Gravitational time dilation for two (or more) masses

[Kate rosser]

6 hours ago, Markus Hanke said:

if you have a region of flat spacetime that is surrounded by a thin massive shell, then you are guaranteed to have a Schwarzschild spacetime in the exterior,

I think my method is right. I found an exact solution to EFE. No solutions were patched together, and there were no boundary conditions, just constants of integration.

The Schwarzschild metric did indeed appear automatically. However its dependence on m, the mass of the shell, fixed one of the constants of integration (like Newtonian gravity fixes the constant of integration m in the Schwarzschild solution).

My method was as follows:

I used Einstein's Field Equations for the static spherical nonvacuum. This takes the form of two simple differential equations;

kappa*rho(r) = 1/g11*r**2 + 1/r**2 - g11'/r*g11**2

kappa*p(r) = 1/g11*r**2 + 1/r**2 + g00'/r*g11*g00

where rho(r) is a spherically symmetric density distribution, and p(r) is the pressure, primes denote differentiation with respect to r, and kappa is -8piG/c**2.  I generalized to the nonvacuum case Dirac's method for the Schwarzschild solution (see General Theory of Relativity, by P.A.M. Dirac, Wiley 1975, p31). Assuming the Equation of State rho=p, at this stage of the calculation, this produces a g00 that looks exactly like the Schwarzschild-de Sitter solution, except that rho(r) is not constant, but appears inside an integral.

Next, I modeled the density function rho(r) for a hollow sphere of radius R as a Dirac delta function: rho(r)=mu*del(r-R), where mu is the mass area density of the shell. This model is an idealization, but the discontinuity can be smoothed out by broadening the delta function in such a way that the mass of the shell remains constant. Taking the integral over rho(r)*r**2, one obtains a step function, which again can be smoothed out by broadening the delta function. Smoothing the delta function would not change the overall result.

The only caveat, as I said in a previous comment, is the Equation of State.

Your method of patching solutions together using boundary conditions is definitely wrong in General Relativity. It is sometimes done, for example in J.A. Wheeler's black hole lattic metric, but it is always an approximation. What I found was an exact solution to EFE.

Do you have any references when you speak of the concensus? If so, that would have been the answer I was seeking when I first entered this forum.

 

[Mordred]

 

4 hours ago, Kate rosser said:

I think my method is right. I found an exact solution to EFE. No solutions were patched together, and there were no boundary conditions, just constants of integration.

The Schwarzschild metric did indeed appear automatically. However its dependence on m, the mass of the shell, fixed one of the constants of integration (like Newtonian gravity fixes the constant of integration m in the Schwarzschild solution).

My method was as follows:

I used Einstein's Field Equations for the static spherical nonvacuum. This takes the form of two simple differential equations;

Non Vacuum? the Schwarzchild metric is a spherically symmetric vacuum solution which must satisfy [latex] R_{\alpha\beta}=0.[/latex] from this you calculate the [latex]A_r =(1-\frac{1-2GM}{c^2r})^{-1}[/latex] and [latex] B_r =-c^2(1-\frac{2Gm}{c^2r})[/latex] both [latex]T_{\alpha\beta}, R_{\alpha\beta}=0[/latex]

The static requirement means for other readers as well that all metric components are independent on the time coordinate. Which I can't tell if your maintaining from what's above so I will assume you are....however for completeness [latex]ds^2=g_{11}dr^2+g_{22}d\theta^2+g_{33}\phi^2+g_{44}dt^2[/latex] which the four components are independent on t

I don't have a copy of that book however I can't see Dirac not employing a vacuum solution. Is it safe to assume you meant vacuum in the quoted section ?

PS have you done latex on other sites if so see here for the latex structure for this site.

https://www.scienceforums.net/topic/108127-typesetting-equations-with-latex-updated/

anyways the density and pressure outside the sphere must be zero for a vacuum solution outside the Schwartzchild radius.

here for the formulism above as I don't have Diracs book to see his methodology

http://physics.gmu.edu/~joe/PHYS428/Topic10.pdf

in terms of what this means on pressure and energy terms.

[latex]\rho=P=0[/latex] for [latex] r> R[/latex] where the stress energy momentum tensor is given by [latex]T_{\mu\nu}=(\rho+p)v_\mu v_\nu)-pg_{\mu\nu}[/latex]

 

 

 

Edited June 13, 2018 by Mordred

[Markus Hanke]

4 hours ago, Kate rosser said:

Your method of patching solutions together using boundary conditions is definitely wrong in General Relativity

The boundary conditions in this case are simply that the spacetime must be asymptotically flat at infinity (i.e. reduces to Newtonian gravity), and smooth and continuous everywhere else, including at the inner and outer boundaries of the shell. That is what it means to have a global spacetime manifold. This continuity condition is crucial - if you have points where spacetime is not continuous and differentiable, then the field equations do not apply there, and the whole thing becomes internally inconsistent. When you account for the continuity condition, the metric constants become fixed automatically.

For an example of how boundary conditions are used in GR to match solutions and ensure continuity, see §23 of Misner/Thorne/Wheeler, which deals with the interior Schwarzschild metric. This way is how I learned to do it, and I don’t see how it could possibly be “wrong”, so long as the result matches the situation in the Newtonian limit, and reproduces the Birkhoff theorem outcome, which it does.

Quote

Do you have any references when you speak of the concensus? If so, that would have been the answer I was seeking when I first entered this forum.

My reference would be the well-understood Newtonian limit. We already know that in the Newtonian case, the gravitational potential in the interior cavity is not the same as the one at infinity. The GR solution must reproduce this boundary case, since the g{tt} component of the metric tensor is directly related to gravitational potential in the Newtonian limit. If time dilation was zero in the cavity, as compared to a reference clock at infinity, you would end up with a contradiction in the Newtonian limit. Therefore, globally speaking, g{tt} cannot be the same inside the cavity and at infinity.

For a more direct confirmation of what the consensus on this is, physicsforums.com would be a good place to ask, since that is where the actual experts in the field go. The general argument though is along these lines: https://www.quora.com/Does-a-hollow-sphere-of-mass-still-cause-GR-time-dilation-inside-it-even-though-there-is-no-net-gravitational-field.

 

[Mordred]

Those boundary conditions are also mentioned here

https://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation

Quote

If the equation is used to model a bounded sphere of material in a vacuum, the zero-pressure condition 

[latex]P=0[/latex] and the condition  [latex]e^\nu=(1-\frac{2GM}{rc^2})[/latex]should be imposed at the boundary. The second boundary condition is imposed so that the metric at the boundary is continuous with the unique static spherically symmetric solution to the vacuum field equations, the Schwarzschild metric:

and Markus is absolutely correct on the requirement of smooth and continuous on both the inner and outer regions but also the transiton between the two regions. The solution should also have no maxima on the inner region.

Edited June 13, 2018 by Mordred

[Kate rosser]

10 hours ago, Mordred said:

Non Vacuum? the Schwarzchild metric is a spherically symmetric vacuum solution ... I don't have a copy of that book however I can't see Dirac not employing a vacuum solution. Is it safe to assume you meant vacuum in the quoted section ?Tμν=(ρ+p)vμvν)−p

No, I meant nonvacuum. Let me try to straighten out this confusion. I derived an exact solution to EFE for the static  (yes, as you defined it) spherical NON-vacuum, with a non-zero mass density distribution rho(r).  The presence of the massive shell of course requires a nonvacuum solution. However, outside the shell, there is a vacuum, and so outside the shell, the total solution looks exactly like the Schwarzschild solution.

To solve the resulting pair of simultaneous differential equations, I used a generalization of the method Dirac used for finding the Schwarzschild solution. The latter is of course a vacuum solution. I generalized Dirac's method to the non-vacuum. The reason I mention Dirac's method is, it is so elegant it is easily generalized. Once he's calculated all the Christoffel symbols and curvature tensors, he derives the Schwarzschild solution in 4 easy steps.

Other than that one point of confusion, everything you said looks perfectly rigorous. Thanks for responding.I highly recommend Dirac's 69 page texbook on GR, which makes it easy to learn to solve Einstein's Field Equations from scratch. You can probably find a used copy.

I will check out your link too.

 

7 hours ago, Markus Hanke said:

The boundary conditions in this case are simply that the spacetime must be asymptotically flat at infinity (i.e. reduces to Newtonian gravity), and smooth and continuous everywhere else, including at the inner and outer boundaries of the shell. That is what it means to have a global spacetime manifold. This continuity condition is crucial - if you have points where spacetime is not continuous and differentiable, then the field equations do not apply there, and the whole thing becomes internally inconsistent. When you account for the continuity condition, the metric constants become fixed automatically.

For an example of how boundary conditions are used in GR to match solutions and ensure continuity, see §23 of Misner/Thorne/Wheeler, which deals with the interior Schwarzschild metric. This way is how I learned to do it, and I don’t see how it could possibly be “wrong”, so long as the result matches the situation in the Newtonian limit, and reproduces the Birkhoff theorem outcome, which it does.

Your point about the continuity of the manifold is well taken. This topic currently being debated at the highest levels of physics. Some theorists freely manipulate singularities in GR, eg for black holes, the big bang, etc. Others argue as you do, that GR breaks down at these singularities. I see useful points on both sides. First, singularities are indeed probably non-physical. So if we can find a cosmological theory, for example, that avoids the big bang singularity, we are likely better off. On the other hand, singularities are an invaluable mathematical technique, and do give some insight.

How this applies to what I have done, I used a spherical Dirac delta function to model the massive sphere. This is an idealization that involves a singularity. Some physicist would argue as you do that GR thus breaks down. However, I would point out that the general solution is well nigh intractible if I had used something else, say a Gaussian, to model the shell. So I took the liberty, as many do, to use a delta function. Now, at the practical level, the delta function can be spread out over some region of r to achieve continuity. This would not affect the solution outside that region.

About authors freely using singularities, note the double Schwarzschild black hole solution I referenced above. The authors used a singular strut to keep the black holes rigid. I agree with you that this singularity can hardly be physical. Yet the method produces interesting results. And ... it was published in Physical Review D. The latter is a very open-minded journal. Which is why I like it.

About matching Newtonian results, yes, this is important for calculating forces etc. The problem here is, the original question was about time dilation, for which Newtonian gravity has no answer. So we're on our own.

[Markus Hanke]

2 hours ago, Kate rosser said:

This topic currently being debated at the highest levels of physics.

To the best of my knowledge, the time dilation in the interior of a shell cavity wrt some clock at infinity is already a well established fact.

2 hours ago, Kate rosser said:

Others argue as you do, that GR breaks down at these singularities

I never mentioned anything about singularities, nor do I claim that GR breaks down anywhere in this type of spacetime. Obviously it doesn’t, since it is globally geodesically complete. 

2 hours ago, Kate rosser said:

The problem here is, the original question was about time dilation, for which Newtonian gravity has no answer.

Time dilation is directly related to differences in gravitational potential, so the Newtonian limit both in the cavity and at infinity are very relevant here. In fact, it is all we need to know - if there is a difference in gravitational potential between these regions, then there will necessarily be time dilation. And we know that there is a difference, even in the weak field Newtonian limit. This difference will increase, the more massive we make the shell.

Edited June 13, 2018 by Markus Hanke

[Janus]

48 minutes ago, Markus Hanke said:

To the best of my knowledge, the time dilation in the interior of a shell cavity wrt some clock at infinity is already a well established fact.

I never mentioned anything about singularities, nor do I claim that GR breaks down anywhere in this type of spacetime. Obviously it doesn’t, since it is globally geodesically complete. 

Time dilation is directly related to differences in gravitational potential, so the Newtonian limit both in the cavity and at infinity are very relevant here. In fact, it is all we need to know - if there is a difference in gravitational potential between these regions, then there will necessarily be time dilation. And we know that there is a difference, even in the weak field Newtonian limit. This difference will increase, the more massive we make the shell.

To illustrate this point we can graph the gravitational potential of a the hollow sphere like this.

 

0 potential is at the top of the graph and potential decreases in the direction of the arrow.  The slope of the line would be the acceleration due to gravity at that point. The red dotted line marks off the boundary of the sphere, with the interior to the left and exterior to the right.  As we move outward from the surface of the shell, both the acceleration due to gravity and the gravitational potential tend towards zero.  Inside the shell,  acceleration due to gravity is zero, but the potential remains the same as that as at the exterior surface of the shell.  And, as pointed out, since time time dilation is tied to difference in gravitational potential, the time dilation inside the sphere will differ from that at a infinite distance from the Sphere.

In order for time dilation to be the same inside the shell as at an infinite distance, the potential would have to look like this.

But this would mean that in order to move from the exterior to the interior of the sphere, you would have to exert energy, the same amount of energy as it would take to lift the same mass from the surface of the sphere to an infinite distance away.

This problem becomes even more apparent if you consider a non-zero thickness for the shell.  Now as you move from exterior surface to interior surface, you have a decrease in potential and a gradual fall off of acceleration due to gravity as you move through the shell material.  But to move across the boundary between shell material and interior of the shell would require jumping from one potential to a higher one.  This makes no sense.  Instead what happens is that as you move through the shell the potential converges towards the potential in the interior of the shell, and the potential inside of the shell is lower than at the exterior surface.

 

[Mordred]

10 hours ago, Kate rosser said:

hNo, I meant nonvacuum. Let me try to straighten out this confusion. I derived an exact solution to EFE for the static  (yes, as you defined it) spherical NON-vacuum, with a non-zero mass density distribution rho(r).  The presence of the massive shell of course requires a nonvacuum solution. However, outside the shell, there is a vacuum, and so outside the shell, the total solution looks exactly like the Schwarzschild solution.

To solve the resulting pair of simultaneous differential equations, I used a generalization of the method Dirac used for finding the Schwarzschild solution. The latter is of course a vacuum solution. I generalized Dirac's method to the non-vacuum. The reason I mention Dirac's method is, it is so elegant it is easily generalized. Once he's calculated all the Christoffel symbols and curvature tensors, he derives the Schwarzschild solution in 4 easy steps.

 

Are not those Christoffels calculated with respect to r and not R ? The Schwartzchild is not a combination of vacuum and non vacuum it is strictly a vacuum solution specifically describing outside a BH or star not the interior. It provides the Schwartchild radius but the metric components become singular at this radius. Ie to an outside observer one will never see a particle cross the EH. Even though a particle will pass the event horizon under the Schwartchild the outside observer will never be able to view such a crossing.

There is a reason why the Schwartzchild metric is invalid inside the sphere itself and you need a different coordinate system. Try Kruskal coordinates, the problem with the Scwartzchild metric is that it has a coordinate singularity at r=2GM/r as well as r=0. the r=0 is the true singularity which no coordinate change can rectify but the coordinate singularity condition can be removed by the appropriate coordinate system change.

Here Sean Carroll has a decent article on this in essence the entire spacetime is represented by 4 regions under Kruskal coordinates.

https://www.preposterousuniverse.com/wp-content/uploads/grnotes-seven.pdf

I was able to find a copy of Dirac on ebooks here he himself specifies the Schwartzchild is only valid for regions outside the EH.

Quote

It is known as the Schwarzschild solution. It holds outside the surface of the body that is producing the field , where there is no matter. Thus it holds fairlyaccurately outside the surface of a star

Edited June 14, 2018 by Mordred

[Mordred]

There is solutions for the interior of 2GM however they require changes of coordinate basis Carroll gives a few examples in the link above but another is to use

Painleve-Gullstrand coordinates see Blackholes II of

http://www.blau.itp.unibe.ch/newlecturesGR.pdf

page 564 however there is a huge section previous describing the problems with Schwartzchild and its coordinate singularity in the previous sections. They amount to the same as the Carroll paper however...

here is another decent article using the latter coordinates in a simpler format

https://arxiv.org/pdf/1008.0470.pdf

just aa side note I seriously doubt a spherical symmetric condition can be preserved for a binary BH system. The two EH's should be elongated towards each other but under approximation for simplicity sake that's ignorable for now

 

 

Edited June 14, 2018 by Mordred

[Markus Hanke]

11 hours ago, Janus said:

Inside the shell,  acceleration due to gravity is zero, but the potential remains the same as that as at the exterior surface of the shell.  And, as pointed out, since time time dilation is tied to difference in gravitational potential, the time dilation inside the sphere will differ from that at a infinite distance from the Sphere.

Precisely. Thanks for your efforts in putting together that excellent post.

[Mordred]

agreed but there is also the problem of role changes for the light cones inside the EH. ie time becomes spacelike and space timelike. Details on that in the links provided.