# Does relativistic mass create gravity?

1. Mar 23, 2005

### eNathan

I know this question was covered here in the past, but I got some mixed answers. I hope somebody can clear this up for me.

2. Mar 23, 2005

### Hurkyl

Staff Emeritus
Energy density (effectively the "same" as relativistic mass) does indeed contribute to gravity1.

However, momentum also contributes to gravity, and in a way different from the Newtonian "pull things together" picture of gravity. Thus, you cannot simply imagine a speedy object as generating the same type of gravitational field as a stationary object.

1: What I really mean is that it affects the geometry of space-time.

3. Mar 23, 2005

### eNathan

Well, thank you for a clear answer :) Someone told me before that it does not contribute to gravity at all. Is there an equation to express its Gravitational effects? (But of course there is, what is it?)

4. Mar 23, 2005

### Crosson

The relativistic equations of gravity are the Einstein Field Equations. This is a set of 16 coupled non-linear partial differential equations.

5. Mar 23, 2005

### pervect

Staff Emeritus
http://lanl.arxiv.org/PS_cache/gr-qc/pdf/9909/9909014.pdf

is still the best reference I've found online.

The clearest statement that kinetic energy does indeed contribute to gravitational mass is found in the abstract of this paper.

The paper is mainly concerned with how the internal kinetic energy of a system with moving parts contributes to it's "gravitational mass" when the momentum of the system as a whole is zero. The guiding result here is that energy and pressure both cause gravity - but, for a closed system, it appears that the virial theorem requires that the appropriate intergal of energy and pressure be equal to the total energy of the system. (This is what I get from reading the paper, I've been meaning to work out some actual examples.)

If you're interested in the gravitational field of a moving object there is an unfortunate problem. As soon as the velocity gets high enough to significantly affect the gravitational field of an object, one cannot consistently view gravity as only a force - the curvature of space itself becomes important. This shows up in the curvature of light, for instance - it curves twice as much as it ought to.

A qualitiative comparison to the electric field of a moving charge can still be made if one does not want exact results. Basically one expects the field to concentrate in a transverse direction rather than to be radially uniform. To really do the problem right requires that one analyze the problem in terms of tidal forces (the Riemann tensor), rather than the "gravitational field".

6. Mar 24, 2005

### Chronos

Under GR, the answer is simply yes. A body in motion posseses kinetic energy plus mass. Energy and mass are equivalent under GR. Particle colliders routinely confirm this prediction. Crash two particles together at relativistic velocities and you create particles with more mass than the sum of the two crashed particles.

7. Apr 12, 2010

### nearc

The ‘speed’ of gravity, like everything else, is also limited to the speed of light. So if two bodies [with mass] are traveling in the same direction with one trailing directly behind the other by some distance and both bodies are traveling at .999 C how does the gravity work between them? For example: If body A is in front and body B is behind. Then A will not feel the gravitational effects for some time. In fact if A is 1 light year ahead of B then it will take 999 years before A will feel the gravitational effects from B, however B will ‘run into’ A gravitational effects in about half a year?

B---> A---> .999 C
|---one light year---|

8. Apr 16, 2010

### cos

In the thread 'Mass dilation' I referred to Colin Ronan's comment in Deep Space that when a particle is accelerated in a cathode ray tube it will curve downwards and must be bought back to its horizontal path using a magnetic field applied beneath the particle.

Ronan's additional comment, that if the particle is accelerated to a greater instantaneous velocity it will curve further down as a result of the planet applying a stronger gravitational force thereby requiring the application of an increased force beneath the particle, has been ridiculed presumably on the basis of the particle's increased inertia i.e. the planet's gravity creates a vertical displacement of 9.8m/sec/sec irrespective of the particle's mass as determined by Galileo.

On the basis that an increase in the particle's velocity-mass exponentially increases the particle's inherent gravitational field strength it seems that Ronan's comment is correct.

The increasing mass-gravitational field strength of the accelerated particle combines with the increasing gravitational field strength of the planet.

So, as Ronan points out, when a particle starts accelerating it requires a certain amount of force applied beneath it in order to maintain a horizontal trajectory but as its velocity increases (and approaches light speed where the gamma factor exceeds 400000) that corrective force has to be proportionally increased in order to overcome the rapidly, exponentially, increasing mutual gravitational attraction?

9. Apr 17, 2010

### Dmitry67

Most of the mass (protons and neutrons) comes from the relativitic mass of quarks.

10. Apr 17, 2010

### Rasalhague

11. Apr 18, 2010

### Jorrie

A similar case has been discussed in the "Long planet and Galileo" thread. At that time I thought that the derivation made by Pervect for Schwarzschild coordinates in this post showed that what you described may be a coordinate independent effect. AFAIK, the issue was not quite resolved in that thread.

As I now understand it, the effect is coordinate dependent and can be 'gauged away', as Carlip wrote in Kinetic Energy
and the Equivalence Principle
(pdf referenced above). It seems that things fall at the same acceleration in any single fame, irrespective of 'horizontal' velocity.

But, I'm not sure that I understand this correctly...

Last edited: Apr 18, 2010
12. Apr 18, 2010

### starthaus

It hasn't been "ridiculed", I simply showed the the equations of motion for a charged particle in a gravitational field do not depend on its relativistic mass (see Rindler's book).
Ronan is confused, the effect is due to the Lorenz force qvxB.
The Lorentz force curves the particle path, this is why the CRTs need to be "degaussed".

Last edited: Apr 18, 2010
13. Apr 18, 2010

### cos

Ronan's comment has been ridiculed in other groups!

Perhaps Ronan was addressing his comment to ignorant types such as myself (in his physics popularization publication) in a fashion that we might more easily understand rather than him talking about the Lorenz force and providing that equation.

You say 'the effect is due to the Lorenz force'. What effect?

Are you talking about Ronan's suggestion that the faster accelerated particle will fall further down on the target screen (i.e. 'an effect')?

It is not the theoretical Lorenz force that curves the particle's path! The particle's path is curved by gravity!

14. Apr 19, 2010

### cos

A point is that the two particle's, having been accelerated to different instantaneous velocities (particle B has been accelerated to a higher speed than particle A), will not have been accelerated at the same rate ergo are not in the same, single, frame either whilst accelerating (at different rates) or when they strike the target at different speeds.

Imagine two identical particle accelerators alongside each other; one of them applies a certain amount of force to make its particle (A) accelerate whereupon that particle strikes the target screen at a point below the horizontal.

The other device applies a much greater force of energy whereupon that particle (B), according to Ronan, strikes the target at a lower point than the first particle due to the fact that when it reaches the target, B possesses a higher instantaneous velocity hence a greater relativistic mass-gravitational field strength than A which, combined with the planet's proportionally increased gravity (perhaps resulting in an increased mutual gravitational attraction) creates that additional deviation.

15. Apr 19, 2010

### starthaus

He's wrong either way.

Charged particles moving in a magnetic field are subjected to a force that depends on the cross product between the field induction (B) and their instantaneous velocity (v). The Lorentz force is responsible for curving the trajectories of charged particles in particle accelerators and in CRTs. This is why CRTs need to be degaussed. This is also why the degaussing is dependent on the CRT location on the Earth surface.

Yes, except that Ronan appears not to understand the effect.

You didn't even know what the Lorentz force is, so how can you claim the above?
You couldn't follow the equations from Rindler's book, so what entitles you to make the above claim? Are you here to learn or to make fringe statements?

16. Apr 19, 2010

### Jorrie

Yes, but we can choose to measure both particle's accelerations in one single inertial frame. In that frame they both fall at the same acceleration, while in the surface (or accelerator) non-inertial frame, they do not. The problem is that the inertial frame must be free-falling towards Earth and can only be momentarily stationary relative to Earth. At the same time your two particles will take time (possibly different times) to travel the distance to the target, complicating the issue.

If the two particles were to strike the target at the same time, the faster one had to depart later, or it had to travel a different distance, with the result that they were on different spacetime geodesics, another complication. If they were 'fired' horizontally at the same place and time, but at different speeds, the faster one would have hit the target earlier and higher than the slower one - by simple orbital principles.

All very confusing...

17. Apr 19, 2010

### Jorrie

Apart from Pervect's equations that I referenced above, I based the latter part of this quote also on the fact that the locally measured circular orbital velocity of a particle in Schwarzschild coordinates, as measured by a momentarily stationary local inertial observer, is given by

$$v_o^2 = \frac{GM}{g_{tt} r}$$

where $g_{tt} = 1-2GM/rc^2$, the time-time coefficient of the metric.

This means the acceleration from a straight line in gravity-free space (centripetal) is

$$a_{cp} = -\frac{v_o^2}{r} = -\frac{GM}{g_{tt} r^2} = -\frac{GM}{r^2}(1+2v_o^2/c^2)$$

This is certainly a 'pseudo acceleration', because an accelerometer will measure zero. However, if we reproduce the same radius and speed in flat spacetime by means of a string in the place of gravity, an accelerometer will measure this centripetal acceleration, not so?

If so, then in a way it is motivation for "horizontally fast moving things fall faster in a gravitational field". (?)

Last edited: Apr 19, 2010
18. Apr 19, 2010

### yuiop

Hi Jorrie,

Not seen you around for a while. If that is the case, welcome back :)

In that old thread, my basic argument was that the Equivalence Principle demands that to first order and locally that the vertical acceleration of a falling particle is independent of its horizontal velocity from the point of view a stationary observer on the surface of the gravitational body. For example, if two particles are initially at the same location and one (A) has zero horizontal velocity and the other (B) has a horizontal velocity of 0.8c, then both will hit the ground simultaneously (from the point of view of the observer at rest with the ground and ignoring orbital effects).

I think it is also important to note that the equal downward acceleration of two particles, does not imply that an equal force of gravity is acting on the same particles.

Now if an observer is moving horizontally at 0.8c, so that he is at rest with particle B, the relativity of simultaneity suggests the two particles will not hit the ground at the same time (and so do not have the same downward acceleration from the point of view of the second observer). This is at odds with Carlip's claim that "things fall at the same acceleration in any single frame" unless the relativity of simultaneity can be "gauged away" by gravity, which seems dubious to me. If I understand (your interpretation of) Carlip's claim correctly, he is in effect claiming that if two spatially separated events are simultaneous in one frame, then they are simultaneous in all inertial frames, which flies in the face of all we know about relativity and using fancy terms like "gauged away" does not hide that flaw in his argument. I suspect the problem lies in your interpretation of what Carlip meant to say, but I have only had a brief look at his paper. It is interesting that in the final paragraph, Carlip states "We can thus tell our students with confidence that kinetic energy has weight, not just as a theoretical expectation, but as an experimental fact."

In other words:
A hot brick weighs more than a cold brick.
A rotating brick weighs more than a non-rotating brick.

19. Apr 19, 2010

### yuiop

This is true when "curvature" is taken into account in the case of a realistic spherical gravitational body when considering orbital motion around the body.

The effective force of gravity per unit mass can be written as:

$$F = \frac{GM}{r^2} + \frac{3GMv^2}{r^2c^2} - \frac{v^2}{r}$$

where v is the instantaneous tangential horizontal or orbital velocity. The first term on the right is the familiar Newtonian equation for gravity and the last term is the familiar Newtonian equation for centrifugal force acting in the opposite direction to gravity for an orbiting particle. The middle term on the right is an additional term unique to GR that can be thought of in terms of the effect of curvature of space (maybe). The above equation can be re-written as:

$$F = \frac{GM}{r^2} + \left(\frac{3GM}{rc^2} -1 \right) \frac{v^2}{r}$$

In this form, it can be seen that when r is less than the photon orbit (r<3GM/c^2) the second term on the right becomes positive and both terms on the right act inwards. At this radial location, the greater the horizontal velocity, the greater the inward force becomes.

Note that outside the photon orbit, the "centrifugal force" acts outwards as we would normally expect in the Newtonian aproximation.

Also note that there is no apparent reference to relativistic mass in the above equations. However it should be remembered that the above equations are effective force per unit mass and there is no direct reference to the mass of the orbiting particle. When the mass of the orbiting particle is taken into consideration, things may be different, but that is another subject for another post or thread.

20. Apr 19, 2010

### Jorrie

Hi Kev, thanks; yes, been away for a while.
I think this is correct only in the case of that "long, flat Earth", where the ground observer is more or less equivalent to an inertial observer. For a realistic body, I think it is only any single free-fall frame that observe those particles falling at the same acceleration (and hit the 'ground' at the same time).

Nope, I interpreted what he said more like: 'the observed accelerations of such free falling particles are coordinate choice dependent'. I don't think there is much doubt that in Schwarzschild coordinates they fall at accelerations that are velocity dependent.

Anyway, to me this is still a somewhat confusing issue and would very much like to hear Pervect's take on it.