Solving GR vs Newton: F = -GMm/R^2 for Relativistic Particles

[lightarrow]

Sorry if you have already covered this subject.
Is it possible to apply the formula F = -GMm/R^2 for a relativistic particle substituting m with its relativistic mass?

 

[pervect]

No, it's not possible to substitute in either relativistic mass or invariant mass into the Newtonian formula and get a correct relativistic answer.

See for example the previous thread https://www.physicsforums.com/showthread.php?t=146245

The electromagnetic analogy (of how the electrostatic force between two charges transforms when one is moving rapidly) is not perfect, but is quite helpful in gaining a general non-quantitative understanding of what happens, including the non-symmetrical appearnace of the tidal gravitational field of a moving mass, and the existence of gravitational equivalents to "magnetic" forces known as "gravitomagnetism".

One should not stretch the electrostatic analogy too far, or it will eventually break, however.

 

[lightarrow]

Thank you for your answer Pervect.
Now I would like to ask about this (different) case:

Inside a laboratory there is a laser which converts a certain amount of energy coming from outside into photons which remain unabsorbed inside. We measure the laboratory's gravitational mass before and after this convertion is made. Do we find any difference?

I would say yes. But, then, how can we sum the gravitational masses? It's not possible to sum the laboratory's grav. mass before with the "photon's gravitational mass", I suppose?

 

[cristo]

I'd say that there would be no difference in the laboratory's mass before and after this conversion, since photons are massless particles.

 

[pervect]

If I am understanding the problem properly, the answer is yes.

You start out with an isolated system. You then allow the system to interact with the outside world, pumping in energy to create photons. (You could just as easily have added photons that originated in the outside world directly). You then isolate the system again. The system as a whole now has more mass. The principle of equivalence guarantees that there is only one mass for the system, that the gravitational mass equals the inertial mass. So it's simplest to just talk about "the mass". You should probably know that when I use the term "mass", I mean "invariant mass" unless I indicate otherwise. (This generally happens on the 7th tuesday of any given month - i.e. not very often :-)).

A simplified variant of this problem can be found in textbooks. A single photon is massless, however, a pair of photons is not. The mass of a system is not the sum of the masses of its components. (This is true for invariant mass, which is as I mentioned above the sort I use.)

You might want to read http://en.wikipedia.org/wiki/Mass_in_general_relativity for more background on the topic of mass in GR. Particularly relevant is the following example:

Imagine that we have a solid pressure vessel enclosing an ideal gas. We heat the gas up with an external source of energy, adding an amount of energy E to the system. Does the mass of our system increase by E/c2? Does the mass of the gas increase by E/c2?

The answers are yes, and no, respectively. Note that when systems are not isolated (or when one talks about the mass of a piece of an isolated system) talking about the mass of such non-isolated system requires a lot of care. The answers to such questions are not observer-independent. (So to make sense of any answer in such cases, one has to specify a lot of details about the observer and the measurement process.)

Mass can be regarded as observer-independent only when the system in question is isolated.

I should probably mention to meet PF guidelines that I'm the primary author of this wikipedia article. While I think it is as useful or more useful than a post, and also has references to the literature, it shouldn't be construed in and of itself as being a peer-reviewed article.