Derivation of geodesic equation from the action - quick question

[binbagsss]

Hi,

I am following this : https://en.wikipedia.org/wiki/Geodesics_in_general_relativity

and all is good except how do we get $\delta g_{uv}=\partial_{\alpha}g_{uv}\delta x^{\alpha}$

Many thanks

 

[binbagsss]

Hi,

I am following this : https://en.wikipedia.org/wiki/Geodesics_in_general_relativity

and all is good except how do we get $\delta g_{uv}=\partial_{\alpha}g_{uv}\delta x^{\alpha}$

Many thanks

Similarly if I want to derive the modified geodesic equation obeyed by a charged particle of some given charge as well as some given mass I need to vary $\delta A_u$ the vector potential, and the first hint given is that first of all you need to explain why:

$\delta A_u = \partial_v A_u \delta x^v$

where $\partial_v = \frac{\partial}{\partial x^v}$

I was perhaps thinking it uses the chain rule and then the product rule, since each term would need be varied wrt $x^u$ but can't really see this working:

$g_{uv}=\partial_{\alpha}g_{uv}dx^{\alpha}$

 

[davidge]

how do we get $\delta g_{uv}=\partial_{\alpha}g_{uv}\delta x^{\alpha}$

Here, $g$ is a function of $x$, i.e. $g_{\mu \nu} = g_{\mu \nu} (x^\alpha)$. $\delta g_{\mu \nu}$ should be regarded as the differential of the function $g_{\mu \nu}$. How do you apply the differential operator $d$ to a function? It's the same thing here.

 

[PeterDonis]

$\delta g_{\mu \nu}$ should be regarded as the differential of the function $g_{\mu \nu}$.

No, it isn't. It's the variation of the function $g_{\mu \nu}$. More specifically, it's the variation of $g_{\mu \mu}$ when you vary the curve $x^\alpha$ that you have chosen. If your question at this point is "what curve", then you need to go back and read, carefully, the section on deriving the geodesic equation via an action.