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I'm looking to gain an intuitive understanding of the geodesic equation (which incorporates the Christoffel symbol) and how it is used to calculate the radial attraction of gravity. In its native form it reads as - $${d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} u^\alpha u^\beta \tag{1}$$

The Christoffel symbol itself $\Gamma^\mu_{\alpha\beta}$ can be read as follows; the two lower indices, $\alpha$ and $\beta$, describe the change in the $\alpha$ basis vector caused by a change in the $\beta$ coordinate. The upper index $\mu$ then gives the specific direction in which the change in the basis vector occurs in.

As such, if we use the geodesic equation for the radial attraction of gravity the equation becomes - $${d^2 r \over d\tau^2} = - \Gamma^r_{tt} u^t u^t \tag{2}$$

The Christoffel symbol becomes - $$\Gamma^r_{tt} = \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) \tag{3}$$ and the geodesic equation as a whole becomes - $${d^2 r \over d\tau^2} = - \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) u^t u^t \tag{4}$$

What I'm trying to understand in $\Gamma^r_{tt}$ - which one these terms $\frac{GM}{c^2r^2}$, $\left(1 - \frac{2GM}{c^2r}\right)$ is the $t$ basis vector, which one is $t$ coordinate, and which one is the upper index $r$ - which gives the specific direction in which the change in the $t$ basis vector occurs in?

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    $\begingroup$ Your question makes no sense. Terms in Christoffel symbols are not equal to basis vectors and $c^2$ definitely appears in the metric. $\endgroup$ Jul 19, 2022 at 11:12
  • $\begingroup$ you are right about the $c^2$ being in the metric. However, I never stated the Christoffel symbol equals the basis vector. I said one of its elements is the basis vector. $\endgroup$
    – Tivity
    Jul 19, 2022 at 11:20
  • $\begingroup$ Which metric you used ? $\endgroup$
    – Eli
    Jul 19, 2022 at 13:41
  • $\begingroup$ The Schwarzschild Metric. $\endgroup$
    – Tivity
    Jul 19, 2022 at 14:50
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    $\begingroup$ The geodesic equation you wrote is incomplete. The Christoffel symbols $\Gamma ^{r}{}_{rr}$, $\Gamma ^{r}{}_{\theta \theta}$ and $\Gamma ^{r}{}_{\varphi \varphi}$ are also non-zero and thus contribute to $\frac{d^{2}r}{d\tau ^{2}}$. Other than that, there is no basis vector in the geodesic equation. The quantity you calculate is a scalar which by construction describes the radial component of the 4-acceleration for an object under the influence of gravitation from a Schwarzschild black hole. $\endgroup$
    – rhomaios
    Jul 19, 2022 at 17:54

4 Answers 4

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As you wrote initially, $\Gamma^{\mu}_{\alpha\beta}$ describes the $\mu^{th}$ component of the change in the $\alpha^{th}$ basis vector when we move it along the $\beta^{th}$ coordinate. Also, we consider torsion-free connections so we have symmetry in the lower two indices, so in the above interpretation, you can interchange the roles of $\alpha$ and $\beta$. Hence, the question of which we consider as the basis vector and which we consider the coordinate along which to move isn't important here. Now, at this stage I should remark that the $\Gamma^{\mu}_{\alpha\beta}$ are functions, they are no longer basis vectors, so your last paragraph doesn't really make sense.

In what follows, I shall set $c=G=1$ for my own sanity. Next, the radial geodesic equation is not as simple as you wrote; there are more terms involved (again, as mentioned in the comments): \begin{align} \ddot{r}+\frac{M}{r^2}\dot{t}^2-\frac{M}{r^2}\frac{1}{\left(1-\frac{2M}{r}\right)}\dot{r}^2-\left(1-\frac{2M}{r}\right)r\dot{\theta}^2 -\left(1-\frac{2M}{r}\right)r\sin^2(\theta)\,\dot{\phi}^2&=0. \end{align} The coefficients of $\dot{t}^2,\dot{r}^2,\dot{\theta}^2,\dot{\phi}^2$ are $\Gamma^{r}_{tt},\Gamma^{r}_{rr},\Gamma^{r}_{\theta\theta},\Gamma^{r}_{\phi\phi}$ respectively. You can obtain this equation by starting from the Schwarzschild metric, and taking the Lagrangian $\mathscr{L}=\frac{1}{2}g_{ab}\dot{x}^a\dot{x}^b$, and writing out the Euler-Lagrange equation for the coordinate $r$: $\frac{d}{d\tau}\left(\frac{\partial \mathscr{L}}{\partial \dot{r}}\right)-\frac{\partial\mathscr{L}}{\partial r}=0$.

Lastly, if you want to really understand the radial motion of geodesics, then simply writing out the radial equation isn't going to be of much help. I realize that sounds counter-intuitive: why is the radial equation alone not helpful to understand the radial behavior of the particle? That's because the geodesic equations are a set of 4 (in this case) coupled ODEs. So, the derivatives of $r$ depend on the other coordinates $t,\theta,\phi$ and their derivatives. Likewise, the derivatives for $\theta$ involves the other coordinates. This makes things difficult. At this point, not all hope is lost. Some of our best friends in physics are symmetries and conservation laws! It turns out the Schwarzschild spacetime is pretty symmetric that we can exploit this quite well. For instance, the metric in Schwarzschild coordinates doesn't depend on $t$ and doesn't depend on $r$. In differential geometry jargon, this says $\partial_t,\partial_{\phi}$ are Killing vector fields. From the perspective of the Euler-Lagrange equations, $t,\phi$ are cyclic coordinates so their corresponding 'momenta' \begin{align} \frac{\partial\mathscr{L}}{\partial \dot{t}}=-\left(1-\frac{2m}{r}\right)\dot{t},\quad\text{and}\quad\frac{\partial\mathscr{L}}{\partial\dot{\phi}}=r^2\sin^2(\theta)\,\dot{\phi} \end{align} are conserved along the geodesics. We denote these constants by $E,L$ respectively (for 'energy' and 'angular momentum' respectively). Next, we can also impose the normalization condition $g_{ab}\dot{x}^a\dot{x}^b=-1$ along the geodesic, and finally due to the spherical symmetry of Schwarzschild, we may always assume that the motion occurs in the $\theta=\frac{\pi}{2}$ equatorial plane. With these conditions, one can actually obtain (see here for references) a much simpler ODE for $r$: \begin{align} \frac{1}{2}\dot{r}^2+\underbrace{\frac{1}{2}\left(1-\frac{2m}{r}\right)\left(\frac{L^2}{r^2}+1\right)}_{:=V(r)}&=\frac{E^2}{2}. \end{align} Now, if you forget about the fact that this particular equation was derived starting from GR and the Schwarzschild metric, then this will be extremely familiar from classical mechanics: we have a 'kinetic energy' $\frac{\dot{r}^2}{2}$ plus some 'effective potential energy' $V(r)$ equal to a constant. So, to understand the behavior of the radial coordinate, you just need to sketch the potential $V(r)$. This is just an elementary exercise in single-variable calculus:

  • $V(r)\to -\infty$ as $r\to 0^+$, and $V(r)\to\frac{1}{2}$ as $r\to\infty$.
  • $V'(r)=\frac{Mr^2-L^2r+3ML^2}{r^4}$, so it has roots at $\frac{L^2\pm\sqrt{L^2(L^2-12M^2)}}{2M}$. In particular, $V'$ has no roots (hence remains positive) if $L^2<12M^2$. Next, $V'$ has a single root if $L^2=12M^2$ (i.e a 'circular orbit'), and $V'$ has two roots if $L^2>12M^2$. You can also go ahead and figure out the second derivatives to determine whether it's a local max/min at each of these points.

Once you have the plot of $V(r)$ for the various cases, you can easily read off the radial behavior of these timelike geodesics, i.e the radial behavior of massive particles under the influence of a large massive object. For a more detailed explanation and calculations, see Wald's text.

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The Christoffel symbol itself $\Gamma^\mu_{αβ}$ can be read as follows; the two lower indices, $α$ and $β$, describe the change in the $\alpha$ basis vector caused by a change in the $\beta$ coordinate. The upper index μ then gives the specific direction in which the change in the basis vector occurs in.

This sounds correct to me. It is essentially this definition of the Christoffel symbols: $$\frac{\partial\mathbf{e}_\alpha}{\partial x^\beta}=\Gamma^\mu_{\alpha\beta}\ \mathbf{e}_\mu$$ or equivalently (written with differentials) $$d\mathbf{e}_\alpha=\Gamma^\mu_{\alpha\beta}\ \mathbf{e}_\mu \ dx^\beta$$

As such, if we use the geodesic equation for the radial attraction of gravity the equation becomes - $${d^2 r \over d\tau^2} = - \Gamma^r_{tt} u^t u^t \tag{2}$$

This looks wrong to me. In the general geodesic equation $${d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} u^\alpha u^\beta \tag{1}$$ you need to be aware that this actually (according to Einstein's summation convention) implies summing over the twice occuring indices $\alpha$ and $\beta$. $${d^2 x^\mu \over d\tau^2} = - \sum_\alpha \sum_\beta \Gamma^\mu_{\alpha\beta} u^\alpha u^\beta$$ So on the right hand side you have a sum with 16 terms, not only one (with indices $tt$) as in your equation (2).

Hopefully this also answers your question at the end, about which one is the $t$ coordinate.

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The metric reads $$ds^2=g_{tt}~c^2dt^2-g_{rr}~dr^2+r^2~d\phi^2.$$ Lets define $$ds\equiv c~ d\tau,~~~\dot{t} \equiv \frac{dt}{d\tau},~~~\dot{r} \equiv \frac{dr}{d\tau}.$$ The four-velocity is then $$u^{\mu}=(\dot{t},\dot{r},0,0).$$Because of energy conservation it holds$$g_{tt}~\dot{t}=k\equiv \frac{E}{m c^2}=const.,$$ and for radial geodesics angular momentum is zero $$\dot{\phi}=0,$$ the geodesics equation can be written as $$g_{tt}~g_{rr}~\dot{r}^2+g_{tt}=k^2.$$ Taking derivative on $\tau$ from both sides of that equation leads to $$\ddot{r}=-\frac{1}{2}~\frac{1}{g_{tt} g_{rr}}\Big((g_{tt} g_{rr})^{'}_{r}~\dot{r}^2+(g_{tt})^{'}_{r}\Big).$$ By comparing above equation with your expression it should be possible to identify your terms. Please check my calculation. I have made it fast. Nice help for such calculations provides the link https://physicssusan.mono.net/9035/General%20Relativity .

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  • $\begingroup$ the $k$ is introduced and dropped without explanation $\endgroup$
    – Yukterez
    Jul 19, 2022 at 20:11
  • $\begingroup$ @Yukterez, you are right, thanks. I have corrected it. $\endgroup$
    – JanG
    Jul 20, 2022 at 7:15
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If the idea is to get an intuitive notion of gravitational attraction expressed by the geodesic equation, it is better to remind that Schwartzschild solution is a tiny correction of Newtonian gravity for ordinary examples. Departing from the equation:

$${d^2 x^\mu \over d\lambda^2} = - \Gamma^\mu_{\alpha\beta} u^\alpha u^\beta \tag{1}$$

We take the curve parameter as linear with time: $\lambda = at + b$. So, $${dx^\mu \over d\lambda} = {dx^\mu \over dt} {dt \over d\lambda} \implies {d^2 x^\mu \over d\lambda^2} = {1 \over a^2}{d^2 x^\mu \over dt^2}$$ $$u^\alpha u^\beta = {dX^\alpha \over d\lambda}{dX^\beta \over d\lambda} = {1 \over a^2}{dX^\alpha \over dt}{dX^\beta \over dt}$$

Now we require that the connection is zero except when $X^\alpha = X^\beta = t$, and $\mu$ is a spatial index. We get:$${d^2 x^i \over dt^2} = - \Gamma^i_{t t}$$

Finally we make: $\Gamma^i_{t t} = \partial_i \phi$, where $\phi = -{GM \over r}$ to get in components, the vectorial equation:$${d^2 \mathbf r \over dt^2} = -{GM\mathbf {\hat r} \over r^2}$$

It is the Cartan approach to Newtonian gravity, using differential geometry. It is amazing that the Riemann tensor, calculated from that connections is not zero, meaning a Newtonian curved spacetime!

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