# The mass of a cube of matter

1. Jul 18, 2006

### quasar987

I wrote something on PF some time ago and nobody said what I wrote was wrong. But now I am almost certain that it is. What I said is that if you have a cube of matter, then its mass is E/c² where E is the total energy of its N constituents:

$$E = \sum_i^N (m_ic^2 + K_i)$$

I would correct that last expression by saying that the total energy of its constituents is the sum of the mass energies added to the total mecanical energy of the system:

$$E=U_{sys} + \sum_i^N (m_ic^2 + K_i)$$

ja?

2. Jul 18, 2006

### pervect

Staff Emeritus
It depends on what you mean by "mechanical energy of the system". For that matter, it depends on what you mean by mi, as well. in GR we usually use the stress-energy tensor to describe a system, that has a well-agreed upon meaning.

3. Jul 18, 2006

### quasar987

By total mechanical energy of the system I mean what I wrote: total kinetic energy of the parts + total potential energy. It's the only kind of "total mechanical energy of the system" I know. Is there another way to interpret those terms?

And by mass I mean rest mass.

4. Jul 19, 2006

### pmb_phy

Just to make your statement clearer; E is not the total (i.e. "sum of") energy of N constituents/particles. The energy of a particle is Ep = Kp + E0p = Kinetic energy of particle "p" + proper energy of particle "p". The potential energy is associated with the electric field which has its particles as the source of charge. You left out the proper energy of the each particle. This is fine if the proper energy is a constant in time.

The total energy, W, of a single particle in an external field" is the sum of kinetic energy of the particle + rest energy of particle + potential energy relating to the particle's position in the field, i.e. W = E + U where E = K + E0. E is sometimes refered to as the "free particle energy" for this reason.

Pete

5. Jul 19, 2006

### quasar987

What do you call "proper energy", and what would be the correct equation for the energy of the cube plz?

6. Jul 19, 2006

### pmb_phy

The proper energy of a particle is the energy as measured in the rest frame in an inertial frame of reference. You might know it as "rest energy." However the later is a confusing term in my humble opinion Since "energy of a particle at rest" will include more terms the kinetic and rest energy. When the particle's 4-momentum 4-vector is expressed in a coordinate system which corresponds to an accelerating frame of reference there is a tendancy for people to still refer to the time component as being proportional to the particle's energy. However it is the time component of the 4-momentum's 1-form that is the energy. When this quantity is evaluated for a particle at rest in a given frame then one can call this "rest energy." However this will be different in general to proper energy. E.g. the rest energy might contain the gravitational potential energy of position.

quasar987 already posted the equation above. See his second equation in his first post.

Pete

7. Jul 19, 2006

### Staff: Mentor

Then you should specify that the $K_i$ in your equation are the kinetic energies of the constituent particles in the reference frame in which the system as a whole is at rest, i.e. its total momentum is zero. That is, you should not include that part of the kinetic energy that corresponds to any net translational motion of the system.

8. Jul 19, 2006

### quasar987

Forgot to mention that the cube was at rest (i.e. its center of mass is at rest, i.e. the sum of the momenta of its constituents is 0)

9. Jul 19, 2006

### pervect

Staff Emeritus
Calculating the exact mass of a large cube in GR is a difficult problem. Probably materials strong enough to hold a large mass in a cube don't exist, anyway. It is only for very large cubes that the fine points in the GR analysis become important.

Let me skip ahead and give the approximate answer, before I put you to sleep:

The total energy of the cube can be approximated by taking the intergal of all sources of non-gravitational energy in the cube (this includes, for example, thermal energy, chemical bond energy - i.e. energy due to the electormagnetic forces between atoms, etc.) over the volume of the cube, and subtracting from this the "gravitational self energy" or "gravitational binding energy" of the cube. The last is not a GR concept, though, it is a Newtonian concept that we are using to approximate the GR calculation.

To get mass, you have to divide the total energy by c^2 if you are using conventional units. (In geometric units, c=1).

In Newtonian theory gravity couples to mass, so we use E/c^2 for the "mass" in calculating the Newtonian binding energy. As far as GR is concerned, we don't really talk about mass, though, we just talk about energy. Gravity couples to all forms of energy. Gravity coupling to mass is a Newtonian concept - gravity copuling to energy is the GR concept.

Here is a rough outline of the full GR procedure which is considerably more complicated than the above.

1) Solve Einstein's equations, $G_{\mu\nu} = 8 \pi T_{\mu \nu}$ for a cubical boundary condition. This is easier said than done - I am not aware of any analytical solutions for a cube.

2) Determine the metric $g_{\mu\nu}$ associated with the Eintein tensor $G_{\mu\nu}$ for the above solution.

3) If (and only if) the cube is stationary and thus has a time-invariant metric, we can use the Komar mass intergal. Otherwise we need to use the ADM mass formula, which doesn't give much physical insight. (We can use it to get a number, though, as long as the cube is in asymptotically flat space-time).

I'm a bit confused about how anything in a cube can be moving. Having moving things in the cube isn't a problem per se, unless it makes the metric a function of time. Unfortuanately, I don't see how you can have a cube with macroscopic moving parts where the metric is not a function of time - the cube just isn't symmetrical enough. Microscopic moving things are OK, though (such as vibarating, rotating, or wiggling atoms) - thermal energy is just fine.

4) Using the Komar formula we can find the mass of the cube as a volume intergal of the stress-energy tensor and the metric coefficients we solved for in the Earlier steps.

The Komar formula is the following volume intergal.
https://www.physicsforums.com/showpost.php?p=1025357&postcount=33

$$\int_{\Sigma} (2-g^{00} g_{00}) T_{00} - g_{00}g^{11} T_{11} - g_{00} g^{22} T_{22} - g_{00} g^{33} T_{33}$$

You can think of $T^{00}$ as being the density of the cube - the total amount of energy of all sorts (except gravitational energy) contained in the volume element dx dy dz. (x,y,z here are coordinates).
This includes the thermal energy (energy due to vibrating, rotating atoms mentioned earlier), chemical bond energy (energy in the electromagnetic fields between atoms), and any other source of energy - except gravitational energy. (I said this before, but it's probably worth repeating :-)).

$T_{00}$ in the above formula is then $g_{00}^2 T^{00}$.

The other terms ($T_{11}$), etc. are related to the stresses in the cube. A trivial consequence of solving Einstein's equation will be to find the complete set of mechanical stresses in the cube.

You see that the metric coefficients we calculated earlier play a strong role in the mass formula.

You will also (if you haven't fallen asleep yet) see that the mass of the cube is not just the intergal of the energy terms. Pressure terms (the stresses I mentioned earler, i.e. $T_{11}, etc$) contribute to the mass as well. In short, "pressure causes gravity".

For a small cube, we can make various approximations to approximate the above complete but very hard calculation. We know that gravity will be nearly Newtonian, we can use Newtonian theory to calculate the gravity, and find the metric coefficients. This leads to the approximate result I described earlier - "total non-gravitational energy" minus "gravitational binding energy".

The contributions of the pressure terms to mass are generally ignored in this approximate answer (total energy - binding energy) BTW - that's part of the whole PPN approximation system. Pressure terms arent terribly significant even in objects as large as the Sun.

Last edited: Jul 19, 2006
10. Jul 19, 2006

### quasar987

Stresses as in F/A?

That was interesting pervect.

11. Jul 19, 2006

### pervect

Staff Emeritus
If F/A is force per unit area, yes - those are the stresses I"m talking about. Note that if you multiply the numerator and deonminator of (force/ area) by distance, you get (energy / volume).

12. Jul 19, 2006

### MeJennifer

Pervect, I try to understand this so bear with me for a moment.

Is not the problem that by taking the sum of the non-gravitational energy over the total volume we discount the space-time curvature? And if so is this because we cannot even map the non-gravitational energy onto curved space-time?

13. Jul 19, 2006

### pervect

Staff Emeritus
To get the Komar mass formula, we've done some tricky things that I've not talked about in this thread.

We start with the idea that we can approximate the far field of a static(*)GR gravitational source by a Newtonian 1/r^2 gravitational field. This far field is in essentially flat space-time. The static metric is our key of converting GR's concept of gravity to a Newtonian concept of "force". We know that F = ma, but how do we measure the accleration a? In arbitrary coordinates there's no general perscription for how to do this, but the existence of a static metric defines a way for us to measure the acceleration. We simply measure the acceleration 'a' relative to a point that's "holding station" with respect to our static metric. Being able to measure a of a "test particle", we can now define the force as m*a, and apply the Newtonian view of "gravity as a force".

Because space-time is essentially flat at infinity, we can apply Gauss's law, and integrate the normal force (defined above) multiplied by the area to get a total gravitational flux, which will be 4 pi G M. Dividing the intergal by 4 pi G (or in geometric units 4 pi, since G=1 in geometric units), we get the enclosed mass M.

Now we do some magical type stuff, and find that we can write the intergal so that we don't have to do it at infinity. By scaling the force by sqrt(|g_00|) (or was it the reciprocal? I'd have to look it up and I don't have the text hand at the moment) which is equivalent to measuring the force "at infinty" as if it were transmitted by a string, we can do the surface intergal in any closed region which entirely surrounds our mass, and get the same constant value for the mass.

When then adopt the above intergal as our defintion of mass for a static gravitating system. This defintion of mass can also be expressed in terms of time translation symmetries of our metric, but I won't go into that now.

For more detail, see Wald's "General relativity" - the emphasis in Wald is on the time translation symmetry approach. While the emphasis is on the time translation approach, Wald goes through the Gauss' law argument as well, and he also demonstrates explicitly that Einstein's field equations guarantee that the surface intergal for the Komar mass formula is constant as long as the surface totally encloses our mass, and that the region outside the surface is a vacuum region.

To convert this surface intergal into a volume intergal (this is not strictly necessary, but that's the form I presented), we use Stoke's theorem (which happens to works for manifolds as well as flat space-time).

http://en.wikipedia.org/wiki/Stokes'_theorem

The equivalence of the volume and surface intergals is also guaranteed by Einstein's field equations (this is also demonstrated in Wald, like the previous demonstration its quite terse).

The net result of all this work is a standard formula for the mass of a static gravitating system called the Komar mass.

There is some related discussion of some of these issues in more elementary terms in

(*) note. While I've said static throughout this thread, I'll have to check the text to see if I really should have said stationary.

14. Jul 20, 2006

### quasar987

Would an even better non-GR approximation of the mass of the cube be given by also adding to the equation the enegy of the electromagnetic field cointainted in the volume of the cube?

$$m_{\mbox{cube}}=\frac{1}{c^2} \left( U_{sys} + \sum_i^N (m_ic^2 + K_i)+\frac{1}{2}\int_{\mbox{cube}}(\epsilon_0E^2+\frac{1}{\mu_0}B^2)d\tau \right)$$

Last edited: Jul 20, 2006
15. Jul 21, 2006

### quasar987

Here's something strange:

Take two electrically neutral spheres of radius R and mass m. Put them together so they touch. Their center is now separated by a distance 2R. Get in a frame where the system is at rest. The potential energy of the system is

$$U_{\mbox{sys}} = -\frac{Gm^2}{2R}$$

So the inertial mass of the system is

$$M=\frac{1}{c^2}\left( 2mc^2 -\frac{Gm^2}{2R}\right) =m\left( 2-\frac{1}{c^2}\frac{Gm}{2R}\right)$$

For a sufficiently large ratio m/R, M is negative.

16. Jul 21, 2006

### pervect

Staff Emeritus
Let's take a look at the Schwarzschild radius of a mass m

r_s = 2 G m / c^2

Substituting this into your formula, we get

$$m (2 - r_s / 4R)$$

Thus we see the expression becomes zero when R = 1/8 rs, i.e. when the spacing betweent the masses m is 1/8 of the Schwarzschild radius of a single mass.

The minimum value of R for the event horizons to be touching is R = 2 r_s, at which point your expression is still positive. It makes no sense physically to have a stationary object inside the event horizon.

The formula will probably not be terribly accurate when R=2_rs, because the forumla itself is just a Newtonian approximation. For the exact answer, you'd have to go through a full-blown GR calculation similar to the one I described. In doing such a calculation it is important to explicitly include the stress-energy tensor of whatever is holding the masses apart (electrical fieldds would be a a convenient choice).

The formula should be ok in the limt when R >> r_s. I'm not sure what the order of the error term is offhand. Note that in the solar system, r_s of the sun is 3 km, whle the sun's radius is somewhere on the order of 700 000 km, thus we'd have R > 200,000 r_s for an object at the surface of the sun.

Last edited: Jul 21, 2006