In summary: Continuing...In summary, this article provides useful equations for static, spherically symmetric spacetimes.
  • #1
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This will be the first of several articles which will provide, for reference, useful equations for static, spherically symmetric spacetimes. This is a common special case that is studied in General Relativity, and it has the advantage of having a general solution for the Einstein Field equation that can be expressed in closed form equations. This first article will focus on those closed form equations for that solution.
First, we should be clear about the set of spacetimes we are talking about. The two key properties of this set of spacetimes can be defined in coordinate-independent terms as follows:
(1) A static spacetime has a timelike Killing vector field which is hypersurface orthogonal.
(2) A spherically symmetric spacetime has a 3-parameter group of spacelike Killing vector fields that satisfy the properties of the Lie group SO(3), which describes rotations in 3-dimensional space.
As noted...

Continue reading...
 
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  • #2
A question from Reddit

Are the gammas with an exponent and two indexes a three variable function ?
 
  • #3
Greg Bernhardt said:
A question from Reddit

I suspect the question is from a person who does not really have the background knowledge to understand the post. Here is a suggested reply:

The gammas are Christoffel symbols [1]; the three indexes are tensor indexes, not variables. The upper index is not an exponent; it's an upper tensor index. (Strictly speaking the Christoffel symbols are not tensors, but they have tensor indexes and are treated like tensors in formulas.) Like all other quantities in the case under discussion, the Christoffel symbols are functions of only one variable, ##r##, because of the staticity and spherical symmetry.

[1] https://en.wikipedia.org/wiki/Christoffel_symbols
 
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  • #4
Hi Peter. An interesting article - thanks. Are you going to go more into the stress-energy tensor and the constraints on it in this case in later articles?
 
  • #5
Ibix said:
Are you going to go more into the stress-energy tensor and the constraints on it in this case in later articles?

I wasn't planning to. One further article will give Maxwell's Equations in a static, spherically symmetric spacetime (and how they couple to the EFE via the stress-energy tensor for an EM field), and another will give equations for an object being slowly lowered in a static, spherically symmetric spacetime.

Did you have a particular question about the SET and constraints on it?
 
  • #6
PeterDonis said:
Did you have a particular question about the SET and constraints on it?
Not really anything specific. The stress-energy tensor seems to me to be something GR texts often seem to assume you understand, somehow. I think I may have missed a lecture somewhere, because it's the bit I feel I have the least grasp of. So I'm always interested in seeing someone talking about it.
 
  • #7
I bit pedantic but shouldn't the definition of spherically symmetric be phrased differently. The Killing fields generate a Lie algebra isomorphic to ##\mathfrak{so}(3)##, the group of isometries is ##\text{SO}(3)##.
 
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  • #8
martinbn said:
I bit pedantic but shouldn't the definition of spherically symmetric be phrased differently. The Killing fields generate a Lie algebra isomorphic to ##\mathfrak{so}(3)##, the group of isometries is ##\text{SO}(3)##.

I think this might be a bit too much detail; I was trying to avoid going into it by saying the Killing fields "satisfy the properties" of ##\text{SO}(3)## (the Lie algebra would be part of the "properties").
 
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  • #9
PeterDonis said:
There seem to be a few errors in the article. ## \partial_{\theta} ## is NOT a killing vector here. I think this is repeated erroneously a few times. Further, ##J(r)## is defined to be the 'redshift factor'. I believe the correct formulation is that the redshift factor is proportional to ##\sqrt{J(r)}## and not ##J(r)##.

It would also be great if you could illustrate the definition of '##s##' in the TOV equation as this seems to have been introduced without adequate precedent.
 
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  • #10
Raychauduri said:
##\partial_{\theta}## is NOT a killing vector here.

You're right, I have updated the article to correct this.

Raychauduri said:
I believe the correct formulation is that the redshift factor is proportional to ##\sqrt{J(r)}## and not ##J(r)##.

It is ##\sqrt{J(r)}##, that's correct. (There is no constant of proportionality required.) I've updated the article to fix this as well.

Raychauduri said:
It would also be great if you could illustrate the definition of '##s##' in the TOV equation as this seems to have been introduced without adequate precedent.

I'm not sure what you mean by "inadequate precedent". The TOV equation is derived assuming that the matter present is a perfect fluid; that means all three spatial diagonal components of the SET must be equal. If you drop that assumption, the radial component can be different from the tangential components, but the two tangential components still need to be the same by spherical symmetry. That's why there are just two functions, ##p## and ##s##.
 
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  • #11
Hi @PeterDonis:

I would much appreciate your taking a look at my calculations regarding a finite static stable universe which I posted in
I started with
FriedmannR.png
.
Since the target model is finite, I used the total mass of the universe M, and substituted for ρ
ρ = M/V​
where
V = 2π2/R3.​

My calculations led to the following results.
Buzz Bloom said:
Λ = 1 / c2R2.​
Combining this with the undifferentiated equation yields
16GM/πc2 - 3R + R = 0.​
Simplifying
R = 8GM/πc2.​
Simplifying
Λ = (1/c2) (π2c4/64G2M2)​
= π2c2/64G2M2.​

CORRECTION MADE June 16
I fixed an error below involving the formula for the 3D volume which is the boundary of a 4D hypersphere.
I find it interesting that the R value for a finite static stable homogeneous and isotropic universe is
1/π times the black hole event horizon radius.

Regards,
Buzz
 
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  • #12
Buzz Bloom said:
I would much appreciate your taking a look at my calculations regarding a finite static stable universe

You seem to be reinventing the Einstein static universe without fully understanding its properties. The feedback you are getting in the thread looks valid to me.

In any case, the form of the EFE you are using in that thread is different from the form I use in the article.
 
  • #13
PeterDonis said:
You seem to be reinventing the Einstein static universe without fully understanding its properties. The feedback you are getting in the thread looks valid to me.
In any case, the form of the EFE you are using in that thread is different from the form I use in the article.

Hi Peter:

I would be very interested to see the form of the EFE you are using. Apparently, in the June 5th article, you have not yet introduced that EFE.

I have learned a few things from the other posts in the thread I cited, but as of yet no one has told me whether my calculations are correct or erroneous. I also do not have any idea about what are properties of the Einstein static universe which I do not fully understand.

CORRECTION June26
I fixed an error below due to the same error if fixed in post #11.
In the most recent post by @George Jones (post #46) George pointed out that it is not sufficient that dR/dt=d2R/dt2=0 for stability. If R is slightly smaller or larger than 2GM/πc2, dR/dt≠0 and d2R/dt2≠0 and both will have the same sign, so R will thereafter either expand or contract. Aside from this improper usage of "stable" on my part, what else about my understanding of this topic (as I have presented it in my posts) needs repair?

Regards,
Buzz
 
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  • #14
Buzz Bloom said:
I would be very interested to see the form of the EFE you are using.

Then you can read my Insights article on the Einstein Field Equation in a static, spherically symmetric spacetime. You know, the one this comment thread is about.

Buzz Bloom said:
Apparently, in the June 5th article, you have not yet introduced that EFE.

Are you referring to the Insights article? What do you think is missing from it?
 
  • #15
Buzz Bloom said:
the thread I cited

Please keep discussion of that thread in that thread.
 
  • #16
Buzz Bloom said:
I also do not have any idea about what are properties of the Einstein static universe which I do not fully understand.

Then I think you would benefit from taking some time to study it. It's a good simple pedagogical model.
 
  • #17
PeterDonis said:
Are you referring to the Insights article? What do you think is missing from it?
Hi Peter:

What would be particularly helpful to my understanding is a description of the physical interpretation of the five functions:
J, m, ρ, p, and s,​
and their units.
Here is what I think I understand about these functions.
1. They are all functions of the radial coordinate r.
2. J is somehow related to the Hubble function H(a), where a is the scale factor variable. If this is correct, I would particularly like to see an equation relating J and H.
3. r(t) is related to a(t) by arbitrarily choosing a particular reference value
r0=r(t0), and​
r=r0a.​
4. ρ is mass density.
5. P is pressure, which Einstein set P=0 for his 1917 model of a dust filled universe.
I do not know what m and s are.

If I am mistaken about this "understanding", I would appreciate being corrected.

I would also like to see a single equation relating H to ρ, m, and s. I would think that such an equation would look something like
FriedmannR.png

or
FriedmannEq.png


Regards,
Buzz
 
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  • #18
PeterDonis said:
Then I think you would benefit from taking some time to study it. It's a good simple pedagogical model.
Hi Peter:

Well, I am trying to understand your article, but I have forgotten almost everything I used to know in my youth, 60 years ago, about tensor calculus. I am looking forward to seeing the other articles in this series.

Regards,
Buzz
 
  • #19
Buzz Bloom said:
What would be particularly helpful to my understanding is a description of the physical interpretation of the five functions

That is discussed in the article.

Buzz Bloom said:
They are all functions of the radial coordinate r.

Yes.

Buzz Bloom said:
J is somehow related to the Hubble function

You are mixing up two different kinds of solutions to the EFE. The kind of solution discussed in the Insights article is for a static, spherically symmetric spacetime. There is no "Hubble function" in such a spacetime.

The kind of solution in which there is a "Hubble function" is an FRW spacetime. FRW spacetimes are not static, and nothing in my Insights article applies to them.

Buzz Bloom said:
r(t) is related to a(t)

The spacetime discussed in the Insights article is static. Nothing is a function of ##t##. There is no scale factor ##a##. See above.

Buzz Bloom said:
ρ is mass density.

More precisely, energy density as measured by a static observer.

Buzz Bloom said:
P is pressure

More precisely, radial pressure as measured by a static observer. And ##s## is the tangential pressure (or stress).

Buzz Bloom said:
Einstein set P=0 for his 1917 model of a dust filled universe

If you are referring to the Einstein static universe, that (unlike all other spacetimes that can be described in FRW coordinates) is indeed a static, spherically symmetric spacetime, so it can be described using the math in my Insights article. However, that article uses coordinates which are very different from FRW coordinates, so a description of the Einstein static universe using the math in my Insights article will look very different from the usual description of it in FRW coordinates.

Buzz Bloom said:
I do not know what m and s are.

Then you didn't read the article very carefully. The meaning of ##m## is discussed near the end. The meaning of ##s## is briefly given in the article, and I gave it again above.

Buzz Bloom said:
I would also like to see a single equation relating H to ρ, m, and s.

There isn't one since there is no ##H## in a static spacetime. See above.
 
  • #20
Hi @PeterDonis:

Thank you very much for your post. I have learned quite a bit from it. I apologize for my previous careless reading of your article. I realized it will take me several readings to grasp all the concepts in it. For the present, I did a quick scan to look for particular topics I am currently trying to learn about. At my age I use the excuse that I fail to see things that are there, and I see things that are not.

I bolded your text from your previous post. I did this because this was more convenient for me than using the Quote feature.

As you have no doubt guessed, the function m(r) can be physically interpreted as the mass inside radial coordinate r.
Does this not mean that m(r) is related to ρ(r)?
ρ(r) = m(r)/V(r)​
where V(r) is the volume of a sphere of radius r. I understand that
V(r) ≠ (4/3)πr3
Let A be the area of a sphere of radius r in a space with curvature.
A(r) = 4πR sin(r/R)2
V(r) = ∫[x= 0 to r] A(x) dx​
= 4πR2 ∫[x= 0 to r] sin2(x/R) dx​
NOTE: ∫sin2(x)dx = x/2 - sin(2x)/4 [ CRC pg 291 Eq 230 ]
Let w = x/R
V(r) = 4πR2 ∫[w=0 to r/R] sin2(w) R dw​
= 4πR3 { (1/2) (r/R) - (1/4) sin(r/R) }​
I did the following just as a check on my math. My first try had an error, so the effort turned out to be useful.
V(πR) = 4πR3 π/2​
= 2π2 R3
The spacetime discussed in the Insights article is static. Nothing is a function of t. There is no scale factor a.
This is a surprise. I did not think that the tensor equations could be specialized for such a static model. Although I probably could not understand an explanation of the GR tensor equations, does the article include a simplification of the tensor equations based on the assumption of a static spacetime? I would like to take a look at that even if I cannot understsand the math. I can still understand the tensor notation.

If you are referring to the Einstein static universe,...
I am. However, I am assuming that the equations
[1] dm/dr = 4πr2 ρ(r)​
[2] (1/2J) dJ/dr = (m+4πr3p)/(r(r-2m))​
[3] dp/dr = -(ρ+p)[(1/2J) dJ/dr) - (2(p-s)/r) ]​
are intended for a universe which may or may not be static, and that you specialize them to be static Is this correct? If not, where in the article was the step where the simplification for static spacetime was explained? I am now getting the message that deriving equations for a static universe from more general equations by assuming
dr/dt = d2r/dt2 = 0​
may not reach the same result as the approach in your article. Is this correct?
Also, I assume for the Einstein 1917 model, pressure p=s=0, so the 2nd and 3rd equations become
[2' -> 4] (1/2J) dJ/dr = m/(r (r-2m))​
[3' -> 5] 0 = -rho (1/2J) dJ/dr​
[4&5 -> 6] 0 = m/(r (r-2m))​
But, this implies
ρ=0 or dJ/dr = 0​
and
m = 0.​
These results (implied by p=s=0) makes no sense to me, so I must have made an error (which I cannot find) in my analysis. Can you help me?

The function J(r) (or more precisely its square root) has the obvious interpretation of being the “redshift factor” for an observer who is static at radial coordinate r.
This "redshift" factor is what suggested to me that it implied a radial speed associated with distance. That is what lead me to think there was a relationship between J and H.

The spacetime discussed in the Insights article is static. Nothing is a function of t. There is no scale factor a.
If there is no function of time, the redshift must be only an effect of seeing an object with a redshift corresponding to it kinetic energy. Is that correct? If so, why is redshift a function of r?

ρ is mass density.
More precisely, energy density as measured by a static observer.
Does this mean that the static observer sees the energy of something as including the energy equivalent of its mass together with its kinetic energy?

p is pressure
More precisely, radial pressure as measured by a static observer. And s is the tangential pressure (or stress).
I confess I do not understand why in a homogeneous and isotropic model the tangential and radial pressure should be different.

Regards,
Buzz
 
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  • #21
Buzz Bloom said:
I did this because this was more convenient for me than using the Quote feature.

I find this difficult to believe since the Quote feature is very easy to use. Please use it. Standard features like that are there for a reason. Making up your own version undermines that.

Buzz Bloom said:
Does this not mean that m(r) is related to ρ(r)?

Yes, and I gave the relationship in the article. It's not the one you are giving here (although later in your post you give the correct one). I'm beginning to wonder if you've actually read the article.

Buzz Bloom said:
I am assuming that the equations
[1] dm/dr = 4πr2 ρ(r)[2] (1/2J) dJ/dr = (m+4πr3p)/(r(r-2m))[3] dp/dr = -(ρ+p)[(1/2J) dJ/dr) - (2(p-s)/r) ]are intended for a universe which may or may not be static. Is this correct?

Now I'm beginning to wonder if you've even read the title of the article. What is the seventh word in it?

(Note, btw, that [1] here is the correct relationship between ##m## and ##\rho##. You can integrate it to get ##m(r)## as a function of ##r##, in terms of an integral of ##\rho## from ##0## to ##r##.)

Buzz Bloom said:
I assume for the Einstein 1917 model, pressure p=s=0

As far as I know, yes, Einstein assumed cold matter for his static universe, which means the only nonzero term in the stress-energy tensor in the formulation I give in the article would be ##\rho##.

Buzz Bloom said:
These results (implied by p=s=0) makes no sense to me, so I must have made an error (which I cannot find)

Then you evidently haven't paid attention very well. See above.

Buzz Bloom said:
This "redshift" factor is what suggested to me that it implied a radial speed associated with distance.

"Redshift" here refers to gravitational redshift. This is standard terminology for a static spacetime.

Buzz Bloom said:
That is what lead me to think there was a relationship between J and H.

That and the fact that you have repeatedly ignored the word "static".

Buzz Bloom said:
If there is no function of time, the redshift must be only an effect of seeing an object with a redshift corresponding to it kinetic energy. Is that correct?

No, it's gibberish. See above on what "redshift" refers to.

Buzz Bloom said:
Does this mean that the static observer sees the energy of something as including the energy equivalent of its mass together with its kinetic energy?

Remember we are talking about the stress-energy tensor of a static spacetime. The stress-energy tensor refers to the source of the gravitational field--a planet, star, whatever, which is static. That means a static observer will be at rest relative to this static object. So the object will not have any kinetic energy relative to the static observer. The energy density ##\rho## is just rest energy density.

Buzz Bloom said:
I confess I do not understand why in a homogeneous and isotropic model the tangential and radial pressure should be different.

Who said anything about homogenous and isotropic? This Insights article is about a static, spherically symmetric spacetime. Such a spacetime does not have to be homogeneous and isotropic, and in the usual cases, a static gravitating body like a planet or star, it obviously won't be.

I have no idea why you persist in thinking I am describing an FRW spacetime when I have repeatedly explained that I am not.

Buzz Bloom said:
This is a surprise. I did not think that the tensor equations could be specialized for such a static model.

Well, then you obviously thought wrong. Also, I have no idea why you would think so.
 
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  • #22
Hi @PeterDonis:

I apologize again, although I did learn a few new things. This time my apology is for my mental denseness wasting your time,. (More senior moments.) I saw "static" and "Einstein" in your article title, and based on my recent interests in the Einstein 1917 model, I fell into a mental trap that your article was coincidentally about the same topic, using different methods. I now, finally, get it. I was completely wrong.

Regards,
Buzz
 
  • #24
What might be helpful to study is the treatment of the TOV (Tolman-Oppenheimer-Volkov) equation for the collapse of a dust cloud in Landau and Lifshitz vol. 2. This is example is (almost) analytically treatable and gives a nice example of the coordinate transformation from the Schwarzschild coordinates to Gaussian (FLRW-like) normal coordinates.
 

What is the Einstein Field Equation?

The Einstein Field Equation is a mathematical equation that describes the relationship between matter and spacetime. It was developed by Albert Einstein as part of his theory of general relativity.

What is a static, spherically symmetric spacetime?

A static, spherically symmetric spacetime is a type of spacetime that is stationary and has a spherical symmetry. This means that the properties of the spacetime do not change with time and are the same in all directions from a central point.

How does the Einstein Field Equation relate to gravity?

The Einstein Field Equation describes how matter and energy in spacetime create curvature, which is what we perceive as gravity. It shows how the presence of matter and energy can warp the fabric of spacetime, causing objects to follow curved paths.

What are some applications of the Einstein Field Equation?

The Einstein Field Equation has many applications in astrophysics and cosmology. It has been used to explain the behavior of black holes, the expansion of the universe, and the effects of gravitational lensing. It is also used in the development of technologies such as GPS and gravitational wave detectors.

Are there any limitations to the Einstein Field Equation?

While the Einstein Field Equation has been incredibly successful in describing gravity, it does have some limitations. It does not take into account quantum effects, and it cannot be used to describe the behavior of extremely small particles. Additionally, it breaks down in extreme conditions, such as at the center of a black hole.

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