# Gravitational time dilation for two (or more) masses

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What is the gravitational time dilation at a point midway between two equal spherically symmetric masses m?

For one mass, the time dilation is given by the tt component g00 of the Schwarzschild metric, and is equal to sqrt(1-2m/r)dt, where m/r is the potential. For two masses, however, I see two intuitive possibilities. First, you could simply take the sum of the potentials to get sqrt(1-4m/r)dt. Second, you could calculate first the time dilation due to one mass, then apply to that dilated clock rate the time dilation due to the other mass. This would be equivalent to taking the product of the two g00 components, with the result (1-2m/r)dt.

These two values for proper time differ by only higher order terms in m/r. But I would think the difference could be measurable, for example in GPS applications. General Relativity does not give the answer explicitly, but I expect simulations, for instance of black hole mergers, would have to take this into account.

Another interesting question is the time dilation inside a hollow massive sphere. Short of emailing a well-known physicist or former professor, I don't know where to find this information. Yet it could be pertinent in developing a modified gravity theory that incorporates Mach's Principle, which is something I would like to do.

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5 hours ago, Kate rosser said:

What is the gravitational time dilation at a point midway between two equal spherically symmetric masses m?

On a conceptual level, a clock placed at such a Lagrange point will be time-dilated by some amount, as compared to some reference clock far away. However, calculating the exact amount of the time dilation is quite a non-trivial problem, because in GR the metric of such a two-body system is not just a linear combination of two (e.g. Schwarzschild) metrics. In fact, I am not aware of any closed analytical solution to the two-body problem, so this would probably have to be done numerically.

Just by thinking about this scenario, I can tell you that the time dilation will not be numerical constant, but a time-dependent function in whatever coordinate system you choose for this scenario. This is because the only way for this system to be stable is if the bodies revolve around a common center of gravity, which means that there is a non-vanishing quadrupole moment, and hence the system will loose energy via gravitational radiation. Therefore the orbits of these bodies will decay over time, affecting the time dilation factor in some non-trivial manner.

Quote

Another interesting question is the time dilation inside a hollow massive sphere.

This one is easier - there will be time dilation compared to a reference clock at infinity, and the dilation factor will be the same at all points within the hollow cavity. This is because spacetime within the cavity is everywhere flat (Birkhoff’s theorem). The time dilation itself depends only on the total mass of the surrounding shell.

You can picture this like a tabletop mountain - the top of the mountain is everywhere flat, but it sits at a different level than the surrounding countryside, and the region in between (i.e. the slopes of the mountain) is curved. Hence a clock will be dilated if compared to a far-away clock, even though spacetime in the cavity has no curvature - but of course, spacetime between the cavity and the far-away clock is curved.

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You're right that the solution is not just a linear combination of two Schwarzschild metrics. If it were, you would get the absurd result that the time dilation is sqrt(2-4m/r)dt, which is not even Minkowski as r approaches infinity. My two examples of possible methods did not of course involve linear combinations of the metric. One method used the sum of the potentials, the other, the product of the metrics. These two methods at least give the correct Minkowski limit.

My question might have seemed more reasonable had I said I was looking for an approximation to the time dilation, since as you point out, the exact time dilation is analytically intractible in GR. To compute the exact time dilation, you would have to plug into the energy-momentum tensor T a density distribution with masses centered at r and -r, and then solve Einstein's Field Equataions for the metric. (Not in my lifetime!)

My question was also seeking an approximate answer in that I deliberately neglected dynamic effects.. It's hard to imagine a case of two static masses, of course. But in many situations, the dynamics would only affect higher order terms. So the static solution would be fairly accurate. An example would be two galaxies unbound to each other gradually receding due to Hubble expansion. Since I'm working on a cosmological model involving Mach's principle, this type of situation is more what I envision.

The Schwarzschild metric, in any event, should give us a clue to the right approximation, since in two limits the Schwarzschild metric must be obeyed exactly---first in the limit as one mass vanishes, second in the limit as the angle between the two masses goes to zero.

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[I would edit my answer above, but can't see how.]

Maybe I've answered my own question at the start of this topic.

By unbelievable coincidence, I just found out that the analytic solution for two static black holes exists. It's called the double Schwarzschild solution, also known as Bach-Weyl or two-center Israel-Khan solution. I stumbled upon it in the April 2018 issue of Physical Review D at the following link: https://journals.aps.org/prd/pdf/10.1103/PhysRevD.97.084020

Now if I can untangle the math for the metric, I should have the gravitational time dilation for two masses! If anyone's interested, please comment and I'll give more details.

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Interesting (though the article is behind a paywall) - I didn’t think of Bach-Weyl spacetime, since this is normally used to describe a massive continuous ring, rather than two separate objects. I had never heard of the Israel-Khan solution, so thank you for pointing this out; a very interesting metric.

I do not at present have access to any CAM system (I’m just an amateur), and the metric is too complicated to perform the calculation by hand, mostly because there are off-diagonal terms in it, so I can’t immediately answer your original question.

I wonder if one could just treat the influence of the second object (approximately) as background curvature to an otherwise standard Schwarzschild metric, in which case the Schwarzschild-deSitter metric could be used as an approximation around the Lagrange point in question. That would not give the precisely correct numerical value, but hopefully a good approximation, and it would be much easier to do. This would give better results the farther the objects are apart from one another.

Mind... Blown

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Interesting idea about the Schwarzschild deSitter metric. I'll look into it.  Here's a quote from the article (but I can't reproduce the equations):

"The double Schwarzschild BH is a static Weyl solution with axial symmetry, featuring two nonrotating, neutral BHs supported in equilibrium by a conical singularity which can be chosen to take the form of either two strings or one strut (see e.g. [32]). The metric can be reduced to the form ds2 ¼ −e2Udt2 þ e−2Uðe2K½dρ2 þ dz2 þ ρ2dφ2Þ; where t, φ are connected respectively to staticity and axial symmetry."

I haven't worked out g00 for this metric yet, but on the face of it, it looks like the time dilation midway between two equal masses is zero! If so, this goes completely against my original intuition.

I parked some screen shots of the equations for the metric at this link (Can't post whole article, APS might get mad at me!):

Sorry this is posted on my rudimentary website. Couldn't think of another way to do it. (I have access to Physical Review D because I bought an individual subscription for \$240.)

No kidding ...

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10 hours ago, Markus Hanke said:

Interesting (though the article is behind a paywall) -

It might be this one: https://arxiv.org/abs/1802.02675

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Yes! Great. You found it. This arxiv version looks just about the same as the one at Physical Review D.

See Section 2 for the equations for the double black hole solution.

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On 6/3/2018 at 4:32 AM, Markus Hanke said:

This one is easier - there will be time dilation compared to a reference clock at infinity, and the dilation factor will be the same at all points within the hollow cavity. This is because spacetime within the cavity is everywhere flat (Birkhoff’s theorem). The time dilation itself depends only on the total mass of the surrounding shell.

You can picture this like a tabletop mountain - the top of the mountain is everywhere flat, but it sits at a different level than the surrounding countryside, and the region in between (i.e. the slopes of the mountain) is curved. Hence a clock will be dilated if compared to a far-away clock, even though spacetime in the cavity has no curvature - but of course, spacetime between the cavity and the far-away clock is curved.

Not depend on the radius (or inner and outer radii) of the sphere also? I am picturing two extremes, two spheres of same mass, one very large and one very small but equal mass. Would the time dilation inside the smaller one not be greater?

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20 hours ago, J.C.MacSwell said:

Not depend on the radius (or inner and outer radii) of the sphere also? I am picturing two extremes, two spheres of same mass, one very large and one very small but equal mass. Would the time dilation inside the smaller one not be greater?

You are correct, it would depend on the radius also. Sorry for not explicitly mentioning that!

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7 hours ago, Markus Hanke said:

You are correct, it would depend on the radius also. Sorry for not explicitly mentioning that!

Thanks

On 6/4/2018 at 1:36 PM, Kate rosser said:

Interesting idea about the Schwarzschild deSitter metric. I'll look into it.  Here's a quote from the article (but I can't reproduce the equations):

"The double Schwarzschild BH is a static Weyl solution with axial symmetry, featuring two nonrotating, neutral BHs supported in equilibrium by a conical singularity which can be chosen to take the form of either two strings or one strut (see e.g. [32]). The metric can be reduced to the form ds2 ¼ −e2Udt2 þ e−2Uðe2K½dρ2 þ dz2 þ ρ2dφ2Þ; where t, φ are connected respectively to staticity and axial symmetry."

I haven't worked out g00 for this metric yet, but on the face of it, it looks like the time dilation midway between two equal masses is zero! If so, this goes completely against my original intuition.

Is it zero also infinitely far away? I would expect it to be above zero at the midway point in that case, in somewhat of a gravity well by comparison.

(not doing any math as you might guess)

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3 hours ago, J.C.MacSwell said:

Is it zero also infinitely far away? I would expect it to be above zero at the midway point in that case, in somewhat of a gravity well by comparison.

Yes, it's zero infinitely far away, and also zero between the two masses, which seems counterintuitive considering the gravity well concept. (By time dilation being zero I mean g00=1).

I did some math on the double Schwarzschild metric, and it has unexpected properties, so maybe it's not valid for answering the question of time dilation between two masses. For example, in the limit as one mass goes to zero, the double S metric approaches g00=1-2m/r for large r, but not for small r. So the double Schwarzschild metric does not approach the Schwarzschild metric in the limit of a single mass. This raises the question of just what IS the double Schwarschild metric. Maybe the singular strut that connect the two masses throws the solution off.

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22 minutes ago, Kate rosser said:

Yes, it's zero infinitely far away, and also zero between the two masses, which seems counterintuitive considering the gravity well concept. (By time dilation being zero I mean g00=1).

How do you get g00=1 between the two masses? There’s a massive rod connecting the two black holes, so on account of this alone one cannot have zero time dilation in or near this region.

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On 09/06/2018 at 11:13 AM, Markus Hanke said:

How do you get g00=1 between the two masses? There’s a massive rod connecting the two black holes, so on account of this alone one cannot have zero time dilation in or near this region.

Good question. I don't understand this double black hole metric very well. How I got g00=1 was by setting the masses equal and doing the algebra for g00 at the origin. (Of course, I may have made a mistake.) For the equations, see Section 2 of the arXiv article at Strange's link above. The authors don't show the singular strut as having mass, nor do they explain what what it really is.

BTW, I looked at your website. Looks like you've done a lot of good work. Impressed

On 03/06/2018 at 12:32 AM, Markus Hanke said:

This one is easier - there will be time dilation compared to a reference clock at infinity, and the dilation factor will be the same at all points within the hollow cavity. This is because spacetime within the cavity is everywhere flat

About time dilation inside a sphere: I managed to solve Einstein's Field Equations for the general static spherical metric with non-zero mass density.  To my surprise (I was expecting something more like your answer), the metric for a hollow shell is exactly Schwarzschild outside the shell, and exactly Minkowski inside the shell. So g00=g11=1 inside the shell. Not only is there no time dilation in the interior of a hollow massive sphere, the potential itself is identically zero, just as it is at infinity. (Unless I made a mistake.) I'm surprised because I agree with the logic of J.C. MacSwell above that the sphere would be deeper in a gravitational well.

I will see about writing an article for viXra to get feedback on my solution, which was exact. The solution was elegant and easy, which also surprised me.

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14 minutes ago, Kate rosser said:

Not only is there no time dilation in the interior of a hollow massive sphere, the potential itself is identically zero, just as it is at infinity. (Unless I made a mistake.) I'm surprised because I agree with the logic of J.C. MacSwell above that the sphere would be deeper in a gravitational well.

My own impression would be that inside the sphere, you are under the influence of two parts of the sphere. If you are not the perfect centre, then what's above you is less massive than what's below you. So you are effectively in two gravity wells, which work against each other. In that case, you could subtract one from the other, to work out the effective gravity well, and calculate the time dilation from that.

At the perfect centre, both would be equal, and the effective gravity well at that point would be zero. So time dilation is zero. But as soon as you move in any direction, your time dilation starts to increase.

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Posted (edited)
18 minutes ago, mistermack said:

At the perfect centre, both would be equal, and the effective gravity well at that point would be zero. So time dilation is zero. But as soon as you move in any direction, your time dilation starts to increase.

Your intuition seems logical. I'm baffled by this too. But my solution to Einstein's Field Equations (if correct) is exact, and it shows everything zero everywhere inside the shell. (g00=-g11=1 for r<R).

One caveat however: To get g00=1, you have to assume the Equation of State that density = pressure (the most common GR Equation of State, and one which is consistent with the Schwarzschild metric). But the question is: Is this EoS valid for a rigid sphere held together say by eletromagnetic, strong and Pauli forces? That's a deeper problem. If the sphere were made of dust, it would not be stable.

Now I need confirmation that my solution is correct. So I plan to post an article on viXra to get feedback. (I'll put a link here, but it may take a few weeks.) The solution is elegant and simple, but there's plenty of room for mistakes with this much algebra.

Edited by Kate rosser
Left out a minus sign

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The only explanation I can think of for everything being zero inside the shell is along the lines of the river model, where gravity is modelled as a flow of inertial frames into the massive object. Outside of the sphere, there is a whole universe of space time, so that a continuous "river" can flow, constantly being replaced from infinity. Inside the sphere, there is no source of replacement frames, so there can be no equivalent "river" of frames into the material of the sphere. No flow means no gravity.

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41 minutes ago, Kate rosser said:

Good question. I don't understand this double black hole metric very well. How I got g00=1 was by setting the masses equal and doing the algebra for g00 at the origin. (Of course, I may have made a mistake.) For the equations, see Section 2 of the arXiv article at Strange's link above. The authors don't show the singular strut as having mass, nor do they explain what what it really is.

BTW, I looked at your website. Looks like you've done a lot of good work. Impressed

About time dilation inside a sphere: I managed to solve Einstein's Field Equations for the general static spherical metric with non-zero mass density.  To my surprise (I was expecting something more like your answer), the metric for a hollow shell is exactly Schwarzschild outside the shell, and exactly Minkowski inside the shell. So g00=g11=1 inside the shell. Not only is there no time dilation in the interior of a hollow massive sphere, the potential itself is identically zero, just as it is at infinity. (Unless I made a mistake.) I'm surprised because I agree with the logic of J.C. MacSwell above that the sphere would be deeper in a gravitational well.

I will see about writing an article for viXra to get feedback on my solution, which was exact. The solution was elegant and easy, which also surprised me.

Assuming a shell of zero thickness, then the gravitational potential everywhere inside of the Shell is the same as that at the exterior surface of the shell.  This means that the time dilation factor everywhere inside of the shell will be the same as that on the surface of the shell. And unless the shell is massless, this will not be 0.

As the thickness of the shell increases (while assuming a constant mass), the potential inside of the hollow becomes lower than that at the surface, and the time dilation factor increases.

For a solid sphere (of uniform density) the gravitational potential is -GM((3R^2-r^2)/R^3) where R is the radius of the sphere and r is the distance from the center of the sphere.  At the exact center this reduces to -GM(3/(2R)).    The potential at the surface is -GM/R.

The lower gravitational potential at the center results in a greater time dilation factor than at the surface even though the net gravitational force is zero.

35 minutes ago, mistermack said:

My own impression would be that inside the sphere, you are under the influence of two parts of the sphere. If you are not the perfect centre, then what's above you is less massive than what's below you. So you are effectively in two gravity wells, which work against each other. In that case, you could subtract one from the other, to work out the effective gravity well, and calculate the time dilation from that.

At the perfect centre, both would be equal, and the effective gravity well at that point would be zero. So time dilation is zero. But as soon as you move in any direction, your time dilation starts to increase.

Time dilation is related to the difference in gravitational potential, Or put another way, the amount of energy per unit mass required to lift something from one point to another against gravity.   It obviously would take energy to lift a mass from the center the Earth to the surface, thus the center of the Earth is at a lower gravitational potential.  As a result, Time dilation at the center would be greater that at the surface and clocks at the center would run slower than those at the surface.

One way of looking at this is by considering the Gravitational time dilation equation: T = t/ sqrt(1-2GM/(rc^2))

One caveat for using this equation is that M is entirely contained within r.    Under these same conditions, escape velocity is found by sqrt(2GM/r).  Note that if you replace this for v in the SR time dilation equation for relative motion, you reproduce the above gravitational time dilation equation.

Now the escape velocity equation is derived by setting the sum of the kinetic energy and the gravitational energy a object to zero like this:

mv^2/2-GMm/r =0

And solving for v.

In the same way, we can find the escape velocity from the center of a uniformly dense sphere using the gravitational potential at the center at I gave above.

m^2/2- 3Gm/(2R)=0

which solves to v= sqrt(3GM/R)

Substituting this for v in the SR time dilation equation gives

T= t/sqrt(1-3GM/(Rc^2)

Where R is the radius of the sphere.

The time dilation at the surface is given by making r=R in the first gravitational time dilation equation I gave.

This gives a greater time dilation factor at the center than at the surface of the sphere.

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1 hour ago, Janus said:

Assuming a shell of zero thickness, then the gravitational potential everywhere inside of the Shell is the same as that at the exterior surface of the shell.  This means that the time dilation factor everywhere inside of the shell will be the same as that on the surface of the shell. And unless the shell is massless, this will not be 0.

I modeled the shell as a Dirac delta function del(r-R). This is equivalent to decreasing the thickness while increasing the density in such a way that the mass m stays constant.

1 hour ago, Janus said:

One way of looking at this is by considering the Gravitational time dilation equation: T = t/ sqrt(1-2GM/(rc^2))

For the Schwarzschild metric, I'm assuming the gravitational time dilation for a stationary clock is ds=sqrt(1-2m/r)dt=g00dt, where s is proper time TAU. So an interval ds of proper time goes to zero at r-2m. In other words, clocks stop at the event horizon.

1 hour ago, Janus said:

Time dilation is related to the difference in gravitational potential,

Yes, so if one is talking about the tt component of the metric in the form g00=(1+2PHI/c**2), then the potential PHI determines the time dilation. (Assuming dr=dOMEGA=0, ie no velocity.)

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2 hours ago, Janus said:

Assuming a shell of zero thickness, then the gravitational potential everywhere inside of the Shell is the same as that at the exterior surface of the shell.  This means that the time dilation factor everywhere inside of the shell will be the same as that on the surface of the shell. And unless the shell is massless, this will not be 0.

I don't think this is correct. Obviously this is an idealized case, but as the thickness approaches zero, the gravitational gradient across the thickness would tend to infinite. Assuming that is correct, the inside and outside of the surface cannot be taken as equivalent unless the shell is massless.

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4 hours ago, J.C.MacSwell said:

I don't think this is correct. Obviously this is an idealized case, but as the thickness approaches zero, the gravitational gradient across the thickness would tend to infinite. Assuming that is correct, the inside and outside of the surface cannot be taken as equivalent unless the shell is massless.

Edit : Rethinking this, the gravitational force is still a finite value on the surface, so that statement of Janus's should be correct.

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14 hours ago, Kate rosser said:

To my surprise (I was expecting something more like your answer), the metric for a hollow shell is exactly Schwarzschild outside the shell, and exactly Minkowski inside the shell.

Yes, this is exactly what I said in my answer, and it is just a result of Birkhoff’s theorem. The local patch inside the shell is Minkowski, as is the asymptotic behaviour of the exterior metric at infinity. However, the global metric that spans the entire spacetime - shell cavity, shell, and exterior vacuum - shows that the locally flat patch inside the cavity still sits at a different potential than the asymptotically flat limit at infinity. Mathematically, if you look at the global situation, all components of the metric tensor are constants at asymptotic infinity, and they are also all constants in the interior cavity of the shell, but the two constants aren’t the same ones. Therefore, in this spacetime you have two regions that are flat (i.e. the metric tensor is isomorphic to Minkowski in those regions), but they nonetheless sit a different gravitational potential. Again, a table mountain - the top of which is as flat as the surrounding countryside, but nonetheless at a different level - is a good analogy for this.

To fully understand how a clock at infinity compares to a clock in the cavity, one must look at the global metric that spans the entire spacetime, not just one local patch, because there is a local gauge freedom in what value the constants in the metric tensor can take. The constants are fixed by the boundary conditions at the shell, as well as by the asymptotic behaviour at infinity. If, at infinity, we choose the metric to be diag{-1,1,1,1), then inside the hollow cavity it will be diag{-a,1,1,1}, with some constant a that is determined by boundary conditions. That is where the global time dilation comes from. Alternatively - as you did - you can reverse this and say that in the interior you want to have diag{-1,1,1,1}, which is fine, but then you will get a different metric constant at infinity. Both metrics are still isomorphic to Minkowski (so they are flat patches of spacetime) - but they are nonetheless different.

14 hours ago, Kate rosser said:

So I plan to post an article on viXra to get feedback.

My advice to you - steer well clear of that site. It’s a repository for cranks and crackpots, with no scientific credibility whatsoever. If you want confirmation of your results, you are better off putting them to the experts on www.physicsforums.com, they will give you instant feedback.

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9 hours ago, Markus Hanke said:

Yes, this is exactly what I said in my answer, and it is just a result of Birkhoff’s theorem. The local patch inside the shell is Minkowski, as is the asymptotic behaviour of the exterior metric at infinity. However, the global metric that spans the entire spacetime - shell cavity, shell, and exterior vacuum - shows that the locally flat patch inside the cavity still sits at a different potential than the asymptotically flat limit at infinity. Mathematically, if you look at the global situation, all components of the metric tensor are constants at asymptotic infinity, and they are also all constants in the interior cavity of the shell, but the two constants aren’t the same ones. Therefore, in this spacetime you have two regions that are flat (i.e. the metric tensor is isomorphic to Minkowski in those regions), but they nonetheless sit a different gravitational potential. Again, a table mountain - the top of which is as flat as the surrounding countryside, but nonetheless at a different level - is a good analogy for this.

To fully understand how a clock at infinity compares to a clock in the cavity, one must look at the global metric that spans the entire spacetime, not just one local patch, because there is a local gauge freedom in what value the constants in the metric tensor can take. The constants are fixed by the boundary conditions at the shell, as well as by the asymptotic behaviour at infinity. If, at infinity, we choose the metric to be diag{-1,1,1,1), then inside the hollow cavity it will be diag{-a,1,1,1}, with some constant a that is determined by boundary conditions. That is where the global time dilation comes from. Alternatively - as you did - you can reverse this and say that in the interior you want to have diag{-1,1,1,1}, which is fine, but then you will get a different metric constant at infinity. Both metrics are still isomorphic to Minkowski (so they are flat patches of spacetime) - but they are nonetheless different.

My results were almost but not exactly the same as yours. I did solve EFE for the global metric that spans the entire spacetime. I understand why you say the metric should be {-1,1,1,1} at infinity and {-a,1,1,1} inside the hollow shell. I considered this situation when I found my solution. It does depened on the constants of integration,  so we follow each other up to that point (which in my experience is a pretty detailed understanding as communication goes). However, when I evaluated the constants of integration, I found the metric to be Minkowski inside the shell, with g00=1, g11=-1 (or in your notation {-1,1,1,1} for rectangular coordinates). If I set the constant of integration differently, I would not have gotten a Schwarzschild metric outside the shell. (I'll check again.) Have you calculated the solution yourself?

So my result has the potential inside the shell as identically zero, as it is at infinity, with no time dilation. I get g00=1-(2m/r)S(R;r), where S(R;r) is a step function S=0 for r<R, S=1 for r>R.

The only way to iron out these mathematical details is to show the entire derivation. Which leads to my next comment:'

10 hours ago, Markus Hanke said:

My advice to you - steer well clear of that site. It’s a repository for cranks and crackpots, with no scientific credibility whatsoever. If you want confirmation of your results, you are better off putting them to the experts on www.physicsforums.com, they will give you instant feedback.

ViXra, like Twitter or the telephone, has no good or bad status in and of itself. It's just a communication vehicle for anyone to publish anything they want, instantly, in beautiful PDF format, at no charge, that is free to read. On viXra, I could show you my full derivation in perfect rigorous notation, and present it as a formal paper.

Cranks? Crackpots? You should read Physical Review D! The wildest ideas of all fly there!

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22 minutes ago, Kate rosser said:

Have you calculated the solution yourself?

As a matter of fact, yes, I have (though it was years ago) - and my result was precisely the one I had described above, which also corresponds to the scientific consensus on this matter. It certainly did not involve any step functions though, I don’t know how you got that. To me that’s a red flag right there, because in general, step functions are not smooth and continuous (and hence not differentiable) at the points where the steps occur.

24 minutes ago, Kate rosser said:

So my result has the potential inside the shell as identically zero, as it is at infinity, with no time dilation.

Then you must have done something wrong, because your result would mean that the energy-momentum of the massive shell has a gravitational effect in one radial direction, but not the other, as seen from the shell. That is obviously unphysical. I’m also pretty sure (but would need to check in more detail) that it would be mathematically inconsistent, since it can’t be asymptotically flat at both infinity and some other point that is not at infinity, while remaining smooth and continuous everywhere in between, in a spacetime that is not globally empty.

Surely you wouldn’t seriously expect to have a mass-energy distribution without any gravitational effect in some region just outside it, would you?

43 minutes ago, Kate rosser said:

If I set the constant of integration differently, I would not have gotten a Schwarzschild metric outside the shell.

Again, this would mean you did something wrong, because 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, due to Birkhoff’s theorem. If you don’t get this, then there is an error in your maths somewhere.

I would say check your boundary conditions at the shell - from what I remember (again, this was years ago, and I don’t know where my notes from that time are gone), these were quite tricky to get right, and I got stuck at that point too. Do remember that the overall global metric has to be both smooth and continuous everywhere, including at the two boundaries, as well as the non-vacuum interior of the thin shell itself. In your case it looks like you have no gravity whatsoever in the interior cavity, and then suddenly gravity in the non-vacuum shell itself, meaning there is a discontinuity in your solution at the interior boundary of the shell. Perhaps that is why you ended up with a step function...?

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