# Relativistic escape velocity

1. Oct 20, 2012

### MLeinenweber

Why is there an inherent presumption imbedded in the field of physics that gravitational force is exempt from the rules of relativity?

The presumption is based on the non-relativistic expression for escape velocity:

v=(2MG/r)^.5

Thus, with a very large mass or a very small radius velocity can exceed the speed of light.

This expression is, of course. derived by setting the expression for kinetic energy equal to gravitational potential as follows:

(mv^2)/2=mMG/r

Our physics textbooks, however, define relativistic potential kinetic energy as:

mc^2(γ-1) where γ=(1-v^2/c^2)^-.5

If, however, you set the relativistic version of potential kinetic energy equal to gravitational potential, after a little algebra you find that:

v=c(1-((rc^2)/(MG+rc^2))^2)^.5

This expression yields nearly equivalent results to the non-relativistic version for observed planets and stars. For example the non-relativistic expression predicts the escape velocity at the surface of the earth is 1.1178839 X 10^4 m/s whereas the relativistic expression predicts 1.1178840 X 10^4 m/s.

For theoretical planets and stars however, the difference between the non-relativistic prediction and the relativistic prediction can be significant. If for example the earth's mass were 5.97 X 10^36 kg versus 5.97 X 10^24 kg the non-relativistic prediction is 1.1178839 X 10^10 m/s (or approximately 37 times the speed of light) whereas the relativistic prediction is 2.9979215 X 10^8 m/s (or just under the speed of light). In fact, no matter how large the mass or how tiny the radius, the relativistic version will never predict a velocity that exceeds the speed of light.

The equation for relativistic escape velocity described above seems to be a very reasonable and natural extension of relativity theory.

Escape velocity predicts the speed that objects and particles must achieve in order to escape the force of gravity. On the other side, however, it also predicts the speed that an object or particle achieves as it is attracted by gravitational force. Since the relativistic form of the equation predicts that gravity cannot accelerate an object or particle to a speed equal to or greater than the speed of light, it also casts a shadow on theories which depend on speeds reaching the speed of light such as black hole theory.

I would be interested in input from anybody on the above analysis.

Last edited: Oct 21, 2012
2. Oct 20, 2012

### willem2

Whenever you get a significant difference with Newtonian gravity, you should use general relativity. In particular this predicts that a mass of 6.97*10^36 of the size of the earth will be a black hole, and you can't escape from it at all from closer than 8.87 * 10^6 km.
Why do you think black hole theory predicts a speed greater than light?

3. Oct 20, 2012

### MLeinenweber

I assume that your calculation of the point of no escape at 8.87 X 10^6 km is based upon the fact that the non-relativistic calculation of escape velocity predicts that an object or particle will achieve the speed of light at such distance. The relativistic calculation, using the formula I described, however predicts that an object or particle will achieve a velocity of about .75 the speed of light at such distance and will never get to the speed of light. Perhaps my statement that black hole theory predicts a speed greater than the speed of light was not well worded, however, it seems to me the conclusion is the same.

4. Oct 20, 2012

### Staff: Mentor

A small point: I assume you mean "kinetic energy". There is no such thing as "potential kinetic energy".

What formula are you using for "gravitational potential"?

5. Oct 21, 2012

### MLeinenweber

Thanks for the correction re kinetic energy.

For gravitational potential I am using:

(mMG)/r

6. Oct 21, 2012

### Staff: Mentor

You are using a non-relativistic formula for potential energy combined with a relativistic formula for kinetic energy. That's why you are getting a different (wrong) answer than the (right) answer given by standard "black hole theory".

7. Oct 21, 2012

### Staff: Mentor

No, it's based upon the fact that the correct, general relativistic calculation of the curvature of spacetime around a massive body leads to the result given for the "point of no escape". It does happen to be the case that the correct calculation arrives at a formula for "escape velocity" which is the same, formally, as the non-relativistic Newtonian formula; but the meaning of the "r" that appears in the correct, general relativistic formula is not the same as the meaning of the "r" that appears in the Newtonian formula, so the resemblance is only on the surface.

8. Oct 21, 2012

### Staff: Mentor

I forgot to respond to this:

There is no such presumption. We have a perfectly good theory of gravity that abides by all the "rules of relativity": general relativity.

9. Oct 21, 2012

### MLeinenweber

Perhaps my opening sentence was overstated.

The point remains, however, that the field of physics has embraced a non-relativistic expression for escape velocity which, in my humble opinion, has led to incorrect speculations such as the existance of black holes.

10. Oct 21, 2012

### pervect

Staff Emeritus
If you check any GR textbook, or some of the trustworthy online GR lecture notes, such as Sean Carrol's http://arxiv.org/abs/gr-qc/9712019 also found at http://www.physics.org/explorelink.asp?id=2086&q=m.&currentpage=1&age=0&knowledge=0&item=4, you'll find the actual GR treatment of black holes, which has little resemblance to your arguments - which are unfortunately "straw men".

You might find such arguments in a few not-so-good popularization - I doubt you'll find them in a serious GR textbook. For instance, here's what Caroll has to say in his lecture notes on GR.

Caroll starts with the Einstein field equation, derives the solution to the EFE in Schwarzschild coordinates, demonstrates some of the problems with the particular coordinate choices (which is easy to solve, but leads to later troubles) , finds some better coordinate choices, and presents the full general-relativistic equations of motion of a test particle / test light beam in well-behaved coordinates, demonstrating that they have the property of not leaving the black hole.

11. Oct 21, 2012

### Staff: Mentor

There isn't. The modern theory of gravity is called general relativity, and as the name implies it is relativistic.

You correctly show that Newtonian gravity is incompatible with relativity, which has been known for over a century. This incompatibility was one of the main motivations for the development of GR, which resolved the issue just under a century ago.

12. Oct 21, 2012

### Staff: Mentor

No, it has embraced a correct, *relativistic* expression which happens to formally resemble the non-relativistic expression. But, as I said in my previous post...

...the resemblance is only formal. If you don't understand what I meant in the above quote, please feel free to ask questions about it; that's much more likely to lead to a fruitful discussion than if you continue to make incorrect statements when we have already given you corrections.

13. Oct 21, 2012

### Bill_K

Let me throw caution to the wind here, and quote the actual general relativistic answer!

If you start off a test particle at rest at infinity and let it fall inward, it will follow a radial geodesic. Its velocity at any point turns out to be (remarkably!) the same expression as for the Newtonian case: dr/dτ = √(2GM/r), where M is the central mass and r is the Schwarschild radial coordinate. This can also be written dr/dτ = c √(2GM/c2r) = c √(rS/r) where rS is the Schwarzschild radius. Thus as you approach the surface of the black hole, the escape velocity approaches c.

What distinguishes the general relativistic result from the Newtonian one is that the time derivative is with respect to τ, the particle's proper time. One might want to also look at the coordinate escape velocity dr/dt. The relationship between the two is dτ/dt = 1 - 2GM/r. This says that "time slows down" as you approach the hole. That is, the particle's proper time runs at a slower rate than coordinate time (= time measured at infinity). Combining, we get dr/dt = (1 - 2GM/r)√(2GM/c2r), which goes to zero at the Schwarzschild surface, where "time is frozen", and the particle appears to an outside observer to no longer move.

14. Oct 21, 2012

### Staff: Mentor

I don't object to throwing caution to the winds, but I have a couple of things to add.

One key clarification: by "velocity" here we mean "velocity relative to a static observer", i.e., relative to an observer who stays at the same value of r. At the horizon, r = 2GM/c^2, and at any smaller value of r, there are no such static observers. This is why the "velocity" can go to c at the horizon, without violating the principle that special relativity is valid locally.

And that the "r" that appears is *not* the physical "radius", i.e., "physical distance from the center", as it is in the Newtonian formula. It is the Schwarzschild r coordinate, which is defined such that a 2-sphere at radial coordinate r has physical area $4 \pi r^2$. Since the spatial geometry in these coordinates is not Euclidean, the radial coordinate r does *not* directly measure "distance from the center".

15. Oct 21, 2012

### pervect

Staff Emeritus
Let me add a note of caution back to this. It's tough sometimes to balance explaining too much vs too little, but I feel impelled not to leave the expalanation at the point it's at, which means expanding it a bit.

I'll start off with saying I agree with everythign Bill K has said so far, it's just that there's even more to the story...

If one considers a fall from a height other than infinity, the equations of motion become

(dr/dtau)^2 = (E^2 - 1) + 2M/r

So depending on E (which is a constant of motion and equal to 1 iif the object is falling from at rest from infinity, and can be thought of as a normalized energy parameter ) dr/dtau can be almost anything.

Furthermore, intervals of dr don't have unit length, because of the metric, ds^2 = dr^2 / (1-2M/r). To get around these issues, one might ask a different question - what is the velocity of the infalling observer relative to a co-located "static observer" who is accelerating to maintain a constant radial coordinate (and a constant distance from the event horizon)? To perform this measurement, the static observer and the falling observer both use their own local clocks and rulers - and they both get the same answer.

After a rather lengthly calculation (https://www.physicsforums.com/showthread.php?p=602558#post602558 posts 29 and 30) one comes up with said relative velocity being

$$\frac{1}{E} {\sqrt{E^2 - (1 - \frac{2M}{r})}}$$

This sort of velocity (which I regard as the most physical of the various sorts computed) approaches uity (i.e. the speed of light) as the object approaches the event horizon - regardless of the value of E.

Static observers don't exist at the event horizon. One can measure relative velocity of the event horizon and an infalling observer , and find it will be equal to 'c', but some caution in interpreting this is needed. The infalling object is not actually reaching the speed of light. What' s happening is that the event horizon is essentially moving outward at the speed of light. Technically, the event horizon can be called a "trapped null surface". You can think of the event horizon as being marked by a trapped beam of light, forever travelling outwards, but never changing it's r coordinate. Because it's like a beam of light, it doesn't have a "frame of reference" of its own. Any newer readers who are still with us at this point might want to consut the FAQ on this point if it isn't obvious.

The free-falling particle does, however, have a frame of reference, and in this frame of reference the "trapped light" moves at the speed of light relative to the particle. Which is only to be expected, as c is always a constant when using local clocks and local rulers.

Last edited: Oct 22, 2012
16. Oct 21, 2012

### Staff: Mentor

Small point: this should be dr^2 / (1 - (2GM / c^2 r) ), correct?

17. Oct 21, 2012

### pervect

Staff Emeritus
I should have pointed out that I used geometric units with c=1 throughout his thread, for the benefit of any newer readers who might be confused by this point - similarly to the way Bill K did. I am rather used to this, but other readers might not be, it's good to mention explicitly.

18. Oct 21, 2012

### Staff: Mentor

Ah, then you probably meant to write ds^2 = dr^2 / (1-2M/r) instead of ds^2 = dr^2 / (1-GM/r).

19. Oct 22, 2012

### pervect

Staff Emeritus
Yes - fixed!