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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?

Níck Aguiar Alves
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aerospace
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1 Answers1

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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
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