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Why in GR does the $g_{\mu\nu}$ describe potential?

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  • $\begingroup$ Related: physics.stackexchange.com/q/33950 $\endgroup$ – Kyle Kanos Jun 13 '15 at 2:20
  • $\begingroup$ Further to agreeing with Timaeus's answer: I'm not sure what your level is- you might be a bit beyond this- but have you read Eintein's 1915 paper, or better still his book "meaning of relativity"? Einstein, unlike many modern treatments, spent a good many words describing the relationship between the metric and Newtonian theory (naturally it was important on the theory's intro to demonstrate the link). There is a most excellent review of this book by Eduardo Guerras Valera,sadly no longer active on this site:check out his wonderful little tale at physics.stackexchange.com/a/61843/26076 $\endgroup$ – Selene Routley Jun 13 '15 at 7:32
  • $\begingroup$ Quote is from which reference? $\endgroup$ – Qmechanic Aug 4 '15 at 12:33
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If you look at the formula for the Christoffel symbol in terms of the metric tensor you see that you need to take derivatives. So ...

Potential $\overset{\text{take derivative}}\rightarrow$ Force $\rightarrow$ Acceleration.

Is like:

Metric $\overset{\text{take derivative}}\rightarrow$ Christoffel Symbol $\rightarrow$ Geodesic Deviation.

I'm not sure any more was meant.

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Although this is an old question, and has been answered satisfyingly, let me add a couple of things that are only tangentially related to the question.

This whole thing about the metric being a gravitational "potential" and the Christoffel symbol being the "field strength" is just an analogy.

The source of the analogy is exactly what you stated, the geodesic equation, in which the Christoffel-symbol looks like a force. Then the metric is the potential because you construct the Christoffel-symbol from the metric's derivatives.

However, if you have nonzero $F$ in electrodynamics (the field strength 2-form), then you have an effective EM-field. On the other hand, you can have nonzero Christoffel-symbol and still not have gravitation fields - all you need is to introduce curved coordinate systems in flat space.

And even if you have curvature, you can still make the Christoffel-symbol vanish along a geodesic (Fermi normal coordinates).

This comes from the fact that the Christoffel-symbol is a part of the Levi-Civita connection, which is not a tensorial field, and both of these strange phenomena have physical meaning attached to them, namely, you can have nonzero Christoffel-symbol in flat space, because if you choose a curved coordinate system, and follow the coordinate curves, then, since they are not geodesics, you will have to experience acceleration (force) to be kept along them, and you can have a vanishing Christoffel-symbol along a geodesic, since if you move along a geodesic, you are in free fall, and thus you don't experience acceleration (force).

The only part of gravity, which cannot be transformed away, is the inhomogenity in gravity, eg. relative acceleration of geodesics, which is governed by the Riemann-tensor. You basically have curvature and so gravity, if the Riemann-tensor is nonzero. So in this respect it would be more natural to see the Riemann-tensor as the gravitational field strength, and then the Christoffel-symbol as the potential.

But the meaning of the Riemann-tensor corresponds more with the tidal force tensor in newtonian gravity, rather than the gravitational acceleration.

A gauge-theoretical formalism of gravity also implies the latter case. In gauge theories, fields correspond to connections on a principal $G$-bundle, with the potentials being the connection 1-forms, and the field strengths being the curvature 2-forms associated with the connection. For GR, the structure group is $\mathrm{SO}(1,3)$, acting on the bundle of orthonormal frames of $TM$, and for these the $\mathfrak{so}(1,3)$-valued connection 1-form is exactly the Christoffel-symbol (when represented on an orthonormal frame, rather than a holonomic coordinate frame), while the curvature 2-form is the Riemann-tensor, which once again implies that the field strength is the Riemann-tensor.

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The relation between the metric and gravitational potential (and between Christoffel symbols and acceleration) is evident in the Newtonian limit of General Relativity. The basic assumptions of this approximation are:

  • Weak gravitational field: the metric $g_{\mu\nu}$ differs from the Minkowski metric $\eta_{\mu\nu}$ only a small amount $$ g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$
  • Non-relativistic speeds: the proper time of the test particle $\tau$ is approximately equal to the coordinate time $t$ (Newton's absolute time)
  • Static metric: the time derivative of the metric is zero.

In the geodesic equation, the only relevant component of the four-velocity is the temporal component: $$ \frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{00} \left(\frac{dt}{ds}\right)^2 = 0$$ The Christoffel symbols that we need are:$$\Gamma^\mu_{00} = \frac{1}{2} g^{\mu\lambda} (g_{\lambda 0,0}+g_{0\lambda,0}- g_{00,\lambda})= -\frac{1}{2} g^{\mu\lambda} g_{00,\lambda} \approx -\frac{1}{2} \eta^{\mu\lambda} h_{00,\lambda} $$ Therefore, the geodesic equation reads: $$ \frac{d^2 x^\mu}{ds^2} -\frac{1}{2}\eta^{\mu\lambda} h_{00,\lambda}\left(\frac{dt}{ds}\right)^2 = 0$$ Since the metric is static, the $\mu=0$ component of the equation is simply $$ \frac{d^2t }{ds^2}=0 $$ while for the spatial components, if we divide by $\left(\frac{d t}{ds}\right)^2$: $$ \frac{d^2 x^i}{dt^2} = \frac{1}{2}h_{00,i}$$ If we compare this equation with its Newtonian analogue $$\vec{a}=\vec{g}=-\vec{\nabla}\Phi$$ then it's natural to identify the [00 component of the] metric with the Newtonian gravitational potential $$ \Phi = -\frac{1}{2}h_{00} $$

We can win further insight if we look at the Einstein field equation. We'll use the trace-reversed version $$R_{\mu\nu} = 8\pi G ( T_{\mu\nu}-\frac{1}{2} g_{\mu\nu}T) $$ where $T_{\mu\nu}$ is the energy-momentum tensor and $T$ its trace. In our Newtonian setting, the only non-negligible component is $T_{00}=- T =\rho $ due to mass density. $$ R_{00}=4\pi G\rho $$ If you compute the Ricci tensor (lef as an exercise for the reader) you get $$ R_{00} = -\frac{1}{2}\nabla^2 h_{00}= 4\pi G\rho$$ And if we use the previous equation for the potential, we recover Poisson's equation: $$ \nabla^2 \Phi= 4\pi G \rho $$

As a conclusion, we can think of the Einstein equation as an upgraded relativistic version of the Poisson's equation. Relativity demands that energy and momentum, in addition to mass, cause gravity, and therefore the source term must be the energy-momentum tensor instead of mass density. Since the new equation has a whole second rank tensor as a source, then the "potential" will also be a tensor: the metric. Of course, this is a too simplistic, semi-handwaving approach to General Relativity, but maybe it's a good start to fill the gap from Newtonian gravity.

References

Albert Einstein: The Meaning of Relativity (1923), lecture IV. Available ar Project Gutemberg
Sean Carroll: Lecture Notes on General Relativity (1997) arxiv:gr-qc/9712019

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