Does Gravity Gravitate? Exploring Mass Measurement

[AdirianSoan]

TL;DR Summary

Trying to ensure my understanding about what Komar Mass refers to is reasonable; namely, that curvature curves curvature, and Komar Mass excludes this effect.

Suppose you have a particle, and you measure the gravitational force one meter away in flat spacetime. Add a second identical particle a negligible distance from the first, and measure the force at the same point one meter in flat spacetime away.

I expect the force to be very slightly less than twice the originally measured force. My internal explanation is that the distance from each particle to the measured point (and every intermediate point) is increased by the curvature of space-time created by the other particle, and it is this curved space that curvature is curving, rather than the imaginary flat space we are using as our basis of measurement.

Thus, Komar Mass is a somewhat more accurate way of measuring mass (if we are concerned with counting the number of atoms).

Is this a reasonable way to think about the situation?

 

[PeterDonis]

As it happens, we have a series of 3 Insights articles on your title topic:

https://www.physicsforums.com/insights/does-gravity-gravitate/

 

[PeterDonis]

Suppose you have a particle, and you measure the gravitational force one meter away in flat spacetime.

You can't; "gravitational force" and "flat spacetime" are inconsistent. If "gravitational force" is present, then spacetime is curved.

You need to rethink your question in the light of the above; as it stands it's unanswerable because the scenario you describe is inconsistent.

 

[AdirianSoan]

My question pertains to the sequence; the flat space referenced there is strictly imaginary.

My current understanding is that "curvature curves curvature, but we already include this effect in how we measure mass". And the way this reduces gravitational mass is by increasing the distance from the mass, to the area under consideration.

Does that make sense? And secondarily, is that a reasonable way to describe what the articles are conveying?

 

[PeterDonis]

the flat space referenced there is strictly imaginary

That doesn't help. Your question is still not well-defined. You can't specify distances in an imaginary space.

My current understanding is that "curvature curves curvature, but we already include this effect in how we measure mass".

This is too vague to know whether it is true or not.

the way this reduces gravitational mass is by increasing the distance from the mass, to the area under consideration.

Increasing the distance compared to what? If your answer is "compared to an imaginary flat space", see above.

 

[PeterDonis]

is that a reasonable way to describe what the articles are conveying?

I can't tell because you haven't yet given a well-defined question that can be answered.

 

[PeterDonis]

Komar Mass is a somewhat more accurate way of measuring mass (if we are concerned with counting the number of atoms).

Are you saying that Komar mass is equivalent to counting the number of atoms? If so, you are mistaken. It's not.

 

[AdirianSoan]

Are you saying that Komar mass is equivalent to counting the number of atoms? If so, you are mistaken. It's not.

Not quite, but I think this is a more useful line of reasoning.

If you were interested in the number of atoms, would Komar mass give you a more accurate estimate than gravitational mass?

 

[PeterDonis]

If you were interested in the number of atoms, would Komar mass give you a more accurate estimate than gravitational mass?

What "gravitational mass" are you referring to?

 

[AdirianSoan]

M, versus the M0 representing Komar mass, in the second in the article series. Mass as measured using orbital speeds and orbital distance from satellites. I don't know what else to call it to differentiate between the two.

(Follow up question, if it does give more accurate estimates; does M0 diverge from M with the total number of atoms, such that adding atoms/mass, holding everything else equal, will make M increasingly inaccurate, for estimating the number of atoms, as compared to M0?)

ETA:
Distance as measured by a sufficiently distant outside observer.

 

[PeterDonis]

M, versus the M0 representing Komar mass, in the second in the article series.

You apparently missed the point that $M_0$ is not an actually observed mass at all. Only $M$ is. For the idealized case I was discussing (a spherically symmetric gravitating body surrounded by vacuum), the Komar mass is $M$, and it is the same as the externally measured mass (from orbital speeds and distances of satellites).

 

[AdirianSoan]

Ah, so I did. I read the first equation and didn't pay attention to the fact that the second equation had a different symbol.

Alright, that clears that up. Thank you!