho}=\partial^{\rho}A_{\mu}$.

Do I now need to figure out an example of $\textbf{E}$ and $\textbf{B}$ fields that leads to a symmetric $\partial^{\mu}A_{\rho}$?

 

[TSny]

Let me try to understand your approach to the problem.

$F^{\nu\rho}$ is always anti-symmetric, so if we can show that $\partial^{\mu}A_{\rho}$ is symmetric for at least one choice of $A_{\mu}$, then we will have shown that the resulting object $F^{\nu\rho}\partial^{\mu}A_{\rho}$ is not always symmetric

If $\partial^{\mu}A_{\rho}$ is symmetric, it does not follow necessarily that $F^{\nu\rho}\partial^{\mu}A_{\rho}$ is not symmetric. That is, $F^{\nu\rho}\partial^{\mu}A_{\rho}$ could be symmetric. Consider a 2 x 2 case.

Let $A = \begin{pmatrix} 0&-1\\ 1&0\\ \end{pmatrix}$ and $B = \begin{pmatrix} 1&0\\ 0&-1\ \end{pmatrix}$

$A$ is antisymmetric and $B$ is symmetric. $AB$ is symmetric.

Can you find an example of fields E and B such that $F^{0 \rho}\partial^{1}A_{\rho} \neq F^{1\rho}\partial^{0}A_{\rho}$?

 

[spaghetti3451]

What if $\textbf{E}=(0,0,0)$ and $\textbf{B}=(0,1,0)$?

 

[spaghetti3451]

If $\partial^{\mu}A_{\rho}$ is symmetric, it does not follow necessarily that $F^{\nu\rho}\partial^{\mu}A_{\rho}$ is not symmetric. That is, $F^{\nu\rho}\partial^{\mu}A_{\rho}$ could be symmetric. Consider a 2 x 2 case.

Let $A = \begin{pmatrix} 0&-1\\ 1&0\\ \end{pmatrix}$ and $B = \begin{pmatrix} 1&0\\ 0&-1\ \end{pmatrix}$

$A$ is antisymmetric and $B$ is symmetric. $AB$ is symmetric.

Can you find an example of fields E and B such that $F^{0 \rho}\partial^{1}A_{\rho} \neq F^{1\rho}\partial^{0}A_{\rho}$?

I'm finding it difficult to find the answer. It would be really helpful if you could provide a small additional hint.

 

[TSny]

What if $\textbf{E}=(0,0,0)$ and $\textbf{B}=(0,1,0)$?

What does the matrix $F^{\mu \nu}$ look like in this case? What does the 4-vector $A^{\mu}$ look like? (You have some freedom in choosing $A^{\mu}$ .)

 

[TSny]

I'm finding it difficult to find the answer. It would be really helpful if you could provide a small additional hint.

We have $T^{\mu \nu} = Q^{\mu \nu} + \frac{1}{4} \eta ^{\mu \nu} F^{\alpha \beta}F_{\alpha \beta}$ where $Q^{\mu \nu} = -F^{\mu \rho}\partial ^{\nu} A_{\rho}$. We're only concerned with the symmetry of $Q^{\mu \nu}$.

Suppose you want to construct an example where, say, $Q^{30} \neq Q^{03}$. Can we do that with a simple example using static E and B fields? For static fields, show that $Q^{30} = 0$. Then write out $Q^{03}$ in terms of components of $\vec{E}$ and/or $\vec{B}$ and components of $A^{\mu}$. Can you find a way to choose components of $\vec{E}$ , $\vec{B}$ , and $A^{\mu}$ so that $Q^{03} \neq 0$?

 

[spaghetti3451]

We have $T^{\mu \nu} = Q^{\mu \nu} + \frac{1}{4} \eta ^{\mu \nu} F^{\alpha \beta}F_{\alpha \beta}$ where $Q^{\mu \nu} = -F^{\mu \rho}\partial ^{\nu} A_{\rho}$. We're only concerned with the symmetry of $Q^{\mu \nu}$.

Suppose you want to construct an example where, say, $Q^{30} \neq Q^{03}$. Can we do that with a simple example using static E and B fields? For static fields, show that $Q^{30} = 0$. Then write out $Q^{03}$ in terms of components of $\vec{E}$ and/or $\vec{B}$ and components of $A^{\mu}$. Can you find a way to choose components of $\vec{E}$ , $\vec{B}$ , and $A^{\mu}$ so that $Q^{03} \neq 0$?

$Q^{30}=-F^{3\rho}\partial^{0}A_{\rho}=0$, since $\partial^{0}A_{\rho}=0$ for static fields ($A^{\mu}$ is constant for static fields).

Now, $Q^{30}=-F^{3\rho}\partial^{0}A_{\rho}=-F^{30}\partial^{0}A_{0}-F^{31}\partial^{0}A_{1}-F^{32}\partial^{0}A_{2}-F^{33}\partial^{0}A_{3}$
$=-E_{2}\partial^{0}A_{0}-B_{3}\partial^{0}A_{1}+B_{1}\partial^{0}A_{3}$

So, let's choose $E_{2}=B_{3}=B_{1}=0$ which means that $\partial_{2}A_{0}=\partial_{0}A_{2}, \partial_{2}A_{3}=\partial_{3}A_{2}, \partial_{1}A_{2}=\partial_{2}A_{1}$.

Now, $Q^{03}=-F^{0\rho}\partial^{3}A_{\rho}=-F^{00}\partial^{3}A_{0}-F^{01}\partial^{3}A_{1}-F^{02}\partial^{3}A_{2}-F^{03}\partial^{3}A_{3}$
$=E_{1}\partial^{3}A_{1}+E_{2}\partial^{3}A_{2}+E_{3}\partial^{3}A_{3}=E_{1}\partial^{3}A_{1}+E_{3}\partial^{3}A_{3}$.

So, for any nonzero value of $E_1$ and $E_3$, we have $Q^{03}\neq 0$.

What do you think?

 

[TSny]

$Q^{30}=-F^{3\rho}\partial^{0}A_{\rho}=0$, since $\partial^{0}A_{\rho}=0$ for static fields ($A^{\mu}$ is constant for static fields).

OK

So, let's choose $E_{2}=B_{3}=B_{1}=0$ which means that $\partial_{2}A_{0}=\partial_{0}A_{2}, \partial_{2}A_{3}=\partial_{3}A_{2}, \partial_{1}A_{2}=\partial_{2}A_{1}$.

Now, $Q^{03}=-F^{0\rho}\partial^{3}A_{\rho}=-F^{00}\partial^{3}A_{0}-F^{01}\partial^{3}A_{1}-F^{02}\partial^{3}A_{2}-F^{03}\partial^{3}A_{3}$
$=E_{1}\partial^{3}A_{1}+E_{2}\partial^{3}A_{2}+E_{3}\partial^{3}A_{3}=E_{1}\partial^{3}A_{1}+E_{3}\partial^{3}A_{3}$.

So, for any nonzero value of $E_1$ and $E_3$, we have $Q^{03}\neq 0$.

How can you guarantee that $\partial^{3}A_{1}$ and $\partial^{3}A_{3}$ aren't both equal to zero?

 

[spaghetti3451]

How can you guarantee that $\partial^{3}A_{1}$ and $\partial^{3}A_{3}$ aren't both equal to zero?

$\vec{E}=-\nabla \phi - \frac{\partial \vec{A}}{\partial t} \implies E_{i} = -\partial_{i}A_{0}-\partial_{0}A_{i}$ so that any given choice of $\partial^{3}A_{1}$ and $\partial^{3}A_{3}$ does not affect the value of $\vec{E}$.

$\vec{B}=\nabla\times \vec{A} = (\partial_{2}A_{3}-\partial_{3}A_{2},\partial_{3}A_{1}-\partial_{1}A_{3},\partial_{1}A_{2}-\partial_{2}A_{1})$ so that any given choice of $\partial^{3}A_{1}$ (and $\partial^{3}A_{3}$) affects only the value of $B_{2}$, but then $B_2$ is not fixed (as shown in the previous post) from the constraint $Q^{30}=0$.

Therefore, with appropriate choices of $A_{1}$ and $A_{3}$ (i.e. make both $A_{1}$ and $A_{3}$ dependent on $z$), we can guarantee that
$\partial^{3}A_{1}$ and $\partial^{3}A_{3}$ aren't both equal to zero.

What do you think?

 

[TSny]

OK. So, as a specific example, you can choose a uniform, static E field, $E_o$, in the x direction and a uniform, static B field, $B_o$, field in the y direction. For the potential, $A^{\mu}$, you can then take $A^0= -E_o x$, $A^1 = B_o z$, $A^2 = 0$, $A^3 = 0$. This choice for $A^{\mu}$ produces the correct E and B fields and also makes $Q^{03} \neq 0$.

For the same E and B fields, you can choose a different $A^{\mu}$ (a different gauge) such as $A^0 = -E_o x$, $A^1 = 0$, $A^2 = 0$, $A^3 = -B_o x$. This choice of $A^{\mu}$ is related to the first choice by a gauge transformation. Now you find $Q^{03} = 0$. This illustrates that $T^{\mu \nu}$ is not gauge invariant.

 

[spaghetti3451]

$Q^{03}=-F^{0\rho}\partial^{3}A_{\rho}=-E_{0}\partial^{3}A_{1}=0$.

Hmm... I see! $Q^{03}$ is not invariant under the given gauge transformation, hence $T^{\mu\nu}$ is not gauge invariant.

For some reason, I have not been able to do this part of the problem even though I've been able to grind through tedious calculations in other parts of the problem.