# Radial geodesic distance with Schwarzschild's solution

• TheMan112
In summary, the geodesic distance between two points with the same angle but at different distances from the center of the body (and field) can be calculated using the Schwarzschild solution. This solution is only valid for points inside of the radius of the field, and creates an insanely complex primitive function inside of that radius.
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.

Last edited:
George Jones said:
You've got a sign wrong.

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

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).

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

Last edited by a moderator:
TheMan112 said:
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.

George Jones said:
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 [tex]I = \int_r2^r_1 ds = r_1 - r_2. Which is the distance in flat spacetime.

Thanks a lot!

## 1. What is the significance of radial geodesic distance in the context of Schwarzschild's solution?

Radial geodesic distance is an important concept in Schwarzschild's solution, which is a mathematical solution to Einstein's field equations that describes the spacetime around a non-rotating, spherically symmetric mass. It represents the distance between two points along a geodesic (the path of a freely falling object) that follows a straight line in the curved spacetime.

## 2. How is radial geodesic distance calculated in Schwarzschild's solution?

The formula for calculating radial geodesic distance in Schwarzschild's solution is given by r = 2GM/c^2 * (1 - 1/sqrt(1 - 2GM/rc^2)), where G is the gravitational constant, M is the mass of the object, c is the speed of light, and r is the distance from the center of the mass.

## 3. What does the value of radial geodesic distance tell us about the behavior of light in a strong gravitational field?

In Schwarzschild's solution, the value of radial geodesic distance decreases as the distance from the center of the mass increases. This indicates that light, which follows a geodesic, will be bent towards the mass in a strong gravitational field. This phenomenon is known as gravitational lensing and has been observed in astronomical observations.

## 4. How does radial geodesic distance change with the strength of the gravitational field?

As the strength of the gravitational field increases, the value of radial geodesic distance also increases. This means that in a very strong gravitational field, objects will experience a greater curvature of spacetime and will follow a more curved geodesic.

## 5. Can radial geodesic distance be used to study black holes?

Yes, radial geodesic distance is an important concept in the study of black holes. In the case of a black hole, the value of radial geodesic distance becomes infinite at the event horizon, which is the point of no return for anything that enters the black hole. This means that even light cannot escape from the black hole once it crosses the event horizon.

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