# Gravitational deviation of matter travelling near c by a massive body

1. Sep 8, 2014

Light traveling transverse to a massive body (e.g. Sun) is deviated by an angle twice the amount predicted by Newtonian gravitational theory. This is predicted by GR and proven experimentally.

What would be the deviation of a matter particle traveling near c transverse to a massive body? Would it also be deviated by an angle nearly twice the amount predicted by Newtonian theory? Or would it much more because its relativistic mass/energy would make its gravitational mass very large? Or would it be deviated by the Newtonian amount?

2. Sep 8, 2014

### A.T.

Yes, because of the spatial geometry:
http://mathpages.com/rr/s8-09/8-09.htm

For nearly c, you get nearly twice the Newtonian deflection. In the limit v << c the deflection tends towards the Newtonian value. That's why we can use Newtonian gravitational theory for most stuff.

3. Sep 8, 2014

Agreed.

This is the part that was bothering me. The relativistic mass of the particle traveling near c should be very large. Does that not increase its gravitational mass exponentially? If it does, then why does it not increase GMm/R^2 proportionately? I am extrapolating from the understanding that the mass increase does take place and increases the gravitational acceleration between the two bodies, as seen in the extra gravitational acceleration between the Sun and Mercury that causes Mercury's orbit to precess. I realize this is a somewhat Newtonian interpretation, but does GMm/R^2 X (relativistic corrections) totally fail here?

4. Sep 8, 2014

### A.T.

I don't think this is a correct understanding. The orbit procession can be explained in the same way as the doubling of the light deflection, by spatial geometry. See the bottom picture here:

http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

5. Sep 8, 2014

### Markus Hanke

No, because the source term of gravity is not relativistic mass, but the stress-energy-momentum tensor. Since this is a fully covariant tensorial quantity, it is invariant under Lorentz transformations, so merely putting an object into relative motion at a very high velocity does not somehow give it a huge gravitational influence.

6. Sep 8, 2014

### jartsa

gravitational force = GMm/R^2

gravitational acceleration = gravitational force / m = GM/R^2

In an accelerating rocket drop a marble and fire a laser gun horizontally at the same time. After a while check where the marble and the laser pulse are. They will be at the same altitude, and they will have the same downwards velocity.

On the surface of the earth drop a marble and fire a laser gun horizontally at the same time. After a very short time check where the marble and the laser pulse are. They will be at the same altitude, and they will have the same downwards velocity.

A hammer, a feather, a photon, a relativistic feather, a relativistic hammer, all accelerate downwards the same way.

7. Sep 8, 2014

### A.T.

He is asking about the total deflection angle, which depends on the speed. So it's not the same for a photon, as for a massive object at less than c. Neither in Newtons theory nor in GR.

Last edited: Sep 8, 2014
8. Sep 8, 2014

9. Sep 8, 2014

### DrStupid

10. Sep 8, 2014

Essentially, we can say that for a transverse velocity $v$, the effective acceleration will be the Newtonian acceleration multiplied by a factor of $(1+v^2/c^2)$. Some of the references in this thread also seem to indicate this, and it is consistent with the known deviation of light as well.