Relativistic mass and space-time curvature

[redtree]

Does relativistic mass curve space-time, i.e., does relativistic mass affect the gravitational field of an object?

 

[mitesh9]

Yes, I suppose!
Do you mean to ask, is the curvature of spacetime greater for relativistic mass, then the rest mass? If so, I would say yes.

 

[dst]

I would love to know. What about say, a neutron star/supernova remnant/whatever on the brink of collapse into a black hole? Does the additional relativistic mass cause it to collapse into a black hole? That would make no sense since in the observer's frame they would have a black hole while from the POV of the black-hole-to-be, it would be just that.

 

[mitesh9]

I think that's what relativity is all about. It is about what would an observer observe.
Say, two space ships are traveling in same line and opposite direction with a relative velocity v. Each will observe other's clock to be slow! (though, I'm yet to understand it fully )
So we can not say, what would have really happened, we can only say, what would an observer observe. Do correct me if I have missed something...

 

[Mentz114]

mitesh9:

So we can not say, what would have really happened, we can only say, what would an observer observe. Do correct me if I have missed something...

If two observers watching the same event, saw different outcomes to the same experiment, that would be a problem. Usually different observers will see events happening on different time scales, but they must see the same outcome or else there would a contradiction.

For instance, the Lorentz transformation 'explains' why different inertial observers always see the same outcome to an EM experiment.

 

[jtbell]

Does the additional relativistic mass cause it to collapse into a black hole?

See the following FAQ:

http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/BlackHoles/black_fast.html

(hmmm, I see that page has been updated recently. It used to be a lot longer.)

 

[mitesh9]

mitesh9:


If two observers watching the same event, saw different outcomes to the same experiment, that would be a problem. Usually different observers will see events happening on different time scales, but they must see the same outcome or else there would a contradiction.

For instance, the Lorentz transformation 'explains' why different inertial observers always see the same outcome to an EM experiment.

Well, In the domain of SR, different observers, according to their position, are bound to see different things. For instance, an observer observes a meter stick to be of 1 meter, and the same meter stick is less then 1 meter for an observer with a relative velocity! As about the EM experiment, they won't notice any difference, just because the speed of light is constant, relative to which everything else is measured! And yet, in an experiment concerning the frequency of light, an observer may observe red shift/blue shift, depending on his position.

 

[pmb_phy]

Does relativistic mass curve space-time, i.e., does relativistic mass affect the gravitational field of an object?

Yes. In fact MTW state that relativistic mass is the source of gravity.

Pete

 

[MeJennifer]

Feel free to quote the chapter and page where you think this is claimed.

Relativistic mass, unlike spacetime curvature, is observer dependent.

 

[peter0302]

Wow, I'm surprised there hasn't been a straight answer yet.

Let me ask this question:

If I am moving toward the Earth at a speed very close to c, and I wish to calculate the instantaneous force of the Earth's gravity on me at any given point in time, do I use the rest mass of the earth, or do I multiply the rest mass of the Earth by my Lorentz factor?

 

[RicFlair]

If you're observing a Newtonian like gravity in a reference frame where you need to account for special relativity then you need to make a relativistic coordinate change from a stationary General Relativity solution. P.C. Aichelburg and R. U. Sexl, Gen. Relativity and Grav. 2, 303 (1971) makes a similar coordinate change. If you make this coordinate change using the Schwarzschild solution then the term that normally reduces to the Newtonian potential becomes the Newtonian potential with the relativistic mass and a contracted coordinate. At distances where a Newtonian type gravity would be used it looks like the extra terms that arise from this coordinate change go to zero. I am not familar enough with General Relativity to determine if this means that you use the relativistic mass but I would speculate that you use a relativistic mass and contract the appropriate coordinate. (Interestingly enough the paper I cited goes on to find the field due to a photon!)