# Solution to Schwarzschild Equation for Constant t,r

• kirkr
In summary, the conversation discusses the Schwarzschild equation and the process of integrating it to find a solution for constant time and radius. The equation is presented in a differential form and the speaker is looking for a way to convert it to an integrated form. The expert provides a generalization of the 2D example to solve the 4D Schwarzschild line element, explaining the process of finding the arc length of a curve on a surface.
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

kirkr said:
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.

renormalize said:
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
renormalize said:
No, arc length is a single, scalar quantity.
kirkr said:
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

strangerep said:
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.)
renormalize said:
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
24.2 KB · Views: 81
kirkr said:
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

OK, I copied the Latex version of the line element formula and added SS in the upper formula but neither one translates to standard mathematical symbols....unless I have to Post the Reply first...we'll see. Thanks for your advice-I have been studying that guide.

kirkr said:
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

OK, I copied the Latex version of the line element formula and added SS in the upper formula but neither one translates to standard mathematical symbols....unless I have to Post the Reply first...we'll see. Thanks for your advice-I have been studying that guide.
Sorry, what does "added SS" mean?

renormalize said:
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

Taking up your response at the formula for scalar arc length,

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

You stated that if there is a function $$\phi_{C}=f(\theta_{C})$$

Then the line element can be expressed as:

$$L_{C}=r\intop_{\theta_{1}}^{\theta_{2}}\sqrt{1+\sin^{2}\theta\left(\frac{df(\theta)}{d\theta}\right)^{2}}d\theta$$
One example of a function that satisfies this requirement is a loxodrome. A loxodrome is defined as a curve that cuts all meridians of a rotating surface at the same angle. Wolfram Mathworld defines it as follows:
A path, also known as a rhumb line, which cuts a meridian on a given surface at any constant angle but a right angle. If the surface is a sphere, the loxodrome is a spherical spiral. The loxodrome is the path taken when a compass is kept pointing in a constant direction. It is a straight line on a Mercator projection or a logarithmic spiral on a polar projection (Steinhaus 1999, pp. 218-219). The loxodrome is not the shortest distance between two points.

Sergio Kos (Rijeka College of Maritime Studies, Croatia) defines the loxodrome as a curve that cuts all meridians of a rotating surface at the same angle. (link to PDF below)
(please print the PDF or have it on hand for reference) (I had trouble translating the formulas for this forum)

He defines the first differential form of the sphere as follows: equation # 6
(see equation 6 in Sergio Kos paper)
The angle between curves on the sphere is defined here equation #7
(see equation 7 in Sergio Kos paper)

The differential equation of a loxodrome id: equation # 8
(see equation 8 in Sergio Kos paper)

The solution and general form of a loxodrome is: equation # 9
(see equation 9 in Sergio Kos paper)

The loxodrome curve is used in maritime navigation. The angle alpha is chosen as a bearing angle and is therefore known and is a constant value.References and Links

Sergio Kos, Differential Equation of a Loxodrome on a Sphere
https://www.researchgate.net/publication/235661334_Differential_Equation_of_a_Loxodrome_on_a_SphereWeisstein, Eric W. "Loxodrome." From MathWorld--A Wolfram Web
Resource. https://mathworld.wolfram.com/Loxodrome.htm
Nord, J. "Mercator's Rhumb Lines: A Multivariable Application of Arclength." College Math. J. 27, 384-387, 1996.Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, pp. 217-221, 1999.

Meridian
A line of constant longitude on a spheroid (or sphere). More generally, a meridian of a surface of revolution is the intersection of the surface with a plane containing the axis of revolution.

https://mathworld.wolfram.com/Meridian.html

## 1. What is the Schwarzschild equation?

The Schwarzschild equation is a mathematical equation that describes the gravitational field around a non-rotating, spherically symmetric mass. It was first derived by Karl Schwarzschild in 1916 as a solution to Einstein's field equations in general relativity.

## 2. What is the significance of finding a solution to the Schwarzschild equation for constant t,r?

Constant t,r refers to a specific set of coordinates in the Schwarzschild metric, where time (t) and radial distance (r) remain constant. Finding a solution to the Schwarzschild equation for these coordinates allows us to better understand the behavior of gravity and spacetime around a massive object, such as a black hole.

## 3. How is the Schwarzschild equation related to black holes?

The Schwarzschild equation is a solution to Einstein's field equations, which describe the effects of gravity on spacetime. This equation is used to model the gravitational field around a non-rotating, spherically symmetric mass, which is a key characteristic of black holes.

## 4. Can the Schwarzschild equation be applied to objects other than black holes?

Yes, the Schwarzschild equation can be applied to any massive object with a spherically symmetric distribution of mass. This includes planets, stars, and other astronomical bodies. However, it is most commonly used to describe the behavior of black holes due to their extreme mass and density.

## 5. Are there any limitations to the Schwarzschild equation?

The Schwarzschild equation is a simplified model that does not take into account factors such as rotation, electric charge, and the effects of other masses in the vicinity. It is also based on the assumption of a non-expanding, non-rotating, and non-charged mass, which may not always be the case in reality. Therefore, it is important to consider these limitations when using the Schwarzschild equation to study gravitational fields.

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