1

I have a Lagrangian of a form:

$$\mathcal{L}=\frac{1}{2}\left (A_\mu g^{\mu\nu}\partial^2 A_\nu-A_\mu \partial^\mu \partial^\nu A_\nu\right ) $$

And I want to show that it is equivalent to the Lagrangian of a form

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

I know that the lagrangians are said to be invariant in a sense that they differ by total derivative:

$$\mathrm{d}(AB)=(\mathrm{d}A)B+A(\mathrm{d}B)$$

$$(\mathrm{d}A)B=-A(\mathrm{d}B)+\mathrm{d}(AB)$$

However, I was unsuccessful in my atempts in calculating in that way (tried various scenarios of taking $A$ and $\mathrm{d}B$ from the Lagrangian, but did not succeed). Could anyone help on how to do it the right way?

Níckolas Alves
  • 17,567
aerospace
  • 47
  • 3
    Chuck the forms and integrate by parts. The forms are meant to help you, not confuse you! – Cosmas Zachos Jan 26 '22 at 15:08
  • @CosmasZachos, that is the part that I am confused about. As integrating by parts is $udv=vu-vdu$, I tried taking $A$ as $u$ and $\partial^2 A$ as $dv$, however, the end result does not make sense to me as it is not the form of a first expression + second expression. – aerospace Jan 26 '22 at 15:17
  • But convince the reader that you can prove this by plain integration by parts, and emphasize that it is your translation to forms that stymies you. – Cosmas Zachos Jan 26 '22 at 15:23
  • Compare with this. – J.G. Jan 26 '22 at 15:26

1 Answers1

1

These two expressions are equal if we can ignore boundary terms (i.e. we can integrate by parts). This means that we can say $A \partial_\mu B = - B \partial_\mu A $.

Your first expression is: $L=\frac{1}{2}(A_\mu \partial_\nu \partial^\nu A^\mu -A_\mu \partial^\mu \partial^\nu A_\nu)$.

Integrating by parts, we get $L=\frac{1}{2}(-\partial_\nu A_\mu \partial^\nu A^\mu + \partial_\nu A_\mu \partial^\mu A^\nu)$.

Then, since all indices are contracted, it does not matter if we switch $\mu$ and $\nu$. Therefore, we symmetrise over $\mu$ and $\nu$: $L=\frac{1}{2}(-\frac{1}{2}(\partial_\nu A_\mu \partial^\nu A^\mu +\partial_\mu A_\nu \partial^\mu A^\nu)+\frac{1}{2}(\partial_\nu A_\mu \partial^\mu A^\nu+\partial_\mu A_\nu \partial^\nu A^\mu))$.

This then factorises into your final expression.

Cameron Gibson
  • 546
  • Thank you very much! I have done it by myself yesterday actually and the main realization was that I fogot to "simplify" the last form in your text to the second one and that is why it made me puzzle for so long as I am not very fluent in those tensor notations. – aerospace Jan 27 '22 at 06:28