Relativity and temperature question

[Physicsguru]

Ok, you got
$$\frac{dt}{dU} \frac{d^2S}{dt^2} = - \frac{dS}{dt} \frac{d}{dt} (\frac{dt}{dU})$$

now what are you trying to show with it? You just go back to standard definition of temperature.

Actually, I'm just trying to answer his question mathematically.

The intuitive answer is this: if the temperature of the vacuum were absolute zero kelvin in reality, and an object immersed in the vacuum had to eventually reach thermal equilibrium with its surroundings, then eventually its temperature would reach zero and truly be constant. On the other hand, the temperature of the vacuum really isn't zero (nor is it constant), so in order for an object in the vacuum to have a constant temperature by eventually reaching thermal equilibrium with the vacuum, the temperature of the vacuum would also have to be constant, and isn't. There are a sea of photons there.

Regards,

Guru

 

[ZapperZ]

Well suppose that (as Crosson said) the definition of temperature for the case of constant mass M (therefore dM/dt=0), and constant volume V (therefore dV/dt=0) is given by:

$$\frac{1}{T} = \frac{dS}{dU}$$

With density $$\rho$$ defined as

$$\rho = \frac{M}{V}$$

Using symbols consistent with the paper cited by Healy.

Now focus on the formula 1/T=dS/dU

Clearly the LHS is undefined for the case of T=0. Most clearly stated, division by zero error must lead to a contradiction, so that the definition above, on its own forbids a temperature of absolute zero from being realized. Not even the vacuum has a temperature of absolute zero, certainly material objects immersed in it won't have a temperature of absolute zero either, in fact they will be hotter than the vacuum. And so, a material object in the vacuum will radiate energy, so that its temperature will drop, and the vacuum local to it will gain energy, assuming the basic laws of thermodynamics are correct. This having been said, consider the RHS, which is dS/dU... the derivative of entropy with respect to energy.

Suppose that U and S, can be related through the variable t, therefore we have two unknown functions U(t), and S(t).

We can take the derivative of S with respect to t, to obtain dS/dt, and we can take the derivative of U with respect to t, to obtain dU/dt. We can then solve for temperature as follows:

$$\frac{1}{T} = \frac{dS}{dU} = \frac{dS}{dt} \frac{dt}{dU}$$

Suppose that the temperature T can be constant. In this case, the derivative with respect to time of the LHS above, will necessarily be zero.

The derivative with respect to time of the RHS is given by:

$$\frac{d}{dt} [\frac{dS}{dt} \frac{dt}{dU} ]$$

Now use the product rule of the differential calculus to obtain:

$$\frac{dt}{dU} \frac{d^2S}{dt^2} + \frac{dS}{dt} \frac{d}{dt} \frac{dt}{dU}$$

So if the temperature can be constant in time, then the formula above must be equal to zero that is:

$$\frac{dt}{dU} \frac{d^2S}{dt^2} + \frac{dS}{dt} \frac{d}{dt} \frac{dt}{dU} = 0$$

From which it follows that

$$\frac{dt}{dU} \frac{d^2S}{dt^2} = - \frac{dS}{dt} \frac{d}{dt} (\frac{dt}{dU})$$

Now, suppose instead that we already do have a formula for entropy in terms of energy U, and that we can just differentiate S with respect to U. In this case, we have again:

1/T = dS/dU

So that if temperature is to be constant then we must have:

$$0 = \frac{d}{dt} \frac{dS}{dU}$$

Realizing that the dt is superfluous, we really just need to look at the differential of the formula for dS/dU, that is the condition for constant temperature is also given by:

$$0 = d ( \frac{dS}{dU})$$

One way for the statement above to be true, is for a system to have a constant entropy. In that case, dS=0, independently of U. Suppose that the entropy of an object must necessarily increase over any two consecutive moments in time. Thus, even though dS is necessarily nonzero, the formula above could still be true if dS must be constant.

Reread your posting above, and count for me the NUMBER of times you said "suppose ... blah blah blah...". You did NOTHING but made one supposition after another, and on top of another! Aren't you the LEAST bit worried that you are dangling from a non-existent branch? Of course you're not!

Note that you have still FAILED to show me any experimental evidence for any of your suppositions... not that I was expecting any.

Here's one last supposition (hey, if you can do it, why can't I?). Suppose that all you're doing is nothing more than blowing smoke? If we go by that, then this thread now belongs in the TD section like the rest of your threads.

Zz.

 

[K.J.Healey]

Actually, I'm just trying to answer his question mathematically.

The intuitive answer is this: if the temperature of the vacuum were absolute zero kelvin in reality, and an object immersed in the vacuum had to eventually reach thermal equilibrium with its surroundings, then eventually its temperature would reach zero and truly be constant. On the other hand, the temperature of the vacuum really isn't zero (nor is it constant), so in order for an object in the vacuum to have a constant temperature by eventually reaching thermal equilibrium with the vacuum, the temperature of the vacuum would also have to be constant, and isn't. There are a sea of photons there.

Regards,

Guru

But, intuitively, if you put a hot item into a hypothetical vacuum, and you left the constraint that the volume of the item must not change, and the volume of the vacuum must not change, and the pressure of the vacuum must not change(to stay vacuum). Then you are left with the fact that the constant pressure heat capacity is 0. So no heat will transfer to the vacuum, becasue there is nothing to transfer it to.
the thermal equillibrium of that object would be its current temperature, without having the possibility of losing heat (thermodynamically).

 

[Physicsguru]

But, intuitively, if you put a hot item into a hypothetical vacuum, and you left the constraint that the volume of the item must not change, and the volume of the vacuum must not change, and the pressure of the vacuum must not change(to stay vacuum). Then you are left with the fact that the constant pressure heat capacity is 0. So no heat will transfer to the vacuum, becasue there is nothing to transfer it to.
the thermal equillibrium of that object would be its current temperature, without having the possibility of losing heat (thermodynamically).

What formulas are you using to draw your basic conclusions?

Regards,

Guru

 

[K.J.Healey]

Cp = T (dS/dT) (with p and N constant)
So for a theoretical vacuum Cp = 0

Otherwise the equations I am using are the equations of state (basic thermo).
Actually, i might be wrong about the above, but it doesn't matter.
Unfortunately a vacuum doesn't exist, 0 temp doesn't exist, so this is moot.
--------
Picture this, you have an object, it is at 160K. The object does not radiate heat. Energy is conserved.

Got it? Ok, now tell me WHY does the temperature have to change with time? I don't see any equations forcing it to. Are you saying that it forces itself to change temperature, like oscillation, and changes the volume proportionally? or pressure?
Why do you insist that this has to happen?

 

[Physicsguru]

Cp = T (dS/dT) (with p and N constant)
So for a theoretical vacuum Cp = 0

Otherwise the equations I am using are the equations of state (basic thermo).
Actually, i might be wrong about the above, but it doesn't matter.
Unfortunately a vacuum doesn't exist, 0 temp doesn't exist, so this is moot.
--------
Picture this, you have an object, it is at 160K. The object does not radiate heat. Energy is conserved.

Got it? Ok, now tell me WHY does the temperature have to change with time? I don't see any equations forcing it to. Are you saying that it forces itself to change temperature, like oscillation, and changes the volume proportionally? or pressure?
Why do you insist that this has to happen?

Pretty much yes, the temperature oscillates. The charge density at the surface (boundary) obeys a wave equation.

 

[K.J.Healey]

Pretty much yes, the temperature oscillates. The charge density at the surface (boundary) obeys a wave equation.

But temperature oscillation doesn't have a spatial dimention, so wouldn't the effect of GR on the charge oscillation be solely time-dependent, and therefor you could just replace your t in wave equations with t' (dialated) and you're done?

 

[Physicsguru]

But temperature oscillation doesn't have a spatial dimention, so wouldn't the effect of GR on the charge oscillation be solely time-dependent, and therefor you could just replace your t in wave equations with t' (dialated) and you're done?

Before I answer this, do you mind if I ask you what it is you do?

Regards,

Guru

 

[K.J.Healey]

I have a degree in applied physics, and I am currently working at sandia national labs, though my area of work is purely crystalline materials modeling.

The reason I bring up the spatial dimention stuff is because if you're thinking that under a non accelerating inertial frame, and whatever property you're observing from stationary happens to be spacially dependent, then the direction of travel would have an effect on the value measured. The effects don't exist perpindicular to the direction of motion.

You see what I'm saying? I hope I'm being clear, I don't use SR or even thermo in my everyday life, so I really could use a refresher, and I could be making a complete mistake, but I like to think I have a little understanding of these topics.


EDIT : I meant SR.

 

[Physicsguru]

I have a degree in applied physics, and I am currently working at sandia national labs, though my area of work is purely crystalline materials modeling.

That sounds very interesting. Is your modelling specifically for the purposes of superconductor development?

The reason I bring up the spatial dimention stuff is because if you're thinking that under a non accelerating inertial frame, and whatever property you're observing from stationary happens to be spacially dependent, then the direction of travel would have an effect on the value measured. The effects don't exist perpindicular to the direction of motion.

You see what I'm saying?

I think so. Suppose that you are studying a property of an object, and the objct is at rest relative to you, but that the measured value of the property which you are trying to measure depends on spatial coordinates in someone elses inertial frame.


For example suppose that you are studying temperature, and that temperature is a function of position (x,y,z) in the universe relative to some fixed point in the universe like the center of mass of the universe (0,0,0), and t in a frame in which the center of mass of the universe is at rest. Therefore, T=T(x,y,z,t), where T is the temperature of your sample.

Now, the center of mass of your sample is located at (x`,y`,z`,t) in your frame, and your frame is moving in the center of the universe CMU frame. So now, if you are at rest relative to the center of mass of the universe, and your sample is at rest relative to you, then the coordinates of the center of mass of the sample arent changing in the center of the universe frame, and therefore have no effect on measurements that you make.

ON THE OTHER HAND, suppose that your velocity in the CMU frame is nonzero. In that case, if temperature is a function of spatial coordinates, then indeed your measurements will be affected by your velocity in the CMU frame. I think that is what you are saying, and if it is then the answer is yes. And yes, obviously what happens to the temperature will depend on what direction you are moving in the frame.

I hope I'm being clear, I don't use SR or even thermo in my everyday life, so I really could use a refresher, and I could be making a complete mistake, but I like to think I have a little understanding of these topics.
EDIT : I meant SR.

I don't use SR, I treat time as absolute, simultaneity as absolute, and use Galilean transformations. But I do understand the mathematical structure of the special theory of relativity.


Kind regards,

Guru

 

[K.J.Healey]

I thought the whole point of this discussion was SR and Thermo?

And no, what you're saying is not what I was trying to say.
If you're in the reference frame of some object, (as if you are the object) then your properties are inconsequential of your non-inertial properties in other reference frames. ALL of your properties.

I believe(that I am right) that if youre traveling at 0.8c, YOU don't see any effects of SR. Everything, including light , from your reference frame, remains intact. If you look at other reference frames, then YES, it does look different, but you still have the same properties. Thats due to the properties being invarient under LT.

 

[selfAdjoint]

I believe(that I am right) that if youre traveling at 0.8c, YOU don't see any effects of SR. Everything, including light , from your reference frame, remains intact. If you look at other reference frames, then YES, it does look different, but you still have the same properties. Thats due to the properties being invarient under LT.

You are right that in your rest frame you see everything as if you were at rest. You ARE at rest, with respect to yourself, and as long as your frame is inertial you will be completely satisfied with the physics of rest. If there is an .8c velocity between you and some particular other observer, that doesn't affect your observations. The other observer sees length contractions and time dilations in your frame but you don't. You see just the same contractions and dilations in her frame. These properties are not invariant under LT, but the LT from your frame to itself, with 0c relative velocity, is the identity.

Notice another thing. You might be interested in just the one observer with the .8c velocity relative to you, but there are in fact an infinite number of possible observers of you with velocities of xc, $$0 \le x < 1$$, relative to you. So you don't have a definite velocity at all, just a set of relative velocities to others.

 

[Physicsguru]

I thought the whole point of this discussion was SR and Thermo?

And no, what you're saying is not what I was trying to say.
If you're in the reference frame of some object, (as if you are the object) then your properties are inconsequential of your non-inertial properties in other reference frames. ALL of your properties.

I believe(that I am right) that if youre traveling at 0.8c, YOU don't see any effects of SR. Everything, including light , from your reference frame, remains intact. If you look at other reference frames, then YES, it does look different, but you still have the same properties. Thats due to the properties being invarient under LT.

The point of this discussion is about SR and thermo, but without choosing sides right away. Specifically I was inquiring whether or not T(t) obeys the Lorentz transformation or not, based upon physical interpretation of the formula after the transformation. For example suppose that in some frame a magnet is floating over a superconductor. If the Lorentz transformation on the formula for T(t) predicts that in the moving frame the temperature of the superconductor is above Tc, then there would be a mistake.

Regards,

Guru

 

[pervect]

I was going to post something more, but then realized that
http://lanl.arxiv.org/abs/gr-qc/9803007
basically has all i needed to know about the topic. If you can read that paper and understand it, your questions will be answered.

I'm finding the paper rather tough going. As nearly as I can tell, density*temperature transforms as one component of a rank-2 tensor, very similar to the stress-energy tensor. This is logical, temperature is just antothre form of energy. The (thermal) energy density depends both on the (thermal) energy/particle, and the number of particles/unit volume (or the energy/unit mass, and the mass/unit volume).

I think the paper is saying that entropy, S, transforms as a 4-vector (similarly to the way the charge-current 4-vector transforms).

[add]
One of the things that still puzzles me somewhat is how the fact that entropy can increase is dealt with. Charge is conserved, so it can't be created or destroyed - entropy, however, can increase (though it can never decrease). I'm used to seeing currents defined from quantites that are conserved, not quantities that can increase.
[end add]

T itself, in isolation, doesn't seem to be a component of any 4-vector or tensor from what I can tell (but as I said before, I'm not really sure I'm following the paper).

 

[Creator]

... suppose that in some frame a magnet is floating over a superconductor. If the Lorentz transformation on the formula for T(t) predicts that in the moving frame the temperature of the superconductor is above Tc, then there would be a mistake.

Not necessarily. IF there is an increase in temperature in the moving frame, it is possible there may also a corresponding increase in the critical temperature. So the Meissner effect is preserved.

Creator

 

[Physicsguru]

Not necessarily. IF there is an increase in temperature in the moving frame, it is possible there may also a corresponding increase in the critical temperature. So the Meissner effect is preserved.

Creator

Yes necessarily. Tc is measured in the rest frame of the object.

Regards,

Guru

 

[K.J.Healey]

Yes necessarily. Tc is measured in the rest frame of the object.

Regards,

Guru

No, he was probably right. Theres a Tc and a Tmax that must both be measured for calculating the visible meisner effect. In the reference frame of the magnet/superconductor, it would be what you would expect. But if you were to observe it from a frame that views the magnet traveling at 0.8c, you would see the temperature of the system changed, as well as the necessary critical temperature.

You can't think of critical values as constants, the only one that is is "c" itself(and therefor mu and epsilon).

 

[gonzo]

I'm not sure how relevant this is here, but from just having read Taylor and Wheeler, isn't this all just handled by the 4-vector momentum-energy, which is always conserved in an isolate system?

Mass is invariant and the same in all inertial frames, and is the magnitude of the 4-vector. Somtimes part of this mass can come from heat. They say this specifically as an example, and even point out that in theory heat should be weighable. They even talk about the mass difference of water at different tempuratures, using celsius.

So, if temperature is just mass, then it would be seen differently in different inertial frames, since the *composition* of the 4 vector is different in different frames, although the magnitude is the same, right? Or maybe not?

They didn't really discuss this aspect in detail. But if you look at an object in a different inertial frame, it has different energy and different momentum, but isn't it's energy just equal to it's rest energy plus it's apparent kinetic energy in frame you are observing from? Which would seem to me to mean that temp. would be the same in all frames observing the object.

However, an isolated system *can* change tempurature. They give the example of an inelastic collision ... to objects smash into each other and stop. After the collision, the two objects are warmer and have greater mass from that. Kinetic energy has been converted into rest energy (which, of course, is just equal to the mass, since this is the case of zero momentum), which I guess is related to temp.

That was a lot of rambling...