Curvature of spacetime inside of stars and planets

[relativityfan]

hi,

how does general relativity work INSIDE stars and planets, since the mass is no longer concentrated within a point, so there are necessarily gravitationnal effects outwards and not only inwards?

 

[George Jones]

hi,

how does general relativity work INSIDE stars and planets, since the mass is no longer concentrated within a point,

Solve Einstein's equation G = 8 Pi T, where T is a reasonable stress-energy tensor for a star. For realistic models, this is done numerically, but Schwarzschild solved Einstein's equation for a constant-density spherical star. See

https://www.physicsforums.com/showthread.php?p=2262718#post2262718

https://www.physicsforums.com/showthread.php?p=1543402#post1543402.

 

[bcrowell]

In the case of ordinary stars and planets (not neutron stars, etc.), I think you can get pretty far without having to solve the Einstein field equations, since this is the weak field limit, which is Newtonian. For instance, suppose you want to find the difference in clock rate between the center of the Earth and the poles. The shell theorem gives a gravitational field g=Mr/R³, corresponding to a potential $\Phi=Mr^2/2R^3$. Then the difference in the gravitational potential is simply $\Phi/c^2=M/2c^2R$, which comes out to be 3.5x10^-10. This is the fractional difference in clock rates.

 

[relativityfan]

thank you but is the schwarzchild interior solution valid? , because i do not understand why time is not dilated by the same factor as length is contracted, where as in the exterior (classical) schwarzchild solution the time dilation factor is the same as the length contraction factor. so both seem to be not compatible with each other

 

[bcrowell]

thank you but is the schwarzchild interior solution valid? , because i do not understand why time is not dilated by the same factor as length is contracted, where as in the exterior (classical) schwarzchild solution the time dilation factor is the same as the length contraction factor. so both seem to be not compatible with each other

There is no fundamental principle that says that these two things have to be equal. In fact, their equality is not even true in any absolute sense for the exterior Schwarzschild metric. If you happen to pick a particular set of coordinates, t and r, where r is the circumference over 2pi, then the metric happens to have g_tt=1/g_rr. If you pick different coordinates, I don't think this is even true.

 

[relativityfan]

i do believe it is true because you can generate the schwarzchild metric just with merely the Lorentz transformations. an object falls with a given speed s, with the right initial conditions, and the gravitationnal field is an acceleration that leads to this speed. from this velocity we deduce the Lorentz factor, and this clearly matches the Schwarzschild metric. so in a different set of coordinates, both should remain reverse because this is physical: the spacetime curvature comes from the Lorentz transformations

 

[relativityfan]

furthermore, this interior Schwarzschild metric has at least one singularity, it would logically mean that there is a black hole inside the sun or even at the center of the earth, i did not know this so far.

 

[George Jones]

i do believe it is true because you can generate the schwarzchild metric just with merely the Lorentz transformations. an object falls with a given speed s, with the right initial conditions, and the gravitationnal field is an acceleration that leads to this speed. from this velocity we deduce the Lorentz factor, and this clearly matches the Schwarzschild metric. so in a different set of coordinates, both should remain reverse because this is physical: the spacetime curvature comes from the Lorentz transformations

furthermore, this interior Schwarzschild metric has at least one singularity, it would logically mean that there is a black hole inside the sun or even at the center of the earth, i did not know this so far.

Did you read the first link that I gave in post #2?

Yes, this solution was found originally by Schwarzschild. It represents a constant density perfect fluid that is spherically symmetric. Derivation of this solution are given in, for example,

General Relativity: An Introduction for Physicists by Hobson, Efstathiou, and Lasenby

Gravitation by Misner, Thorne, and Wheeler.

$r = R$ represents the surface of the spherical body. If

$$R > 2M \frac{9}{8},$$

then the solution has no singularities. At

$$R = 2M \frac{9}{8},$$

the solution develops a pressure singularity at its centre.

This solution is an "almost realistic" toy model for a spherical body.

I think that you are confusing two very different solutions to Einstein's equation:

1) inside the event horizon of the Schwarzschild black hole solution, where there "is a singularity";

2) the non-singular solution inside a spherically symmetric, constant density massive object.

 

[JesseM]

furthermore, this interior Schwarzschild metric has at least one singularity, it would logically mean that there is a black hole inside the sun or even at the center of the earth, i did not know this so far.

The interior solution George Jones was referring to is a different solution that the vacuum one, in this interior solution there is no singularity at the center (nor an event horizon). For a spherically symmetric nonrotating star or planet the spacetime geometry at radii past the surface is the same as the spacetime geometry around a nonrotating Schwarzschild black hole of the same mass, but at radii less than the radius of the surface, the geometry is different.

 

[Dale]

furthermore, this interior Schwarzschild metric has at least one singularity, it would logically mean that there is a black hole inside the sun or even at the center of the earth, i did not know this so far.

This is not correct, there is no singularity in the interior Schwarzschild metric.

EDIT: oh, double-scooped here. I must be getting slow.

 

[relativityfan]

Did you read the first link that I gave in post #2?


I think that you are confusing two very different solutions to Einstein's equation:

1) inside the event horizon of the Schwarzschild black hole solution, where there "is a singularity";

2) the non-singular solution inside a spherically symmetric, constant density massive object.

Yes i did read your post but I misread r=2M with R=2M, sorry. it was just that I do not completely trust that solution because if we describe spacetime curvature as space that is falling, then it would be necessarily because of Lorentz transformations.

 

[Dale]

The Schwarzschild solutions (interior and exterior) are derived from the Einstein field equation, not the Lorentz transform.

 

[pervect]

The metric in the interior of a spherical body with constant density is (for r<R, R being the radial coordinate of the massive body, i.e. R would be its radius if space-time was flat) is:

$$- \left( \frac{3}{2} \left(1 - 2M/R\right) ^{\frac{1}{2}} - \frac{1}{2} \left(1-2Mr^2/R^3\right)^{\frac{1}{2}} \right) ^2 dt^2 + \frac{dr^2}{1-2M/r} + r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2$$

Assuming I typed it correctly - the source is MTW's "Gravitation", pg 608-610. It is in geometric units, so the factors of G and c^2 are not present.

Deriving this isn't a matter of fiddling with the Lorentz transforms, it requires solving Einstein's equations. This would be what would be called the "interior Schwarzschild metric", it was derived originally by Karl Schwarzschild. Outside the massive body, for r>R, one replaces the expression for dt^2 with the exterior solution -(1-2M/r) dt^2

 

[relativityfan]

thank you but one can derive the original schwarzchild metric without the einstein equations, with Lorentz transforms. both give the same result. so i believe other metrics should be consistent too with other Lorentz transforms

 

[bcrowell]

thank you but one can derive the original schwarzchild metric without the einstein equations, with Lorentz transforms. both give the same result. so i believe other metrics should be consistent too with other Lorentz transforms

No, this is incorrect. You may have convinced yourself that you derived the Schwarzschild metric just from the Lorentz transformations, but that doesn't mean your derivation is actually correct. It's not correct, because it's leading you to other conclusions that aren't correct.

If you want to get yourself unstuck mentally from some of the confusion you've gotten yourself into about GR, a couple of good books to start with might be Gardner, Relativity Simply Explained and Taylor and Wheeler, Exploring Black Holes.

 

[Dale]

thank you but one can derive the original schwarzchild metric without the einstein equations, with Lorentz transforms.

I don't think this is correct. The Schwarzschild metric is curved and the Lorentz transform is only for a flat Minkowski metric. I don't see how you can get a curved metric from a flat one.

Can you demonstrate this?

 

[pervect]

thank you but one can derive the original schwarzchild metric without the einstein equations, with Lorentz transforms. both give the same result. so i believe other metrics should be consistent too with other Lorentz transforms

Well, discounting the fact that the metric I gave above was written by Schwarzschild, I think you are mistaken when you say "one can derive the Schwarzschild metric" (assuming you mean the exterior metric) "without the Einstein equations"..

Do you have a reference for this derivation?

 

[Passionflower]

Well, discounting the fact that the metric I gave above was written by Schwarzschild, I think you are mistaken when you say "one can derive the Schwarzschild metric" (assuming you mean the exterior metric) "without the Einstein equations"..

Do you have a reference for this derivation?

Take a look at this:

http://arxiv.org/abs/gr-qc/0309072

 

[PAllen]

Take a look at this:

http://arxiv.org/abs/gr-qc/0309072

Of course, to paraphrase the author, this is not a derivation of the Schwarzschild solution. It heuristic based on Newton and SR that accidentally yields the Schwarzschild solution.

 

[relativityfan]

to write shortly:
-G."M" /r^2 =dv/dt, (-G."M" /r^2) .dr=(dv/dt).dr , dv.dr/dt=v.dv, G."M" /r =v^2/2, from this we get the Lorentz factor gamma, that is a hyperbolic rotation, and we apply it to the flat metric on g_tt and g_rr , to get the Schwarzschild metric.

well i ca

 

[relativityfan]

No, this is incorrect. You may have convinced yourself that you derived the Schwarzschild metric just from the Lorentz transformations, but that doesn't mean your derivation is actually correct. It's not correct, because it's leading you to other conclusions that aren't correct.

If you want to get yourself unstuck mentally from some of the confusion you've gotten yourself into about GR, a couple of good books to start with might be Gardner, Relativity Simply Explained and Taylor and Wheeler, Exploring Black Holes.

thank you but i have read some books too, and even if can do mistakes, general relativity can do mistakes too.

 

[Passionflower]

Of course, to paraphrase the author, this is not a derivation of the Schwarzschild solution. It heuristic based on Newton and SR that accidentally yields the Schwarzschild solution.

I do not think there are accidents in scientific formulas but that is just me.

As relativityfan mentioned the free falling observer from infinity, who travels at escape velocity, will observe that the Schwarzschild radial coordinate $dr$ is equal to $d\rho$ which is the proper distance. His velocity v(r) will Lorentz contract the proper distance to the Schwarzschild radius with respect to a static observer o(r) in such a way that it simply becomes r.

For instance if we take a Schwarzschild radius of 1 and we measure the proper distance to the EH we use the following integrand for a static observer:

$$(1-r^{-1})^{-1/2}$$

The velocity v(r) for an observer free falling from infinity is:

$$v = \sqrt {{r}^{-1}}$$

Hence his Lorentz factor with respect to a static observer is:

$$\gamma = (1-v^2)^{-1/2} = (1-r^{-1})^{-1/2}$$

But this is exactly the integrand to obtain the proper distance for a static observer!

To obtain the length contracted proper distance we have to divide the integrand by gamma, thus the integrand becomes:

$$dr$$

Of course the proper acceleration a(r) for a stationary observer is simply Newton's law times gamma of the escape velocity v(r) of a free falling observer from infinity.

Thus:

$$0.5\,{r}^{-2}(\sqrt {1-{r}^{-1}})^{-1}$$

 

[bcrowell]

-G."M" /r^2 =dv/dt, (-G."M" /r^2) .dr=(dv/dt).dr , dv.dr/dt=v.dv, G."M" /r =v^2/2, from this we get the Lorentz factor gamma, that is a hyperbolic rotation, and we apply it to the flat metric on g_tt and g_rr , to get the Schwarzschild metric.

The last step is the problem. There is no physical principle that says you can derive a metric in GR in this way. It gives the wrong answer in the case of the field inside a uniform sphere. Note that Visser's heuristic method gives a result (11) that is correct, but that violates the condition g_rr=1/g_tt that you claim is universally valid.

 

[PAllen]

I do not think there are accidents in scientific formulas but that is just me.

The author defined what is meant by accident: the procedure is not a valid general procedure, and when applied to other situations, gives the wrong answer. That makes it a 'motivation' not a 'derivation'.

While I can't recall any at the moment, over the years of following physics I recall a number of situations where a heuristic derivation, with logical flaws, yields the correct answer for a simple case. This doesn't make it useless - it remains interesting as a motivation that the result is not unexpected.

 

[Dale]

Interesting. Thanks for posting the heuristic.

relativityfan, any question of the form "how does general relativity work in ..." can be answered using the Einstein field equation. That is the central equation of the theory.

 

[bcrowell]

Another fairly simple way to see that the method given in #20 doesn't really make sense is that for r less than the Schwarzschild radius, it involves a Lorentz transformation with v>c, so you end up with a metric having imaginary components. This is a symptom of the fact that it starts from the motion of a test particle in Galilean relativity, with a Newtonian gravitational force -- but Galilean relativity and Newtonian gravity aren't valid.

 

[relativityfan]

Another fairly simple way to see that the method given in #20 doesn't really make sense is that for r less than the Schwarzschild radius, it involves a Lorentz transformation with v>c, so you end up with a metric having imaginary components. This is a symptom of the fact that it starts from the motion of a test particle in Galilean relativity, with a Newtonian gravitational force -- but Galilean relativity and Newtonian gravity aren't valid.

well we can say in general relativity that "a black hole is a place where space falls faster than light"

http://jila.colorado.edu/~ajsh/insidebh/waterfall.html

furhermore, time and space change role, which is consistent with the imaginary Lorentz factor

 

[relativityfan]

I believe that stating that the energy density is constant is not really compatible mathematically with general relativity(further gravity means further density near the center), and this would explain why g_tt is so different than g_rr.

i do not say that g_tt must always be similar with g_rr, but for a symmetrical static solution i believe this should be the case

 

[pervect]

I would say that g_tt could be anything you like in GR, because one has a complete freedom to choose arbitrary coordinates. Since the numerical value of g_tt simply reflects one's coordinate choice, and the coordinate choice is arbitary, the numerical value of g_tt doesn't have any physical significance at all.