Sorry, my question may be feel dumb but still I am finding it hard to understand & calculate, how can one calculate $s$ from Schwarzschild metric equation $ds^2=-c^2(1-M/r)dt^2+r^2(d\theta^2+sin^2\theta d\phi^2)+\frac{dr^2}{1-M/r}$ from say $r=10, \theta=0, \phi=0, t=0$ to $r=2, \theta=\pi, \phi=\pi, t=10$ with $M=1$ , taking the shortest geodesic path?

i.e. how can one integrate the given equation, should one integrate each term & then take a square of each term and add or should one doubly integrate each term and add, e.g. if one go by the former method than is this relation correct, $s^2=-(ct)^2(1-M/r)+r^2(\theta^2+sin^2\theta\phi^2)+\left(\int\frac{dr}{1-M/r}\right)^2$ ?

Please provide the complete integrated expression and the value of $s$ for limits specified above. Thank you.

  • $\begingroup$ The interval depends on the path. It is the integral of $ds$ along the path taken from the initial to the final point. So it is not possible to answer your question unless you specify what path was taken. $\endgroup$ – John Rennie Aug 29 at 9:53
  • $\begingroup$ Sorry, take the shortest geodesic path between the points specified. $\endgroup$ – rim Aug 29 at 15:39
  • $\begingroup$ You have to solve the geodesic equation and integrate over that geodesic path. $\endgroup$ – G. Smith Aug 29 at 16:42

For a general metric, $ds^2 = g_{ij}dx^i dx^j$. If we specify a path and a parameterization, so $x^i=x^i(\lambda)$, then this becomes $ds^2=g_{ij}\frac{dx^i}{d\lambda} \frac{dx^j}{d\lambda} d\lambda^2$. Taking the square root and integrating yields the path length:

$$s=\int ds = \int \sqrt{g_{ij} \frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}} d\lambda$$

Consider 2D polar coordinates $(r,\theta)$, where $g_{ij}=\pmatrix{1&0\\0&r^2}$. In this case,

$$s = \int \sqrt{\left(\frac{dr}{d\lambda}\right)^2 + r^2 \left(\frac{d\theta}{d\lambda}\right)^2}d\lambda$$

Consider now the path given by $r(\lambda) = \lambda^2$ and $\theta(\lambda)=4\lambda$ for $\lambda\in[0,4]$. That's a spiral path which looks like this:

enter image description here

In this parameterization $\frac{dr}{d\lambda} = 2\lambda$ and $\frac{d\theta}{d\lambda} = 4$, and so

$$ds = \int_0^4 \sqrt{(2\lambda)^2+(\lambda^2)^2\cdot (4)^2} d\lambda = \int_0^4 \sqrt{4\lambda^2+16\lambda^4}d\lambda$$ Now we're just left with a standard integral, which WolframAlpha informs me is equal to $\frac{65\sqrt{65}-1}{6}$.

So that's how you integrate along some arbitrary specified path. In 4D and with the Schwarzschild metric, this is far more involved algebraically, but the principle is precisely the same.

If you want to integrate over a geodesic path, then you need to find the geodesic path first. To do this, you have to solve the geodesic equation:

$$\frac{d^2 x^i}{d\lambda^2} +\Gamma^i_{jk} \frac{dx^j}{d\lambda} \frac{dx^k}{d\lambda} = 0$$

This is it's own world of difficulty, in general. There are references all over the place about the solutions to this equation for the Schwarzschild metric - for example, here or here. Once you've figured out the geodesic path you're looking for, you can plug it into the expression for $ds$ and integrate to find the geodesic distance.

You shouldn't expect it to work out nicely, though. The calculation that you are asking somebody to do for you is going to be a nightmare which will presumably have to be worked out numerically.

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  • $\begingroup$ Thank you J Murray, you have basically explained the gist of differential geometry & cleared all my confusion on the subject. So, I just need to get the parametric path equations for $dr, d\theta, d\phi$ with $dt$ as the parameter using geodesic equations and then I can calculate $s=\int{\sqrt{{-c^2(1-M/r)}+r^2{(\frac{d\theta}{dt}})^2+r^2sin^2\theta({\frac{d\phi}{dt}})^2+\frac{1}{(1-M/r)}({\frac{dr}{dt}})^2}dt}$ , right!? please correct me if I am wrong at any point. Can anyone suggest what the parametric equations for $r, \theta, \phi$ taking $t$ as the parameter will be? $\endgroup$ – rim Aug 30 at 11:14
  • $\begingroup$ @rim Yes, that's the idea. Again, those equations are second order, coupled, nonlinear differential equations given by the geodesic equation I wrote above. The first link I provided below the geodesic equation will give you the explicit expressions for the Christoffel symbols. $\endgroup$ – J. Murray Aug 30 at 16:04

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