# Special relativity treatment of gravity?

• Peeter
In summary, the conversation revolves around the topic of gravity and its relationship with special relativity and general relativity. The participants discuss the differences between Newtonian gravitation and relativistic gravitation, with one participant seeking a simplified approach to understanding general relativity. They also address the issue of how gravity affects the weight and acceleration of objects on the surface of a neutron star. The conversation ends with a question about the bending of light near a gravitational body.

#### Peeter

In high school we learn Coulomb's law for electrostatics, and later in University find that the correct version of this equation is in field form as part of maxwell's laws where the set of these equations together are lorentz invarient. In fact in "Principles of Electrodynamics" (Dover, by Schwarz) maxwell's equations are "derived" this way by Lorentz transformation and symmetry arguments starting with the statics equation.

Does anybody know a good treatment of gravity (online references preferred) that is consistent with special relativity without introducing the complexities of GR which I am not yet mathematically equipt to deal with (have no hardcore physics background, only engineering, and am studying for fun).

My expectation is that one could do the same Lorentz invarient treatment for div g = \rho in gravitation as for div E = \rho in the electrodynamics book, and come up with the retarded time potential equations and other relativistically corrected versions of the Newtonian GmM/r^2 "law" (probably as first order linear approximations of GR). I have a guess of what a Lorentz invarient formulation of div g = rho would look like, but figured I was reinventing the wheel. If somebody can point me in the right direction for reading I'd appreciate it.

http://en.wikipedia.org/wiki/Gravitomagnetism" [Broken]deals with such a similarity, but is based on GR.

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Peeter said:
Does anybody know a good treatment of gravity (online references preferred) that is consistent with special relativity without introducing the complexities of GR which I am not yet mathematically equipt to deal with (have no hardcore physics background, only engineering, and am studying for fun).
Incorporating a special relativistic correction to Newtonian gravitation without incorporating general relativistic corrections (e.g., frame dragging) will yield results that are less accurate than good old Newtonian gravitation. The reason is that frame dragging and the finite speed of transmission nearly cancel each other. To look at gravitational relativistically you have no choice but to go whole-hog. Weak field approximations of the full GR equations can yield perturbation corrections to Newton's theory of gravitation.

Peeter said:
Does anybody know a good treatment of gravity (online references preferred) that is consistent with special relativity …

Hi Peeter!

Didn't Einstein try that, and find he was a factor of 2 out?

tiny-tim said:
Didn't Einstein try that, and find he was a factor of 2 out?

I'm not the one to ask that question of:) I basically just want a clue of the big picture while I slowly learn enough differential calculus and tensor math to understand GR. I naively assumed an SR approximation would be simple and easier to understand.

Is that factor of two the frame dragging issue mentioned above?

I have a very simple question that might help put things in perspective here (for me).

If Newtonian physics predicts that a mass (m) will weigh GMm/R^2 on the surface of the of a very massive gravitational body such as a neutron star, does GR predict a different value for what a local observer on the surface with a set of bathroom scales would get for m (by reading the result he directly sees on the scales)?

kev said:
I have a very simple question that might help put things in perspective here (for me).

If Newtonian physics predicts that a mass (m) will weigh GMm/R^2 on the surface of the of a very massive gravitational body such as a neutron star, does GR predict a different value for what a local observer on the surface with a set of bathroom scales would get for m (by reading the result he directly sees on the scales)?

I have another closely related question. If Newtonian gravity predicts that the acceleration due to gravity at the surface of the neutron star is GM/R^2 is that what a local observer will measure by timing the fall of object over a very short distance (say a meter) or does GR predict the local observer will measure something very different from the Newtonian prediction? (Assume a significantly strong gravitational field and that the neutron star is not spinning).

tiny-tim said:
Hi Peeter!

Didn't Einstein try that, and find he was a factor of 2 out?

If you just consider a photon to obey Newtons laws and have a mass from E=mc^2, then work out how it is deflected by an object then you get an answer that is only half the correct value predicted by GR.
This was the experimental test of General Relativity by Eddington.

kev said:
I have a very simple question that might help put things in perspective here (for me).

If Newtonian physics predicts that a mass (m) will weigh GMm/R^2 on the surface of the of a very massive gravitational body such as a neutron star, does GR predict a different value for what a local observer on the surface with a set of bathroom scales would get for m (by reading the result he directly sees on the scales)?

kev said:
I have another closely related question. If Newtonian gravity predicts that the acceleration due to gravity at the surface of the neutron star is GM/R^2 is that what a local observer will measure by timing the fall of object over a very short distance (say a meter) or does GR predict the local observer will measure something very different from the Newtonian prediction? (Assume a significantly strong gravitational field and that the neutron star is not spinning).

Even a simple yes or no answer would be better than nothing, or does no one know the answer?

kev said:
Even a simple yes or no answer would be better than nothing, or does no one know the answer?

Hi kev!

The local gravitational potential is 1/(r - 2GM), giving a local g of -1/(r - 2GM)².

Newton uses 1/r instead of 1/(r - 2GM), giving g = -1/r².

The difference is negligible provided the radius of the star is a lot more than 2GM (the Schwarzschild radius).

But the mathematical treatment of the bending of light which grazes the surface (tangential motion) is a lot different from the treatment of falling objects (radial motion).

tiny-tim said:
Hi kev!

The local gravitational potential is 1/(r - 2GM), giving a local g of -1/(r - 2GM)².

Newton uses 1/r instead of 1/(r - 2GM), giving g = -1/r².

The difference is negligible provided the radius of the star is a lot more than 2GM (the Schwarzschild radius).

But the mathematical treatment of the bending of light which grazes the surface (tangential motion) is a lot different from the treatment of falling objects (radial motion).

tiny-tim said:
Hi kev!

The local gravitational potential is 1/(r - 2GM), giving a local g of -1/(r - 2GM)².

Newton uses 1/r instead of 1/(r - 2GM), giving g = -1/r².

The difference is negligible provided the radius of the star is a lot more than 2GM (the Schwarzschild radius).

But the mathematical treatment of the bending of light which grazes the surface (tangential motion) is a lot different from the treatment of falling objects (radial motion).

Hi Tim,
can you give any background or references to the equations you posted as I am still a bit unsettled about a couple of things?

For example in this post https://www.physicsforums.com/showpost.php?p=1526798&postcount=25 (I think) I showed that the coordinate orbital period and photon orbit radius of relativity is equivalent to the Newtonian (& Keplarian) equivalents suitably adjusted by the gravitational time dilation factor. The caveat as pointed out by Jorrie here https://www.physicsforums.com/showthread.php?t=190028&highlight=shapiro&page=3 are that it only applies to circular orbits, as obviously the time dilation varies with radius so elipitical orbits are more complex to analyse.

Now if I pad out your statement "Newton uses 1/r instead of 1/(r - 2GM), giving g = -1/r²." to Newton uses -GM/r for potential which when integrated with respect to radius gives an acceleration factor of GM/r^2, shouldn't it follow that if the GR definition of potential is -GM/(r-2GM), then the integral with respect to radius should give an acceleration of -(GM log(R-2GM))?

The local potential you quoted seems to be the Newtonian potential multiplied by the square of the gravitational gamma factor. Whatever the local acceleration is, it should be greater than the coordinate gravitational acceleration (that measured by an observer at infinity) by a factor of (1-2GM/R)^(-1.5) because vertical distance is length contracted by (1-2GM/R)^(0.5) and time is dilated by (1-2GM/R)^(-0.5). Distance (d) is proportional to $\frac{at^2}{2}$ so acceleration is proportional to 2d/t^2 so coordinate acceleration is less than local acceleration by a factor of gamma cubed. Agree?

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kev said:
Now if I pad out your statement "Newton uses 1/r instead of 1/(r - 2GM), giving g = -1/r²." to Newton uses -GM/r for potential which when integrated with respect to radius gives an acceleration factor of GM/r^2, shouldn't it follow that if the GR definition of potential is -GM/(r-2GM), then the integral with respect to radius should give an acceleration of -(GM log(R-2GM))?

Hi kev!

GM/r for potential which when differentiated with respect to radius gives an acceleration factor of GM/r^2;

so GM/(r-2GM) gives GM/(r-2GM)^2.
The local potential you quoted seems to be the Newtonian potential multiplied by the square of the gravitational gamma factor. Whatever the local acceleration is, it should be greater than the coordinate gravitational acceleration (that measured by an observer at infinity) by a factor of (1-2GM/R)^(-1.5) because vertical distance is length contracted by (1-2GM/R)^(0.5) and time is dilated by (1-2GM/R)^(-0.5). Distance (d) is proportional to $\frac{at^2}{2}$ so acceleration is proportional to 2d/t^2 so coordinate acceleration is less than local acceleration by a factor of gamma cubed. Agree?

Sorry … that's doing my head in!

tiny-tim said:
so [noparse][taking the gradient of][/noparse] GM/(r-2GM) gives GM/(r-2GM)^2.
Note well: tiny-tim is using units in which the speed of light is one. In systems of units where this is not the case (e.g., SI units),

$$\Phi = \frac{GM}{r-2\frac{GM}{c^2}} = \frac{GM}{r-r_s}$$

where $r_s \equiv 2\frac{GM}{c^2}$ is the Schwarzschild radius.

kev said:
I have a very simple question that might help put things in perspective here (for me).

If Newtonian physics predicts that a mass (m) will weigh GMm/R^2 on the surface of the of a very massive gravitational body such as a neutron star, does GR predict a different value for what a local observer on the surface with a set of bathroom scales would get for m (by reading the result he directly sees on the scales)?

The GR (Schwarzschild) coordinate acceleration has the same expression as Newtonian acceleration, but an accelerometer carried by a hovering observer would actually read

$$\left(1 - \frac{2Gm}{c^2 r} \right)^{-\frac{1}{2}} G\frac{m}{r^2}$$

tiny-tim said:
Hi kev!

GM/r for potential which when differentiated with respect to radius gives an acceleration factor of GM/r^2;

so GM/(r-2GM) gives GM/(r-2GM)^2.

Hi Tim,

You are absolutely right. It should be differentiate. I was away from my PC when I suddenly realized my error and rushed back hoping I would be able to correct my mistake before someone else spotted it, but too late! oops

Anyway, I still think local acceleration should be the coordinate acceleration multiplied by gamma cubed based on:

$$a = \frac{d}{2t^2}$$

$$a ' = \frac{d'}{2t '^2} = \frac{d \gamma^{-1}}{2t^2 \gamma^{2}} = \frac{a}{\gamma^3}$$

where $$\gamma = \frac{1}{\sqrt{1-\frac{2GM}{Rc^2}}}$$

Sorry if the rest of the stuff is a lot to get your head round :tongue: but there are some important issues/principles at stake here. I can not find anything on the internet that specifically addresses what the local gravitational acceleration would be, which is why I was hoping you could point to a reference on the subject. :shy:

It also seems worth mentioning that the acceleration that acts on a stationary object is different to the acceleration acting on a moving onject. For example a falling photon is actually de-accelerating as it falls according to a distant observer using coordinate measurements.

kev said:
Even a simple yes or no answer would be better than nothing, or does no one know the answer?
Imagine using the thrust of a rocket to hover at a constant radius R outside a black hole of mass M. How much thrust would the rocket of mass m need to exert? The required thrust is

$$\left(1 - \frac{2Gm}{c^2 r} \right)^{-\frac{1}{2}} G\frac{m}{r^2}$$

You can find the answer in Hartle's book: Gravity, page 261

kahoomann said:
Imagine using the thrust of a rocket to hover at a constant radius R outside a black hole of mass M. How much thrust would the rocket of mass m need to exert? The required thrust is

$$\left(1 - \frac{2Gm}{c^2 r} \right)^{-\frac{1}{2}} G\frac{m}{r^2}$$

You can find the answer in Hartle's book: Gravity, page 261

Tim has given local acceleration as $$\frac{GM}{R^2}\gamma^4$$

and kahoomann has given the local acceleration as $$\frac{GM}{R^2}\gamma$$

Clearly there is a large discrepancy.

Earlier I stated that $$a_{{LOCAL}} = a_{(COORD)}*\gamma^{3}$$

which is consistent with the equation given by Tim as being the local acceleration (as he claimed) and the equation given by kahoomann as being the coordinate acceleration. However kahoomann claims that his equation is the local acceleation and this is backed up by Hartle's book. Kahoomanns claim is also backed up by this http://www.mathpages.com/rr/s7-03/7-03.htm" [Broken]

Any ideas how we resolve this?

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kev said:
Tim has given local acceleration as $$\frac{GM}{R^2}\gamma^4$$

and kahoomann has given the local acceleration as $$\frac{GM}{R^2}\gamma$$

Clearly there is a large discrepancy.

Earlier I stated that $$a_{{LOCAL}} = a_{(COORD)}*\gamma^{3}$$

which is consistent with the equation given by Tim as being the local acceleration (as he claimed) and the equation given by kahoomann as being the coordinate acceleration. However kahoomann claims that his equation is the local acceleation and this is backed up by Hartle's book. Kahoomanns claim is also backed up by this http://www.mathpages.com/rr/s7-03/7-03.htm" [Broken]

Any ideas how we resolve this?

This Wikpedia article http://en.wikipedia.org/wiki/Proper_acceleration#Acceleration_in_.281.2B1.29D states that

"In the unidirectional case i.e. when the object's acceleration is parallel or anti-parallel to its velocity in the spacetime slice of the observer, proper acceleration α and coordinate acceleration a are related[6] through the Lorentz factor γ by α=γ3a. "

which is exactly what I claimed. Later on http://en.wikipedia.org/wiki/Proper_acceleration#Surface_dwellers_on_a_planet it states that the local proper gravitational acceleration is proportional to $$\frac{GM}{R^2}\gamma$$ which is supported by kahoomann, Hartle's book and mathpages. Put the two facts together and the coordinate gravitational acceleration is

$$\frac{GM}{R^2}\frac{\gamma }{\gamma^3} = \frac{GM}{R^2}\left(1-\frac{2GM}{Rc^2}\right)$$

The significance of that equation is that the coordinate acceleration when the radius is less than the Schwarzschild radius is negative and accelerates particles outwards from the centre of a vacuum solution black hole towards the event horizon. The proper acceleration below the event horizon is imaginary while the coordinate acceleration is real indicating the coordinate acceleration is the physical solution while the traditional picture of a black hole as central singularity of infinite mass is not. The coordinate solution shows there is in fact infinite acceleration outwards at the centre of a black hole indicating that it impossible for a true singularity to form. This agrees with another thread where we showed that the interior Schwarzschild solution shows that a region of negative time forms inside a gravitational body before it has collapsed to the density required to form a black hole and that this would be likely to prevent a central singularity forming. This shows that a black hole is in fact a hollow shell that is marginally larger than the Schwarzschild radius with no mass inside the event horizon. The radius of the shell is asymptotically collapsing toward the Schwarzschild radius but never quite reaches it.

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Warning: I am only a beginner in GR, so I stand to be corrected by an expert.

kev said:
Earlier I stated that $$a_{{LOCAL}} = a_{(COORD)}*\gamma^{3}$$

which is consistent with the equation given by Tim as being the local acceleration (as he claimed) and the equation given by kahoomann as being the coordinate acceleration. However kahoomann claims that his equation is the local acceleation and this is backed up by Hartle's book. Kahoomanns claim is also backed up by this http://www.mathpages.com/rr/s7-03/7-03.htm" [Broken]
$$a_{proper} = \gamma^3 a_{coord}$$​

is, I think, true for a local inertial coord system (in which SR is approximately true), but I believe it's not generally true in other coords, due to gravitational time dilation and length contraction.

kev said:
The significance of that equation is that the coordinate acceleration when the radius is less than the Schwarzschild radius is negative and accelerates particles outwards from the centre of a vacuum solution black hole towards the event horizon. The proper acceleration below the event horizon is imaginary while the coordinate acceleration is real indicating the coordinate acceleration is the physical solution while the traditional picture of a black hole as central singularity of infinite mass is not. The coordinate solution shows there is in fact infinite acceleration outwards at the centre of a black hole indicating that it impossible for a true singularity to form. This agrees with another thread where we showed that the interior Schwarzschild solution shows that a region of negative time forms inside a gravitational body before it has collapsed to the density required to form a black hole and that this would be likely to prevent a central singularity forming. This shows that a black hole is in fact a hollow shell that is marginally larger than the Schwarzschild radius with no mass inside the event horizon. The radius of the shell is asymptotically collapsing toward the Schwarzschild radius but never quite reaches it.
To my inexpert ears, this sounds nonsensical. Is there a GR expert to confirm this?

Are you aware that the t and R coordinates of the Schwarzschild solution represent the time and space according to a distant observer (theoretically at infinity), and do not represent the time and space of a local observer at the event being measured. Thus coordinate acceleration d2R/dt2 does not represent the acceleration measured by a "stationary" local observer. And are you aware that inside the event horizon, t is proportional to local distance and R is proportional to local time?

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DrGreg said:
Warning: I am only a beginner in GR, so I stand to be corrected by an expert.
So am I, so expert opinions are welcome

DrGreg said:
$$a_{proper} = \gamma^3 a_{coord}$$​

is, I think, true for a local inertial coord system (in which SR is approximately true), but I believe it's not generally true in other coords, due to gravitational time dilation and length contraction.
That equation comes directly from the Schwarzschild solution so it relates the measurements of a observer at infinity to that of a local observer. The Schwarzschild solution takes gravitational time dilation and length contraction into account. In fact it defines it.

kev said:
The significance of that equation is that the coordinate acceleration when the radius is less than the Schwarzschild radius is negative and accelerates particles outwards from the centre of a vacuum solution black hole towards the event horizon. The proper acceleration below the event horizon is imaginary while the coordinate acceleration is real indicating the coordinate acceleration is the physical solution while the traditional picture of a black hole as central singularity of infinite mass is not. The coordinate solution shows there is in fact infinite acceleration outwards at the centre of a black hole indicating that it impossible for a true singularity to form. This agrees with another thread where we showed that the interior Schwarzschild solution shows that a region of negative time forms inside a gravitational body before it has collapsed to the density required to form a black hole and that this would be likely to prevent a central singularity forming. This shows that a black hole is in fact a hollow shell that is marginally larger than the Schwarzschild radius with no mass inside the event horizon. The radius of the shell is asymptotically collapsing toward the Schwarzschild radius but never quite reaches it.
DrGreg said:
To my inexpert ears, this sounds nonsensical. Is there a GR expert to confirm this?
I do not take offence that you think it is nonsense, because it is not the mainstream conclusion of numerous books on the subject. On the other hand, it is much easier to sell books describing a black hole as singularity that may be the portal to other universes than the description of black hole as a boring hollow shell of matter that remains in this universe.

More seriously, coordinate systems like those of Eddington-Finkelstein and Kruskal seem to prove that a free falling observer reaches the centre of the black hole in finite time. However, these alternative coordinate systems are derived from the vacuum Schwarzschild solution so they assume a priori that a black hole is a singularity of infinite density so they can not be used to prove that a black hole is a stable singularity of infinite density. None of the alternative coordinate systems seem to take the interior Schwarzschild solution into account. They also seem to assume that proper local measurements are more valid than coordinate measurements. Further on in this post, I hope I have shown that is not necessarily true.

DrGreg said:
Are you aware that the t and R coordinates of the Schwarzschild solution represent the time and space according to a distant observer (theoretically at infinity), and do not represent the time and space of a local observer at the event being measured.

Yes and yes, but technically the Schwarzschild solution contains both local and coordinate measurements. The dtau in the equation is local proper time and dt is the time measured by an observer at infinity. In fact we do not need to be limited to a binary choice of just local observers or an observer at infinity. The gravitational time dilation derived from the Schwarzschild solution can be expressed as:

$$dt ' = dt \frac{\sqrt{1-R_s/R_o}}{\sqrt{1-R_s/R}}$$

where Rs is the Schwarzschild radius, Ro is the location of the observer and R is radius where the time dilation is being calculated for. It is easy to see that when Ro=infinity that $dt ' = dt/\sqrt{1-R_s/R}$ and when Ro=R that dt'=dt. The beauty of this formulation is that the observer can be above or below where the measurement is being made so for example when Ro=4Rs and R=5Rs the observer sees the clock above him running faster (blue shift) and when Ro=5Rs and R=4Rs the observer sees the clock below him running slower (red shift).

Similarly length contraction can be expresses as:

$$dr ' = dr \frac{\sqrt{1-R_s/R}}{\sqrt{1-R_s/R_o}}$$

$$dr '/dt ' = dr/dt \frac{(1-R_s/R)}{(1-R_s/R_o)}$$

$$d^2 r'/dt^2' = d^2r/dt^2 \frac{(1-R_s/R)^{(1.5)}}{(1-R_s/R_o)^{(1.5)}}$$

The above expressions for velocity and acceleration do not tell us what the local velocity or acceleration actually is unless we know the coordinate value or vice versa.

We know that the local speed of light is always c so we can say with confidence that:

$$c ' = c\frac{(1-R_s/R)}{(1-R_s/R_o)}$$

and from considerations of escape velocity the velocity of a particle with mass falling from infinity is

$$v' = c\sqrt{Rs/R} \frac{(1-R_s/R)}{(1-R_s/R_o)}$$

For acceleration the local gravitational acceleration (from sources quoted earlier in this thread) is:

$$a=\frac{GM}{R^2}\frac{1}{\sqrt{1-R_s/R}}$$ so the trasformation is:

$$a'=\frac{GM}{R^2} \frac{(1-R_s/R)}{(1-R_s/Ro)^{(1.5)}}$$

Below the event Schwarzschild radius the above expression for gravitational acceleration inverts and anything that happens to be below the horizon is accelerated outwards according to an observer at infinity. To the local observer it appears as if the acceleration is towards the centre of the black hole. Most of the literature gives priority to the measurements of the local observer (the proper measurements) but I challenge that. While in Special Relativity the measurements made by one observer of a system moving relative to him are ambiguous the measurements made at a distance in General Relativity are not. For example, an observer low down in a gravitational weel see the clock odf an observer above him as running fast while the upper observer see the clock of the observer below as running slower. No contradiction or ambiguity of measurements there. If the lower observer moves up to the higher observer he finds that his clock has indeed recorded less elapsed time than the higher observer. The measurements made at a distance in General Relativity are accurate and unambiguous. Now the lower observer measure the rate of his own clock as running at one second per second and so does the upper observer. Obviously local proper measurements are not very useful. Indeed when the lower observer realizes that his clock is running slower than that of the upper observer (which he can easily prove to himself by meeting with upper observer and comparing elapsed times) he soon realizes his measurement of the local speed of light lower down in the gravity well as c, is in fact an illusion. The above arguments should make clear that coordinate measurements made by comparing the measurements of various observers are the "big picture" and represent physical reality in so far as there is any such thing ;)

A surprising useful conclusion falls out of the Schwarzschild coordinate solution. Run the universe backwards towards the big bang and you eventually get to a time when the scale factor of the universe is such that the mass contained within the universe is contained within the Schwarzschild radius and the universe is effectively a black hole. I think I worked this out to be about a radius of 300 times the size of the Milky Way galaxy.Some people would object to this view and say the universe was never really bigger or smaller but just differently scaled. On the other hand I think everyone would have to agree that the universe must have had a greater density in earlier epochs and that at one time the density exceeded that required for a black hole. Now if we run the universe forward again we have a potential problem because the classical view is that nothing can escape a black hole which implies we must now still be in a black hole. Observations do not seem to support that conclusion. One solution to this problem would be for the universe to gradually seep out the primordial black hole in the form of Hawking radiation but that would mean the universe started as a big wimper rather than a big bang. A simpler solution is provided directly by General Relativity and the Schwarzschild solution. The equation I gave above for the gravitational acceleration shows that if all the mass of the universe was contained within say a radius of one Planck length then there would be extreme acceleration outwards which would be not unlike the rapid inflation that was supposed to occur early on.

DrGreg said:
Thus coordinate acceleration d2R/dt2 does not represent the acceleration measured by a "stationary" local observer. And are you aware that inside the event horizon, t is proportional to local distance and R is proportional to local time?

As mentioned above, I do not believe the "proper" measurements made by a local observer are the best represention of reality. It is also worth noting that the proper acceleration measured by the local observer below the event horizon is the sqare root of a negative number and therefore imaginary, while the same measurement by the observer at infinity is real.

This paper here http://prola.aps.org/abstract/PRD/v25/i12/p3191_1 seems to support my argument as far as repulsive gravity is concerned.

"To the distant observer, who uses measuring instruments not affected by gravity, gravitational repulsion can occur anywhere in the Schwarzschild field. It depends on the relationship between the transverse and radial Schwarzschild velocities. On the other hand, local observers, whose measuring instruments are affected by gravity, cannot detect a positive value for the acceleration of gravity."

I am aware of the view that time and distance coordinates swap over below the Schwarzschild radius but I am also aware that view was bitterly disputed in a thread in this forum. I am not sure who won ;) Personally I don't think you can shift coordinates around like that, as you would have to define another coordinate system to shift (or rotate) the original coordinate system relative to.

As I mentioned in another thread ,once it is accepted that a black hole is a hollow shell of mass just outside the Schwarzschild radius, the Information Loss Paradox, location of entropy and source of thermal radiation problems all automatically go away and do not require exotic explanations involving holograms, virtual particles and multiple universes.

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1. "Proper", "local" and "coordinate"

To clear up any possible misunderstanding:

"Coordinate acceleration" is just d2x/dt2 in whatever coordinates are currently under consideration.

"Local acceleration" is measured in local coordinates i.e. by an inertial observer at the event where the measurement is made, using local clocks and local rulers which behave as if SR were approximately valid in a small region near the observer.

"Proper acceleration" is a special case of local acceleration, where the object being measured is momentarily stationary in the local coordinate system being used. It's what an accelerometer attached to the object would measure.

The equation

$$a_{proper} = \gamma^3 \, a_{local}$$​

is always true (I think), even in Special Relativity (where all inertial coordinates are "local"). I don't know what the equation is for "remote" acceleration measured by a distant observer.

2. Hollow black holes

The problem with the interior of a hollow shell is that the Schwarzschild solution doesn't apply there. In Newtonian theory there's no gravity at all inside a uniform shell. I suspect this is also true in GR, in the sense that there's no curvature, though I'm not 100% sure. But it must be approximately true in some sense, and the interior Schwarzschild solution looks nothing like the right answer.

3. Imaginary acceleration

The local "acceleration due to gravity" is measured by an observer who hovers a fixed distance from the centre, and is that observer's own proper acceleration. As it is impossible to hover inside the event horizon, it's not surprising any formula for proper acceleration fails there.

Hi DrGreg,

DrGreg said:

1. "Proper", "local" and "coordinate"

To clear up any possible misunderstanding: ...
Your definitions of the various forms of acceleration seem reasonable.

DrGreg said:
2. Hollow black holes

The problem with the interior of a hollow shell is that the Schwarzschild solution doesn't apply there. In Newtonian theory there's no gravity at all inside a uniform shell. I suspect this is also true in GR, in the sense that there's no curvature, though I'm not 100% sure. But it must be approximately true in some sense, and the interior Schwarzschild solution looks nothing like the right answer.
I have not heard anyone prove the interior Schwarzschild solution is wrong or even suggest that. Maybe your concern is that the interior solution does not appear to show that the gamma factor within the hollow shell is uniform? The interior solution as I presented it, is the simplified form that assumes a uniform density in the body as a whole and so is not directly applicable to the hollow shell. The more general form of the solution that allows for non uniform distribution of matter in the body is:

(Eq1) $$\frac{dtau}{dt} = \frac{3}{2}\left(1-\frac{2GM}{R_mc^2}\right)^{0.5}-\frac{1}{2}\left(1-\frac{2GM}{Rc^2}\frac{p(4/3)\pi R^3}{p(4/3)\pi R_m^3}\right)^{0.5}$$

where p is the density of the gravitational body.

For uniform density the expression reduces to that given in the earlier post:

(Eq2) $$\frac{dtau}{dt} = \frac{3}{2}\left(1-\frac{2GM}{R_mc^2}\right)^{0.5}-\frac{1}{2}\left(1-\frac{2GM}{c^2}\frac{R^2}{R_m^3}\right)^{0.5}$$

For density increasing towards the centre according to p/R the expression becomes:

(Eq3) $$\frac{dtau}{dt} = \frac{3}{2}\left(1-\frac{2GM}{R_mc^2}\right)^{0.5}-\frac{1}{2}\left(1-\frac{2GM}{c^2}\frac{R}{R_m^3}\right)^{0.5}$$

For either distribution of matter in the body, it can be seen that a event horizon appears at R=0 when Rm=9Rs/8. As the shell collapses below 9Rs/8 time starts to run backwards at the centre of the body even when the mass is not enclosed within the Schwarzschild radius and the only way to prevent that happening is for the mass to redistribute itself outwards. Time running backwards is an unstable condition and immediately acts to undo that condition. Earlier I suggested repulsive gravity is the mechanism that causes the mass to redistribute itself. It should be remembered that negative gravitational curvature is not ruled out in general Relativity on a cosmological scale. The authors of this paper http://odarragh.astro.utoronto.ca/Schwarzschild.pdf [Broken] suggest that the "pressure" distribution inside the body acts to prevent the negative time zonewithin the body. Either way, the formation of a singularity of infinite density seems to an unphysical proposition. (Quantum physics would tend to agree).

Eq1 can also be more conveniently expressed as:

(Eq4) $$\frac{dtau}{dt} = \frac{3}{2}\left(1-\frac{2GM}{R_mc^2}\right)^{0.5}-\frac{1}{2}\left(1-\frac{2GM}{Rc^2}\frac{(Enclosed Mass)}{(Total Mass)}\right)^{0.5}$$

from which it is easy to see that the gravitational gamma factor is constant everywhere inside the cavity of a hollow shell. Any body falling inside the shell dynamically changes the situation.

DrGreg said:
3. Imaginary acceleration

The local "acceleration due to gravity" is measured by an observer who hovers a fixed distance from the centre, and is that observer's own proper acceleration. As it is impossible to hover inside the event horizon, it's not surprising any formula for proper acceleration fails there.

I agree. The equation I gave before just states that if an object is stationary it would be accelerated outwards. To state that it impossible to hover inside an event horizon means that you have to present equations that show the proper acceleration everywhere inside a black hole is infinite and you have to prove whether the inside of a black contains all the matter in a singularity at the centre, so that you can use the exterior solution, or some other distribution of matter (including the hollow configuration) that requires the interior solution. The question becomes a bit circular. However it is possible to show that a particle falling from infinity reaches the speed of light at the event horizon and exceeds the local speed of light below the event horizon. The only way to ensure that the speed of light is always c relative to a local observer is to assume that any matter that passes through the event horizon is turned into light energy. The relativistic velocity addition formula shows that the velocity of anything relative to a photon is c and so the postulate of relativity is maintained. The imaginary acceleration result is because a photon is not a valid inertial observer in the normal sense so the proper observation is not "real".

It is obvious that the acceleration acting on a particle with velocity is different from the gravitational acceleration acting on a stationary particle. This can be seen in the fact that the coordinate velocity of a falling photon de-accelerates as it falls while a particle that is stationary or almost stationary accelerates. From a straight comparison of the acceleration of photons and particles with rest mass it seems that the greater the velocity towards the centre of the black hole the less the acceleration and at a certain critical velocity at any given height the acceleration becomes negative. It would be nice if someone could supply a equation that computes coordinate acceleration and takes initial velocity into account. :)

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DrGreg said:
...
3. Imaginary acceleration

The local "acceleration due to gravity" is measured by an observer who hovers a fixed distance from the centre, and is that observer's own proper acceleration. As it is impossible to hover inside the event horizon, it's not surprising any formula for proper acceleration fails there.

One additional thought. There IS one place inside the event horizon that a hypothetical observer would be stationary. That is at the centre. It is safe to assume that in the vacuum solution of a black hole that the observer at the centre of the black hole is going nowhere (until he is slowly emitted as Hawking radiation) so the exterior solution for acceleration:

$$a '=\frac{GM}{R^2} \frac{(1-R_s/R)}{(1-R_s/Ro)^{(1.5)}}$$

applies, which according to the observer at Ro=infinity, the coordinate acceleration at R=0 is infinite and directed outwards.

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kev said:
For uniform density the expression reduces to that given in the earlier post:

(Eq2) $$\frac{dtau}{dt} = \frac{3}{2}\left(1-\frac{2GM}{R_mc^2}\right)^{0.5}-\frac{1}{2}\left(1-\frac{2GM}{c^2}\frac{R^2}{R_m^3}\right)^{0.5}$$

Oops..the earlier post I mentioned, where the symbols are defined, was in another thread here: https://www.physicsforums.com/showpost.php?p=1767802&postcount=17

kev said:
I have not heard anyone prove the interior Schwarzschild solution is wrong or even suggest that. Maybe your concern is that the interior solution does not appear to show that the gamma factor within the hollow shell is uniform?
I may have misused the terminology. When I referred to "the interior Schwarzschild solution", I meant inside the event horizon but outside the central mass, so the standard ("exterior") Schwarzschild metric still applies (albeit with spacelike t and timelike r). My concern was that you appeared to be using that metric to claim outward acceleration (by the way, outward acceleration need not imply outward velocity) but then you talk of a shell of matter which invalidates that metric inside the shell. I still believe spacetime is flat inside a uniform shell.

I'm not sure where all the equations you quote come from. Are they from a book or other reliable source, or have you derived them yourself? If the latter, I'd want to know how you got them, as there might be some flaw in your logic.

I can quote some other published equations (I can't pretend I fully understand their derivations):

For a radially free-falling particle of unit mass, in units where G = c = 1, the following quantities E (energy) and L (Lagrangian) are conserved (constant over time):

$$E = \left(1 - \frac{2M}{r} \right) \frac{dt}{d\tau}$$
$$2L = \left(1 - \frac{2M}{r} \right) \left( \frac{dt}{d\tau} \right)^2 - \frac{1}{1 - 2M/r} \left( \frac{dr}{d\tau} \right)^2$$​

References:

1. Woodhouse, N.M.J. (2007), General Relativity, Springer, London, ISBN 978-1-84628-486-1, pages 100, 107, 124.

2. Woodhouse, N.M.J. (2003), http://www2.maths.ox.ac.uk/~nwoodh/gr/index.html [Broken] on which the book (1) was closely based, pages 54-55 (Lecture 12), p59 (L13), p73 (L15).

These sources also confirm Hartle's formula for "the proper acceleration due to gravity" quoted in post #17 (1, p.99) (2, p.54, L12) (3, p.230)

3. Rindler, W. (2006 2nd ed), Relativity: Special, General and Cosmological, Oxford University Press, Oxford, ISBN 978-0-19-856732-5.

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## 1. What is the concept of special relativity in relation to gravity?

Special relativity is a fundamental theory in physics that explains how the laws of physics are the same for all observers, regardless of their relative motion. It also describes how space and time are intertwined in what is known as spacetime. In the context of gravity, special relativity shows that the effects of gravity can be described as a curvature of spacetime.

## 2. How does special relativity explain the phenomenon of gravitational time dilation?

According to special relativity, time passes at different rates for observers in different frames of reference. In the presence of a gravitational field, time appears to pass slower for objects closer to a massive object, such as a planet or star. This is because the massive object curves the surrounding spacetime, causing time to flow at a slower rate in its vicinity.

## 3. Can special relativity explain the bending of light in a gravitational field?

Yes, special relativity can explain the bending of light in a gravitational field. According to the theory, the curvature of spacetime causes the path of light to bend as it travels through a gravitational field. This was famously confirmed by observations of starlight bending near the sun during a solar eclipse, providing strong evidence for the theory of relativity.

## 4. How does special relativity differ from general relativity in its treatment of gravity?

Special relativity only considers the effects of gravity in flat, unchanging spacetime, while general relativity includes the curvature of spacetime caused by massive objects. General relativity provides a more complete understanding of gravity and has been extensively tested and confirmed by experiments.

## 5. Can special relativity be used to explain the behavior of gravity on a quantum level?

No, special relativity is a classical theory and does not take into account the principles of quantum mechanics. To fully understand the behavior of gravity on a quantum level, a theory of quantum gravity is needed. This is an area of ongoing research and has not yet been fully developed.