# Why is the covariant derivative of the metric tensor zero?

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 $$\Gamma ^{\sigma}_{\mu \beta}$$

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

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

• 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. May 22, 2017 at 16:51
• 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 Jun 13, 2018 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.

• 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 $v,w,t.$ Dec 30, 2012 at 11:14
• Thanks for the wonderful answer. I will try to go through the wald book. Dec 30, 2012 at 11:18
• 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 Dec 30, 2012 at 11:27
• @Christoph yes, that's important, I should have mentioned it. Dec 30, 2012 at 11:42

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.

• doesnt the first line imply already that $Dg_{ik} = 0$...? Jun 20, 2022 at 18:18

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}

• That is a good question I'd like to have answered too. But it is actually the essence in classical GR. Jun 13, 2018 at 21:34
• 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. Jun 14, 2018 at 11:45
• Why do you assume that a locally flat coordinate system exists in the real universe?
– user32023
Oct 26, 2018 at 1:03
• Why do you ask me? Ask Einstein; that was his general assumption. Oct 26, 2018 at 16:48
• The dimension of the tangent space is exactly equal to the dimenion of the manifold. This is on purpose so that it is a suitable place to do linear approximations to the manifold. But the existence of geodetic coordinates is a mathematical consequence of a Riemannian metric. And they have no physical significance, they merely simplify calculations. So they or their use or their formulas are not the consequence of any additional physical assumptions except that there is a metric. Dec 7, 2020 at 7:43

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.

• 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. Apr 10, 2014 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.

• Why do you assume this? What evidence do you have that there is any place in the universe where the acceleration is zero?
– user32023
Oct 26, 2018 at 1:02