Relativity and temperature question

[Physicsguru]

Since the other thread got moved for a reason that I do not understand, let me try to ask a question which belongs in this category.

Suppose that the temperature of some object is T(t), where t is the time coordinate in the objects rest frame. Now, in reality nothing has a perfectly constant temperature, rather there are tiny fluctuations about some average temperature Te.

Suppose the temperature of some object in its rest frame is given by

$$T(t) = T_e + T_0 sin(\omega t)$$

Where Te is the average temperature of the object, and T0 is the maximum amplitude of the temperature flux, and $$\frac{dT_0}{dt} = 0$$, and $$\frac{d\omega}{dt} = 0$$

What does relativity say that the temperature is in a reference frame moving at a relative speed of v?

Regards,

Guru

 

[dextercioby]

Ignore relativity:what does that $T=T(t)$ mean,PHYSICALLY SPEAKING...?

Daniel.

 

[Physicsguru]

Ignore relativity:what does that $T=T(t)$ mean,PHYSICALLY SPEAKING...?

Daniel.

That the temperature is a function of time.

Regards,

Guru

 

[ZapperZ]

Since the other thread got moved for a reason that I do not understand,

Considering that a bunch of your threads got moved to the TD section, you appear to be a very slow learner, or you simply refuse to learn.

Can you point out one, JUST ONE, example of such time variation of temperature that is in a thermal equilibrium? And if you point out that this occurs in a superconductor once again via a supposition, I will repeat my recommendation.

Zz.

 

[dextercioby]

That the temperature is a function of time.

Regards,

Guru

Sorry,this the mathematical interpretation... PHYSICS,please...

Daniel.

 

[Physicsguru]

Considering that a bunch of your threads got moved to the TD section, you appear to be a very slow learner, or you simply refuse to learn.

Can you point out one, JUST ONE, example of such time variation of temperature that is in a thermal equilibrium? And if you point out that this occurs in a superconductor once again via a supposition, I will repeat my recommendation.

Zz.

I want to watch someone do a Lorentz transformation on a very simple formula.

 

[russ_watters]

Is this another one of those 'if time dilation slows down motion, doesn't that decrease the temperature of an object?' misunderstandings? Hint: Temperature is measured in your frame. Time dilation is something that happens in the other guy's frame. Time dilation does not affect temperature. (measurements of radiated energy of moving objects, however...)

 

[Physicsguru]

Sorry,this the mathematical interpretation... PHYSICS,please...

Daniel.

Mathematically to say that something is a function of time t, means many things, the main one being that if we differentiate that function with respect to t, that we will get a formula, rather than zero.

Physically, to say that temperature is a function of time, would be to take out statistical analysis, and say that there is some precise reason why the temperature must fluctuate. Is that the kind of answer you were looking for?

Regards,

Guru

 

[dextercioby]

And to what kind of "fluctuations" would you refer to...?

Daniel.

 

[ZapperZ]

I want to watch someone do a Lorentz transformation on a very simple formula.

Fine. However, if you start invoking any "magical" superconductivity stuff without bothering to actually learn what it is, you can bet that I'll be all over it like a cheap suit.

Zz.

 

[Physicsguru]

And to what kind of "fluctuations" would you refer to...?

Daniel.

Well, a body can emit photons, and absorb photons. Those photons either carry away energy (in the form of heat) or supply energy (in the form of increased speed of electrons). So to say that there is a "temperature" fluctuation, it suffices to say that there is a "thermal energy" fluctuation.

Regards,

Guru

 

[dextercioby]

So what chapter of physics could account for your formula...?

Assume i know nothing and i ask you what discipline of (more or less) modern physics explains $T=T_{equil}+Amplit\cdot\sin\omega t$

Daniel.

P.S.And promiss Zapper you won't bring superconductivity into discussion...

 

[russ_watters]

P.S.And promiss Zapper you won't bring superconductivity into discussion...

No, see, since electrons are moving "fast," they get cold due to time dilation, thus: superconductivity!

 

[Physicsguru]

So what chapter of physics could account for your formula...?

Assume i know nothing and i ask you what discipline of (more or less) modern physics explains $T=T_{equil}+Amplit\cdot\sin\omega t$

Daniel.

P.S.And promiss Zapper you won't bring superconductivity into discussion...

Quantum mechanics covers the emission and absorption of photons from a hydrogen atom. For atoms of higher atomic number, the mathematics is too complex to find a precise formula for the wavefunction. This would be the 'modern physics' answer.

According to quantum physics, a photon has an energy given by:

$$E = hf$$

Where f is the frequency of a photon, and h is Planck's constant, which is presumed to be a fundamental constant of nature. In other words, h is NOT a function of time, so that dh/dt=0.

In quantum mechanics, the frequency f of a photon is proportional to omega we have:

$$\omega = 2\pi f$$

Which appears in the quantum mechanical wavefunction:

$$\psi (x,y,z,t) = e^{i(k\vec r - \omega t)}$$


Less modern, would be to try and use thermodynamics/statistical mechanics, founded by Boltzmann in the late 1800's. One could also try to derive h from classical electromagnetism, thermodynamics, and statistical mechanics, by trying to connect stephan's constant $$\sigma$$ to quantum mechanics, and electrodynamics, but that would be difficult, if not impossible.

Regards,

Guru

 

[pervect]

I've always seen temperature defined in the rest frame of the fluid when speaking relativistically.

For an ideal gas/ideal fluid, temperature is (physically), proportional to the kinetic energy of the particles in the fluid in the fluid's rest frame, as dexter was hinting.

For more complex fluids such as actual gasses, one has to consider factors such as rotational and vibrational degrees of freedom -- every degree of freedom has an energy of kT/2 because of the equipartition theorem.

 

[dextercioby]

Quantum mechanics covers the emission and absorption of photons from a hydrogen atom.

Incorrect.

This would be the 'modern physics' answer.

An incorrect answer...

According to quantum physics, a photon has an energy given by:

$$E = hf$$

Where f is the frequency of a photon, and h is Planck's constant, which is presumed to be a fundamental constant of nature. In other words, h is NOT a function of time, so that dh/dt=0.

Yes.

In quantum mechanics, the frequency f of a photon is proportional to omega we have:

Incorrect.

$$\omega = 2\pi f$$

Which appears in the quantum mechanical wavefunction:

$$\psi (x,y,z,t) = e^{i(kx - \omega t)}$$

Incorrect.

Less modern, would be to try and use thermodynamics/statistical mechanics, founded by Boltzmann in the late 1800's.

Why would you say that...?

One could also try to derive "h" from classical electromagnetism, thermodynamics, and statistical mechanics, by first finding out how stephan's constant $$\sigma$$ connects to quantum mechanics, and electrodynamics, but that would be messy.

"h" cannot be derived."h" is postulated.Period.

Daniel.

 

[Physicsguru]

Incorrect.



An incorrect answer...



Yes.



Incorrect.

$$\omega = 2\pi f$$



Incorrect.




Why would you say that...?



"h" cannot be derived."h" is postulated.Period.

Daniel.

Then by all means, suggest to me your corrections, and let's see if i correct them or not.

Kind regards,

Guru

 

[K.J.Healey]

Hi all, I am slightly interested in this question so I thought I'd jump aboard.
Questions I have before further thought on this:
By temperature do you mean the moving objects rate at which it emits radiation?
Or do you mean it has a temperature that, if it weren't moving, would remain constant. It emits nothing?

Are you asking what the emmitted radiation would do under Lorentzian transformation?

Why does temperature have to be a function of time? Can it not be constant?

Are you trying to imply that the rate of change of temperature(emmitted radiation) is a function of velocity?

Thanks.

 

[Physicsguru]

I've always seen temperature defined in the rest frame of the fluid when speaking relativistically.

For an ideal gas/ideal fluid, temperature is (physically), proportional to the kinetic energy of the particles in the fluid in the fluid's rest frame, as dexter was hinting.

Mmm hmm, U = 3/2 kT, in the rest frame of the fluid, using the Mawellian speed distribution formula. The derivation of the formula quite long, but rather interesting.

Regards,

Guru

 

[dextercioby]

The reason why i put that "incorrect" word after almost every phrase of yours is because you don't seem to grasp the object of quantum mechanics...Specifically,you do not know that in the field of quantum mechanics,the word photon doesn't have any meaning...And Planck's constant is postulated...

Do you insinuate that,by performing a LT over the formula giving internal energy vs.equilibrium temperature,the time dependence of temperature would become explicit...?

Daniel.

 

[Physicsguru]

Questions I have before further thought on this:
By temperature do you mean the moving objects rate at which it emits radiation?
Or do you mean it has a temperature that, if it weren't moving, would remain constant. It emits nothing?

By temperature, I mean that temperature is a measure of the internal kinetic energy of the parts which make up the object, in some frame where the center of mass of the system is at rest. A balloon contains helium gas, the gas molecules are all inside of it, and the balloon has a center of mass. Consistent with what Pervect wrote.

Regards,

Guru

PS:
Operational definition: Put your hand on an electric stove which is off, and your hand will not get burned. Turn on the stove, wait five minutes, and repeat the experiment. You will be on your way to the hospital, unless you performed the experiment with your eyes open. After five minutes, you will see that the metal is glowing red hot. Common sense will then tell you not to put your hand on it, because you will get a severe burn. What has changed, is the speed of the parts in the metal.

 

[ZapperZ]

By temperature, I mean that temperature is a measure of the internal kinetic energy of the parts which make up the object, in some frame where the center of mass of the parts is at rest. Consistent with what Pervect wrote.

Regards,

Guru

In all of this, there seems to be something missing here. Are you trying to develop a theory for including relativistic effects in thermodynamics? I mean, aren't you aware of the development of relativistic thermodynamics, or did you miss that completely (considering your track record of ignoring already-established theory, I wouldn't put this pass you completely). There are already tons of stuff out there on Relativistic thermodynamics[1], even a text on it![2] Are you disputing these works, or are you trying to come up with new ones?

Zz.

[1] http://lanl.arxiv.org/abs/gr-qc/9803007
[2] R.C. Tolman "Relativity, Thermodynamics and Cosmology" (Dover, 1987)

 

[dextercioby]

Zz,is the 1934 edition fundamentally different than the one republished in 1987...?

Daniel.

 

[Physicsguru]

The reason why i put that "incorrect" word after almost every phrase of yours is because you don't seem to grasp the object of quantum mechanics...Specifically,you do not know that in the field of quantum mechanics,the word photon doesn't have any meaning...And Planck's constant is postulated...

You don't postulate a constant of nature, you measure it. What do you mean by "field of quantum mechanics"? There is quantum physics, quantum theory, quantum mechanics, quantum electrodynamics, quantum chromodynamics?

Do you insinuate that,by performing a LT over the formula giving internal energy vs.equilibrium temperature,the time dependence of temperature would become explicit...?

Daniel.

The way I have phrased the question, starts off with time dependence of temperature being 'explicit', in the rest frame of an object. The question then is, what is the temperature of the object as measured by an observer moving at a speed v relative to the object.

Regards,

Guru

 

[dextercioby]

You don't postulate a constant of nature, you measure it. What do you mean by "field of quantum mechanics"? There is quantum physics, quantum theory, quantum mechanics, quantum electrodynamics, quantum chromodynamics?

1.Yes,u do postulate it...
2.Sorry,poor wording:THE OBJECT/DOMAIN of quantum mechanics...
"Quantum theory" is an extremely vague term... Don't use it...


Daniel.

 

[Physicsguru]

In all of this, there seems to be something missing here. Are you trying to develop a theory for including relativistic effects in thermodynamics? I mean, aren't you aware of the development of relativistic thermodynamics, or did you miss that completely (considering your track record of ignoring already-established theory, I wouldn't put this pass you completely). There are already tons of stuff out there on Relativistic thermodynamics[1], even a text on it![2] Are you disputing these works, or are you trying to come up with new ones?

Zz.

[1] http://lanl.arxiv.org/abs/gr-qc/9803007
[2] R.C. Tolman "Relativity, Thermodynamics and Cosmology" (Dover, 1987)

First off, I am not trying to develop a theory, let me be clear on that. At this stage, I just want to know if formulas for temperature transform correctly from one frame to another using Lorentz transformations, and by correctly I mean that statements made in those frames are consistent with facts that are necessarily frame independent.

Regards,

Guru

 

[ZapperZ]

First off, I am not trying to develop a theory, let me be clear on that. At this stage, I just want to know if formulas for temperature transform from one frame to another using Lorentz transformations, and if the conclusions are consistent with facts which are frame independent.

Regards,

Guru

Then read the book! [There appears to be COMMON theme here]

Zz.

 

[Physicsguru]

Hi all, I am slightly interested in this question so I thought I'd jump aboard.
Questions I have before further thought on this:

Are you asking what the emmitted radiation would do under Lorentzian transformation?

No, I am asking whether or not the temperature of an object T(t) transforms correctly using a Lorentz transformation.

Why does temperature have to be a function of time? Can it not be constant?

The condition for an object to have a temperature which is independent of time (i.e. constant temperature), is for its temperature to be absolute zero degrees kelvin. This is impossible, therefore temperature must be a function of time.

From experiments, we know that an object will reach thermal equilibrium with its surroundings, if left alone for a long enough time $$\delta t$$, where t is the time coordinate in the objects rest frame. Thermodynamics was designed to say that two objects placed in thermal contact will reach an equilibrium temperature... that this is the most 'probable' result. This is most interesting in the case of an object in the vacuum.

Regards,

Guru

 

[Crosson]

As long as you agree that the number of particles and the total volume in a given reference frame will not change, the definition of temperature is the following:

1/T = dS/dU "one over temperature is the derivative of entropy with respect to energy".

So to solve your problem, you would first calculate entropy (the number of microstates corresponding to a given macrostate) as a function of energy, so that du/ds = T + Tcos(wt). Once you had that (physically unrealistic) entropy function, you could make relativistic corrections, and take the derivative with respect to energy to find the temperature in a given lorentz frame.

 

[Physicsguru]

As long as you agree that the number of particles and the total volume in a given reference frame will not change, the definition of temperature is the following:

1/T = dS/dU "one over temperature is the derivative of entropy with respect to energy".

So to solve your problem, you would first calculate entropy (the number of microstates corresponding to a given macrostate) as a function of energy, so that du/ds = T + Tcos(wt). Once you had that (physically unrealistic) entropy function, you could make relativistic corrections, and take the derivative with respect to energy to find the temperature in a given lorentz frame.

Ok suppose that I had the entropy as a function of energy S(U). I could then take the derivative of S with respect to U, to obtain dS/dU. By definition, this is equal to the inverse of the absolute temperture T. Clearly T=0 is impossible, division by zero error. Now, invert the formula for dS/dU, to obtain dU/dS, and we have a formula for temperature (expectedly one that depends on the time coordinate t in the frame). My question started off with:

$$T(t) = T_e + T_0 sin(\omega t)$$

Equating the expression above with dU/dS we have:

$$\frac{dU}{dS} = T_e + T_0 sin(\omega t)$$

Now suppose I multiply both sides of the equation above by the differential of entropy, to obtain:

$$dU= [ T_e + T_0 sin(\omega t) ] dS$$

Now, suppose that we integrate both sides of the equation above:

$$\int_{U_1}^{U_2} dU= \int_{S_1}^{S_2} [ T_e + T_0 sin(\omega t) ] dS$$

Where U1,S1 are simultaneous values at time coordinate t=t1, and U2,S2 are simultaneous values at time coordinate t=t2. The LHS is easy enough to integrate so that we have:

$$U_2 - U_1 = \int_{S_1}^{S_2} [ T_e + T_0 sin(\omega t) ] dS$$

But in order to integrate the RHS, i need time as a function of entropy don't I?

Regards,

Guru

 

[K.J.Healey]

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.

 

[Physicsguru]

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 briefly perused the paper, and he is discussing entropy as if it's a current. Does that make sense?

Regards,

Guru

 

[ZapperZ]

The condition for an object to have a temperature which is independent of time (i.e. constant temperature), is for its temperature to be absolute zero degrees kelvin. This is impossible, therefore temperature must be a function of time.

This is bogus. Again, you have made a statement without justification. Show me exactly where it has been shown that what you said is true. The whole idea of thermodynamic equilibrium IS to establish a time-independent equation of state, which is perfectly valid as far as experimental observations have shown - this is way more than what you have.

Please take note that PF is populated not just by theorists, but also by experimentalists. If you are bold enough to make supposition/guesswork such as this, be prepared to prove that what you are suggesting has been verified experimentally. If not, then (i) consider yourself warned (ii) how can you build your knowledge on top of something unverified? Does this not give you cause for concern, or do you not care if what you believe is valid or not?

Zz.

 

[Physicsguru]

Originally Posted by Physicsguru
The condition for an object to have a temperature which is independent of time (i.e. constant temperature), is for its temperature to be absolute zero degrees kelvin. This is impossible, therefore temperature must be a function of time.

This is bogus. Again, you have made a statement without justification. Show me exactly where it has been shown that what you said is true.

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.

 

[K.J.Healey]

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.