I am taking an introductory course in general relativity, and we are following Sean Carroll's book Spacetime and Geometry. I've found that he writes his tensor equations without specifying at which point the tensors are to be evaluated. For example, he writes $g_{\mu \nu}$ and not $g_{\mu \nu}(x)$ or $g_{\mu \nu}(x_0)$. I find myself constantly making mistakes because I don't know how to apply the equations without knowing how they are evaluated.

My question: How do I know how to evaluate tensors, in particular the metric tensor or the Christoffel symbol when applying the equations? Often it does not become clear to me from the context.

For example, suppose that you want to parallel transport a vector $V^\mu$ on $S^2$, Carroll writes up the equation of parallel transport $$ 0 = \frac{\mathrm d}{\mathrm d \lambda} V^\mu + \Gamma^\mu_{\sigma\rho} \frac{\mathrm dx^\sigma}{\mathrm d\lambda} V^\rho $$ According the Carroll, this gives me a DE for the continuation for the vector. However, the type of DE that I obtain is dependent on whether $\Gamma^\mu_{\sigma\rho}$ is a constant with respect to $\lambda$, or not. By constant, I mean that there exist an $a^\mu_{\sigma\rho} \in \mathbb{R}$ such that $\Gamma^\mu_{\sigma\rho}(x^\nu(\lambda)) = a^\mu_{\sigma\rho}$ for all $\lambda$. An example of non constant $\Gamma^\mu_{\sigma\rho}$, in contrast, is $\Gamma^\mu_{\sigma\rho}\circ x^\nu$ is an element of the set of smooth functions on the reals with non-compact support. Initially I thought he meant the latter.

Through trial and error, I discovered that what he means is that $\Gamma^\mu_{\sigma\rho}$ is actually $\Gamma^\mu_{\sigma\rho}(a)$, where $a$ is the point at which the initial value of $V^\mu$ is specified.

  • $\begingroup$ What? This is no different than $f$ vs. $f(x)$ in analysis. $\endgroup$ – Ryan Unger Mar 5 '17 at 20:40
  • $\begingroup$ Sorry, I'm still not sure what you mean. What does 'constant with respect to lambda' mean? What's the difference in how you evaluate the equation in either case? $\endgroup$ – knzhou Mar 5 '17 at 21:39
  • $\begingroup$ @0celo7 I apologize that my question wasn't clear enough. I have added an example. Since you have taken analysis you know that the equations of uniform continuity and regular continuity are identical. However the interpretation is context dependent. I find that Carroll has minimal context, and therefore unclear what he means. I hope I made myself more clear now. $\endgroup$ – Mikkel Rev Mar 5 '17 at 21:44
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    $\begingroup$ Your edit still does not make sense. Everything is being evaluated along the curve. That is clear from context. $\endgroup$ – Ryan Unger Mar 5 '17 at 23:23

The objects in question normally live in some bundle over spacetime and need to be evaluated at the same point for operations to be well defined.

For your particular example, we have

$$ 0 = \frac{d}{d \lambda} V^\mu(\lambda) + \Gamma^\mu_{\sigma\rho}(x^\nu(\lambda)) \; \frac{dx^\sigma}{d\lambda}(\lambda) \; V^\rho(\lambda) $$

with function arguments restored.


I'm doubt this has anything to do with general relativity. If I'm not mistaken, the core confusion can also be found in Newton's second law, $$F = m \frac{d^2 x}{dt^2}.$$ Your question would then translate to asking at what time $F$ should be evaluated. But the force that matters at time $t$ is clearly, uh, the force at time $t$. That is, in really pedantic notation, we mean $$F(t=t_0) = m \frac{d^2 x(t)}{dt^2} \bigg|_{t=t_0}.$$ The same thing is going on with the equation you wrote. The connection coefficients play the role of $F$, and if the particle is at point $x_0^\mu$, then the only thing relevant is the connection coefficients at the point, $\Gamma(x = x_0^\mu)$.

  • $\begingroup$ Suppose $m$ is not a constant for all time, but it is a function of $t$. Then $F(0) = m a(0)$ would drive you crazy, because you wouldn't know when to evaluate $m$ $\endgroup$ – Mikkel Rev Mar 5 '17 at 22:03
  • $\begingroup$ In Carroll not clear that if the particle is at point $x^μ_0$, then the only thing relevant is the connection cofficients at the point, $Γ(x=x^μ_0)$. $\endgroup$ – Mikkel Rev Mar 5 '17 at 22:09

Although you have chosen an answer, a good one, I would like to add a little from the point of view of actually solving the equation analytically or numerically. You can think of the Christoffel symbol as a position dependent force term. In general you need to know the path you are integrating along to make sense of the parallel transport equation. Geodesics, or auto-parallels, obey the same equation with V = dx/dl.

In this case you can't even say in theory that you have a well defined x(l) along which you can evaluate these terms. Your goal is to identify the particular x(l) that obeys the geodesic equation from among all possible curves on M.

You might set up the system the same way you would any other ODE system, p = dx/dl, and get

d(p^mu)/dl = -Cristoffel(x)^mu_ab * (p^a) * (p^b)

d(x^mu)/dl = p^mu

in 3+1 dim space-time an 8-dim state vector. In the force equation you cannot and should not try to evaluate Cristoffel at a particular point x, or parameter l. The solution should provide you with that x(l) which satisfies the equations.

For parallel transport you would assume x(l) to be known and you really do have an ODE in one parameter. This is how you might propagate a viel-bein along a geodesic.

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    $\begingroup$ We do have MathJax here that makes your equations look not-horredous. You can search 'notation' in help center for details, if you aren't familiar with it. $\endgroup$ – Kyle Kanos Dec 21 '18 at 19:09

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