Does mass increase in SR mean higher gravity in GR?

1. Dec 23, 2004

I812

It has been years since I have thought of relativity, but I came across a book by Jerome Drexler in which he attempts to support his view that ultra high energy protons (cosmic rays) are what dark matter is all about in the galactic halos. In spite of the fact that it seems to me these protons would interact and emit radiation (hence be detectable), he argues that the mysterious gravitational influences from these photons is due to the relativistic mass increase due to their high velocities. This doesn't make sense to me. If this were true, it seems like I could tie a ball to a string and twirl it around fast enough to create enough relativistic mass to bring the moon crashing down to Earth! Obviously this is crazy, but I think the idea is the same. Does anyone have any explanations or thoughts as to what gravitational effects fast moving objects would have. It does not make sense to me that they would have any at all.

2. Dec 24, 2004

Andrew Mason

You would need a very strong string. And alot of energy. But in theory, you are correct.

According to SR a mass that experiences a gain of kinetic energy (eg. by absorption of a photon) experiences an increase in its mass (relativistic mass). GR says that relativistic mass contributes to gravitation the same as rest mass. I am not adept enough in GR mathematics to elaborate, but I think that is essentially the principle.

AM2

3. Dec 24, 2004

Chronos

Incorrect. The energy used to accelerate an object must be subtracted from the apparent relativistic mass gain. The net effect is zero.

4. Dec 24, 2004

pervect

Staff Emeritus
Well, I've tried to work that particular problem (the gravitational field of a fast moving body) out two different ways - and I've gotten two different answers!

They agreed qualitatively, but not quantitatively :-(. Somoene else here has worked on the problem, too, but I don't really believe his answer either.

I do have a paper which works out the results for low velocity, though.

Qualtiatively, though, the field will increase in the transverse direction, i.e if X marks the "spot" (particle).

X----> direction of motion
|
|
\/ max field

There will also be a "magnetic" like field that will affect moving particles (but it won't matter to the force on a stationary particle) this is known as "frame dragging", or also as gravitomagnetism.

For the electric field of a relativistically moving charge (which will be generally similar, but not exactly the same as the gravitational result) see

java applet

for this java applet, note that the field strength is equal to the density of the field lines.

and if you'd rather see the equations, this link has them

equation form

5. Dec 24, 2004

Andrew Mason

Not if you add energy to the system - which you would obviously have to do in order to give the rotating mass such a huge relativistic mass that it causes the moon to fall out of its orbit. The energy added would have to be much greater than the energy contained in the mass of the earth (ie. earth mass all converted into energy). As I say, you would need a strong string.

AM2

6. Dec 25, 2004

pmb_phy

Nobody said the object was being accelerated. The question was about the mass and gravity as a function of speed. One can simply change their frame of reference to observer the effects. That way even a black hole can be observered from two different frames of references with merely a change in speed of the observer ... which requires much less energy than trying to accelerate a black hole, :rofl:

Pete

7. Dec 25, 2004

Chronos

No object can change speed without being accelerated. The only relevant factor is the reference frame from which you measure it.

Last edited: Dec 25, 2004
8. Dec 25, 2004

pmb_phy

If there is a star at rest in my frame of reference and I move to a new frame of reference which is moving relative to the old one with a speed v = 0.9c. I will now measure the speed of the star to be v = 0.9c. How much work did I do on the star? Where did that energy come from?

However if the person swinging the ball got energy from outside the Earth to swing the ball then the gravitational field of the Earth/ball system would increase.

Pete

Last edited: Dec 25, 2004
9. Dec 25, 2004

gonzo

Actually, I would be really happy if someone could give me a good explanation of how kinetic energy works in special relativity.

I mean, if a 1 kg rock hits the earth at 100 m/s, you usually calculate the energy of that collision by using the mass of the rock. Why don't you use the mass of the Earth? Why isn't the Earth moving towards the rock instead?

10. Dec 25, 2004

yogi

Einstein concluded in one of his papers that a particle's increase in inertial mass due to motion would result in an increase in its gravitational mass As Andrew Mason has said. I don't know of any actual reported experiments however - anyone have a reference on this. It does seem to raise some paradoxical questions- but it also seems that it must follow from the equivalence principle.

11. Dec 26, 2004

Andrew Mason

Of course, the measure of kinetic energy is frame dependent. I can give the entire earth more rotational energy in my frame of reference by travelling west in my car along the equator at a very high speed. But I can only collide with a tree once and that energy is lost.

The gravitational effect, however, is the same in both reference frames. Since gravity is the product of both masses, it doesn't matter which mass the $\gamma$ factor applies to. It is still $F \propto \gamma m M/R^2$

AM

12. Dec 26, 2004

pervect

Staff Emeritus
See either of the two links I posted (the java applet, or the equations) for why the electric field of a charge is NOT just qE1 E2 / R^2 when the charge is moving at a relativistic velocity.

Columb's law does not work for relativistically moving charges. Similarly, Newton's law of gravitation will not work for relativistically moving masses.

13. Dec 26, 2004

pervect

Staff Emeritus
OK, here's the good concise explanation of the problem you asked about.

The physically significant quantity in a collision between two masses is the energy in the center-of-mass frame.

If a 1kg rock hits the earth at 100m/s, the center of mass frame is essentially that of the Earth.

You can acutally work the problem of the rock hitting the earth out in any frame you desire, but the center of mass frame is the simplest, and the most standard.

Note that this is why particle accelerators go to the bother of creating anti-protons to slam into protons with an equal and opposite velocity. The energy that counts is the energy in the center-of-mass frame, and much of this energy is lost when a hyper-relativistic particle strikes a stationary target. You get well over double the energy in a "head-on" collision.

14. Dec 26, 2004

pmb_phy

That is an incorrect expression for the gravitational force in GR.

The active gravitational mass of a moving body is not $\gamma M$, its $\gamma M(1+\beta^2)$.

See - http://www.geocities.com/physics_world/gr/moving_body.htm

Pete

15. Dec 26, 2004

pervect

Staff Emeritus
I still don't see where you get eq 3a) in your derivation - I was hoping that the paper you cited by Harris would answer this question, but it does not.

Harris ("Analogy between general relativity and electromagnetism .......") gives the formula

$$m\frac{d^2 x^u}{d \tau^2} + m\Gamma^u{}_{ab}(\frac{dx^a}{d \tau})(\frac{dx^b}{d \tau})$$

his eq 1a), as the formula for deriving "force", which I do believe, at least as long as one is in a coordinate basis (the last point may seem picky, but it was the source of at least some of my earlier difficulties).

16. Dec 26, 2004

pmb_phy

17. Dec 26, 2004

pervect

Staff Emeritus
I should add, that by applying the above definition, for the metric for Newtonian gravity ($$U = - \Phi$$)

(1+2U)dx^2 + (1+2U)dy^2 + (1+2U)dz^2 - (1-2U) dt^2

with a 4-velocity
$$(t,x,y,z) = (\gamma, \beta \gamma, 0, 0)$$

$$\beta = v/c, \gamma = 1/\sqrt{1-\beta^2}$$

I'm getting

$$\frac{d^2 x}{d \tau^2} = \frac{\frac{\partial U}{\partial x} - 2 \gamma^2 \frac{\partial U}{\partial t}}{1+2U}$$

the main concern is the term in the partial of U with repsect to t (as U is not necessarily indpedent of time depending on the coordinate system used).

also
$$\frac{d^2 y}{d \tau^2} = \frac{1+\beta^2}{(1+2U)(1-\beta^2)} \,\frac{\partial U}{\partial y}$$

which also didn't seem compatible

Last edited: Dec 26, 2004
18. Dec 26, 2004

pervect

Staff Emeritus
http://www.geocities.com/physics_world/gr/eq_gif/gr01-eq-08.gif

conflicts with Harris's article (Anology between General Relativity and electromagnetism for slowly moving paticles in weak gravitational fields) and MTW, which give

$$m\frac{d^2 x^u}{d \tau^2} + m\Gamma^u{}_{ab}(\frac{dx^a}{d \tau})(\frac{dx^b}{d \tau})$$

instead of what you wrote, so that the equivalent 4-force is

$$-m\Gamma^u{}_{ab}(\frac{dx^a}{d \tau})(\frac{dx^b}{d \tau})$$

If we use the above formula to calculate the 'x' component of the 4-acceleration with a 4-velocity $$\beta$$ in the x direction, we get

$$\frac{d^2x}{d \tau^2} = -\frac{\partial U}{\partial x} / (1 + 2U)$$

which reduces to

mx / R^3 in the limit where R >> m, R being of course x^2+y^2+z^2

(This is because the metric at this point is static, we consider the mass stationary and the particle to be the only thing that is moving).

This compares very well with my "method two" which was based on consideration of the formula (for a Schwarzschild geometry)

$$rdot = \frac{dr}{d \tau} = \sqrt{E^2-(1-2*M/r)*(1+L^2/r^2)}$$

We can use this to calculate $$\frac{d^2 r}{d \tau}^2$$ as follows

$$\frac{d^2 r}{d \tau^2} = \frac{d rdot}{d r} \frac{dr}{d \tau} = \frac{d rdot}{dr} rdot$$

and when L = 0, for an radially infalling/outgoing particle, we get $$\frac{d^2 r}{d \tau^2} = -M/r^2$$ regardless of the value of rdot, which is the same as the previous result.

Furthermore, I realized the "simple" way to work this problem is to work entirely within the simple static metric, then convert the 4-acceleration to a new set of moving coordinates after the 4-acceleration is computed in the "mass centered" coordinates via a final Lorentz boost.

I still get rather strange-looking results with a factor of (1-3v^2) in them for the x-acceleration after doing the final boost, but I'm beginning to believe them, though my explanation for the change in sign of the force at v^2 = 1/3 (!!) is a bit strained at this point.

19. Dec 27, 2004

pmb_phy

I don't recall MTW using a definition of force and I lost my copy of Harris.

The definition I used is dP_u/dt where u is a covariant index and let u = k = 1,2,3. Had I not used this definition the gravitational force would not reduce Newton's expression in the Newtonian limit. Had I use proper time rather than coordinate time then a moving body would not weigh more than the same body at rest, which it must. Both Moller and Mould use this definition. Ohanian uses dP_u/d(tau)

Pete

20. Dec 27, 2004

pervect

Staff Emeritus
Last edited: Dec 27, 2004