# Questions regarding mass /= volume x density

1. Jan 7, 2016

### Buzz Bloom

I was amazed to learn the following from a post by bcrowell:
"You can't get the total mass of the system by adding up the masses of all its parts."​
The reason for this is that the GR equations regarding mass are nonlinear.
Obviously for small volumes, like in a laboratory,
M = V × ρ​
works just fine. The first question that occurs to me is how big does the volume have to be before the nonlinear aspects of mass makes this equation noticeably false. That is, if the above equation is only a linear approximation, perhaps the actual mass in a given volume might have a power series representation, maybe something like the following:
Mlinear = V × ρ
M = Σ (i = 1 to ∞) ci Mlineari
where c1 = 1.
If this is plausible, then a question might be:
which terms in the summation have non-zero coefficients?​
That is, are all powers present in the summation, or perhaps only odd powers, like the sine?
To calculate the first nonlinear term one needs to know both its exponent and the coefficient.

If the following simplification would be useful, it would be OK to assume spherical volumes with all with the same uniform distribution of dust particles per unit volume.

Does anyone know if such a calculation, or a similar one, has been published somewhere? If not, can someone post the equation(s) that need to be solved in order to make this calculation?

I would much appreciate any help anyone can offer.

Regards,
Buzz

2. Jan 7, 2016

### James Nelson

I am not an expert but I think I have something to give you a start. The way that energy curves spacetime is likely what creates the problem with this. For example, energy is what causes the effects of mass in terms of gravity, not necessarily actual mass. To solve Einstein's field equations you need things like momentum, energy flux, the "folding of space," and density, which are not accounted for by simply adding rest masses. But in a frame at rest relative to those masses, where spacetime is locally (kind of) linear, the equation you posted makes sense, so GR is functionally compatible.

3. Jan 7, 2016

### Staff: Mentor

You don't need general relativity for this. It's true in special relativity, too.

Consider a box with mass $m_{box}$ that contains N identical gas molecules each with mass m. The sum of the masses of the individual components of this system is $m_{box} + Nm$.

The mass, energy and momentum of each individual component are related by $E^2 = (mc^2)^2 + (pc)^2$. Similarly, the mass, energy and momentum of the system as a whole are related by $E_{sys}^2 = (p_{sys}c)^2 + (m_{sys}c^2)^2$, that is, $$(m_{sys}c^2)^2 = E_{sys}^2 - (p_{sys}c)^2$$

$E_{sys}$ is the total energy. Assuming the box is at rest: $$E_{sys} = m_{box}c^2 + Nmc^2 + NK_{avg}$$ where $K_{avg}$ is the average kinetic energy of the individual molecules.

$p_{sys}$ is the magnitude of the total vector momentum of the system. The box is at rest, so its momentum is zero. The molecules are moving around randomly in different directions, so their momemtum vectors add up to zero: $p_{sys} = 0$.

Therefore $$m_{sys}c^2 = E_{sys} \\ m_{sys}c^2 = m_{box}c^2 + Nmc^2 + NK_{avg} \\ m_{sys} = m_{box} + Nm + NK_{avg}/c^2$$ which is greater than the sum of the masses of the components by the amount $NK_{avg}/c^2$. That is, the kinetic energy of the molecules contributes to the mass of the system, but not to the masses of the individual molecules.

4. Jan 7, 2016

### pervect

Staff Emeritus
Let's suppose we have an ideal gas. One simple way of modelling such an ideal gas is as a swarm of idealized particles, that interact by bouncing off each other when they contact, but don't otherwise interact.

Suppose we take a small volume dV of a container containing such a pressurized gas / swarm of particles We want to know that the energy, momentum, and mass of the small volume element dV. We want a relativistic expression for all of the above that works in any frame of reference, not just a frame of refrence at rest with respect to the gas.

We start to run into issues. The container of gas, and the individual volume elements, Lorentz contract in a moving frame. So we need to make sure we can account for that, and all other relativistic issues, properly.

If the volume element of gas was an isolated system, we'd know that m^2 = E^2 - p^2. But we run into a problem here, too. When we read the fine print of the math (for instance in Taylor & Wheler's SR textbook "Space-time Physics", we find that E^2 -p^2 is constant if we have a single particle or an isolated system. But our element of pressuized gas is neither - it's a swarm of particles, and they are not isolated, they're interacting with the other particles in other volume elements of the gas in the pressurized container.

It turns out that a calculation of E^2 - p^2 depends on the frame of reference - for example, a frame of reference moving with respect to the gas will find different values for E^2 - p^2 than a frame of reference with respect to the gas. This can be verified by carrying out a calculation of the scenario with only the basic tools of SR, but I don't have a specific textbook reference that carries out this calculation. I have done it myself in a PF post for a box-of-particles, though I don't think it was typo-free, unfortunately. If there's enough interest I could try and find it and post a link. Hopefully it suffices to say that if you ignore the box walls, and consider only the contests contained within the walls of the box, a calculation of E^2 - p^2 for a box of relativistic particles turns out to depend on the frame of reference rather than being frame independent.

The solution to all these problems is well-documented in textbooks, though the explanation is not particularly intuitive, and most of the treatments are of necessity rather advanced. The solution turns out to be the Stress-energy tensor. See for instance http://web.mit.edu/edbert/GR/gr2b.pdf, which derives the stress-energy tensor for our idealized gas made up of a swarm of interacting particles.

The end result of this is that the total energy-momentum (which is a 4-vector) in the volume element dV is equal to the rank 2 Stress-Energy tensor T, multipled by the 4-velocity u of the volume element, multipled by the scalar value of the volume.

As for mass, the mass of the volume element dV is no longer frame-independent. The mass per unit volume is given by the value of the 0,0 component $T^{00}$ of the stress-energy tensor T. T transforms as a second rank tensor. This describes concisely how things work, but unfortunately it may not be too helpful without a working knowledge of tensors. Given that the end result for how we calculate the energy-momentum in a unit volume winds up being expressed in terms of a tensor, it's rather hard to get into the details unless one already know what a tensor is :(.

5. Jan 8, 2016

### Buzz Bloom

Hi James and jt:

I appreciate your posts and your efforts to help. I don't think that "momentum, energy flux, the 'folding of space' are relevant to my question, but I might be mistaken especially regarding the distortion of space. I also pretty confident that kinetic energy is not relevant either. My assumptions for making a calculation include that the particles are dust, and I might add that the temperature is zero. I think pervect's post is closer to what I am looking for.

Regards,
Buzz

6. Jan 8, 2016

### Buzz Bloom

Hi @pervect:

Thank you for your post. I used to have some facility with tensors many decades ago when I took a course as an undergraduate using the 1951 textbook by Sokolnokoff Tensor Analysis Theory and Applications. Unfortunately, doing nothing with tensors for so long, I no longer have my former facility, but I still have the textbook. If we assume the pressure is zero, and there is no radiation accompanying the dust particles, what form does T00 take in spherical coordinates?

I vaguely remember that the spherical symmetry behavior of gravity is the same in GR as it is in Newtonian gravity. That is: (1) a spherically symmetric mass distribution acts the same out side the boundaries of the spherical distribution as a point mass at its center of the same mass; and (2) there is no gravitational field inside a spherical shell. Is that correct?

If so, then I am guessing that the spherically symmetric mass inside a radius R is not effected by a spherical shell of inner radius R and thickness dR, but the shell would be affected by the internal spherical mass so as to change the mass of the shell. Is that correct?

I assume that the mass is composed of non-moving dust with zero pressure and temperature, and no radiation.

If I am right about the above spherical symmetry affects, then I would much appreciate your help in finding an equation that shows how the non-linearity of T00 causes the change in the mass of the shell due to a central point mass?

Regards,
Buzz

Last edited: Jan 8, 2016
7. Jan 8, 2016

### pervect

Staff Emeritus
$T^{00}$ is equal to the density $\rho$ in the rest frame of the dust. If we consider a moving frame, we can compute the tensor by doing an appropriate Lorentz boost for a rank 2 tensor. The result is that in a moving frame, $T^{00}$ becomes $\rho / \sqrt{1-(v/c)^2}$ where v is the velocity of the boost. Furthermore other components become non-zero - if the boost is in the "1" direction, for instance, $T^{01} = T^{10}$ becomes non-zero after the boost.

If I'm understanding you correctly, that's correct in the rest frame of the matter assuming it has one. But if the sphere is moving or rotating, this picture may be misleading. For instance, if you have a static spherical shell, you can regard the gravitational force as zero, but if the shell is rotating, you have frame dragging effects. If you have a moving shell of matter that's spherical in it's rest frame, Lorentz contraction will make the shell non-spherical in the moving frame.

I suspect this is not a part of your question, but a concern for completeness leads me to remark that it's definitely wrong to assume that the gravitational field of a moving point mass is spherically symmetrical.

It depends to a large extent on what you mean by effect. Gravity in GR has geometrical effects that are not a part of Newtonian theory, for example time dilation. If you regard time dilation as an "effect", then since there is time dilation inside a hollow sphere as compared to the outside, it has an "effect" even though the Newtonian field is zero.

With those assumptions, and a few additionally assuming weak gravitational fields, the Gravitoeletrromagnetic model is useful. See for instance the wiki article https://en.wikipedia.org/w/index.php?title=Gravitoelectromagnetism&oldid=677033222

An informal summary which I hope isn't too misleading - there is a gravitational effect rather similar to magnetism when you have moving masses.

Mass gets tricky in GR, but for static geometries there is a notion of the total energy of the system (which can be regarded as its mass times c^2). Note though that in order to have a static system, you do have to include pressure - a shell of dust without pressure will be collapsing, not static. The pressure has gravitational effects in GR, unlike in Newtonian theory. Trying to ignore the gravitational effect of pressure will give rises to inconsistencies in the full theory. An informal statement of the significance of the stress-energy tensor is that mass, momentum, and pressure all have gravitational effects in GR, as opposed to Newtonian theory where mass is the sole source of gravity.

Using this notion of energy of a static system (sometimes called Komar mass), by subtracting the assembled mass of the system from the disassembled mass. For instance, you might think of the system as being composed of blocks of stone, you have a mass for the assembled system, then you dissassemble it by lifting the blocks with a crane so that they are very far apart (which requires worse). If you assume the blocks are also incompressible, he difference between the assembled mass and the disassembled mass is the binding energy, the amount of work needed to take the mass apart. If you don't assume the blocks are incompressible, you have to divide the energy up into 3 parts - rest energy, internal energy (due to compression/heating of the blocks), and gravitational binding energy.

It may not be terribly helpful to you, but MTW in their book "Gravitation" does have a formula for the gravitational binding energy on pg 604 in terms of $\rho(r)$ which you can regard as the density of the incompressible blocks we are assuming the spherically symmetric mass is made of. MTW's analsys also computes the internal energy if the blocks are conmpressible, something I've omitted. The results are:

$$m(r) = \int_0^r 4 \pi r^2 \rho(r) dr \quad \Omega = \int_0^r \left( \frac{1}{\sqrt{1-2m(r)/r}} -1 \right) 4 \pi r^2 \rho(r) dr$$

Here r is the Schwarzschild radial coordinate, $\rho(r)$ is the density of the blocks (which is $T^{00}$), m(r) is the "mass inside radius r" for the assembled system (as a function of $\rho(r)$), and $\Omega$ is the gravitational binding energy.

MTW also has a few numerical examples as well later in the chapter, and a comparison to the Newtonian limit (which turns out to agree for "small" masses).

8. Jan 8, 2016

### Buzz Bloom

Hi @pervect:

Thank you very much for your post. I expect I will have some further questions after I fully digest it.

I recently borrowed the MTW book from our library network for three weeks, while I was absorbing answers to my questions (in the other thread I cited) regarding calculating the total mass of a closed universe as M = V × ρ, which MTW says is impossible. Unfortunately I had to return MTW, but I can probably borrow it again.

If I assume a closed static universe, zero pressure and temperature, and the metric for the spherical coordinates are changed in the your equations for m(r) and Ω to adjust for the curvature, is the total mass-energy in a sphere of radius r equal to the sum of the integrals given for m(r) + Ω/c2? I want to also assume ρ is a constant. By static universe I mean that the value of the constant term ΩΛ in the Friedmann equation

is adjusted so that H = 0. That is:
a = 1, and
ΩΛ = -ΩMk.​

I can see this is wrong since the Ωi are densities over a critical density, and the critical density is zero if H = 0. So, the Ωi variables have to be replaced by corresponding static densities. Well, I think you get the general idea of what I am trying to calculate.

BTW, in the integral you have for Ω, do you really want the integral to include the space inside the event horizon at r = 2m(r)? Or is 2m(r)/r always < 1?

If ρ is a constant, then m(r) = (4/3) π r3 ρ
2m(r)/r = (8/3) π r2 ρ which is not always < 1.

Regards,
Buzz

Last edited: Jan 8, 2016
9. Jan 8, 2016

### pervect

Staff Emeritus
Well, if you borrow MTW, there's a very short section in it on the "total mass of a closed universe". The textbook answer, as you'll find if you borrow the book again, is that as far as we know there isn't any such concept in general, though certain special cases (for instance, the Einstein universe) one might be able to apply the notions of the Komar mass that I mentioned earlier.

The basic issue is that a definition of "mass of the universe" should be independent of one's choice of coordinates. We know how to make things independent of coordinates via tensor notation, but there isn't any tensor with the proper properties to represent mass.

There are three main notions of mass in General relativity, and some others that are not so popular. The three main ones are Bondi mass, ADM mass, and Komar mass. The first two require asymptotic flatness, the last requires a stationary universe (one that isn't expanding or contracting). Our universe doesn't match any of these three. The third notion is the one used in the calculation of binding energy that I quoted from MTW.

There was a fourth notion of mass that wasn't particular popular that caught my interest, unfortunately I can't recall the name. It involved setting a particular gauge condition. The consensus view is probably that the necessity for a gauge condition makes the definition very artificial, even though the gauge choice is a fairly natural one (Harmonic gauge). I thought the idea was interesting though alas, the name has slipped my mind. In any event, if you talk about "the mass" of the universe without any qualifiers, you will get at best a polite question of "what exactly do you mean by that" as there are some definite known and currently unsolved problems with finding a satisfactory defintion.

10. Jan 8, 2016

### PAllen

The gauge condition was the Harmonic aka the De Donder gauge. The mass was an integral based on an energy pseudo-tensor. You and also get momentum and angular momentum this way. Nakanishi was the author of papers discussed in that thread. Our advisor Samalkhait posted a bit about this, in that thread. However, while the special gauge choice combined with the pseudo-tensor gives you the ability to talk about quasilocal contribution of gravity (esp. GW) to energy, you don't get any notion of global mass without assuming that you have asymptotically Lorentz transforms at infinity - which obviously amounts to the same thing as asymptotic flatness.

11. Jan 8, 2016

### Buzz Bloom

Hi @pervect:

Thanks again for your explanations. I expect I will have some more questions after I think about all this some more.

I was adding stuff to my previous post while you were making your post. Please take a look at my questions regarding your integral for Ω having zero as a lower bound.

Regards,
Buzz

Last edited: Jan 8, 2016
12. Jan 9, 2016

### pervect

Staff Emeritus
I don't think we can assume all of that while being consistent with GR, at least if we are also assuming the cosmological principle (that there is a set of observers to whom the universe looks homogeneous and isotropic on a large scale).

If the only non-zero term in the stress energy tensor is $T^{00}$, usually denoted as $\rho$, and if $\rho$ is nonzero, if we imagine a spherical arrangment of test particles experiencing no external forces other than gravity, we know from Einstein's equations that the second derivative of the volume of test particles will be nonzero. If $\rho$ is positive, the second derivative of the volume will be negative. Or in other words, the self-gravity due to the positive matter density will cause a ball of matter to contract, making it shrink. See for instance http://math.ucr.edu/home/baez/einstein/ "The Meaning of Einstein's equation".

The only way to avoid contraction is to have $\rho$ equals zero. Which is a rather boring empty universe. (You might be able to make it only slightly non-boring if you assume that there is a positive distribution of $\rho$ due to matter, and a negative effective $\rho$ due to some non-standard form of the cosmological constant, but the sum will still be zero, you'll basically have an empty universe.

An empty universe would typically be open rather than closed - I suppose you could artifically make it closed by fiat with a suitiable (and not particularly physically motivated) "cutting" and "gluing" of the edges. While you do get a sensible result that says an empty universe has no mass, this doesn't help us come up with any concept of mass for a non-empty universe.

I'd still recommend digging up the textbook and reading it, though. Let me know if I should dig up the page number or not.