# Relativistic balance

1. Jun 28, 2006

### bernhard.rothenstein

I read in “N. David Mermin, It’s about time, (Princeton University Press, 2005) p.144:
”Furthermore, even when you put together identical bricks, it turns out (in the relativistic case) that the mass of the object you construct depends on how you put the bricks together.”
In order to test the way in which the students understand the statement I proposed them the following problem:
Consider a high sensitivity balance with equal arms. Put on its left pan a cube of volume a3. Put on its right pan a sphere of radius r made from the same material (density ). The balance is located in a uniform vertical gravitational field (g). The balance being in a state of mechanical equilibrium and in thermal equilibrium with the surrounding, determine the relationship between a and r.
what is the solution you propose?

2. Jun 28, 2006

### clj4

You said " uniform vertical gravitational field (g)". This is key.

Surprisingly enough, if one calculates the forces applied to the cube and to the sphere thru volume integration, one finds that the forces have the application point in the symmetry centers of the two respective objects, directed perpendicular to the surfaces of the two balance platters.
This is something that I believe was taught in 9-th grade physics. So the problem reduces to a simple algebraic equation in a and r.

$$a^3=4/3*pi*r^3$$

What does this have to do with relativity?

Last edited: Jun 29, 2006
3. Jun 29, 2006

### bernhard.rothenstein

You are not able to answer without to offend? In accordance with Mermin's statement the potential energies of the cube and of the sphere are different!
sine ira et studio

4. Jun 29, 2006

### clj4

Doubt it. Not in a "uniform vertical gravitational field". I may be mistaken but you can calculate the volume integrals and you will find out. Either way, what does this have to do with relativity?

Last edited: Jun 29, 2006
5. Jun 29, 2006

### pervect

Staff Emeritus
The solution I propose is probably too advanced for the class :-(.

First off, I would propose replacing the sphere and the cube with a cube (1x1x1) and a non-cubical box - (4x.5x.5) for example, with the first number being the height.

Then we ask: given the metric of a "uniform gravitational field", what are the equations of continuity for the stress-energy tensor?

(modern notation: $\nabla_a T^{ab}$ - semicolon notation $T^{ab}{}_{;a}$)

i.e.

given the metric dx^2 = -(1+gz)dt^2 + dx^2 + dy^2 + dz^2

let us assume the stress-energy tensor is

$T^{tt}$ = rho(z)
$T^{zz}$ = P(z)

Physically, we want to make (1+gz)^2*rho(z) constant to represent a density that appears constant in the orthonormal basis of any observer regardless of height, z, i.e. we want

rho(z) = rho_0 / (1 + gz)^2

[note that this doesn't include the stress terms due to the self-gravity of the box, only the stress terms due to the "uniform gravitational field"]

We compute $\nabla_a T^{ab}$

There is only one non-trivial equation in the result. Non-relativistically, we expect that z component of pressure would be proportional to depth.

i.e. non-relativistiaclly

g*rho(z) + dP/dz = 0

So non-realtivistically, the pressure * area of the tall box (equal to the force exerted on the scale) would be the same as the pressure*area of the cube, because the solution for pressure vs depth would be linear.

But relativistically, we get instead from the above

g (1+gz) * rho(z) + g*P(z)/(1+gz) + dP(z)/dz = 0

in terms of rho_0 this becomes
g * (rho_0 + P(z) ) /(1+gz) + dP(z)/dz = 0

Thus, we get an exponential solution for pressure vs depth and not a linear solution. Therfore pressure * area is not equal for the tall box and the cube - their weights are different.

The way I would describe this is that "pressure causes gravity"

Note that I've used geoemtric units. In standard units, pressure gets divided by c^2, so the contribution of pressure to gravity is very small.

This is only a toy version of the problem, but I don't think there's any simple way to find the stress-energy tensor (and the associated metric!) if the self-gravity of the box is included.

It probably would be more to the spirit of the original problem if we assumed that most of the pressure was due to the self-gravity of the system, and that the acceleration was very small, rather than my assumption that most of the pressure is due to the "uniform gravitational field". But I'm not sure how to set up that case easily.

This also illustrates how the pressure changes associated with the problem setup (the "uniform gravitational field" affect the system. The case that would probably best illustrate Mermin's original point is one in which the acceleration is very very low, minimizing the effects of the pressure changes and disturbing the system as little ias possible. In this case, the balance should essentially measure the Komar mass - which won't be equal for the different shapes.

The pressure terms though, are one good way of explaining why the intergal of the energy density / c^2 does not give the mass. Note that the discrepancy is very small.

Last edited: Jun 29, 2006
6. Jul 1, 2006

### clj4

Ahh, how could I have missed that this is another one of your not-so subtle self adverisments? You sent me the link :

http://arxiv.org/abs/physics/0604145

Now, it is all pefectly clear. Why do you ask questions when you appear to know the answer, just to point back to your papers?

Now, since you pointed me to the paper, I am going to say that , though you claim that it will be published in Am. Journal of Physics, it is also wrong.
-I gave a classical explanation (based on my understanding of "uniform gravitational field"),
-pervect gave another one, base on advanced GR

I think that yours is incorrect. Let me explain why: you resort to a (mis)application of the famous Einstein equation $$\Delta E=\Delta mc^2$$.
If you look at the famous 1905 Einstein short paper you will see that the above has been derived for a very specific situation of em radiating substances. In your paper you chose an inellastic collision instead but is a highly unrealistic experiment to conduct on a balance: the two moving bodies must be perfectly symmetrical wrt the platter in order for the experiment to work, otherwise the platter holding them will tilt.
Come on, are you expecting to conduct a collision experiment in a precision balance while maintaining equilibrum? Even for a gedank experiment, it is farfetched. AmJPhys publishes pedagogical stuff (no original research) but I have to question the common sense of your gedank.

I much prefer the original Einstein derivation because , as Einstein points out, it lends itself to experimental verification. Yours doesn't.

For the Einstein paper, see here:

http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

At a second glance, the paper is not only difficult to digest but also seems to use some of the conclusion in the reasoning:

-How do you know that after the collision the balance is still in equilibrum as you state in the text and in the caption for fig b?

-Also, right from the beginning, eq 1 , while not wrong is pedagogically wrong in that it is already expressed in the form mc^2 (which is what you do not know yet since this is what you are trying to establish!)

The whole logic of the paper seems to be going in circles.

On a lesser note, eq 2 is wrong. (probably a typo)

Last edited: Jul 2, 2006
7. Jul 12, 2006

### Creator

Bernhard....
I think you should realize that in your article (http://arxiv.org/abs/physics/0604145) to which this post is making reference, you are making the implicit assumption that gravitational and inertial mass are equivalent at finite temperature, an assumption which has been shown to be invalid.

See for ex.; Donoghue & Holstein, 'Principle of Equivalence at Finite Temperature', Gen. Relativity & Gravitation, 17, no.3,p 207, 1985.

The violation of equivalence principle at finite temperature is extremely small. However, so also is the temperature-mass effect you are proporting to demonstrate in your 'imaginary experiment'.
Since your demonstration of 'mass equivalence of temperature' is based on equivalence principle being exact at non-zero temperature, your methodology is ambiguous. You should re-evaluate your methodology in light of Donoghue & Holstein.

Creator

Last edited: Jul 12, 2006
8. Jul 13, 2006

### pervect

Staff Emeritus
Is this paper online? It seems to contradict, for instance

http://arxiv.org/abs/gr-qc/9909014

9. Jul 13, 2006

### MeJennifer

Does the energy required to maintain the structure (binding of the material) of the cube and sphere not influence the mass as well?

10. Jul 13, 2006

### bernhard.rothenstein

I think it is included in the mass with which the start

11. Jul 13, 2006

### MeJennifer

So if I were to compare an identical mass of a flat slab and a sphere with the same start mass in a uniform gravitational field for say 60 million years they would still be the same?
The force to maintain the structure of the sphere for instance, doesn't that cost more energy than the force that maintains the slab in such a gravitational field?

12. Jul 13, 2006

### bernhard.rothenstein

13. Jul 13, 2006

### bernhard.rothenstein

I do not understand what you say. If y put a body on a balance I measure all the energies involved with its existence? How does time appare in the problem?
sine ira et studio

14. Jul 13, 2006

### actionintegral

I realize you folks are way over my head, but upon re-reading bernhard's original post, I think the question is whether the identical bricks are gently brought together or smashed together. That is, how much energy went into their cohesion.

On the other hand, I do NOT think that Mermin was referring to the geometric shape of the conglomerate when he said "how you put the bricks together". So I don't think sphere or cube hits the heart of Mermin's illustration.

15. Jul 13, 2006

### bernhard.rothenstein

but I think that the problem has some in common with Mermin's statement!

16. Jul 15, 2006

### Creator

Hello, Pervect
I figured you'd be the one to catch the implications and have questions.

The Donoghue-Holstein EP violation is not widely known at large, but it is generally accepted as valid in the GR community, ( and I am not aware of any refutation).

The effect is so tiny as to make it easy to ignore (or at least making it more palatable), the amount of the violation being :

(alpha)(pi)T^2/3m^2
(where T = temp., and m is mass of particle.)
....
Thus it is not experimentally accessible, (IOW, it is below the Eotvos type detectability limits).

The violation occurs through additional contributions to the particle mass that arise from finite temperature radiative corrections.
Another way to appreciate it, that may be more to your understanding, is that there arises a 'non-covariant temperature dependent component' in the energy momentum tensor.

IOW, as pointed out by Holstein, 'it appears to occur because of the lack of Lorentz invariance of the finite temperature vacuum'.
(The analysis is done in the weak field, low temperature limit).

However, fear not, surprisingly, GR suffers no damage as you may suspect, a point also made by Holstein; (probably of neccessity - gads look where it is being published.- the journal "Gen. Relativity & Gravitation" ) .....since one can measure absolute velocity & accel. relative to a heat bath, the conditions under which EP was formulated are not met for finite temperature.

I suggest you get a copy, as I think you will find it very interesting. As pointed out by Holstein the results not only show a connection between Lorentz covariance and EP, but also demonstrates that there exists physical situations in which equivalence principle is violated.

Creator

BTW... You may be able to locate another version (which may be more to your liking) printed in Physical Review D: under then heading...
"Renoramlization of the Energy-momentum tensor and the Validity of Equivalence principle at finite Temperature", Holstein, Donoghue, Robinett, Physical Review D, Vol. 30, pp 2561, (1984).

--Procrastinators: leaders of the future. --

Last edited: Jul 15, 2006
17. Jul 15, 2006

### pervect

Staff Emeritus
I'm not sure I totally grasp the claims being made yet.

These sound like interesting enough articles to shell out  for. Do you know how many pages are there in each article? Our particular public library charges $.10 per page for long articles (much cheaper than online, though a longer lead time), so knowing the page count in advance is helpful. On another topic, I can describe what I'd expect, this would help in seeing where these articles are claiming something new, or if they are just using scary language to describe ideas I've run across before. Here's what I would expect from what I know: some of these statements may disagree with the paper you cite, if so it would be helpful to know which ones: I would expect that E^2 - p^2 of a system in a temperature bath (or a pressure bath, for that matter) to depend on the frame chosen - so that sqrt(E^2 - p^2), the SR defintion of the system mass, would not be a Lorentz invariant and would depend on the frame. I would also expect that the energy and momentum contained per unit volume in the radiation field representing a temperature bath would transform covariantly. (In fact, that assumption is what I'd start with to analyze the problem). I would definitely expect that the radiation field would have a "preferred frame". 18. Jul 16, 2006 ### Creator These certainly aren't 'new' claims as they were established in the 1980's; but like I said before most are simply unaware. Sorry I forget to address your question whether there is internet access:: Article General Relativity and Gravitation Publisher: Springer Science+Business Media B.V., Formerly Kluwer Academic Publishers B.V. ISSN: 0001-7701 (Paper) 1572-9532 (Online) DOI: 10.1007/BF00760243 Issue: Volume 17, Number 3 Date: March 1985 Pages: 207 - 214 Research Articles The principle of equivalence at finite temperature John F. Donoghue1, Barry R. Holstein1 and R. W. Robinett1 (1) Department of Physics and Astronomy, University of Massachusetts, 01003 Amherst, Massachusetts Abstract : We demonstrate that the equivalence principle is violated by radiative corrections to the gravitational and inertial masses at finite temperature. We argue that this result can be attributed to the Lorentz noninvariance of the finite temperature vacuum. Research supported in part by the National Science Foundation. This essay received the fourth award from the Gravity Research Foundation for the year 1984-Ed. -------------------------------------------------------------------------------- You may also access an equivalent article in European Journal of Physics: http://www.iop.org/EJ/abstract/0143-0807/8/2/006/ J F Donoghue and B R Holstein Dept. of Phys., Massachusetts Univ., Amherst, MA, USA Abstract. According to the weak form of the equivalence principle all objects fall at the same rate in a gravitational field. However, recent calculations in finite-temperature quantum field theory have revealed that at T>0 heavier and/or colder objects actually fall faster than their lighter and/or warmer counterparts. This unexpected result is demonstrated using elementary quantum mechanical arguments. Print publication: Issue 2 (April 1987) Creator "Never eat more than you can lift" -- Ms. Piggy Last edited: Jul 16, 2006 19. Jul 16, 2006 ### pervect Staff Emeritus Thanks for the references - at 7 pages, I can order this for just$2.00 at our local public library. Ordering this online would be \$30.00 (which is just not going to happen). Of course, it will be a while before I have a chance to read it (I probably won't even place the order until I go in to pick up my last order, which should be arriving Real Soon Now).

Is there one particular version that you'd recommend?

None of the things I mentioned in my previous response would make a colder object fall faster than a hotter one, so you can safely ignore my previous remarks.

According to the abstract, this would appear to be some sort of quantum mechanical effect, rather than a classical GR effect.

Last edited: Jul 16, 2006
20. Jan 7, 2007

### bernhard.rothenstein

When I posted that thread I did not suppose that it it will receive such sophisticated answers many of them harsh enough. I restate it.
Consider N identical bricks and a balance as sensitive as a thought experiment admits. The balance is located in an uniform vertical gravitational field. Put all the bricks in contact with the surface of the left pan and let M1 be the mass I should put on the right pan in order to establish the equilibrium. Put the same bricks one over the other and let M2 the mass I should put on the right pan in order to establish the equilibrium. Does (M2-M1)cc equate the difference between the potential energies of the two bricks configurations on the pan,
I get accustomed to the harsh style used by many contribuitors so I do not mention sine ira et studio.