The way I understand it, temperature is the average kinetic energy of the particles of a system. However, kinetic energy depends on the velocity of the particles, which is relative. If a cup of coffee were traveling at 99% the speed of light relative to an observer, its particles would have much higher kinetic energy compared to a cup of coffee at rest. Would it be hotter to this observer then? Would this observer disagree on the temperature with another observer traveling along side the cup of coffee?


2 Answers 2


This is an open problem. There is no definitive answer.



Ott’s result means that a moving body appears hotter, opposite to the Einstein-Planck temperature transformation...

The existence of two opposite solutions to the same problem created a controversy on the correct temperature transformation that led to a significant number of publications during the decades of 1960 and 1970. Among those studies a new interesting theory appeared on 1966 and 1967 when Landsberg published two articles questioning the results of Einstein and Planck on the temperature transformation, and proposing that the temperature is a Lorentz invariant.

Landsberg then proposed a new definition of temperature that ensures relativistic invariance and a different generalization of the definition of temperature in relativistic thermodynamics given by

$\frac{1}{T} = \frac{1}{\gamma} (\frac{\partial S}{\partial E})_{V,P}$

This result implies that the internal energy is a Lorentz invariant, and since TdS is also an invariant, then we can define the temperature as

$\frac{1}{T} = (\frac{\partial S}{\partial U})_{V,P}$

in all frames of reference. His view was supported by a number of authors...

Since Newburgh, several authors came to the conclusion that different definitions of a thermometer lead to different temperature transformations, and hence all these works have supported the position that it is not possible to find a general relativistic transformation for temperature...

In recent years, and with the controversy over relativistic transformations for a thermodynamical system still open after more than a century, new tools and computational capabilities have led a to a series of more complex numerical experiments and solutions.

Recently, Dunkel et al. pointed out that the main reason for the existence of the controversy is the number of different definitions of heat and work, all equally plausible, that lead to different conclusions and transformations of the thermodynamic quantities. In their study they defined the thermodynamic quantities with respect to the backward-lightcone of an observation event, and showed that it is possible to obtain both Planck-Einstein, and Ott formalisms by taking different choices in their definitions. Dunkel also suggested that, while Ott (and later Van Kampen) results seem to be more reasonable, it is almost impossible to tell which one is the correct one, given the current impossibility to perform an experiment which could shed light into this topic. Within this context, we then conclude that the long-standing controversy on the construction of a theory of relativistic thermodynamics is mainly based on the initial assumptions, which need to be tested in the future in order to discern which set of Lorentz transformations is correct for quantities such as temperature and pressure.

  • $\begingroup$ But is it really that hard to accept that a moving body looks blue-shifted from one end, red-shifted from another and that it therefor does not look like it is in thermal equilibrium as seen from an observer moving relative to it? $\endgroup$ Sep 25, 2022 at 5:46
  • $\begingroup$ That is interesting. I don't know. but if you're measuring it using entropy, which i believe is the fundamental definition of temperature, i don't see how moving the whole system will change those measurements. $\endgroup$
    – jelly ears
    Sep 25, 2022 at 6:08
  • $\begingroup$ +1 but I'm curious; "...but if you're measuring it using entropy..." How does one measure temperature of a cup of coffee traveling at 0.99c using entropy? $\endgroup$
    – uhoh
    Sep 25, 2022 at 9:53
  • 1
    $\begingroup$ @uhoh you're right. i don't know how to measure the fundamental definition of entropy. in terms of counting states but it seems to me that this is real thing. that shouldn't vary with movement. anyway, this topic is probably over my head and i can't answer. the best i can do is cite this article that i found interesting. $\endgroup$
    – jelly ears
    Oct 2, 2022 at 8:33
  • $\begingroup$ I'm certainly all for "real things" in Physics :-) $\endgroup$
    – uhoh
    Oct 2, 2022 at 9:20

Well let's take the case of you in your spaceship with a cup of hot water in your hand, and a thermometer in the cup registering "90 Celsius". You speed up to 99% of the speed of light and go zooming right past the rest of us who are watching nearby. As this occurs you hold the cup up to a porthole so we can read the thermometer.

It reads "90 Celsius". But one of us spectators has a magic telescope that allows us to watch the water molecules jiggling around in the cup of water and measure their average speed. "Gadzooks!" she exclaims. "Those water molecules are HARDLY jiggling!"

How can that be? How can those water molecules be almost frozen in time, but the thermometer tells us the cup of water containing them is at 90 C?

This suggests that we have to know something about the frames of reference in which the hot object resides and in which we observe thermometers- and that if we do not shift our experiences between those frames according to the rules of relativity, we'll get the wrong answers and draw the wrong conclusions from what we are observing.

  • $\begingroup$ In other words, temperature is not a relativistic scalar. $\endgroup$ Sep 25, 2022 at 4:08
  • $\begingroup$ @FlatterMann Some say it is and others say it isn't. This is still unknown. $\endgroup$
    – jelly ears
    Sep 25, 2022 at 5:43
  • $\begingroup$ @jellyears I was just trying to reduce your answer to one sentence. I don't have a strong opinion on it. A moving body in thermal equilibrium does not look like it's in thermal equilibrium because of Doppler shift, but if we are making a closed system out of it by surrounding it with a reflective sphere, then it would be in thermal equilibrium with the radiation inside the sphere, again, would it not? So it's more of a problem of defining how we measure the temperature of the object, isn't it? $\endgroup$ Sep 25, 2022 at 5:52
  • $\begingroup$ @FlatterMann If a blackbody was moving toward you, I don't believe you would see the graph of intensity shift from the red to something like the blue.en.wikipedia.org/wiki/Black-body_radiation#/media/… I don't think it would look like a hotter blackbody. it would be something else. but I could be wrong $\endgroup$
    – jelly ears
    Sep 25, 2022 at 6:39

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