Radial geodesic distance with Schwarzschild's solution

[TheMan112]

Given we have a spherically symmetric gravitational field around a spherically symmetric body of mass M.

How can I calculate the actual (geodesic) distance between two points with the same angle but at different distances from the center of the body (and field).

I came immediately to think of the Schwarzschild solution which reduces to:

$$ds^2 = -(1-r_s/r)^{-1} dr^2 = \frac{1}{(\frac{r_s}{r}-1)} dr^2$$

The geodesic distance would then be:

$$\int ^{r_1}_{r_2} ds = \int ^{r_1}_{r_2} \frac{1}{\sqrt{r_s/r-1}}dr$$

Which can't be used outside $$r_s$$, and creates an insanely complex primitive function inside $$r_s$$.

What should I do? I need help urgently.

 

[George Jones]

You've got a sign wrong.

Care is needed when talking about spatial distances in general relativity.

See https://www.physicsforums.com/showthread.php?t=215488".

 

[TheMan112]

You've got a sign wrong.

Care is needed when talking about spatial distances in general relativity.

See https://www.physicsforums.com/showthread.php?t=215488".

Thank you, although I don't see how you came to that integral.

This is what I get by automatic integration: (where of course x = r)

-------------------------
You have entered: f (x) = ((1-((2m)/(x))))^(-1/2).

$$\int f(x)dx= \sqrt{1 - {2m \over x} }x +m \log \left( \sqrt{1 - {2m \over x} } +1 \right) -m \log \left( \left| \sqrt{1 - {2m \over x} } -1 \right|\right)$$
-------------------------

 

[George Jones]

Thank you, although I don't see how you came to that integral.

This is what I get by automatic integration: (where of course x = r)

-------------------------
You have entered: f (x) = ((1-((2m)/(x))))^(-1/2).

$$\int f(x)dx= \sqrt{1 - {2m \over x} }x +m \log \left( \sqrt{1 - {2m \over x} } +1 \right) -m \log \left( \left| \sqrt{1 - {2m \over x} } -1 \right|\right)$$
-------------------------

I used Maple, but I previously have done this integral by hand. My answer and your answer are equivalent; your answer differs from what is inside my square brackets by an additive constant. This is OK for indefinite integration.

The stuff inside your absolute value bars is always negative (why), so take them off and change the sign of the stuff inside. After this, combine the logs to show the equivalence between our answers.

If you want further hints, just ask.

 

[TheMan112]

I used Maple, but I previously have done this integral by hand. My answer and your answer are equivalent; your answer differs from what is inside my square brackets by an additive constant. This is OK for indefinite integration.

The stuff inside your absolute value bars is always negative (why), so take them off and change the sign of the stuff inside. After this, combine the logs to show the equivalence between our answers.

If you want further hints, just ask.

I think I've got it now. I've also checked that my results correspons with the Newtonian approximation by setting m = 0 and integrating at $$lim_{x \rightarrow \infty}$$ giving $$I = \int_r2^r_1 ds = r_1 - r_2. Which is the distance in flat spacetime.<br /> <br /> Thanks a lot!$$