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In general relativity, when deriving the field equation using the variational principle we use $\hat{g}_{\mu\nu}=g_{\mu\nu}+\delta{g_{\mu\nu}}$.

Does $\delta{g_{\mu\nu}}$ mean the measurement of how $g_{\mu\nu}$ changes when we change the form of the equations in the components of ${g_{\mu\nu}}$ by changing the coordinates or doing some other thing that changes their form, Or does it measure how $g_{\mu\nu}$ changes when we translate $r$ by $\triangle{r}$?

MrDi
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    ...what does $\delta\phi$ mean to you for other fields? – ACuriousMind Jul 23 '15 at 22:58
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    It means neither. It means that we assume a perturbation of the $g_{\mu\nu}$ at which the action is staionary. – Sebastian Riese Jul 23 '15 at 22:58
  • @ACuriousMind, it means the variation of $\phi$ by a small amount, but doesn't that mean that we are changing the variables that $\phi$ depends on by a small amount. – MrDi Jul 23 '15 at 23:02
  • These variations are variations of situations in some region but situation is the same at the boundaries. For this case $\delta g_{\mu\nu}(x)$ is difference in (metric) field configuration in some region but field configuration is fixed to be the same at the boundaries. – Saksith Jaksri Feb 21 '17 at 13:05

2 Answers2

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Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. The metric $g$ can be viewed as a classical field of this kind.

OP is asking about finding the Euler-Lagrange equations. In that case, the variations are vertical.

There are other applications where variations are not necessarily vertical, e.g. Noether's theorem, cf. this Phys.SE post.

Community
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Qmechanic
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Usually, the variation of a filed $\delta\phi$ is defined to be

\begin{equation} \delta\phi\left(x\right)=\phi^{'}\left(x\right)-\phi\left(x\right) \end{equation}

where the new field $\phi^{'}$ and old field $\phi$ are evaluated at the same point, if we take a active transformation point of view.

So I think $\delta g_{\mu\nu}$ should be

\begin{equation} \delta g_{\mu\nu}\left(x\right)=\hat{g}_{\mu\nu}\left(x\right)-g_{\mu\nu}\left(x\right) \end{equation}

where they are evaluated at the same point.

So if you make a translation, $x_{\mu}\rightarrow x_{\mu}+\epsilon_{\mu}$, then $\hat{g}_{\mu\nu}\left(x+\epsilon\right)=g_{\mu\nu}\left(x\right)$.

But \begin{equation} \hat{g}_{\mu\nu}\left(x+\epsilon\right)=\hat{g}_{\mu\nu}\left(x\right)+\epsilon^{\alpha}\partial_{\alpha}g_{\mu\nu}\left(x\right) \end{equation}

So by our definition, we have

\begin{align*} \delta g_{\mu\nu}\left(x\right) & =\hat{g}_{\mu\nu}\left(x\right)-g_{\mu\nu}\left(x\right)\\ & =\hat{g}_{\mu\nu}\left(x+\epsilon\right)-\epsilon^{\alpha}\partial_{\alpha}g_{\mu\nu}\left(x\right)-g_{\mu\nu}\left(x\right)\\ & =-\epsilon^{\alpha}\partial_{\alpha}g_{\mu\nu}\left(x\right) \end{align*}

Nahc
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  • You say "new field", and the rest of the answer looks as if this "new" field must be obtained from the old one by some transformation on spacetime. That is not how the variation is supposed to work. $\phi'$ may be any field configuration "close" to $\phi$ in a suitable sense. – ACuriousMind Jul 24 '15 at 13:50
  • A suitable definition of "close" is given by the "$C^0$ open topology" on the bundle of symmetric tensors, cf. e.g. Hawking & Ellis, The large scale structure of space-time (1973). – Ryan Unger Jul 24 '15 at 13:56