# Euler-Lagrange Tensor Equations

• A
• binbagsss
In summary, when varying the term in the Lagrangian given by ##\frac{1}{3}A^{\mu} \partial_{\mu}\Phi##, we get the corresponding term in the equations of motion as ##\frac{1}{3}\partial^m\partial^n\Phi-\frac{1}{4}\eta^{\alpha \beta} \partial^f \partial_f \Phi##.

#### binbagsss

I need to vary w.r.t ##a_{\alpha \beta} ##

##\frac{\partial L}{\partial_{\mu}(\partial_{\mu}{a_{\alpha\beta}})}-\frac{\partial L}{\partial {a_{\alpha \beta}}}## (1)

I am looking at varying the term in the Lagrangian of ##\frac{1}{3}A^{\mu} \partial_{\mu}\Phi ##

where ##A^{\beta}=\partial_k {a^{k\beta}} ##

##\frac{\partial L}{\partial_{\mu}(\partial_{\mu}{a_{\alpha\beta}})}-\frac{\partial L}{\partial {a_{\alpha \beta}}}##

My working

(Expect it to yield a corresponding term in the EoMs as: ##\frac{1}{3}\partial^m \partial^{n} \Phi-\frac{1}{4}\eta^{\alpha \beta} \partial^f \partial_f \Phi##  )

I don't think I've ever done this properly, so my apologies, but my guess is want to keep things as general as possible with the indicies in (1) and those used on the ##a_{\alpha \beta} ## tensor so (also, with the first term in (1)- is the index ##\mu## are they both supposed to be the same or is it better to keep it more general, something like:##\frac{\partial L}{\partial_{\nu}(\partial_{\mu}{a_{\alpha\beta}})}##?) :

## \frac{1}{3}A^{\mu} \partial_{\mu}\Phi= \frac{1}{3}\partial_c a^{c \mu} \partial_{\mu} \Phi, ##

So if I say I am varying it w.r.t ##\partial_{k}a^{mn}## for the first term of (1), (obviously second term of (1) not relevant):

first I lower the indices on : ##\frac{1}{3}\partial_c \eta^{wc} \eta^{zu}a_{wz}\partial_{\mu} \Phi ##

then, (this is the bit I'm more unsure of- get deltas from requiring the indices on the derivative and tensor to match...)when varying wrt ##\partial_{k}a^{mn}## :

##\frac{1}{3}\delta_{c,k}\delta_{m,w}\delta_{n,z}\eta^{wc}\eta^{zu}\partial_{\mu}\Phi ##

##=\frac{1}{3}\eta^{mk}\eta^{n\mu}\partial_{\mu}\Phi ##

so for ##\frac{\partial L}{\partial_k(\partial_{k}{a_{mn}})}## get:

##\partial_k\frac{1}{3}\eta^{mk}\eta^{n\mu}\partial_{\mu}\Phi ##
##= \frac{1}{3}\partial^m\partial^n\Phi ##

And so I have got the first term of  but not the second term. :(

Thanks.

Last edited:

Hello! It looks like you have made some progress in your calculations, but there are a few things that need to be clarified. First, in the first term of (1), the index ##\mu## should be the same as the one in the derivative of the Lagrangian, so it should be ##\frac{\partial L}{\partial_{\mu}(\partial_{\mu}{a_{\alpha\beta}})}##.

Secondly, when you vary wrt ##\partial_{k}a^{mn}##, the indices should match, so it should be ##\frac{\partial L}{\partial_{k}(\partial_{m}{a_{n\mu}})}##. Then, when you perform the variation, you should get:

##\frac{\partial L}{\partial_{k}(\partial_{m}{a_{n\mu}})} = \frac{1}{3}\partial^m\partial^n\Phi - \frac{1}{3}\delta_{mk}\delta_{n\mu}\partial^f\partial_f\Phi ##

The second term in this expression is the one that matches the second term in .

I hope this helps clarify things for you! Keep up the good work.