# Absolute Zero and time travel?

1. Jun 19, 2008

### campal

I've been told that as a person approaches the speed of light, time relative to others being viewed slows. when you make the speed of light ( if possible, I’m aware of distance change and mass increase and of the immense amount of energy needed to possibly reach this speed to push that mass) time for others stops and past the speed of light time goes back wards. Does this huge mass increase make an objects gravitational force so immense that it drops through space time and can continue traveling at higher speeds then light? Such could be possible in the even of a star becoming too heavy and dropping through space time to create a black hole. BUT what happens as you slow down? As in all the particles in your body slow down to absolute zero, where your temperature and kinetic motion would be zero. Would this result in forward time travel? Does mass decrease?

2. Jun 19, 2008

### Mentz114

Hi Campal,
you've been told wrong. All motion is relative, and there's no absolute velocity. So it does not make sense to say ' as a person approaches the speed of light' unless you say who is measuring them. Relative to someone in a distant galaxy you may be travelling at a huge velocity, but it makes no difference to you, you are stationary in your own frame.

M

3. Jun 26, 2008

### Xezlec

There's also another problem with what you're saying, at least based on my limited understanding of SR.

You can't reach the speed of light or pass it (at least not in SR... I've heard claims you can do it in GR under ludicrously bizarre circumstances, but I'm pretty sure this isn't what you mean). It's not that there are immense requirements, it's that the requirements are infinite, that is, impossible. Nothing with mass can travel at the speed of light; it would violate all kinds of rules. Going faster than light wouldn't necessarily mean you go back in time, it would just mean the known laws of physics are wrong and therefore we can't say anything at all anymore.

And "forward time travel" is what we are all doing right now.

As Mentz114 said, whenever you talk about velocity (or almost anything else) in relativity, you have to add the phrase "relative to ..." because your velocity relative to me might be 200 Mm/s but your velocity relative to you is always zero.

4. Jun 27, 2008

### mtworkowski@o

He does have a good question there though. You have to admit.....BTW Einstein did talk about all this crazy stuff and it's been discussed by all sorts of people, so it's not out of line or anything is it? Not in my book.

5. Jun 27, 2008

### Xezlec

OK, I'm not trying to be snotty, but if you can explain his good question to me I'll answer it.

Are you talking about the temperature thing? If I interpret his question to be "does something moving the speed of light have an internal temperature of zero", I believe the answer is technically yes, depending on how you define things, but thermo isn't my area of expertise so I'll let someone correct me if I'm wrong there. Since objects going the speed of light are massless, I'm not sure how the definition of temperature would be affected.

As for "forward time travel" and "mass decrease", I don't know what he's referring to.

6. Jun 27, 2008

Staff Emeritus
No, it's not. Ignoring the fact that an extended, massive object (i.e. something with a temperature) cannot travel at c, the question can be recast as "What is the Lorentz transformation of Temperature?". On dimensional grounds, one would think that it grows as gamma (i.e. would get hotter as it goes faster), but a more careful analysis shows that temperature does not have a Lorentz transformation - i.e. a body that is in thermodynamic equilibrium with itself in one frame is not necessarily in thermodynamic equilibrium with itself in all frames.

7. Jun 28, 2008

### Xezlec

So your answer is that temperature is ill-defined for things with no mass? The other part of your answer I don't think I understand. I don't really think the Lorentz transformation is needed here. We are talking about something going at c. So everything is in the well-known limiting case where time is stopped, etc. Nice easy invariants, no need to resort to any math.

8. Jun 28, 2008

Staff Emeritus
That is correct. You need an ensemble of particles to have a temperature, and an ensemble of even massless particles in thermal equilibrium has a mass.

But your "something" going at c has mass. SR doesn't let something with mass travel at c.

9. Jun 28, 2008

### Xezlec

Explain to me how an ensemble of massless particles at 0 K (remember, that was the claim) has a mass.

No it doesn't. I think I've been pretty explicit about that.

10. Jun 28, 2008

Staff Emeritus
Do you perhaps mean "Please, explain to me..."? Surely that's you meant.

Consider an ensemble of N photons in a massless conducting box. The conducting box imposes boundary conditions such that there is a maximum photon wavelength, and thus a minimum photon energy. Now place it in its minimum energy state and consider it in its center of momentum frame. It's energy is greater than 0; it's momentum is 0, and E^2 - p^2 = m^2. Therefore, m > 0.

Yes, you've been saying that over and over. Nevertheless, you cannot create an ensemble of particles with a temperature such that the mass of the ensemble is zero. If you think you can, please describe it.

11. Jun 28, 2008

### Xezlec

My sincerest apologies for that offense.

Not sure how you can have a massless conducting box, nor how you can have any photons "bouncing around" (to use classical terminology) in a reference frame moving at c. A reference frame moving at c (with respect to any other frame) has no time dimension, unless I'm greatly mistaken. Nothing moves.

The only way I know of that you can have anything resembling an "ensemble" of photons at all that are moving together is for them all to be moving in the same direction. But why restrict it to photons? Let's say we have a group of photons, gluons, and gravitons or something all moving in the same direction at c. That's about the most general scenario I can think of that actually involves a collection of distinct objects moving together at c. And in this case their velocities relative to one another are, of course, zero.

The point at issue here was whether this collection of particles, stopped and timeless in their own reference frame (if you can even call it a reference frame) can technically be referred to as having a temperature of 0 K. Since the only description I've ever heard for "stuff at 0 K" is "all the particles in the ensemble have zero velocity in the ensemble's center-of-momentum frame", I thought that description sounded relatively accurate. But maybe speed-of-light reference frames are specifically excluded from this picture somehow; I wouldn't know.

I didn't really have a problem with the answer "there's no temperature at all", I just said I didn't think Lorentz transformations had much to do with the question. But after this discussion, I wonder, are "0 K" and "not having a temperature" the same thing? If not, would you kindly be so kind as to enlighten me about the difference?

12. Jun 28, 2008

Staff Emeritus
What I am trying to say is that there are two classes of objects: one is "objects with a well defined temperature" and the second is "objects that move at c" (or alternatively "objects with zero mass"), I am then arguing that there are no objects that fall into both classes.

That's a very oversimplified picture. For example, consider a system of many simple harmonic oscillators, all in their ground states. It is at 0K because the oscillators are at their state of minimum motion, but they most certainly are not stopped.

13. Jun 28, 2008

### Mentz114

I'm not sure that is true. 0oK can't be achieved precisely because the ground state is not zero.

Check out 'Bose-Einstein condensation' ( assuming you haven't done so already).

14. Jun 29, 2008

Staff Emeritus
Well, ignoring for the moment that 0K can't be achieved in the real world in general, it's not true that the ground state having an offset makes a different here. I'm sure it's worked out in e.g. Reif, but it's simple to show for N identical oscillators:

$$\overline{E} =N \hbar \omega \left( \frac{1}{2} + \frac{1}{e^{\hbar \omega / kT}-1} \right)$$

One can see that at T = 0, this works out to the energy of the system such that each oscillator is in its ground state.

I have a PhD in physics. I've heard of it. It's not necessary (or in my opinion, to further complicate this discussion with spin statistics) when people still misunderstand the basics.

15. Jun 29, 2008

### Xezlec

And what I am trying to say is that my failure to understand stems from my (continuing) lack of knowledge of any definition of the former class. At this point, I have actively searched around and not found no such definition online that clearly excludes this scenario.

And thank you for providing your example of a changing system at 0 K. I hadn't thought about it that much but I can now see how my description was oversimplified. Unfortunately though, it still doesn't help me see the flaw in my reasoning about the photons.