# Active Gravitational Mass in GR

1. Nov 20, 2005

### pmb_phy

In the past I've spoken about mass a kagillion times. There was one aspect that I had to hold off on until I taught myself the details and meaning to a good level of understanding. I've now come to that level and have made another web page to describe it. The aspect I'm speaking of is "active gravitational mass." The page is at

http://www.geocities.com/physics_world/gr/active_grav_mass.htm

You'll note that active gravitational mass is identical to inertial mass (what some people call "relativistic mass") and that, for a perfect fluid, it is a function of the fluid pressure.

For those of you who have Schutz's new book "Gravity from the Ground Up" will want to look on page 194 box 15.4 - "How pressure resists acceleration"

Pete

2. Nov 21, 2005

### pervect

Staff Emeritus
The answer is right, at least for weak fields and low velocities. Some of the details of your derivation still leave me scratching my head, though.

It's a bit of a challenge to find a system where the intergal of the pressure terms contributes to the mass of the system. A pressurized sphere won't do it, for instance - the pressure inside the sphere is balanced by the tension in the walls, so the total intergal of the pressure terms is zero.

Offhand, the only two "simple" cases I can think of where pressure should contribute are in an expanding spherical gas shell, or a rotating object.

Unfortunately I haven't seen detailed relativistic solutions for either of these.

With a detailed solution, we could compare the mass defined in terms of pseudotensors (which is equivalent to the ADM mass) to your conception to make sure they are the same.

The expanding spherical shell should have an exterior Schwarzschild solution, of course, but we would need the interior solution too to get anywhere with a comparison.

Last edited: Nov 21, 2005
3. Nov 22, 2005

### pmb_phy

Please clarify and I'll try to modify the page to correct it.
You're thinking of the mass of the sphere+fluid system and not the fluid itself. The mass of the system is a function only on the inertial energy of the system.
A star comes to my mind. The gravity holds the fluid together. That the energy of the gravitational field is left out is what makes this an approximation and the resulting equation linear.

Schutz goes over a similar calculation in his new text "Gravity from the ground up" with regards to the inertia of pressure. Do you have this text?

Pete

4. Nov 22, 2005

### pervect

Staff Emeritus
An interesting point. My initial reaction was - is this measurable? Thinking about this, I suppose the "mass" (Komarr mass) of the fluid should be measurable separable from that of the shell, in theory, by taking an accelerometer and measuring the acceleration of gravity at the surface of the fluid and using the Komarr surface intergal (basically Gauss's law, with some time dilation correction terms). This relies on the fact that the gravitational field of a hollow spherical shell is zero.

Probably the biggest thing that I think is missing from your webpage is that you are concerned with some definition of "mass" that is only an approximation.

The question is - where exactly will your approximation work? For comparision, we know that the Komarr mass works in any static metric, and that the ADM mass works in any asyptotically flat space-time. It's a little unclear to me exactly where your approximations will (and won't) work.

The usual treatment of the Newtonian limit (PPN formalism, for instance) actually neglects the pressure terms, because they are of order (v/c)^2. It's probably counter-productive to include some terms of order (v/c)^2 if one is not going to include all terms of order (v/c)^2 - it's just extra work.

Nope. I could order it via interlibrary loan, but it would cost a couple of bucks.

One final techncial point. When you talk about mass as a function of the "inertial energy of the system", I assume that you are talking about mass as some intergal of the stress-energy tensor.

Note that the ADM mass is not formulated in this manner at all!

So it seems to me that your mass can best be viewed as an approximation to the Komarr mass, the concept of mass that's associated with any static space-time. AFAIK the Komarr mass is the only form of mass that can be associated with an intergal of the stress-energy tensor.

5. Nov 23, 2005

### pmb_phy

The definition is exact. It is indentical to inertial mass - the ratio of the magnitude of momentum to speed.
Same place as where the approximation to Einstein's field equation given in Eq. (3) holds. I followed the derivation to Eq. (3) in Peacock some time back but I forget the approximation made. But it holds for a relativistic fluid.

If you'd like I can scan Peacock and you can follow his derivation of Eq. (3) from Einstein's equation. Just PM your e-mail address or let me know if you'd prefer that I place the scans online.

Pete

6. Nov 23, 2005

### kleinwolf

I had once a course about GR, and it was asked : because of the equivalence mass-energy, is a rotating sphere attracting a mass the same as if that sphere was not rorating....indeed, in GR, the curvature depends on energy, and if the sphere is rotating, energy should be higher, hence space-time more curved, which could be stated as : the mass is higher...is it like that ? Is a rotating sphere more inertial than a non rotating one to external forces ?
(I suppose this could be solved if the Scwazschild metric leads to the Kerr (i'm not sure if it's that one, for a spinning/rotating space-time) just by changing the mass...)

Last edited: Nov 23, 2005
7. Nov 23, 2005

### pervect

Staff Emeritus
I signed up for an email at physicsforums.com under my userid here (pervect@.....), but it didn't look to me like it was working when I checked just recently :-(

I'll certainly take a look at a scan if you post it online.

I would guess from what you said that the assumption you are making is that you can find the momentum of a system by adding up all the momentum of its pieces.

This should be OK in flat space-time, and not-OK in curved space-time (the parallel transport will depend on the path taken). So if space-time is "flat enough", I suppose the approximation should work, I'll have to think about this some more.

One other quick question, if I may - given a stress energy tensor T_ij, is the momentum you are calculating equal to $\int_V T_{0j} \, dV$?

Last edited: Nov 23, 2005
8. Nov 26, 2005

### pervect

Staff Emeritus
I think I've come up with a very simple argument that shows that the formula for mass $\int \rho+3P$ can NOT work in the most general (non-static) case. It only works for problems where a static metric exists.

Consider a hollow sphere, pressurized with an ideal gas. As we have discussed earlier, because of the pressure terms P in this intergal, the interior of the sphere has a higher mass than $\int \rho$, while the shell of the sphere appears to have a lower mass, as the shell has a negative pressure (tension) which exactly balances the positive pressure in the interior of the sphere.

This re-distribution of mass is detectable in principle by measuring the acceleration of gravity at both the inside and outside surfaces of the shell and performing the appropriate intergal to determine the Komar mass.

The total mass of the shell + sphere, however, is equal to $\int \rho$ over the entire system, because the pressure terms average out to zero.

Now we ask - what happens when we blow up the sphere? Lets assume that we can cause the highly stressed sphere to fail and break up into pieces, without going overmuch into the details of the mechanism as to how we accomplish this (perhaps it is a large number of distributed timed explosive charges, for instance).

When the sphere fails, the tension terms in the shell disappear. However, a short time after the explosion, the pressure terms P and the density $\rho$ of the gas will not have had time to change. We are thus led to the conclusion that the negative tension terms disappear, while the positive (rho+3P) terms remain. If the formula $\rho+3P$ worked, we would be led to the conclusion that the mass of the system had just increased.

But this is of course not possible. Them actual explanation for what happens is that when the sphere explodes, the concept of the Komar mass of the system becomes undefined, because the system is no longer static.

The ADM mass of the sphere (assuming that that this is all happening in an asymptotically flat space-time) remains defined throughout the explosion, as it can be derived from the metric coefficients at infinity.

However, the ADM mass can not be used to localize "where" the mass is, it only gives the total mass of the system. The ADM mass also cannot be expressed as an intergal of the stress-energy tensor, it is only a function of the metric coefficients at infinity.

Thus the Komar mass formula must fail and $\int \rho+3P$ cannot give the correct answer for the mass of the exploding sphere.

I have assumed in this argument that the sphere is perfectly rigid, which means that no work is done in going from a state of tension to a state of no tension. This is not a realistic assumption, but it does not appear to me that including this effect will allow the problem to have a different resolution. If the shell is elastic, it will expand when pressurized, this will cool the gas in the interior down slightly via expansion. The amount of expansion of the shell, and the amount of energy transfered, will depend on the stiffness (Young's modulus) of the material of which the shell is composed.

This energy transfer from the gas to the sphere, whereby the gas does work on the sphere (force*distance), complicates the analysis of the problem, but I do not believe that it introduces any fundamentally new factors. We can reduce the amount of this energy transfer by using as stiff a material as possible, and by making the sphere thicker.

9. Nov 26, 2005

### pmb_phy

Who said otherwise?

Pete

10. Nov 26, 2005

### pervect

Staff Emeritus
As you may recall (or look back in the thread), I was asking you about when your equation for mass applied, and you said that the definition was exact, and applied to "a relativistic fluid".

I thought this answer was a bit vague, but I took it as implying that you thought your defintion would apply to a small enough sphere of expanding fluid, even if it were expanding at relativistic velocities.

This question of whether or not this equation would give the right answer to this particular case has been on my mind for some time, as evidenced by the following remarks:

Your response to this question did not enlighten me about under what conditions your formula would or would not work, though it did raise some other interesting points, so it was an interesting exchange of views.

Another relevant quote:

Previously, I was in the position of doubting whether the equation would necessarily give the right answer to the case of an expanding sphere, for the reasons I mentioned.

Now I'm in the position of beliving it's probably giving an incorrect answer (a stronger statement).

I'm going to digress a bit and point out that the for the two specific examples mentioned, the second case example (a rotating object that's symmetrical about its axis of rotation) should satisfy the criterion of having a static metric, and thus the equation should apply to this case - as long as the rotating object is not "too big" in which case the exact Komar mass formula should be used.

11. Nov 26, 2005

### pmb_phy

Take a look at exactly the statement that I was responding to. You said
And it was this comment I was responding to when I said "The definition is exact. It is indentical to inertial mass - the ratio of the magnitude of momentum to speed." I still hold to that. You must have assumed that when I said that then it meant that the "equation" must be exact. Its not. Recall the question that you posed
and to this I responded - "Same place as where the approximation to Einstein's field equation given in Eq. (3) holds."
I'll tell you what I'm going to do my dear friend pervect. I'll post the derivation of the equation as it is obtained from Einstein's field equations. Then youy can make note of the approximations and assumptions made. I can scan and e-mail Peacock if you want but he's not as clear as I will be when I make the web page.
Is the rotating sphere described by the stress-energy-momentum tensor for a perfect fluid? The answer is no.

Pete

12. Nov 26, 2005

### pervect

Staff Emeritus
That might give me a better idea of what you are thinking about. As is, I'm just not following your line of thinking.

Ooops, you're right, your formula won't in fact calculate the mass of a rotating sphere, because you've assumed the pressure is isotropic when you've assumed that the medium is a perfect fluid.

A fairly trival correction (replacing rho+3P by rho+P_xx+P_yy + P_zz) removes this assumption and allows the formula to work properly (for a small sphere).

13. Nov 27, 2005

### pmb_phy

Alrighty then! All done. I've scanned Peacock and placed that part online which pertains to this topic. Its a scan of pages 24-25 and placed in a PDF file. Go to http://www.geocities.com/physics_world/gr/gr.htm and then click on "Peacock Scan" and it should download. Peacock does a lot of handwaving there so I plan on filling in those steps someday. Maybe tonight or this week.

Pete

14. Nov 27, 2005

### pervect

Staff Emeritus
Peacock appears to be calculating the mass of a stationary Newtonian fluid.

So the answer to my original question was that you were assuming that the metric was stationary in order to define your mass. (This is fine, but it's important to point this out!).

Last edited: Nov 27, 2005
15. Nov 27, 2005

### pmb_phy

The fluid is not Newtonian. This is a relativistic fluid, i.e. we cannot assume that $p << \rho c^2$
Nothing is assumed when defining mass. At best you're making assumptions on the solution. The definition of mass is exact while the resulting equation is not.

Pete

16. Nov 27, 2005

### pervect

Staff Emeritus
Peacock starts out by computing the mass of a stationary Newtonian fluid. He then goes on to relax the assumptions a bit to compute the mass of a stationary non-Newtonian fluid. Peacock never relaxes the assumption that the metric is stationary.

You are assuming that momentum exists and is a conserved quantity in order to use your definition. In order that momentum exists and is conserved, certain conditions must be met. I've mentioned this once or twice :-) before on this forum.

The conditions that momentum exist and is conserved are the same conditions that energy exists and is conserved (unsurprisingly, since momentum and energy are unified in GR).

These conditions are either an asymptotically flat space-time, or a static-space time. Peacock takes the later choice, by assuming the metric is static. You may have missed this on your first read, or you may be stubborn (in the way that only you can be stubborn) and refuse to ever admit that Peacock is making this assumption, but this assumption is very clearly there if you look.

17. Dec 2, 2005

### pervect

Staff Emeritus
I've been taking a sanity break from this whole thread, but there is something more I want to add.

Volume intergals of the "gravitational mass" of a system exist in terms of the Lanadu-Lifschitz pseudotensors. These are not true tensors, but if one does not mind coordinate dependent physics, they should be very useful. (As has been discussed elsewhere, relativistic mass is coordinate dependent physics, so the psuedo-tensors are likely to fit in reasonably well with this approach - IMO anyway.)

An approach based on these pseudo-tensors should work in any asymptotically flat space-time, regardless of whether or not the space-time is stationary. Which is what I think is desired here - being static makes the problem much simpler, but purely static solutions are not as general as the non-static but asymptotically flat case.

There is some discussion of this in MTW, 464-466, Wald 84-85, and the original source appears to be "Classical Theory Of Fields" (1962) by (of course) Landadu & Lif****z, which I have not had the pleasure of reading, but which is referenced by both MTW and Wald.

I'm not familiar enough with these entities yet to actually compute with them or say anything terribly useful about them other than that they look very interesting.