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What's the difference? $$\nabla_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho~\text{ and }~\partial_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho~?$$

In John Dirk Walecka's book 'Introduction to General Relativity', He told us that we can use connection to describe the change of basis vectors: $$\partial_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho$$ Then the difference of basis vector $e_\mu$ between points $x_0$ and $x_0+\delta x^\mu$ is: $\Gamma_{\mu \nu}^\rho e_\rho \delta x^\mu$

Then conbine with the definition of metric field: $$g_{\mu \nu}=e_\mu\cdot e_\nu$$ We will get the metric compatible equation: \begin{align} \partial_\rho g_{\mu\nu}&=(\partial_\rho e_\mu)\cdot e_\nu +e_\mu\cdot (\partial_\rho e_\nu)\nonumber\\ &=\Gamma^\lambda_{\rho\mu }g_{\lambda\nu}+\Gamma^\lambda_{\rho\nu }g_{\mu\lambda} \end{align}

But in many books, we find another formula: $$\nabla_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho$$ Then what is the difference between the two formulas? And from this formula can we know the difference of the basis vector between $x_0$ and $x_0+\delta x^\mu$?

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    $\begingroup$ "But in many books, we find another formula:" -- what book(s)? What page(s)? $\endgroup$
    – Andrew
    Commented Jul 2 at 4:41
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    $\begingroup$ Which page in Walecka's book? $\endgroup$
    – Qmechanic
    Commented Jul 2 at 5:16
  • $\begingroup$ Without specifying the exact source, we can not really tell the difference. Nonetheless, $\partial_\mu$ is usually used as the extension of the three-dimensional $\nabla$ into the $(3+1)$-spacetime. Meanwhile, $\nabla_\mu$ is the covariant derivative used when curved spacetime is involved. $\endgroup$
    – hendlim
    Commented Jul 2 at 5:20
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    $\begingroup$ John Dirk's book page39, 3.64 the first formula. The second formula appears in Misner's Gravitation page259 10.13 $\endgroup$ Commented Jul 2 at 7:14
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    $\begingroup$ Eq. (3.64) uses infinitesimal displacement $d$ not partial derivative $\partial_{\mu}$. $\endgroup$
    – Qmechanic
    Commented Jul 2 at 7:59

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The expression $\partial_\mu \mathbf{e}_\nu$, at face value, does not really make sense in curved space due to the vectors being in different tangent spaces. In order to make it work, vectors need to be transported from one tangent space to another, which is why a connection is needed. A derivative operator using this connection is often given a new notation. In the canonical case of the Levi-Civita connection, this becomes the usual covariant derivative, denoted by $\nabla$.

In flat space, $\nabla$ reduces to the "standard" derivative which is what is meant by $\partial$ in the first place before the addition of curvature. But, as explained in the first sentence, this concept doesn't work in curved space. So, when incorporating the Levi-Civita connection, the notation is usually changed from $\partial$ to $\nabla$. However, I have come across a small number of texts where the author does not change the notation and continues using $\partial$. So I suspect that the two notations really mean the same thing and $\partial$ does not refer to a separate notion of the partial derivative different from $\nabla$.

The expression $\nabla_\mu \mathbf{e}_\nu = \Gamma^\rho_{\mu\nu} \mathbf{e}_\rho$ cannot really be derived but rather is a definition of the Christoffel symbols. For more information, see this post.

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  • $\begingroup$ It's a definition, so the problem is how to explain it, in John Dirk's book, $\Gamma_{\mu\nu}^\rho e_\rho \delta x^\mu$ seems to be explained as the small change of basis $\delta e_\nu$ when moves from $x$ to point $x+\delta x^\mu$, So in the definition $$\nabla_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho$$ how to explain the linear combination of basis vector $\Gamma_{\mu \nu}^\rho e_\rho$, Can we give a different explanation different from Dirk's? $\endgroup$ Commented yesterday
  • $\begingroup$ @Jianbingshao It can be written as a linear combination of the basis precisely because the basis is a basis which means it spans the tangent space. $\endgroup$ Commented 8 hours ago

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