# Why can't gravity accelerate an object past c?

1. Jul 16, 2008

### Sbratman

The fact that it is impossible to accelerate an object to (or past) the speed of light is typically explained as connected to the relativistic increase in inertial mass that occurs with increasing velocity; eg., for any given force, the acceleration produced by that force will decline as the the object approaches the speed of light. But what if the force is gravity? Since gravitational force exactly matches inertial mass, wouldn't gravity exert a constant rate of acceleration? I would suppose that the answer lies in general relativistic "corrections" to the Newtonian gravitational force law, but I've never seen it explained.

2. Jul 16, 2008

### haushofer

Well, properly spoken gravity isn't a force in general relativity, because of the equivalence principle. A gravitational field can locally always be transformed away by going to an accelerating frame and applying special relativity. This is the whole idea of space-time being a manifold.

You say that "gravitational force matches inertial mass", but what does that mean? The fact that the mass which causes inertia and the mass which causes a gravitational field are the same doesn't imply that an object can reach a speed v>c. Why would it?

You also speak of "general relativistic corrections to Newton's gravitational law", so I think you mean that we should somehow extend the scalar gravitational potential in a general relativistic way? This turns out to give nonsense, and eventually we turn to the notion of gravity being space-time geometry. The scalar potential of Newton ( with respect to 3D Cartesion space ) is hereby replaced by the metric tensor ( of 4D space-time ).

So the problem is that general relativistically we can't write down a four-acceleration due to a gravitational field :)

3. Jul 16, 2008

### Mentz114

Welcome, Sbratman.

Inertial mass is a property of matter that is not observer dependent, so this is a poor explanation. The rules are built into special relativity where the hyperbolic nature of the space-time keeps everything travelling below c. General relativity inherits this to a degree.

Its interesting that if we caculate the velocity of a test body falling from infinity towards a Schwarzschild black hole, that the velocity of the body would reach c exactly on the event horizon. Neat.

M

4. Jul 16, 2008

### yuiop

This raises the question of what happens if the initial velocity of the test particle at infinity is a significant fraction (say 0.99) of the speed of light, what will its terminal velocity be at the event horizon? It is difficult to find a published equation that takes the initial velocity of the falling onject into account but it easy to make some observations that the acceleration due to gravity is velocity dependent. The coordinate acceleration of gravity acting on a stationary body is proportional to GM/R^2(1-R_s/R) and as R tends towards R_s the coordinate acceleration tends towards zero. The coordinate speed of a falling photon is given by c'=c(1-R_s/R) so the final coordinate velocity of a falling photon at the event horizon is zero so the photon is obviously being decelerated as it falls. This suggests that the coordinate acceleration is velocity dependent and above a critical velocity at a given height, gravity decelerates rather than accelerates falling objects.

Some people prefer to work with local measurements as made by a stationary observer local to where the measurement is being made. The local observer always measures the velocity of the falling photon to be c and the local acceleration due to gravity is proportional to GM/(R^2*sqrt(1-R_s/R)) which goes to infinite acceleration at the event horizon. The difficulty with local measurements is that if the acceleration at the event horizon is infinite then there can not be a stationary observer at the event horizon. Below the event horizon the local acceleration becomes imaginary (but the coordinate acceleration remains real).

5. Jul 16, 2008

### Sbratman

Obviously, I'm missing a great deal, but it's commonly stated (including by Feynman in those old books of his) that while the structure of bMInkowski space itself demonstrates that a moving object can't reach the speed of light, this has to work out in a practical way for a spaceship trying to accelerate itself to the speed of light. From the perspective an inertial frame through which a spaceship is trying to accelerate, its nice ion drive capable of inducing a constant 1 G acceleration at rest begins to produce a lower rate of acceleration as the spaceship grows more massive, and so light speed can't be reached.

However, suppose the spaceship has gotten up to .9999 times the speed of light, and then approaches, say, a neutron star whose escape velocity is .9999 times the speed of light. These numbers are chosen a bit arbitrarily, but, anyway, it's easy to come up with numbers so that the spaceship exceeds light speed if F=GMm/r^2 continued to hold. The essential difference is that, as opposed to the ion drive, the gravitational force follows the weak principle of equivalence (gravitational mass = inertial mass), and so there is no diminution in the ability of gravity to accelerate the object.

But maybe even the idea of a spaceship failing to keep accelerating due to mass increase is simply a heuristic example Feynman used, and not at all representative of the real situation. (I'm not in any way doubting the conclusion, just looking for a qualitative understanding.)

Detail of example: The (Newtonian) escape velocity from a planet of radius r and mass M is the square root of 2GM/r; conversely this is the velocity gained by dropping something from infinity onto that planet. The velocity gained by dropping an object from twice the radius down to the planet surface is the square root of GM/r. So if the object being dropped is negligible in size, it will increase in velocity by (almost) the escape velocity / sqrt2. Or .7 C in the case of a very heavy neutron star whose surface escape velocity is very near to the speed of light.

For example, the famous "oh my God particle" discovered a few years back, a proton going 1 − (5×10−24)] times c, would, unless some other effect takes over, have been accelerated past the speed of light by the earth itself.

6. Jul 16, 2008

### DaveC426913

No. You're forgetting time dilation. From the spaceship's point of view, there is no loss in acceleration.

7. Jul 16, 2008

### Sbratman

I probably need a more qualitative explanation. :-) The "weak equivalence principle" is that gravitational mass = inertial mass. Eg, this is the reason why heavy bodies fall as fast as light bodies. So far as I understand it, Einstein explained this "weak equivalence principle" with a much stronger equivalence principle, that gravity is, in a sense, a geometric pseudoforce equivalent to acceleration; this not only explains the weak equivalence principle, it lead to unique field equations for general relativity. However, I don't understand general relativity at all at the mathematical level. At some point f=ma and/or F= GMm/r^2 is going to have to break down when gravity is the "force," else a body being accelerated by gravity will pass light speed under various scenarios. The problem here is the weak equivalence principle -- other forces besides gravity are unable to accelerate an object to light speed "because" the the mass of the object increases; gravity itself, though, would have no such problem.

I suspect that the whole idea of decreased acceleration due to increased mass is simply a heuristic one used by Feynman and others. But it's a bit odd that the apparent gravitational exception to this heuristic explanation is left out. In essence, does the weak equivalence principle cease to hold as velocity approaches light speed? (Again, please remember that though I understand special relativity fairly well mathematically, my understanding of general relativity is only qualitative.)

8. Jul 16, 2008

### Sbratman

That's why I said "from an inertial frame observing the spaceship." But I'm really just quoting the heuristic explanation all kinds of people (including Feynman) give for why one can't accelerate to the speed of light. Actually, he (and others) use it as a lead-in to relativistic mass; the idea is that since it becomes harder and harder to accelerate a spaceship as light speed nears, this is very much like the spaceship gaining inertial mass. He goes from there to properly derive the energy/momentum four vector. Maybe the heuristic example is just a bunch of handwaving. Alternatively, maybe the "weak equivalence principle" (that gravitational mass = inertial mass) breaks down near light speed. Or something else entirely?

Please remember that I only understand general relativity qualitatively; I have at least a basic understanding of special relativity mathematically.

Thanks ....

9. Jul 16, 2008

### Sbratman

If the acceleration due to gravity is velocity dependent at some point, that would solve the problem. But note that one doesn't need a black hole to cause acceleration past C (using the Newtonian force law, and the weak equivalence principle, which obviously must fail somehow, but I don't know how.) In one of my other posts I mentioned what would happen if one had a neutron star massive enough to have an escape velocity of, say, .999c. It's easy to come up with scenarios that cause a velocity > c. (see my post that mentions the "oh my god particle," that famous proton going ridiculously near the speed of light)

10. Jul 16, 2008

### Mentz114

SBratman, Kev,

thinking of gravity as 'accelerating' anything in GR is straining the analogy. It is space-time geometry and not force that causes the changes in motion. We don't observe things reaching the speed of light because it is barred by the geometry. If you try to account for this with Newtonian F=Ma thinking - you'll go mad.

SBratman:
Me too.

Kev:
One can see from the geodesic equation that coordinate acceleration is proportional to velocity2, and the factor that connects them is a Christoffel symbol ( affine connection).

M

11. Jul 16, 2008

### kahoomann

The problem with this explanation is that, by itself, relativistic mass does not prevent the observation of superluminal objects. For it does not imply the addition of velocity formula and therefore two observers in two IRFs moving in opposite directions may observe the other to be traveling at a speed greater than c. Thus it is not a sufficient reason for the existence of a fundamental speed limit for any object.

12. Jul 17, 2008

### MeJennifer

Gravity does not accelerate objects, at least not in an proper acceleration sense, however it does maniipulate the distances between objects and depending on how you chart it it can appear that objects move faster than light in relation to each other. For instance in the case of an expanding universe using a comoving coordinate chart.

13. Jul 23, 2008

### methane

When electron and positron meet and explode it produce energy. Space absorbed that energy. but what happened then? Was Einstein's cosmological constant wrong?

14. Jul 23, 2008

### DaveC426913

The energy is emitted as gamma ray photons.

15. Jul 23, 2008

### shamrock5585

awesome explanation!!! ;-)

16. Jul 27, 2008

### methane

what is Geon? does space have any characteristics of its own beyond mass?

17. Jul 28, 2008

### yuiop

An intuitive alternative explanation to relativistic mass being the reason that objects with rest mass can not be accelerated to the speed of light is time dilation. As the rocket gets faster relative to an observer, time dilation slows down chemical and nuclear reactions in the rocket so that that propellants are ejected slower and slower rate as the rocket's relative velocity approaches c. However, there is a big problem with rejecting the concept of relativistic mass altogether and that is consideration of elastic collisions of particles.

Imagine a particle of mass m1=1 and velocity u1=0.8c colliding with a stationary particle of mass m2=2 in a straight head on elastic collision. Newtonian calculations predict that the larger mass will move away with a velocity of 0.53333c while the smaller mass will rebound with velocity -0.26667. By taking into consideration the relativistic masses of the particles the final velocities of the small and large masses are -0.32432c and 0.64234c respectively. Actual experiments show that the final velocities agree with the assumption of relativistic mass. It is difficult to explain the results of these elastic collisions without the use of mv/sqrt(1-v^2/c^2) instead of just mv. Suggestions that it is the velocity rather than the mass that is increased by a factor of gamma is easily proven to be false by timing the particle over a given distance. Can you show how the final velocities of -0.32432c and 0.64234c are obtained with assuming relativistic mass?

Relatistic mass does not need to imply the relativistic addition formula as the formula is explained by length contraction of rulers and time dilation of clocks and the relativity of simultaneity. Relativistic mass ensures that no object moves at greater than c relative to any single inertial observer. The relativistic addition formula by itself does not always guarantee that the relative velocity of one observer relative to another is always less than c. For example if observer A is moving with velocity vA=2c and observer B is moving with velocity vB=-0.1c the velocity of A relative to B is 1.75c. By ensuring that neither observer moves faster than c in a given IRF in the first place, relativistic mass ensures that the relativistic addition or subraction formula always gives a result that is less than c.

18. Jul 28, 2008

### yuiop

I have found a reference that shows gravity IS velocity dependent here: http://www.mathpages.com/rr/s6-07/6-07.htm

They give the velocity dependent gravitational acceleration equation for a particle falling from infinity as:

$$a = \frac{-GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)\left(3\left(1-\frac{2GM}{rc^2}\right)\left(1-\frac{V^2}{c^2}\right)-2\right)$$

where $$V^2/c^2$$ is the initial downward velocity of the particle at infinity. For a photon, V/c=1 at infinity and the equation reduces to:

$$a = \frac{2GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)$$

Note that the equation for the acceleration of a falling photon is always positive for r>2GM/c^2 meaning that a falling photon is decelerated all the way from infinity to the event horizon starting with a velocity of c and finishing with a coordinate velocity of zero. This is consistent with the velocity of a falling photon derived directly from the Schwarzchild metric:

$$c' = \frac{dr}{dt} = c\left(1-\frac{2GM}{rc^2}\right)$$

As can be seen from this diagram http://www.mathpages.com/rr/s6-07/6-07_files/image017.gif the greater the initial velocity the greater the deceleration of gravity acting on it bringing it to zero velocity in coordinate terms (or c in local terms) at the event horizon. The velocity never exceeds c locally no matter how fast the intial velocity. This is due to a antigravity effect that is not often mentioned. Mathpages makes this antigravity effect clear in this quote:

" Notice that the value of (d2r/dt2) / (-m/r2) is negative in the range from r = 2m to r = 6m/(1 + 4m/R), where d2r/dt2 changes from negative to positive. This signifies that the acceleration (in terms of the r and t coordinates) is actually outward in this range."

and in this quote too:

"This shows that the acceleration d2r/dt2 in terms of the Schwarzschild coordinates r and t for a particle moving radially with ultimate speed V (either toward or away from the gravitating mass) is outward at all radii greater than 2m for all ultimate speeds greater than 0.577 times the speed of light. For light-like paths (V = 1), the magnitude of the acceleration approaches twice the magnitude of the Newtonian acceleration – and is outward instead of inward. The reason for this outward acceleration with respect to Schwarzschild coordinates is that the speed of light (in terms of these coordinates) is greater at greater radial distances from the mass.

By accelerating I mean that the velocity is increasing over time or the distance traveled per unit time is increasing. In this context it is not important whether the cause of the acceleration is gravity or geometry. If a person jumps off a high tower the situation can be explained by the person being stationary and the tower and the Earth attached to it that is actually accelerating upwards to meet him (as can be proven by an accelerometer) or you could say he is being accelerated downward by gravity (ridiculous, I know :rofl: ) or you could say he is simply moving along a geodesic. Either way you know it is going to hurt. The point is that a the velocity of a falling object (whether caused by gravity or geometry) is not always increasing as it falls towards a gravitational mass. If you look very carefully at the graphs on the Mathpages page you will see that an object shot outwards from a radius of less than 2Rs can increase it velocity as it moves AWAY from the gravitational mass and only start slowing down when the radius exceeds 3Rs.

Mathpages obtains the parameter K from the affine parameter on this page http://www.mathpages.com/rr/s6-04/6-04.htm and uses the parameter K interpreted as:

$$K=\sqrt{1-V^2/c^2}$$

or

$$K=\frac{1}{\sqrt{1-2GM/(Rc^2)}}$$

that he uses in the equations used on the page I discussed above http://www.mathpages.com/rr/s6-07/6-07.htm