# Special relativity treatment of gravity?

1. Jun 5, 2008

### 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.

2. Jun 5, 2008

3. Jun 5, 2008

### Staff: Mentor

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.

4. Jun 5, 2008

### tiny-tim

Hi Peeter!

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

5. Jun 5, 2008

### Peeter

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?

6. Jun 5, 2008

### yuiop

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)?

7. Jun 5, 2008

### yuiop

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).

8. Jun 5, 2008

### mgb_phys

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.

9. Jun 6, 2008

### yuiop

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

10. Jun 6, 2008

### tiny-tim

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).

11. Jun 6, 2008

### yuiop

12. Jun 8, 2008

### yuiop

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?

Last edited: Jun 8, 2008
13. Jun 8, 2008

### tiny-tim

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.
Sorry … that's doing my head in!

14. Jun 8, 2008

### Staff: Mentor

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.

15. Jun 8, 2008

### kahoomann

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

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

16. Jun 8, 2008

### yuiop

Hi Tim,

You are absolutely right. It should be differentiate. I was away from my PC when I suddenly realised 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.

17. Jun 8, 2008

### kahoomann

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

18. Jun 8, 2008

### yuiop

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 mathpages link.

Any ideas how we resolve this?

19. Jun 9, 2008

### yuiop

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 Schwarschild radius but never quite reaches it.

20. Jun 12, 2008

### DrGreg

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

$$a_{proper} = \gamma^3 a_{coord}$$​

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.

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?