1
$\begingroup$

Suppose that you want to take a covariant derivative $$\nabla$$ of some arbitrary (2,0)-tensor $$T^{\mu\nu}$$such that $$\nabla_{\alpha}T^{\mu\nu}=\partial_{\alpha}T^{\mu\nu}+\Gamma^{\mu}_{\alpha\beta}T^{\beta\nu}+\Gamma^{\nu}_{\alpha\beta}T^{\mu\beta}.$$ In order to calculate the Christoffel symbol we must use metric, such that $$\Gamma^{\mu}_{\alpha\beta}=\frac{1}{2}g^{\mu\sigma}(\partial_{\beta}g_{\sigma\alpha}+\partial_{\alpha}g_{\sigma\beta}-\partial_{\sigma}g_{\alpha\beta}).$$ My question is; is there any other method to calculate $\Gamma^{\mu}_{\alpha\beta}$ (without resorting to computers) or is the metric the only way?

$\endgroup$

2 Answers 2

4
$\begingroup$

There are other methods. I'll show you one of my favorites and provide references to other techniques. This one is particularly useful when the metric is diagonal on the coordinate system you are using. Take, for example, the metric on the unit sphere, $$g_{\mu\nu} = \begin{pmatrix} 1 & 0 \\ 0 & \sin^2\theta \end{pmatrix}.$$

As you know, the geodesic equation is $$\frac{\textrm{d}^2 x^{\mu}}{\textrm{d} \tau^2} + \Gamma^{\mu}{}_{\nu\sigma} \frac{\textrm{d} x^{\nu}}{\textrm{d} \tau}\frac{\textrm{d} x^{\sigma}}{\textrm{d} \tau} = 0.$$

We also know the geodesic equation can be obtained from the Lagrangian $$L = \frac{1}{2} g_{\mu\nu} \frac{\textrm{d} x^{\mu}}{\textrm{d} \tau}\frac{\textrm{d} x^{\nu}}{\textrm{d} \tau},$$ as shown, e.g., on pp. 43–45 of Bob Wald's General Relativity and also discussed on a recent question you and I are familiar with: Different Lagrangian formulas for geodesics.

Let us consider the particular case for the sphere. The Lagrangian becomes $$L = \frac{\dot{\theta}^2}{2} + \frac{\sin^2\theta \dot{\phi}^2}{2},$$ where the dots denote differentiation with respect to the affine paremeter $\tau$, which here we use just to help with the calculation. The Euler–Lagrange equations are $$\ddot{\theta} - \sin\theta \cos\theta \dot{\phi}^2 = 0$$ and $$\ddot{\phi} + 2 \cot\theta \dot{\theta}\dot{\phi}= 0.$$

If we open up the components of the geodesic equation, it reads $$\ddot{\theta} + \Gamma^{\theta}{}_{\theta\theta}\dot{\theta}\dot{\theta} + 2\Gamma^{\theta}{}_{\phi\theta}\dot{\phi}\dot{\theta} + \Gamma^{\theta}{}_{\phi\phi}\dot{\phi}\dot{\phi} = 0$$ and similarly for $\phi$. Hence, comparing the Euler–Lagrange equations and the geodesic equation (which must coincide), we find that $$\Gamma^{\theta}{}_{\phi\phi} = - \sin\theta \cos\theta, \quad \text{and} \quad \Gamma^{\phi}{}_{\phi\theta} = \cot\theta$$ and all other Christoffel symbols vanish. This is precisely what you would get if you went through the painful calculation.

This is my favorite technique, and works quite well for diagonal metrics, which lead to simple Euler–Lagrange equations. It also works for other metrics, but it gets more cumbersome.

There are still other techniques for computing curvatures. Wald's book discusses a tetrad-based technique on Sec. 3.4b, which I don't remember well enough to discuss here. He also gives an example calculation when dealing with the Schwarzschild metric on Chap. 6. Still on Sec. 3.4b (p. 52, to be exact) he also mentions yet another approach based on spinors, and provides references to it.

$\endgroup$
2
  • $\begingroup$ Thanks for the information! Now to actually solve the 2 EL equations would we use range-kutta method? $\endgroup$
    – aygx
    Nov 23, 2021 at 4:19
  • $\begingroup$ @aygx If you want to solve the geodesic equation, that would be a possibility, but to find the Christoffel symbols it is just a matter of algebraic manipulation. Notice that you don't have to solve the equations: it suffices to find the EoM, write them in a fashion that resembles the geodesic equation and read the Christoffel symbols by comparison between the EoM you computed from the Lagrangian and the usual form of the geodesic equation $\endgroup$ Nov 23, 2021 at 4:39
2
$\begingroup$

@NickolasAlves has pointed out the most practically useful way to calculate the symbols. I'd point out another method that is of limited practical use but is conceptually fun. In fact, it is very closely related to the method pointed out in the other answer in that what it relies on is exploiting the appearance of the symbols in the geodesic equation.

We know that the geodesic equation in a generic coordinate system reads \begin{align} \frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau} = 0 \end{align} Now, let's say that the local inertial coordinates $\{\xi^{\mu}(x)\}$ are known which can be inverted to write $\{x^{\mu}(\xi)\}$. This is precisely the reason as to why you won't ever actually use this method to calculate the symbols -- because what would be usually explicitly known is the metric components in a given coordinate system -- not how the local inertial coordinates at each point relate to the given coordinates. Of course, you can find them out but it would require computing the Christoffel symbols in the first place (or you can do something equivalent but even more computationally heavy) :) Anyway, in principle, one can imagine that we know $\{\xi^{\mu}(x)\}$ (for example, if we cooked up the non-inertial coordinates (such as the Rindler coordinates) as functions of the inertial coordinates).

Then, we know that in the local inertial frame, the equation of motion ought to read

\begin{align} \frac{d^2\xi^\mu}{d\tau^2} &= 0\\ \implies \frac{d}{d\tau}\bigg(\frac{d\xi^\mu}{d\tau} \bigg)&= 0\\ \implies \frac{d}{d\tau}\bigg(\frac{\partial\xi^\mu}{\partial x^\alpha}\frac{dx^\alpha}{d\tau}\bigg) &= 0\\ \implies \frac{\partial^2\xi^\mu}{\partial x^\beta\partial x^\alpha}\frac{dx^\beta }{d\tau}\frac{dx^\alpha}{d\tau} +\frac{\partial \xi^\mu}{\partial x^\alpha}\frac{d^2x^\alpha}{d\tau^2} &= 0\\ \implies \frac{\partial x^\nu}{\partial \xi^\mu}\frac{\partial^2\xi^\mu}{\partial x^\beta\partial x^\alpha}\frac{dx^\beta }{d\tau}\frac{dx^\alpha}{d\tau} +\frac{\partial x^\nu}{\partial \xi^\mu}\frac{\partial \xi^\mu}{\partial x^\alpha}\frac{d^2x^\alpha}{d\tau^2} &= 0\\ \implies \frac{\partial x^\nu}{\partial \xi^\mu}\frac{\partial^2\xi^\mu}{\partial x^\beta\partial x^\alpha}\frac{dx^\beta }{d\tau}\frac{dx^\alpha}{d\tau} +\delta^\nu_{\alpha}\frac{d^2x^\alpha}{d\tau^2} &= 0\\ \implies \frac{d^2x^\nu}{d\tau^2} + \frac{\partial x^\nu}{\partial \xi^\mu}\frac{\partial^2\xi^\mu}{\partial x^\beta\partial x^\alpha}\frac{dx^\beta }{d\tau}\frac{dx^\alpha}{d\tau} &= 0\\ \implies \frac{d^2x^\mu}{d\tau^2} + \bigg[\frac{\partial x^\mu}{\partial \xi^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}\bigg]\frac{dx^\alpha }{d\tau}\frac{dx^\beta}{d\tau} &= 0 \end{align}

where in the last step, I have simply renamed a bunch of indices. Now, comparing this to the geodesic equation for the generic coordinate system, we can read off

$$\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial \xi^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}$$

So, given the $\xi^\mu(x)$ (and the inverse), we can simply calculate the symbols using the expression above. In fact, what we have done is (almost) derive the geodesic equation in light of the equivalence principle. This is the route that Weinberg takes towards introducing the symbols. He only then goes on to derive the relation between the metric tensor and the symbols as a result following from the above definition of the symbols.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.