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Q & A: constancy of mass in gravitational field

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Most recent answer: 04/18/2013
Q:
Can you tell me what experiments have been done to show that mass is the same at different heights in a gravity field, and to what accuracy? Thank you very much.
- David Martin (age 41)
Phoenix, Arizona
A:
There's a bit of a complication in interpreting your question. Since in standard reference frames general relativity gives us that frequencies depend on position in a gravitational field, and frequency is really just energy (E=hf) and energy is really just mass (m=E/c2) there's a sort of intrinsic dependence of mass on height. This gravitational frequency shift has often been measured, to good precision.

So I think what you're asking is whether there's an additional effect, say on rest mass, that would cause various noticeable effects locally. The relative rates of different clocks (vibrating reed, atomic, etc. ) would change if rest masses changed by some factor, leaving other physical constants (h, c, e) unchanged. This would violate the equivalence principle, the basis of general relativity. So the many experimental tests of GR in effect are testing whether such effects occur. ()

For example, the clocks on satellites used for GPS would be sensitive to those mass changes. If there are any changes in their local rates due to mass changes they must be quite small compared to the basic clock rate effects caused by general relativity, otherwise they would have been noticed.

Mike W.

(published on 04/01/2013)

Follow-Up #1: gravity and mass variation

Q:
Reply to the answer headed "Constancy of mass in gravitational field". Thank you for your reply Mike. When you say "there's a sort of intrinsic dependence of mass on height", you seem to mean that it's linked to the gravitational redshift effect. In that case, this might be the effect I was wondering about (which I arrived at via another route), because the ratio between the mass values at two heights in the field would be the same as the ratio between the time rates there, ie sqrt(1 - [2GM/r[1]c^2]) / sqrt(1 - [2GM/r[2]c^2]) . That's a very small effect. The factor difference in mass at the height of a GPS orbit would be 1 + 5e-10. There's another effect that might be the same effect or a different one, perhaps you can tell me. The gravitational potential energy, being energy, should have mass. So that might make a difference to the mass at different radii. But the grav potential is - GM/r. If that's energy, when you convert to mass via the mass-energy equivalence, you get - GM/rc^2. If the mass ratio between radii is (GM/r[1]c^2) / (GM/r[2]c^2) , then surely that would make much larger variation in mass values across the field. Perhaps you can disentangle these possible effects? Thanks very much.
- David Martin (age 41)
Phoenix, Arizona
A:
Very perceptive question. The two effects you're discussing are the same. If you look at that sqrt, in the limit of small effects (about all I can handle), it's just a factor of 1-GM/rc2. If you think about the potential energy you're including in the second term, that's just a correction to the energies, including the rest mass energy, and gives the same factor.

As you say, these factors are small. For GPS systems, they're still very important. Over a day (~105 s) your little fractional effect would give an error of ~5*10-5 s. On your wrist watch, that wouldn't be a big deal. When you're figuring out location by triangulation using lengths given by delay times times c, that's around a 15km error in a length.  Not good for a GPS! So they all include GR corrections.

Things get interesting, and would require more expert help, when you start looking at the GR effects on the gravitational energy itself.

Mike W.



(published on 04/17/2013)

Follow-Up #2: more on gravity, mass, and time

Q:
Thanks Mike. That makes sense. I didn't know that the grav potential involves a correction to the rest mass energy, but I can see what you mean, I guess (for the factor difference), at infinity the rest mass is 1. That means less mass as you approach the centre of the field. I knew that if you remove the 2 and the sqrt box from the grav redshift expression sqrt (1 - [2GM/rc^2]) you get a close approximation in weak gravity. They're very similar. But the question arises, which is the correct expression? Einstein got to the expression sqrt (1 - [2GM/rc^2]) via the equivalence principle, he used SR and the Doppler effect together to get there. So that should be right - it gives the time rate changes as well. So does that mean that the formula arrived at via the energy is just an approximation? It seems so, and that near a black hole, the formula for both would be the GR one for the grav reshift. Is that right? Thanks very much.
- Anonymous
A:
Hi David- I hope I didn't give the impression that the gravitational potential gave a special correction to the rest mass. It affects equally all rates, as seen from a distance, and, as argued before that means all energies, including the big part we call the rest mass.

I'm very weak at GR calculations, but your sqrt (1 - [2GM/rc^2])  is indeed the exact expression, in Schwarzschild coordinates, according to standard sources. I should caution here that when you worry about terms beyond the lowest order correction, you have to be very careful about the meaning of the terms. "R" here is (1/2π)*distance around the spherical object as measured with standard local rulers. That's not the same as half the distance across as measured with local rulers. In the case of a black hole, the difference is a factor of infinity. Likewise "M" here is the sum of all the locally measured masses in the spherical ball. If you start trying to use some remote definition of M then you run into problems. For a black hole, with its infinite redshift, those problems are really severe.

If this sounds a bit fuzzy, that's because of my knowledge here is shallow.  I usually only think to lowest order, in which these issues don't come up and the difference between
(1 - [2GM/rc2])1/2 and (1 - GM/rc2) is neglected.

Mike W.

(published on 04/18/2013)

Follow-up on this answer.