Equation of motion Chern-Simons

[Lapidus]

The Lagrangian (Maxwell Chern-Simons in Zee QFT Nutshell, p.318)

has as equation of motion:

Where does the 2 in front come from?

Thank you very much

 

[Orodruin]

You have two $a$ in the $\gamma$ term.

 

[Lapidus]

Right. And a derivative in front of one a.

Do I get one term from the RHS and one from the LHS of equation of motion and then I add them together?

 

[Orodruin]

Right. And a derivative in front of one a.

Do I get one term from the RHS and one from the LHS of equation of motion and then I add them together?
View attachment 240456

Why don’t you try it and see what you get?

 

[Lapidus]

I get the equation but without the 2 in front. I do not see how the 2 comes about. How to sum over the indices. I find the indices confusing. Hence my question.

 

[Orodruin]

I get the equation but without the 2 in front. I do not see how the 2 comes about. How to sum over the indices. I find the indices confusing. Hence my question.

We cannot help you unless you show what you actually did. Otherwise we have no way of knowing where you went wrong.

 

[Lapidus]

I differentiate γε^μνλa_μ∂_νa_λ w.r.t. a_μ and I get γε^μνλ∂_νa_λ

and γε^μνλa_μ∂_νa_λ w.r.t. ∂_νa_λ which gives γε^μνλa_μ.

Thus γε^μνλ∂_νa_λ - γε^μνλa_μ = 0 , which is not 2γε^μνλ∂_νa_μ = 0.

 

[Orodruin]

You forgot to take the derivative with respect to $x^\nu$ of the derivative with respect to $\partial_\nu a_\lambda$.

Edit: Also, if you want the equation of motion for $a_\mu$, you must take the derivative with respect to $\partial_\nu a_\mu$, not $\partial_\nu a_\lambda$.

 

[Lapidus]

L = a_μ∂_νa_λ

∂L/∂a_μ - ∂_ν (∂L/∂(∂_νa_μ)) = ∂_va_μ - ∂_v ?

I do not know what and how to differentiate in the second term. Also, I need to add two identical terms to get the factor two. But there is a minus sign.

 

[Orodruin]

I suggest you read this: https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

Edit: Also, this:

L = aμ∂νaλ

is not what L is.

 

[Lapidus]

Orodruin, I really appreciate your time and effort. I read carefully all your posts in this thread and the link you gave. But unfortunately, I still can not answer my initial question. Maybe I try somewhere else. Thank you!

 

[Orodruin]

I am sorry you don't feel you have enough. I think you would benefit significantly from showing your computations in more detail instead of just stating what you get and by thinking of each step in terms of what I said in the Insight. I could of course just give you the derivation, but I doubt you would learn as much from that as you would if you present it, clarify exactly in which steps your confusions lie, and get help in seeing how to resolve them.

In particular, I think you are guilty of #8 in the Insight and that this is causing you trouble. However, it is difficult to tell since you have not given us your explicit stepwise computations, just what you think each term should be.

 

[samalkhaiat]

Where does the 2 in front come from?
Thank you very much

Do not differentiate with respect to repeated indices. Write $$\mathcal{L} = \epsilon^{\sigma\rho\nu} \ A_{\sigma} \ \partial_{\rho}A_{\nu} + A_{\sigma}J^{\sigma}.$$ Now, differentiate with respect to $A_{\mu}$, and use $\frac{\partial A_{\eta}}{\partial A_{\mu}} = \delta^{\mu}_{\eta}$ to get $$\frac{\partial \mathcal{L}}{\partial A_{\mu}} = \epsilon^{\mu\rho\nu} \ \partial_{\rho} A_{\nu} + J^{\mu} . \ \ \ \ \ (1)$$ Next, differentiate $\mathcal{L}$ with respect to $(\partial_{\tau}A_{\mu})$ and use the identity $$\frac{\partial (\partial_{\rho}A_{\eta})}{\partial (\partial_{\tau}A_{\mu})} = \delta^{\tau}_{\rho} \ \delta^{\mu}_{\eta} ,$$ to obtain $$\frac{\partial \mathcal{L}}{\partial (\partial_{\tau}A_{\mu})} = \epsilon^{\sigma\tau\mu} \ A_{\sigma} = - \epsilon^{\mu\tau\sigma} \ A_{\sigma}.$$ Thus $$\partial_{\tau} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\tau}A_{\mu})} \right) = - \epsilon^{\mu\tau\sigma} \ \partial_{\tau}A_{\sigma} = - \epsilon^{\mu\rho\nu} \ \partial_{\rho}A_{\nu} . \ \ \ (2)$$ Now $(1) – (2) = 0$ is the E-L equation. It gives you $$2 \epsilon^{\mu\rho\nu} \ \partial_{\rho}A_{\nu} + J^{\mu} = 0.$$

 

[Lapidus]

Fantastic! Thanks Samalkhaiat! Kronecker Deltas from derivatives and indices swapping in the Levi-Civita. Got it.

I must study a little tensor calculus..