Can index swapping be applied to relativistic Lagrangian equations?

bananabandana
Messages
112
Reaction score
5

Homework Statement


Show that

$$ \mathcal{L} = -\frac{1}{4}F^{\mu \nu}F_{\mu \nu} = - \frac{1}{2}\partial^{\mu}A^{\nu}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}) $$

Where $$ F^{\mu \nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu} $$

Homework Equations

The Attempt at a Solution


$$ \mathcal{L} = -\frac{1}{4} F^{\mu \nu}F_{\mu \nu} = -\frac{1}{4}(\partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu})(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}) $$
Which expands out to:
$$ -\frac{1}{4} \bigg( \partial^{\mu}A^{\nu}\partial_{\mu}A_{\nu} - \partial^{\mu}A^{\nu}\partial_{\nu}A_{\mu} -\partial^{\nu}A^{\mu}\partial_{\mu}A_{\nu}+\partial^{\nu}A^{\mu}\partial_{\nu}A_{\mu} \bigg) $$

So if I just exchange indices on half of the terms, and then take out a factor, I get to the result I want... question is, how am I allowed to do that??
 
Physics news on Phys.org
It does not matter what you call summation indices, in general
$$
\sum_\mu V_\mu W^\mu = \sum_\nu V_\nu W^\nu.
$$
The only difference in your expression is that the summation convention is being used and therefore we do not write out the sums explicitly but always sum over repeated indices.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top