# Solution to Schwarzschild Equation for Constant t,r

• kirkr

#### kirkr

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'll guess the reason that no one else has answered so far is that your question does not make sense....

What do you mean by "...solve the Schwarzschild solution"? What are really trying to derive? The equation of motion of test particle in a Schwarzschild field? That's just a matter of computing the associated geodesic equation (although restricting to constant is likely to turn out rather trivial).

If not that, then... what?? (Btw, you should probably learn how to write equations in Latex in this forum.)

• Grelbr42, martinbn and milkism
Thank You for your reply. It is good to know how my question is being viewed and interpreted. I am working on latex and will revise my question when I get a good translation. Meanwhile, I hope you will cope with my current version. I have pasted the shortened version of the Schwarzchild equation below.
dt =0, dr=0
Euclidean space
d l^2 = r^2*(d(theta)^2+sin^2(theta)*d(phi)^2)
I have seen the Schwarzchild equation discussed on various websites; but I have never seen a solution that showed a value for arc length (l). That is, the equations are always in differential form and the integrated form is never shown. My question is a purely mathematical one. What is the proper technique to integrate the equation (line element)?
Euclidean space
d l^2 = r^2*(d(theta)^2+sin^2(theta)*d(phi)^2)

My efforts, so far, have yielded the two equations shown below:

Phi (1)*r*cos(theta)-Phi(2)*r*cos(theta) + integration constant (C)
r*theta(1) - r*theta(2) + integration constant (C')

This form suggests vector valued quantities (?)

Any guidance here would be appreciated. Kirk

This form suggests vector valued quantities (?)
No, arc length is a single, scalar quantity.

Consider your 2D example of the surface of a sphere of radius ##r## with line element ##ds^{2}=r^{2}\left(d\theta^{2}+\sin^{2}\theta\:d\phi^{2}\right)##. Draw any curve ##C## on the surface (see diagram), geodesic or otherwise. Since ##C## is 1D, positions along the curve can be written in terms of a single parameter ##p## that ranges in value from the start of the curve ##p_{1}## to its end ##p_{2}##. The 2D surface coordinates of the curve are then given by ##\theta_{C}=\theta(p),\;\phi_{C}=\phi(p)##. Hence, on the curve the line element becomes:$$ds^{2}=r^{2}\left(\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}\right)dp^{2}$$Taking the square root and integrating over ##p## from ##p_{1}## to ##p_{2}## gives the (scalar) arc length ##L_{C}## of ##C##:$$L_{C}=\intop_{p_{1}}^{p_{2}}ds=r\intop_{p_{1}}^{p_{2}}\sqrt{\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}}dp$$And if you can write a function relating the coordinates along the curve ##C##, such as ##\phi_{C}=f(\theta_{C})##, then the arc length simplifies to:$$L_{C}=r\intop_{\theta_{1}}^{\theta_{2}}\sqrt{1+\sin^{2}\theta\left(\frac{df(\theta)}{d\theta}\right)^{2}}d\theta$$From this 2D example you can readily generalize to the 4D Schwarzschild line element. Last edited:
• Grelbr42
Thank you for your reply. If I'm not mistaken, this solution looks something like a parameterization using p values on a one dimensional line. I had been trying to figure out how to make that conversion. I'll study this and get back to you with comments or possibly more questions. This is basically what I have been searching for.

No, arc length is a single, scalar quantity.

Consider your 2D example of the surface of a sphere of radius ##r## with line element ##ds^{2}=r^{2}\left(d\theta^{2}+\sin^{2}\theta\:d\phi^{2}\right)##. Draw any curve ##C## on the surface (see diagram), geodesic or otherwise. Since ##C## is 1D, positions along the curve can be written in terms of a single parameter ##p## that ranges in value from the start of the curve ##p_{1}## to its end ##p_{2}##. The 2D surface coordinates of the curve are then given by ##\theta_{C}=\theta(p),\;\phi_{C}=\phi(p)##. Hence, on the curve the line element becomes:$$ds^{2}=r^{2}\left(\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}\right)dp^{2}$$Taking the square root and integrating over ##p## from ##p_{1}## to ##p_{2}## gives the (scalar) arc length ##L_{C}## of ##C##:$$L_{C}=\intop_{p_{1}}^{p_{2}}ds=r\intop_{p_{1}}^{p_{2}}\sqrt{\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}}dp$$And if you can write a function relating the coordinates along the curve ##C##, such as ##\phi_{C}=f(\theta_{C})##, then the arc length simplifies to:$$L_{C}=r\intop_{\theta_{1}}^{\theta_{2}}\sqrt{1+\sin^{2}\theta\left(\frac{df(\theta)}{d\theta}\right)^{2}}d\theta$$From this 2D example you can readily generalize to the 4D Schwarzschild line element
Hi, I pasted in the latex code you had written and I can't get it to translate. I must be missing something.....

View attachment 324248
No, arc length is a single, scalar quantity.
Thank you for your reply. If I'm not mistaken, this solution looks something like a parameterization using p values on a one dimensional line. I had been trying to figure out how to make that conversion. I'll study this and get back to you with comments or possibly more questions. This is basically what I have been searching for.

Consider your 2D example of the surface of a sphere of radius ##r## with line element ##ds^{2}=r^{2}\left(d\theta^{2}+\sin^{2}\theta\:d\phi^{2}\right)##. Draw any curve ##C## on the surface (see diagram), geodesic or otherwise. Since ##C## is 1D, positions along the curve can be written in terms of a single parameter ##p## that ranges in value from the start of the curve ##p_{1}## to its end ##p_{2}##. The 2D surface coordinates of the curve are then given by ##\theta_{C}=\theta(p),\;\phi_{C}=\phi(p)##. Hence, on the curve the line element becomes:$$ds^{2}=r^{2}\left(\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}\right)dp^{2}$$Taking the square root and integrating over ##p## from ##p_{1}## to ##p_{2}## gives the (scalar) arc length ##L_{C}## of ##C##:$$L_{C}=\intop_{p_{1}}^{p_{2}}ds=r\intop_{p_{1}}^{p_{2}}\sqrt{\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}}dp$$And if you can write a function relating the coordinates along the curve ##C##, such as ##\phi_{C}=f(\theta_{C})##, then the arc length simplifies to:$$L_{C}=r\intop_{\theta_{1}}^{\theta_{2}}\sqrt{1+\sin^{2}\theta\left(\frac{df(\theta)}{d\theta}\right)^{2}}d\theta$$From this 2D example you can readily generalize to the 4D Schwarzschild line element.

View attachment 324248

I'll guess the reason that no one else has answered so far is that your question does not make sense....

What do you mean by "...solve the Schwarzschild solution"? What are really trying to derive? The equation of motion of test particle in a Schwarzschild field? That's just a matter of computing the associated geodesic equation (although restricting to constant is likely to turn out rather trivial).

If not that, then... what?? (Btw, you should probably learn how to write equations in Latex in this forum.)
No, arc length is a single, scalar quantity.

Consider your 2D example of the surface of a sphere of radius ##r## with line element ##ds^{2}=r^{2}\left(d\theta^{2}+\sin^{2}\theta\:d\phi^{2}\right)##. Draw any curve ##C## on the surface (see diagram), geodesic or otherwise. Since ##C## is 1D, positions along the curve can be written in terms of a single parameter ##p## that ranges in value from the start of the curve ##p_{1}## to its end ##p_{2}##. The 2D surface coordinates of the curve are then given by ##\theta_{C}=\theta(p),\;\phi_{C}=\phi(p)##. Hence, on the curve the line element becomes:$$ds^{2}=r^{2}\left(\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}\right)dp^{2}$$Taking the square root and integrating over ##p## from ##p_{1}## to ##p_{2}## gives the (scalar) arc length ##L_{C}## of ##C##:$$L_{C}=\intop_{p_{1}}^{p_{2}}ds=r\intop_{p_{1}}^{p_{2}}\sqrt{\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}}dp$$And if you can write a function relating the coordinates along the curve ##C##, such as ##\phi_{C}=f(\theta_{C})##, then the arc length simplifies to:$$L_{C}=r\intop_{\theta_{1}}^{\theta_{2}}\sqrt{1+\sin^{2}\theta\left(\frac{df(\theta)}{d\theta}\right)^{2}}d\theta$$From this 2D example you can readily generalize to the 4D Schwarzschild line element.

View attachment 324248

I am not certain in what format I have to enter formulas in this forum. I used the Overleaf latex editor to translate the line element formula that you used in your response. Then I was unable to paste it back to this forum. I attached an example and I guess I'm going to have to put together a response and use the attach file method.

#### Attachments

• Forum Script 40923.docx
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I am not certain in what format I have to enter formulas in this forum.
See the PF Latex Help Guide.

It's located under the "help" tab at the bottom of most (all?) PF screens. (I also have trouble finding it sometimes.)

(This info appeared when you read the PF guidelines when creating your account. You did read the guidelines, right? )

line element formula
$$L_{C}=\intop_{p_{1}}^{p_{2}}ds=r\intop_{p_{1}}^{p_{2}}\sqrt{\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}}dp$$

L_{C}=\intop_{p_{1}}^{p_{2}}ds=r\intop_{p_{1}}^{p_{2}}\sqrt{\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}}dp

$$L_{C}=\intop_{p_{1}}^{p_{2}}ds=r\intop_{p_{1}}^{p_{2}}\sqrt{\left(\frac{d\theta(p)}{dp}\right)^{2}+\sin^{2}\theta(p)\left(\frac{d\phi(p)}{dp}\right)^{2}}dp$$