# Thermal effect on rest mass

1. Aug 6, 2010

### DaTario

Hi All,

According to the basic notions of Relativity, in the reference frame, in which a certain body is at rest, we are allowed to measure the rest mass of this body. Nevertheless, we know from thermodynamics that at temperature T there is a certain amount of disordered microscopic movement inside any body. This microscopic movement is to have relativistic implications, as well. So my question is:
1) When we say that the rest mass of an object is 1 kg did we also have to say that this measurement was done in a lab at 250 K, for instance ?

Thank you all

Best Regards,

DaTario

2. Aug 6, 2010

### Jonathan Scott

The rest mass already includes all forms of internal energy, so it takes that into account.

3. Aug 6, 2010

### DaTario

So, when the lab's temperature increases, so do the rest mass, although it remains at rest. Is it correct?

Best wishes

DaTario

4. Aug 6, 2010

### Passionflower

According to the definition yes. On a subatomic level there are very little things at rest, sometimes movements at these levels reaches even sub-relativistic speeds. For instance consider the electrons in a gold atom.

I find it amusing that on this forum some people promote rest mass to the best thing since sliced bread to the detriment of relativistic mass; as if rest mass is some kind of perfect representation of mass.

Last edited: Aug 6, 2010
5. Aug 6, 2010

### bcrowell

Staff Emeritus
Yes, heating increases both the gravitational mass and the inertial mass.

I think you're oversimplifying or misunderstanding what people have said.

6. Aug 6, 2010

### DaTario

So it seems correct to say that rest mass is a rather idealized concept, as it would only be possible at 0 K.

Otherwise one is to write an expression for the rest mass which is a function of temperature.

Is it a reasonable statement?

Best wishes

DaTario

7. Aug 6, 2010

### bcrowell

Staff Emeritus
No. If you define rest mass as $\sqrt{E^2-p^2}$, then it's not necessary to talk about temperature. For a macroscopic physical object whose temperature is kept fixed, $\sqrt{E^2-p^2}$ is a constant, and you can say it's the rest mass for the object at this temperature.

If it was necessary to get rid of internal degrees of freedom in order to talk about rest mass, then rest mass would be a completely useless concept. For example, a proton is made out of quarks, and we don't even know with any decent precision how much of its mass comes from the kinetic energy of the quarks. Nevertheless a proton has a fixed, exact value of $\sqrt{E^2-p^2}$.

8. Aug 6, 2010

### JesseM

Also, as far as anyone knows the simplest particles like electrons (which unlike protons aren't thought to be made up of any smaller particles) don't have internal degrees of freedom that would allow them to have a changing temperature, so the "rest mass" for such particles would be a constant of nature.

9. Aug 6, 2010

### DaTario

First question: what happens to $\sqrt{E^2-p^2}$ when the temperature increases ?

Second question: How would you define this E in $\sqrt{E^2-p^2}$ ?

Thank you

Best Regards,

DaTario

10. Aug 6, 2010

### bcrowell

Staff Emeritus
That's true, but in 1950 we would have said exactly the same thing about the proton. If string theory turns out to be right, then an electron actually does have internal degrees of freedom. IMO this example actually strengthens my point. Any definition that required knowing about all possible internal degrees of freedom would be a useless definition.

It increases. For example, if I heat a rock while keeping it at rest (p=0), then m=E, and both m and E increase.

It's the total mass-energy of the object. It's the thing that's conserved in relativity (as opposed to mass and energy, which are not separately conserved).

11. Aug 6, 2010

### DaTario

We must agree that this "mass depending on the reference frame" has to do with Lorentz (relativistic) corrections on the electromagnetic fields inside matter. By correcting these fields to another reference frame, the energy due to these fields change, and so do the mass according to E = m c c.

12. Aug 6, 2010

### bcrowell

Staff Emeritus
No, that's incorrect. For example, a neutrino's inertial and gravitational mass depend on the frame of reference. In general, electromagnetic fields do not have any special, foundational role in relativity. It would be easy to get that impression from early papers on relativity, but you have to keep in mind that in 1905 the electromagnetic field was the only known fundamental field.

13. Aug 6, 2010

### Ken Natton

Hmmm. Even at zero Kelvin, subatomic particles do not stop. If they did, it would be a violation of the uncertainty principle.

And. Temperature is all about motion. Thus 'thermal effect on rest mass' is oxymoronic.

14. Aug 6, 2010

### DaTario

I assume this misconception. But, in trying to survive in the world of ideas I would ask you if my statement would have some chance of being valid if instead of electromagnetic fields I would have put internal fields.

Best Regards,

DaTario

15. Aug 6, 2010

### DaTario

I agree with you. It is a provocative title in deed. It is nice to have you here discussing with us. But I disagree with you when you seems to believe that zero Kelvin would be an achievable temperature.
BTW, do you think rest mass, as a physical concept, is independent of the temperature of the lab?

Best wishes

DaTario

16. Aug 7, 2010

### Ken Natton

Temperatures within a fraction of a degree of absolute zero have been achieved. Exactly absolute zero has not, and maybe never will be achieved. But that doesn’t mean that it is not known that subatomic particles would not stop at absolute zero, for exactly the reason I stated. There is a minimum value on the uncertainty, and it does place interesting constraints on some phenomena. Unfortunately, I haven’t got access to the text that I took this from at the moment. But I will do in a few days and I’ll tell you some more about it then, if you wish.

Clearly there are others who can give you a better assessment of ‘rest mass’ than me. But my understanding is that ‘rest mass’ is an abstract, one that makes no consideration of temperature. Certainly we all know about the relationship between mass and energy, and temperature is just energy of a form – but this I think is ‘relativistic mass’. ‘Rest mass’ I think, does not vary as a function of temperature.

17. Aug 7, 2010

### DaTario

As you said, temperature is all about motion. 0 Kelvin implies zero kinetic energy, and therefore, zero momentum. And it is not a situation where we may talk about zero momentum in average, for it has come from zero kinetic energy, which has velocity squared inside. Thus, UP implies infinite dispersion in position, which IMO is an idealized concept-situation. From this follows that, whithin the framework of this theory, zero Kelvin is a non-achievable temperature.

Best wishes

DaTario

18. Aug 7, 2010

### JesseM

I don't think that's true about string theory. String theory is still a quantum theory so I assume the exclusion principle would apply to fermions like electrons, which means if the electrons had internal degrees of freedom beyond the ones we know about, two electrons which were in identical states for all the variables we do know about would not exclude one another in the way we predicted if they had different states for the extra internal variables. Of course, in theory you could construct some kind of hidden-variable theory where the exclusion principle would only apply to the observable properties and not to the hidden properties, but I don't think string theory is like that. My understanding is that in string theory different particles are treated as strings vibrating at different frequencies, so all particles of a given type are identical strings vibrating in an identical way.

Also, note that even though we may think of strings as continuous 1-dimensional objects that should have an infinite number of possible shapes in space, string theory apparently doesn't allow for a finite region of space to have an infinite number of degrees of freedom, as the holographic principle which is thought to hold in string theory implies the Bekenstein bound on the amount of information that can be contained in a finite region. This article has a good summary of these ideas, and it's written by one of the physicists who played a main role in coming up with these holographic ideas.

19. Aug 7, 2010

### DrGreg

No. Rest mass does vary with temperature. As has been said earlier, it is given by

$$m = \sqrt{E^2 - |\textbf{p}|^2}$$​

(in units where c=1), where p is the sum of all the particles' momenta and E is the sum of all the sources of energy within the object, which includes each particle's rest mass, each particle's kinetic energy as well as any potential energy. Measured in the inertial frame where the object is at rest, p is zero, but E increases with temperature (and, for that matter, with angular rotation).

When we talk of a macroscopic body (or even more so when we talk of a collection of unrestrained particles) we sometimes say "invariant mass" or "system mass" instead of "rest mass", because it might seem odd to talk of the "rest" mass of an object comprised of lots of particles that aren't at rest.

The idea behind the invariant mass m of a macroscopic object is the object, in many ways, "behaves like" a particle of rest mass m.

20. Aug 7, 2010

### bcrowell

Staff Emeritus
I don't think your argument can be correct, because if it were correct it would apply to protons as well. Protons have internal structure, but that doesn't lead to violations of the expected statistical behavior of protons as fermions. Protons are just composite fermions. Composite fermions generally just behave as law-abiding fermions, and likewise for composite bosons. A good example of this is the different behavior of superfluid 3He and superfluid 4He. It doesn't matter that 3He is composite, or that its nucleus is composite, or that the protons and neutrons in the nucleus are composite. All that matters is that 3He behaves as a (composite) fermion. Another good example is that when you do nuclear scattering experiments with identical target and projectile nuclei, you get distinctive interference patterns that are exactly what you expect from their fermionic or bosonic nature. It doesn't matter that they're composite.

In general, it would be too good to be true if compositeness caused different statistical behavior. If it did, then we wouldn't need particle accelerators. We would just look for anomalous statistical behavior at low energy that would allow us to infer the small-scale composite structure. We would have known about quarks in the 1930's.

21. Aug 7, 2010

### JesseM

Well, is the internal structure such that different protons can have different values for some internal variable, or are the quarks in every proton exactly identical? If the latter that would suggest there's no way a proton could have the kind of internal degrees of freedom that would allow us to speak of different protons having different "temperatures" or any other kind of variation in energy, so we could still speak of the rest mass of a proton being a basic constant of nature even though it's a composite particle.

22. Aug 7, 2010

### bcrowell

Staff Emeritus
There's a spectrum of baryon resonances like the $\Delta^+$. These are the excited states, and the neutron and proton are the ground state.

Temperature is a concept that works best with Avogadro's number worth of particles, but it can also be applied as an approximation to smaller numbers of particles. In nuclear physics, we routinely talk about "hot" nuclei, and even the temperatures of nuclei (which are usually measured in MeV). This is typically with systems of ~100 nucleons. The smaller the number of particles, the worse you do trying to apply concepts like thermal equilibrium. It would be silly to talk about the temperature of a 3-quark system, but I don't think it's relevant to the present discussion whether or not thermodynamic concepts apply or not.

23. Aug 7, 2010

### JesseM

So would the excited states behave in an observably different way from what we call "protons"? If someone didn't know about quarks, would they be able to tell the difference, maybe just treating the excited states as entirely different particles?

In terms of the exclusion principle, for our imaginary physicist who has access to modern experimental equipment but has no theory of quarks and assumes protons are fundamental particles, would there be any sense in which there could be two protons that seem to him to be in the same "state" as far as all measurable variables are concerned, but someone who knows about quarks would be able to figure out a way to deduce they were in different states? If not I don't see how the proton example is relevant to disproving my notion that you can use the exclusion principle to judge if two fermions might have additional internal "degrees of freedom" that you don't know about.

24. Aug 7, 2010

### Ken Natton

As ever, I have no option but to defer to those with greater expertise, but you surprise me Dr Greg. Are you telling me that if I have a 1kg lump of iron, say, in front of me at 20ºC, and I then heat it up to 50 ºC it is now more than 1 kg? And if I cool it down somewhere close to 0ºK its mass steadily decreases as I cool it?

25. Aug 7, 2010

### JesseM

Yes, that's what the theory predicts, although the change in inertial mass (or weight) would probably be far too small to measure. But see this paper for an attempt to look at data on the weight of materials and the kinetic energies contained in their atoms to find evidence for the claim that kinetic energy does contribute to weight.