- #1

kirkr

- 10

- 1

- Homework Statement
- How can the Schwarzschild solution be solved in a spherical space with a constant radius and constant time?

- Relevant Equations
- General Schwarzchild equation

ds2 = (1−rG/r)c2dt2 −(1/(1−rG/r)dr2 −r2(d(theta)2 +sin2(theta)d(phi)2)

In 1916, Karl Schwarzschild was the first person to present a solution to Einstein's field equations. I am using a form of his equation that is presented in Tensors, Relativity and Cosmology by Mirjana Dalarsson and Nils Dalarsson (Chapter 19, p.205).

I am approaching what may be the simplest problem related to this equation. That is finding a solution for constant time and radius (t=constant, radius = constant).

I have written my results for this simplification below. I have two questions. One, am I doing the integrations correctly and two if the integrations are correct, how are the results combined and interpreted?

For time (t) = constant and r constant then the three dimensional space met-

ric becomes the metric of a two-dimensional sphere or radius r embedded in the

Euclidean space

d l^2 = r^2*(d(theta)^2+sin^2(theta)*d(phi)^2)

Taking square root of both sides :

dl = r*sqrt((d(theta)^2+sin^2(theta)*d(phi)^2))

Question : can this be solved by first setting theta =

constant, d(theta) = 0 and then setting phi = constant, d(phi) = 0 .d(theta) =

0? Result for d(theta) I think??

dl^2 = r*2*sqrt(sin*2(theta)*d(phi)^2) dl = r*sin(theta)*d(phi)

I am approaching what may be the simplest problem related to this equation. That is finding a solution for constant time and radius (t=constant, radius = constant).

I have written my results for this simplification below. I have two questions. One, am I doing the integrations correctly and two if the integrations are correct, how are the results combined and interpreted?

For time (t) = constant and r constant then the three dimensional space met-

ric becomes the metric of a two-dimensional sphere or radius r embedded in the

Euclidean space

d l^2 = r^2*(d(theta)^2+sin^2(theta)*d(phi)^2)

Taking square root of both sides :

dl = r*sqrt((d(theta)^2+sin^2(theta)*d(phi)^2))

Question : can this be solved by first setting theta =

constant, d(theta) = 0 and then setting phi = constant, d(phi) = 0 .d(theta) =

0? Result for d(theta) I think??

dl^2 = r*2*sqrt(sin*2(theta)*d(phi)^2) dl = r*sin(theta)*d(phi)