I've consulted several books for the explanation of why

$$\nabla _{\mu}g_{\alpha \beta} = 0,$$

and hence derive the relation between metric tensor and affine connection $\Gamma ^{\sigma}_{\mu \beta} $

$$\Gamma ^{\lambda } _{\beta \mu} = \frac{1}{2} g^{\alpha \gamma}(\partial _{\mu}g_{\alpha \beta} + \partial _{\beta} g_{\alpha \mu} - \partial _{\alpha}g_{\beta \mu}).$$

But I'm getting nowhere. May be I've to go through the concepts of manifold much deeper.

  • 1
    $\begingroup$ As a side note; to show that $g_{\alpha\beta;\sigma}=0$ all we have to do is show that it is zero in a locally inertial frame (which it trivially is) and therefore it must be in all frames. $\endgroup$ – Quantum spaghettification May 22 '17 at 16:51
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    $\begingroup$ intuitively speaking, the interpretation is trivial: the metric tensor is the ruler used to measure how fields change from place to place. It makes sense that the ruler does not change as measured by the ruler $\endgroup$ – lurscher Jun 13 '18 at 21:06

The connection is chosen so that the covariant derivative of the metric is zero. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection $\Gamma^{\sigma}_{\mu \beta}$. You could in principle have connections for which $\nabla_{\mu}g_{\alpha \beta}$ did not vanish. But we specifically want a connection for which this condition is true because we want a parallel transport operation which preserves angles and lengths.

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    $\begingroup$ Good answer. Some details are given in Wald section 3.1. We want the inner product $(v,w) = g_{ab} v^a w^b$ to remain constant under parallel transport along a curve with tangent $t^c$, which gives rise to the condition $t^c \nabla_c (g_{ab} v^a w^b) = 0.$ But (using parallel transport), this is the same as $t^c v^a w^b \nabla_c g_{ab} = 0$ and this should be true for all $v,w,t.$ $\endgroup$ – Vibert Dec 30 '12 at 11:14
  • $\begingroup$ Thanks for the wonderful answer. I will try to go through the wald book. $\endgroup$ – Aftnix Dec 30 '12 at 11:18
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    $\begingroup$ also note that the condition $\nabla g = 0$ is not enough to specify a unique connection - another condition (eg vanishing torsion) is necessary for that $\endgroup$ – Christoph Dec 30 '12 at 11:27
  • $\begingroup$ @Christoph yes, that's important, I should have mentioned it. $\endgroup$ – twistor59 Dec 30 '12 at 11:42
  • $\begingroup$ See also Schrödinger's "Time-Space structure". @twistor59's answer is correct, but for details you need to learn more about (generally independent) concepts of connection and metric. $\endgroup$ – Murod Abdukhakimov Dec 30 '12 at 13:09

It can be show easily by the next reasoning. $$ DA_{i} = g_{ik}DA^{k}, $$ because $DA_{i}$ is a vector (according to the definition of covariant derivative). On the other hand, $$ DA_{i} = D(g_{ik}A^{k}) = g_{ik}DA^{k} + A^{k}Dg_{ik}. $$ So, $$ g_{ik}DA^{k} + A^{k}Dg_{ik} = g_{ik}DA^{k} \Rightarrow Dg_{ik} = 0. $$ So, it isn't a condition, it is a consequence of covariance derivative and metric tensor definition.

The relation between Christoffel's symbols and metric tensor derivations can be earned by cyclic permutation of indexes in the covariance derivative $g_{ik; l}$ expression, which is equal to zero.


This is only meant to supplement the first answer.

If we think physically, then we live in one particular (pseudo-)Riemannian world. In this world, there is only one metric tensor (up to scalar) and it can pretty much be measured. If I found it here, and if an alien measured it, and we compared our answers, they would be scalar multiples of each other (choice of Parisian metre stick for me, choice of Imperial foot for the alien, or, vice versa..). There is precisely one connection, and it can be calculated from the metric.

So I quarrel with the word used by @twistor59, «chosen». There is no choice. Given a metric, the connection is determined. I agree with the rest of the answer, but would like to see the word «chosen» replaced by «given». I would rather say,

given a metric, the connection is determined by the metric.

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    $\begingroup$ Given a metric, the Levi-Civita connection is determined by the metric. We often find it convenient to choose the Levi-Civita connection over other possible choices. $\endgroup$ – Timothy Wofford Apr 10 '14 at 8:10

Consider the analogy with Newtonian gravity. In Newtonian gravity, we have a potential $\Phi$, and differentiating that gives the gravitational field.

In GR, the metric plays the role of the potential, and by differentiating it we get the Christoffel coefficients, which can be interpreted as measures of the gravitational field.

Now in GR we have the equivalence principle (e.p.), and one way of stating the e.p. is that we can always choose a local frame of reference such that the gravitational field is zero. Therefore coordinates exist such that $\nabla_\alpha g_{\mu\nu}=0$. But $g$ is a tensor, and the whole point of the covariant derivative $\nabla$ is that it's a tensor (unlike the partial derivatives with respect to the coordinates). And a tensor that's zero in one set of coordinates is zero in all other coordinates. Therefore we must have $\nabla_\alpha g_{\mu\nu}=0$ in whatever set of coordinates we choose.

  • $\begingroup$ Why do you assume this? What evidence do you have that there is any place in the universe where the acceleration is zero? $\endgroup$ – Donald Airey Oct 26 '18 at 1:02

Here is another straight forward calculation, but assuming the existence of locally flat coordinates $\xi^i\left(x^\mu\right)$. Then \begin{align} D_\rho g_{\mu \nu} &= \partial_\rho g_{\mu \nu} - g_{\mu \sigma} \Gamma_{\nu\rho}^{\sigma} - g_{\sigma\nu} \Gamma_{\mu\rho}^{\sigma} \\ &= \partial_\rho \left( \frac{\partial \xi^i}{\partial x^\mu}\frac{\partial \xi^i}{\partial x^\nu} \right) - g_{\mu \sigma} \frac{\partial x^\sigma}{\partial \xi^i} \frac{\partial^2 \xi^i}{\partial x^\nu \partial x^\rho} - g_{\sigma \nu} \frac{\partial x^\sigma}{\partial \xi^i} \frac{\partial^2 \xi^i}{\partial x^\mu \partial x^\rho} \\ &= \frac{\partial^2 \xi^i}{\partial x^\rho \partial x^\mu}\frac{\partial \xi^i}{\partial x^\nu} + \frac{\partial \xi^i}{\partial x^\mu}\frac{\partial^2 \xi^i}{\partial x^\rho \partial x^\nu} - \frac{\partial \xi^j}{\partial x^\mu}\underbrace{\frac{\partial \xi^j}{\partial x^\sigma} \frac{\partial x^\sigma}{\partial \xi^i}}_{\delta^j_i} \frac{\partial^2 \xi^i}{\partial x^\nu \partial x^\rho} - \frac{\partial \xi^j}{\partial x^\sigma}\frac{\partial \xi^j}{\partial x^\nu}\frac{\partial x^\sigma}{\partial \xi^i} \frac{\partial^2 \xi^i}{\partial x^\mu \partial x^\rho} \\ &= 0 \end{align}

  • $\begingroup$ That is a good question I'd like to have answered too. But it is actually the essence in classical GR. $\endgroup$ – Diger Jun 13 '18 at 21:34
  • $\begingroup$ Actually the above calculation is also valid if you consider a higher dimensional flat space $i,j=1,...,N$ where the variety is embedded $\mu,\nu,\rho,\sigma=1,...,M$ with $M<N$. Then the Christoffel-symbols can still be defined as long as $$ \frac{\partial^2 \xi^i}{\partial x^\mu \partial x^\nu} = \Gamma_{\mu\nu}^\rho \frac{\partial \xi^i}{\partial x^\rho} $$ The inverse used above is not really necessary. $\endgroup$ – Diger Jun 14 '18 at 11:45
  • $\begingroup$ Why do you assume that a locally flat coordinate system exists in the real universe? $\endgroup$ – Donald Airey Oct 26 '18 at 1:03
  • $\begingroup$ Why do you ask me? Ask Einstein; that was his general assumption. $\endgroup$ – Diger Oct 26 '18 at 16:48
  • $\begingroup$ and your job as a scientist is to never question your base assumptions? I missed that lecture. $\endgroup$ – Donald Airey Oct 26 '18 at 18:23

protected by Qmechanic Jun 10 '15 at 10:07

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