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The covariant derivative is

$$\nabla_r V_m=\delta_r V_m - \Gamma{^{t}_{r,m}} V_t$$

What is the meaning of the indexes $t$, $r$ and $m$ of the Christoffel symbol in this equation? I saw Leonard Susskind's video GR3, and I understood that $r$ corresponds to the observer's frame, $m$ corresponds to the observed frame and $t$ is an arbitrary orthonormal frame, but comparing my interpretation with textbooks, I do not come to the same result.

What is the correct, real sense of $t$, $r$ and $m$?

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  • $\begingroup$ $t$ is a dummy index that’s being summed over, so it’s not something you get to choose. $\endgroup$
    – Ghoster
    Nov 9, 2022 at 18:22
  • $\begingroup$ Link to video? Which minute? $\endgroup$
    – Qmechanic
    Nov 9, 2022 at 19:58

2 Answers 2

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You can start from the very definition of Christoffel symbols, as the components of the derivatives of the vectors of a natural basis w.r.t. the coordinates, or as the second derivative of the position w.r.t. the coordinates to describe the space $\mathbf{r}(q^j)$, given

$\mathbf{b}_i(q^j) = \dfrac{\partial \mathbf{r}}{\partial q^i}(q^j)$,

$\dfrac{\partial^2 \mathbf{r}}{\partial q^k \partial q^i}(q^j) = \dfrac{\partial \mathbf{b}_i}{\partial q^k}(q^j) := \Gamma_{ik}^{\ell}(q^j) \ \mathbf{b}_{\ell}(q^j)$

I'll try here to give some interpretation to the indices in terms of observer/observed quantities to a covariant derivative, that is nothing more than a component of the gradient of a (tensor) field. Let's do it for a vector field $\mathbf{v}(\mathbf{r})$, that can be written as , $\mathbf{v} = v^m \mathbf{b}_m$, using the local natural basis induced by the coordinates $q^i$ chosen to describe the system, $\mathbf{b}_i(q^j) = \frac{\partial \mathbf{r}}{\partial q^i}(q^j)$. The gradient of the vector field can be written as

$\nabla \mathbf{v} = \mathbf{b}^r \otimes \mathbf{b}_m \underbrace{\left[ \dfrac{\partial v^m}{\partial q^r} + \Gamma^{m}_{rt} v^t\right]}_{\nabla_{/r} v^m}$.

Now, I'd say that the observed quantity is the vector field $\mathbf{v}$. This observed vector field has been written using the local natural basis, that is induced by the choice of coordinates $q^r$, and thus by the one who observes/ describes the problem.

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  • $\begingroup$ My question does not concern the maths, with my question I want to understand the physical signification of $l$, $i$ and $k$ in your equation. $\endgroup$
    – Moonraker
    Nov 9, 2022 at 17:09
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    $\begingroup$ All the quantities, if not specified otherwise, are referred to the same set of coordinates, that induces the natural base vectors. There is no need for different reference frames, besides the RF locally induced by the choice of the coordinates $\endgroup$
    – basics
    Nov 9, 2022 at 17:15
  • $\begingroup$ I am no mathematician, but as far as I understand, this does not answer my question. $\endgroup$
    – Moonraker
    Nov 9, 2022 at 17:31
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    $\begingroup$ Let's wait for other answers, then $\endgroup$
    – basics
    Nov 9, 2022 at 17:51
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My question does not concern the maths, with my question I want to understand the physical signification of l, i and k in your equation. – Moonraker 1 hour ago

According to Leonard Susskind's lecture, the covariant derivative equation of a covariant vector component $$\nabla_rv_m=\partial_rv_m-\Gamma_{rm}^tv_t$$ is written as a combination of the change in the vector components $v_m$ over a given dimension $r$ summed/subtracted with the change in the basis $\pmb{e}_m$ within your curvilinear coordinate system.

Recall that: $$\nabla_r\pmb e_m=\frac{\partial\pmb e_m}{\partial x^r}+\Gamma_{rm}^t\pmb e_t\quad where\quad \frac{\partial\pmb e_m}{\partial x^r}=0$$ given that basis vectors, by definition, are always tangential to our coordinate lines leaving us with: $$\nabla_r\pmb e_m=\Gamma_{rm}^t\pmb e_t$$ This value is non-zero whenever the coordinate system that the basis vectors correspond to is curvilinear, implying a change in their orientation over position. Knowing this, a comparison can be made where the only difference between $\nabla_rv_m$ and $\nabla_r\pmb e_m$ (besides the "$-$" sign convention) is the addition of a change in the vector component's over position $x^r$ that basis vectors obviously lack.

To interpret the Christoffel symbol $\Gamma_{rm}^t$ in the equation above, the $t$ index is a dummy index meant to be summed over each dimension $x^r=(x^0,x^1,x^2,x^3)$ that the basis vector travels along since Christoffel symbols represent the component coefficients of the resulting derivative vector you get by comparing the orientation of a vector at two different events in spacetime due to the curvature of a coordinate system. $m$ and $r$, on the other hand, are free indices with $m$ representing a specific vector component and $r$ representing the chosen dimension to transport the vector along.

The order that $r$ and $m$ are written does not matter since Christoffel symbols are symmetric with respect to their lower indices. $$\Gamma_{mr}^t=\Gamma_{rm}^t$$

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