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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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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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