# Relativity of temperature paradox

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## The imagined scenario:

Part A:

From special relativity we know that velocity is a relative physical quantity, that is, it is dependent on the frame of reference of choice. This means that kinetic energy is also relative, but this does not undermine the law of conservation of energy as long as we are consistent with our choice of frame. So far so good.

Part B:

On the other hand, from statistical mechanics, we know that the average kinetic energy of a system and its temperature are directly related by the Boltzmann constant $$\langle E_k \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{3}{2} k_B T$$ which leads to conclude that when the notion of temperature in physics is expressed in terms of a system's kinetic energy, then it too ought to be a relative quantity, which is a bit mind-boggling, because I had always thought of temperature as absolute.

Part C:

Furthermore, we know that all objects at non-zero temperature, radiate electromagnetic energy with a wavelength given as a function of the body/object's temperature, this is the Blackbody radiation. Thus in principle, I am able to infer an object's temperature (i.e. the temperature in its own rest-frame of reference) by measuring its emitted radiation, regardless of the frame I find myself in. But this seems to violate the previously expected relativity of temperature as defined by average kinetic energy.

## Proposed resolutions:

The resolutions that I imagine to this paradox are:

• a) Depending on the frame of reference from which I measure the emitted blackbody radiation of the object, the radiation will undergo different Doppler blue/red-shifts. Thus the relativity of the temperature in the context of blackbody radiation, is preserved due to the Doppler effect.

• b) I suspect that treating temperature as nothing but an average kinetic energy does not in general hold true, and to resolve this paradox, one should work with a more general definition of temperature (which I admit I do not know how in general temperature ought to be defined, if not in terms of state of motion of a system's particles).

Case a) resolves this hypothetical paradox by including the Doppler effect, but does not contradict the relativity of temperature.

Case b) on the other hand, resolves the problem by challenging the definition that was used for temperature, which in the case that we define temperature more generally, without relating to kinetic energy, may leave temperature as an absolute quantity and not relative to a frame.

## Main question:

• But surely only one can be correct here. Which begs to ask: what was the logical mistake(s) committed in the above scenario? In case there was no mistake, which of the two proposed resolutions are correct? If none, what is then the answer here? Very curious to read your input.
This post imported from StackExchange Physics at 2016-06-18 20:38 (UTC), posted by SE-user user929304
asked Jun 13, 2016
retagged Jun 18, 2016
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The average kinetic energy is just that. it describes the energy content of the heated molecules. As the post below explained, it's best to look at the heated body in its own reference frame and apply relativistic doppler shift for the change in the radiation due to relativistic effects. If you tried to analyse the relativistic effect of the heated body traveling relative to you, all you would be doing is applying the same velocity vector to each molecule and arrive at the same place anyway. It would fall out as a constant.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Peter R
The temperature can be related to kinetic energy in the bulk flow rest frame, as I mention at http://physics.stackexchange.com/a/218643/59023. The issue of a relativistic distribution was raised at http://physics.stackexchange.com/q/216819/59023.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user honeste_vivere
The temperature can be related to kinetic energy in the bulk flow rest frame, as I mention at http://physics.stackexchange.com/a/218643/59023. The issue of a relativistic distribution was raised at http://physics.stackexchange.com/q/216819/59023.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user honeste_vivere
@PeterR I see what you mean. So if I understand correctly, the point here is that: looking at the object under study at a molecular level, one cannot meaningfully even define an inertial frame in which the molecules would all be at rest, not even their average motion. But where does that leave us? is this to say that temperature is not a relative quantity? sorry I'm still confused...

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user user929304
There is another issue. The conversion to temperature requires some equation of state, which is not trivial in relativistic fluids.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user honeste_vivere
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@JohnRennie: Which other question?

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier
@ArnoldNeumaier: the same one that you linked above on June 18th

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user John Rennie

## 13 Answers

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All kinds of weird things happen if you try to define temperature in a moving object. The paradox to me (not a generalized accepted answer) resolves by realizing that temperature should only be defined as measured when the object is stationary. Not only is not a scalar but it is not even well defined for areference frame in relative motion. Is temperature a Lorentz invariant in relativity?

Another problem to the ones you mentioned is , for a moving object that is at thermal equilibrium on its own reference frame, that as seen in the reference frame where the object is moving, the distribution of particles' velocities no longer follows the boltzmann distribution, not even relative to the center of mass: particles moving perpendicular to the motion of the object will not change its average velocity, but those moving parallel to the object will. In addition, because the compositions of velocities is non-linear, this deviation will neither be symmetrical between those moving in the same direction of the center of mass and those moving opposite to the center of mass.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Wolphram jonny
answered Jun 17, 2016 by (30 points)
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I think you can avoid all these troubles if you define the temperature as proportional to the variance of velocity, i.e.

$$E[(v-\overline{v})\cdot(v-\overline{v})]=E(v\cdot v)-\overline{v}\cdot\overline{v}$$

Here $E$ means expected value, $v$ ranges over the velocities of the individual particles, and $\overline{v}=E(v)$.

Clearly this is frame-independent, because a change of frame adds some constant vector $v_0$ to each $v$ and to $\overline{v}$, leaving their difference unchanged.

I have a friend who likes to describe cold windy days by saying that on average, the air molecules have too much velocity and not enough speed. Clearly the measure of velocity --- that is $\overline{v}\cdot \overline{v}$ --- is frame-dependent: You don't feel the wind if you move along with it. But the measure of speed --- or, more accurately, of speed-minus-velocity --- that is, the displayed expression above --- is frame-independent, and I believe it's what is measured by your thermometer.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user WillO
answered Jun 17, 2016 by (10 points)
what is measured by your thermometer is always measured in the rest frame of the thermometer. Thus it is meaningless to ask for a quantity that is frame independent, unless there are independent reasons for the latter.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
@ArnoldNeumaier : But if two thermometers are in uniform motion with respect to each other when they both encounter the same cloud of gas, it still makes sense to ask whether they'll show the same readings --- and it seems to me that if the answer is yes, we'd want to call that a frame-independent measurement, no?

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user WillO
The question is meaningful but whether they would show the same reading is not clear. Wouldn't the high-speed flow across the thermometer heat it up, and the amount depends on the relative speed and on how long the measurement takes? So it is questionable whether the temperature is meaningfully defined through such a measurement.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user Arnold Neumaier
Thanks, I m not so sure about your last paragraph, i mean isn't speed just another name for the norm of the velocity vector? So when you speed minus velocity, i dont really understand. Would you a bit more clarity?

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user user929304
@user929304: The speed of a particle is the norm of its velocity vector. The average speed of a particle is the sum of the speeds of all the particles, divided by the number of particles. The average velocity of a particle is the sum of the velocities of all the particles, divided by the number of particles. When it's very windy, the average velocity is high. When it's very cold, the average speed is low.

This post imported from StackExchange Physics at 2016-06-18 20:39 (UTC), posted by SE-user WillO
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The question may be about the covariance or otherwise of temperature, in which case have a look here. As well, have a look at the paper "Temperature in special relativity" by J. Lindhard, Physica Volume 38, Issue 4, 5 June 1968, Pages 635-640.

This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user jim
answered Jun 19, 2016 by (40 points)

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