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Why the variation of the Lagrangian density has a term which is it's 4-gradient if it doesn't explicitly depend on $x^{\mu}$

+ 1 like - 0 dislike
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The Lagrangian density, as told in the excerpt below, isn't explicitly dependent on $x$. Why isn't $\partial _{\mu}\cal{L}$ isn't $0$?

There is no more information about this in the book.

Thanks!

asked Jul 2, 2015 in Theoretical Physics by pniau7 (25 points) [ no revision ]
recategorized Jul 2, 2015 by Dilaton

How do you want us to know it if there is no more information about this in the book?

Because $\partial_{\mu}$ is the total (explicit plus implicit) derivative with respect to $x^{\mu}$. Implicit means the dependence through the fields. A field theory is translational invariant, i.e. there are no special points in space-time (apart from boundary/initial conditions), if and only if the Lagrangian density doesn't depend on $x$ explicitly. This is what that book calls "a basic hypothesis of Field Theory", a sentence I don't like at all. Note that an external field (external means that its dynamics it is not described by your theory, but given as an input) makes the Lagrangian explicitly dependent on $x$ and breaks invariance under translations.

The result arrived in the excerpt is quite general (as you can see in Peskin & Schroeder, for example). The reason I show it and not another is because it shows the most general variation possible (with fixed end-points), but it could be any other book. The question is still pertinent, why $\partial_{\mu}\cal{L}\neq 0$ if $\cal{L}$ doesn't explicitly depend on $x^{\mu}$?

@Drake Thanks for your answer. How do you see that it's the total derivative? The notation I'm used to is $\partial_{\mu}=\left(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$ which, in other words, is the 4-gradient.

Should $\partial_{\mu}\cal{L}=\frac{d\cal{L}}{dx^{\mu}}=\frac{\partial\cal{L}}{\partial \phi}\frac{\partial\phi}{\partial x^{\mu}}+\frac{\partial\cal{L}}{\partial \left(\partial_{\nu}\phi\right)}\frac{\partial\left(\partial_{\nu}\phi \right)}{\partial x^{\mu}}$ be the right notation?

I know it's the total derivative because otherwise it would be wrong and also because it's standard notation (although perhaps unfortunate). Your last identity is right, provided that there is no explicit dependence, which is not right in the presence of external fields. 

1 Answer

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Actually the $\partial_\mu$ is a "total derivative", otherwise it wouldn't be a total divergence and we would not be able to get rid of it during the derivation of Lagrange equations

$$\partial_\mu \mathcal{L} \equiv \frac{\partial \mathcal{L}}{\partial \phi} \partial_\mu \phi + \frac{\partial \mathcal{ L}}{\partial ( \partial _\nu \phi)} \partial_\mu \partial _\nu \phi$$

This is a slightly confusing but in the end somewhat practical convention in classical field theory. For instance, you can see that this very same derivative is used in the Lagrange equations

$$\partial_\mu ( \frac{\partial \mathcal{ L}}{\partial ( \partial _\mu \phi)}) - \frac{\partial \mathcal{L}}{\partial \phi} = 0$$

etc. This notation is possible only thanks to the fact that the "real partial derivative" of $\mathcal{L}$ is postulated to be always zero and that there cannot thus be any ambiguity in what is meant by $\partial_\mu \mathcal{L}$.

Some authors prefer to use notation such as $ d/dx^\mu$ or $D/dx^\mu$ to underline the "totalness" but this always feels like notation abuse to me. My opinion is that if anything should change then it is the discussion of the "omitted pullback" $\mathcal{L}(\phi,...) \to \mathcal{L}(\phi(x^\mu),...)$, what does $\partial/\partial (\partial_\mu \phi)$ really mean in term of the pullback and so on.

answered Jul 3, 2015 by Void (1,505 points) [ no revision ]

That "postulate" is only sensible if there are no external fields. It is actually not a postulate, but a consequence of the absence of external fields or a requirement in order to have translational invariance. 

@drake In the case of external fields you simply have the Lagrangian as a pullback of the external field $\mathcal{L}(x,...) = \mathcal{L}(\phi_{\rm ext} (x),...)$. The $\partial_\mu$ then works in the same chain-rule way (the "true partial derivative" is zero) and the only difference is that one does not allow the $\delta \phi_{\rm ext}$ variation in determining the equations of motion. Of course you could say the external field is hard-wired into the Lagrangian but I don't find that particularly instructive.

@Void That notation is non-sense because the external field is a GIVEN (EXPLICIT) function of $x$. You can't consistently treat external fields and degrees of freedom in the same way. I understand what you mean, but  it's an inconsistente notation. 

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