I Missing proof of the Shell theorem in General Relativity

[Bosko]

TL;DR Summary

Is there a proof for the shell theorem used in Swartzchild's solution?

In the classical Newtonian theory of gravity, the shell theorem holds. ( https://en.wikipedia.org/wiki/Shell_theorem )

...
Isaac Newton proved the shell theorem and stated that:

1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center.
   ...

In the beginning of the derivation of the Schwarzschild solution, the spherically symmetric object is replaced by a point mass.
The proof that this can be done in curved space-time is missing.

If the proof of this statement were to use the Schwarzschild solution directly or indirectly, it would not be logically valid. It would be so called "circular reasoning".

Is there a proof of the shell theorem for GR independent of the Schwarzschild solution?

 

[PeterDonis]

In the beginning of the derivation of the Schwarzschild solution, the spherically symmetric object is replaced by a point mass.

No, it isn't. That's not how the Schwarzschild solution is derived.

 

[PeterDonis]

Is there a proof of the shell theorem for GR independent of the Schwarzschild solution?

The second part of the shell theorem holds in GR, in the form that the sptacetime geometry of a vacuum region inside a spherically symmetric matter distribution is flat.

The first part isn't even meaningful in GR: there is no meaningful GR solution that I'm aware of for the spacetime geometry surrounding a point mass. Nor is that what the Schwarzschild solution describes (or how it's derived, as I noted in my previous post).

 

[Bosko]

No, it isn't. That's not how the Schwarzschild solution is derived.

How then?
Is there a point mass?

 

[weirdoguy]

How then?

Where did you get that it is? What are your sources?

 

[PeterDonis]

How then?

Where have you looked to find out? Most GR textbooks discuss how the Schwarzschild solution is derived.

Is there a point mass?

In the Schwarzschild solution? No.

 

[Bosko]

The second part of the shell theorem holds in GR, in the form that the sptacetime geometry of a vacuum region inside a spherically symmetric matter distribution is flat.

The first part isn't even meaningful in GR: there is no meaningful GR solution that I'm aware of for the spacetime geometry surrounding a point mass. Nor is that what the Schwarzschild solution describes (or how it's derived, as I noted in my previous post).

The first part can be derived from the second part .
E.g. The array of shells with the same mass and with 1/n radius . By subtracting results of the consecutive elements of the array you can get equivalence with the point mass.
Of course it is not the physical object any more but the mathematical model of the spherically symmetric mass.

 

[Bosko]

Where did you get that it is? What are your sources?

Any.
Some are explicit with the point mass as the represent of the spherically symmetric object
and other are using the equivalent form in one of the steps in derivation.

 

[Bosko]

In the Schwarzschild solution? No.

It might be better to ask an equivalent question:
Let two spherically symmetric objects of the same mass of radius $r_1 \lt r_2$ be given.

Is the spacetime for $r \gt r_2$ the same in both cases?

So a larger sphere can be replaced by a smaller ... etc ... arbitrarily smaller ... the point mass.

Edit:
Is there a proof of this claim for GR independent of the Schwarzschild solution?

 

[PeterDonis]

The first part can be derived from the second part .

Perhaps it can in Newtonian physics. That doesn't mean it can in GR.

E.g. The array of shells with the same mass and with 1/n radius . By subtracting results of the consecutive elements of the array you can get equivalence with the point mass.

Not in GR, you don't. Long before you get to that point, you will have a configuration that violates Buchdah's Theorem and therefore is not stable.

Of course it is not the physical object any more but the mathematical model of the spherically symmetric mass.

As I've already said, there is no consistent mathematical model of a point mass with a gravitational field in GR. GR is not Newtonian physics.

a larger sphere can be replaced by a smaller ... etc ... arbitrarily smaller ... the point mass.

Not in GR, it can't. See my comment about Buchdahl's Theorem, above.

 

[PeterDonis]

Moderator's note: Thread moved to the relativity forum as the topic is well within standard GR.

 

[PeterDonis]

Any.
Some are explicit with the point mass as the represent of the spherically symmetric object
and other are using the equivalent form in one of the steps in derivation.

Sorry, but this isn't a responsive answer. You need to cite a specific chapter/section/page in a specific textbook or peer-reviewed paper that gives such a derivation. If you can't do that, this thread will be closed as there is no valid basis for discussion.

 

[PAllen]

Any.
Some are explicit with the point mass as the represent of the spherically symmetric object
and other are using the equivalent form in one of the steps in derivation.

Ok then it would be easy to provide one. For example, I have about 10 GR texts, spanning publication dates from 1921 to 2015. All derive the Schwarzschild solution and none mention a mass point - because it is impossible. The interior of the horizon of the Schwarzschild solution contains(ends with) an axial singularity, not a mass point.

 

[PeterDonis]

The interior of the horizon of the Schwarzschild horizon is best described as an axial singularity, not a mass point

And note that the singularity (the locus $r = 0$ in the Schwarzschild geometry) is spacelike, not timelike, so it's "axial" only in the sense that it's a one-dimensional line, not in the sense that it's a line sitting in space and existing for some length of time.

 

[PAllen]

It might be better to ask an equivalent question:
Let two spherically symmetric objects of the same mass of radius $r_1 \lt r_2$ be given.

Is the spacetime for $r \gt r_2$ the same in both cases?

So a larger sphere can be replaced by a smaller ... etc ... arbitrarily smaller ... the point mass.

Edit:
Is there a proof of this claim for GR independent of the Schwarzschild solution?

Up to the stability issue raised by @PeterDonis (Buchdahl's theorem), there are many sources that prove this, even in, e.g. the case of a radially vibrating spherically symmetric mass. That is, outside the outer bound of vibration, the solution is uniquely determined to be Schwarzschild with some mass parameter. One example is chapter 7 of J. L. Synge's book "Relativity: The General Theory".

More generally, search for "Birkhoff's Theorem", and several rigorous presentations in arxiv may be found.

 

[PAllen]

And note that the singularity (the locus $r = 0$ in the Schwarzschild geometry) is spacelike, not timelike, so it's "axial" only in the sense that it's a one-dimensional line, not in the sense that it's a line sitting in space and existing for some length of time.

Never intended to imply that. Adjusted the wording slightly, hopefully for the better ...

 

[Orodruin]

TL;DR Summary: Is there a proof for the shell theorem used in Swartzchild's solution?

In the beginning of the derivation of the Schwarzschild solution, the spherically symmetric object is replaced by a point mass.

No it isn’t. The usual derivation starts by assuming spherical symmetry and a vacuum solution and then proceeds to solve the EFEs under those assumptions.

 

[Bosko]

No it isn’t. The usual derivation starts by assuming spherical symmetry and a vacuum solution and then proceeds to solve the EFEs under those assumptions.

There is no parameter of a spherically symmetric object in the solution. For example, the radius .
The solution does not depend on the size of the object.
Let's find some detailed explanation of the formula and see when the radius of the object vanishes .

Is this from Wikipedia good for analysis?
https://en.wikipedia.org/wiki/Derivation_of_the_Schwarzschild_solution

I'm looking for a detailed derivation available on the internet.

Can we agree that the solution does not depend on the radius of a spherically symmetric object?
I think that this is some form of the shell theorem.

 

[PeroK]

Can we agree that the solution does not depend on the radius of a spherically symmetric object?
I think that this is some form of the shell theorem.

The Schwarzschild solution is valid outside an object, as long as the radius of the object is greater than its Schwarzschild radius. For a fixed mass M, therefore, there is a minimum radius for which the solution is valid. If the mass is contained within the Schwarzschild radius, then you have an unstable collapsing scenario.

The full vacuum solution has a characteristic mass, but that mass itself is not part of the spacetime. Instead, you have a singularity, which is not a point in space.

In any case, in GR you cannot consider a gravitating body as a point mass - although you may consider test particles as point masses.

 

[A.T.]

In the beginning of the derivation of the Schwarzschild solution, the spherically symmetric object is replaced by a point mass.

Can we agree that the solution does not depend on the radius of a spherically symmetric object?
I think that this is some form of the shell theorem.

First you claimed that it is used at the beginning of the derivation, then you say it follows from the solution. These are two different claims.

As others said: That a solution doesn't depend on a parameter, doesn't necessarily imply that it is valid for any value of that parameter.

 

[Bosko]

The Schwarzschild solution is valid outside an object,

In other words, for any $r_1$ and $r_2$ the solution is the same outside of $max(r_1,r_2)$.
Am I right, even if one of $r_1$ or $r_2$ is smaller than $r_s$ (Schwarzchild radius)

 

[Bosko]

As others said: That a solution doesn't depend on a parameter, doesn't necessarily imply that it is valid for any value of that parameter.

That's exactly what I thought.
Edit: That is valid

 

[Bosko]

Moderator's note: Thread moved to the relativity forum as the topic is well within standard GR.

Thanks. I put the topic in "Beyond the Standard Models" so as not to violate the forum rules in case someone offers non-standard thinking.

 

[PeroK]

In other words, for any $r_1$ and $r_2$ the solution is the same outside of $max(r_1,r_2)$.
Am I right, even if one of $r_1$ or $r_2$ is smaller than $r_s$ (Schwarzchild radius)

If $r_1 < r_s$, then you don't have a Schwarzschild solution. Note that for $r < r_2$, the $r$ is a time-like parameter. It's not a spatial radius (or anything like it). Note that for $r > r_s$, $r$ is technically not a Euclidean radius, but an areal radius.

 

[Bosko]

What are $r_1$ and $r_2$?

Two objects of the same mass but different radii.

 

[PeroK]

Two objects with the same mass but different radii

Sorry, I didn't read the earlier post. See my updated reply. $r_1 < r_s$ is not valid as the radius of an object.

 

[Bosko]

Sorry, I didn't read the earlier post. See my updated reply. $r_1 < r_s$ is not valid as the radius of an object.

Do you think that for $r_1 < r_s$ the mathematical model does not correctly represent physical reality?

 

[PeroK]

Do you think that for $r_1 < r_s$ the mathematical model does not correctly represent physical reality?

For $r < r_s$, the coordinate $r$ is timelike. Just because you use $r$ as a coordinate doesn't mean it's a radius. The Schwarzschild coordinates have a coordinate singularity at $r = r_s$.

The region of spacetime represented by $0 < r < r_1 < r_s$ for any $\phi, \theta$ and fixed $t$ is not a spatial volume. It cannot be the volume of a massive object.

There was another student confused about this recently. See this thread:

H

Thread 'Volume of a black hole using the Schwarzschild metric'

Feb 25, 2025

Hi,

I'm wondering if it is possible to calculate the volume of a black hole using the Schwarzschild metric.

After computing the volume I get the follow integral:

$$V = 4 \pi \int_0^r \frac{1}{\sqrt{ (1- \frac{r_s}{r'})}} r'^2 dr$$

This integral diverges at the upper bound. The only way I found to compute the integral analytically is if $$r \gg r_s$$, then I can perform a Taylor's expansion.

However, in this case the volume is not really that of a black hole.

Is there something I don't understand?

Thank you

• happyparticle
• Black hole General relativity Schwarzschild
• Replies: 34
• Forum: Advanced Physics Homework Help

 

[PeterDonis]

I put the topic in "Beyond the Standard Models" so as not to violate the forum rules in case someone offers non-standard thinking.

This is not a good way to choose a subforum. "Non-standard thinking" in this context means personal speculation, which is off limits here. The issue we are discussing in this thread is standard GR and does not require going beyond that.

 

[PeterDonis]

There is no parameter of a spherically symmetric object in the solution.

You clearly do not understand how the Schwarzschild solution is derived. I suggest reviewing the derivations in Wald or MTW. Both of those clearly state the assumption of spherical symmetry and what it implies mathematically.

The solution does not depend on the size of the object.

You clearly do not understand what the Schwarzschild solution we are discussing represents. It represents a vacuum solution. For a spacetime containing a spherically symmetric mass, it represents the vacuum region outside the mass. The geometry of the vacuum region does not depend directly on the radius of the surface of the mass (it does depend on the mass itself), but that radius still has to be larger than the limit imposed by Buchdahl's Theorem, which is 9/8 of the Schwarzschild radius corresponding to the mass, i.e., outside the horizon.

I think that this is some form of the shell theorem.

No, it isn't. Nothing I said above has anything to do with the shell theorem.

 

[PeterDonis]

The Schwarzschild solution is valid outside an object, as long as the radius of the object is greater than its Schwarzschild radius.

Actually it has to be greater than 9/8 of the Schwarzschild radius, per Buchdahl's Theorem.

 

[PeterDonis]

Do you think that for $r_1 < r_s$ the mathematical model does not correctly represent physical reality?

That's not the issue. The issue is that you don't understand what physical reality the mathematical model represents for values of $r$ less than $r_s$. For such values of $r$ the model represents the interior of a black hole. It does not represent the vacuum region outside an ordinary gravitating object with that radius.

 

[PeterDonis]

For $r < r_s$, the coordinate $r$ is timelike.

This is true for Schwarzschild coordinates, but there are other charts for which it is not true. You give a much better description of the actual problem (which can be stated in a form that's invariant, so it doesn't actually depend on your choice of coordinates) here:

The region of spacetime represented by $0 < r < r_1 < r_s$ for any $\phi, \theta$ and fixed $t$ is not a spatial volume. It cannot be the volume of a massive object.

 

[PAllen]

There is no parameter of a spherically symmetric object in the solution. For example, the radius .
The solution does not depend on the size of the object.
Let's find some detailed explanation of the formula and see when the radius of the object vanishes .

Is this from Wikipedia good for analysis?
https://en.wikipedia.org/wiki/Derivation_of_the_Schwarzschild_solution

I'm looking for a detailed derivation available on the internet.

Can we agree that the solution does not depend on the radius of a spherically symmetric object?
I think that this is some form of the shell theorem.

You assume nothing but spherical symmetry, and vacuum outside some r. You assume nothing about inside r (except spherical symmetry throughout). You derive that the solution is unique up to one parameter - a mass parameter associated with everything inside r. Thus you have derived, not assumed, that nothing about the nature or size of what is inside r can matter. There is simply no place for it in the solution. The shell theorem in GR is thus derived, not assumed. It is also different from the Newtonian statement ( you can’t talk about a mass point). The GR statement is: if all vacuum outside of r, nothing about the inside matters except one parameter describing total mass.

 

[Orodruin]

There is no parameter of a spherically symmetric object in the solution. For example, the radius .

There is no object at all in the Schwarzschild spacetime. It is a vacuum solution. The assumptions are spherical symmetry of the solution and vacuum.