# Help required regarding deriving E-L equations for EM

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## Main Question or Discussion Point

I am a retired High School teacher trying to use tensors in getting the Euler-Lagrange equations from the em lagrangian density.

I attached a document in my post since I am not fluent in writing LaTex.

Can anyone, please check my work.

Thanks.

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jedishrfu
Mentor
Hi @grzz I looked at your handwritten notes and part of it is cut off and what remains is somewhat hard to read.

We really prefer Latex input because then we can copy paste parts of your expressions in our responses.

where I found these useful tools:

https://www.codecogs.com/latex/eqneditor.php

Here's another that might be better:

https://arachnoid.com/latex/

The arachnoid site has examples that you can view and tease out what you need

The last equation in the list is the Einstein field equation:

$\displaystyle R_{\mu v} - \frac{1}{2} R g_{\mu v} + \Lambda g_{\mu v} = \frac{8 \pi G}{c^4} T_{\mu v}$

which is basically this latex expression bracketted by double # signs

Code:
\displaystyle R_{\mu v} - \frac{1}{2} R g_{\mu v} + \Lambda g_{\mu v}  = \frac{8 \pi G}{c^4} T_{\mu v}
and here's the Schrodinger's equation complete with an invisible cat :-) jk

$\displaystyle i \hbar\frac{\partial \psi} {\partial t}= \frac{-\hbar^2}{2m} \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \psi + V \psi.$

with the Latex code here:
Code:
\displaystyle i \hbar\frac{\partial \psi}
{\partial t}=
\frac{-\hbar^2}{2m}
\left(\frac{\partial^2}{\partial x^2}
+ \frac{\partial^2}{\partial y^2}
+ \frac{\partial^2}{\partial z^2}
\right) \psi + V \psi.
Our site has a more detailed Latex guide here:

https://www.physicsforums.com/help/latexhelp/

Once you get the hang of it you'll wonder why you didn't learn it sooner.

Last edited:
Thank you 'jedishrfu'.
First I will have some practice using LaTex. Then I intend to repeat my post in a better format!
Thanks again.

I am following the suggestion by 'jedishrfu' and repeating my original post but removing the pdf attachment and trying LaTex instead.

The langragian for an electromagnetic field with sources is,
$$\mathcal{L}= - \frac {1}{4} F_{αβ}F^{αβ} + j^μA_μ.$$
Hence,
$\mathcal{L} = - \frac {1}{4} F_{αβ}η^{ασ}η^{βρ}F_{σρ} + j^μA_μ = - \frac {1}{4}(∂_αA_β - ∂_βA_α)η^{ασ}η^{βρ} (∂_σA_ρ - ∂_ρA_σ) + j^μA_μ$

Therefore,
$\frac {∂\mathcal{L}}{∂(∂_μA_ν)} = - \frac {1}{4} [η^{μσ}η^{νρ} (∂_σA_ρ - ∂_ρA_σ) - η^{νσ}η^{μρ} (∂_σA_ρ - ∂_ρA_σ) +(∂_αA_β - ∂_βA_α)η^{αμ}η^{βν} - η^{αν}η^{βμ} (∂_αA_β - ∂_βA_α) ]$
$= - \frac {1}{4} [(∂^μA^ν - ∂^νA^μ) - (∂^νA^μ - ∂^μA^ν) + (∂^μA^ν - ∂^νA^μ) - (∂^νA^μ - ∂^μA^ν) ]$
$= - \frac {1}{4} ( F^{μν} - F^{νμ} +F^{μν} - F^{νμ} )$
$= F^{νμ}$ since the $F^{μν}$ is an anti symmetrical tensor.

Also $\frac {∂\mathcal{L}}{ ∂A_μ} = j^μ.$

Hence the Euler-Langrange equations,
$$\frac {∂\mathcal{L}}{ ∂A_μ} - ∂_ν\left( \frac {∂\mathcal{L}}{ ∂(∂_νA_μ)} \right) = 0$$
for the fields $A_μ$ are,
$$j^μ - ∂_ν(F^{νμ} ) = 0$$
$$∂_ν(F^{νμ} ) = j^μ .$$

That is the Euler-Langrange equations give the two source Maxwell's Equations while the other two Maxwell's Equations are not given since the latter are not dynamical equations but are just mathematical identities.

Help is appreciated if somebody checks my work and perhaps suggest some shorter method.

Thanks.

Last edited:
jedishrfu
Mentor
Perhaps @Orodruin, one of our science advisors can comment on your equations.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
You can clean it up a bit by first noting that
$$\partial F_{\alpha\beta}F^{\alpha\beta} = 2F_{\alpha\beta} \partial F^{\alpha\beta} = 2F^{\alpha\beta} \partial F_{\alpha\beta},$$
where $\partial$ is any first derivative. This follows directly from the Leibniz rule. Then you can just note that
$$\frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)} = \delta^\mu_\alpha \delta^\nu_\beta - \delta^\nu_\alpha \delta^\mu_\beta = 2 \delta^{\mu}_{[\alpha} \delta^{\nu}_{\beta]},$$
where the brackets denote anti-symmetrisation with respect to the enclosed indices. Given these relations, you easily find that
$$\frac 14 \frac{\partial F_{\alpha\beta} F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)} = F^{\alpha\beta} \delta^\mu_{[\alpha} \delta^\nu_{\beta]} = F^{\mu\nu},$$
since $F$ is anti-symmetric.

Thanks 'Orodruin".

I am not so fluent with tensors and hence,
$$\partial F_{\alpha\beta}F^{\alpha\beta} = 2F_{\alpha\beta} \partial F^{\alpha\beta} = 2F^{\alpha\beta} \partial F_{\alpha\beta},$$
is not so obvious to me and I have to work it out.

\begin {align} ∂(F_{αβ}F^{αβ}) & = F^{αβ}∂(F_{αβ}) + F_{αβ}∂(F^{αβ}) \nonumber\\ &= F^{αβ}∂(F_{αβ}) + η _{ασ}η_{βρ}F^{σρ}∂(η^{αλ}η^{βν}F_{λν}) \nonumber\\ &= F^{αβ}∂(F_{αβ}) + δ^λ_σδ^ν_ρF^{σρ}∂(F_{λν}) \nonumber\\ &=F^{αβ}∂(F_{αβ}) + F^{λν}∂(F_{λν})\nonumber \\ &= 2F^{αβ}∂(F_{αβ}).\nonumber \end{align}

Is this working required for a beginner or is it to be left out?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Is this working required for a beginner or is it to be left out?
What do you mean by required? If you are doing an exam you might speak of requirements. Otherwise what is "required" is just that you convince yourself that it is true. Note that there is no need to pull the metrics out of the $F_{\alpha\beta}$ in your second term. The indices will be raised by the metrics from the $F^{\alpha\beta}$ once you pull them out of the derivative.

What do you mean by required?
I think that it all depends on with whom I share my work.
Perhaps if I share it with a beginner like me, it is good to show all steps.