# Is temperature a Lorentz invariant in relativity?

If an observer starts moving at relativistic speeds will he observe the temperature of objects to change as compared to their rest temperatures? Suppose the rest temperature measured is $T$ and the observer starts moving with speed $v$. What will be the new temperature observed by him?

This is a very good question. Einstein himself, in a 1907 review (available in translation as Am. J. Phys. 45, 512 (1977), e.g. here), and Planck, one year later, assumed the first and second law of thermodynamics to be covariant, and derived from that the following transformation rule for the temperature: $$T' = T/\gamma, \quad \gamma = \sqrt{1/(1-v^2/c^2)}.$$ So, an observer would see a system in relativistic motion "cooler" than if he were in its rest frame.

However, in 1963 Ott (Z. Phys. 175 no. 1 (1963) 70) proposed as the appropriate transformation $$T' = \gamma T$$ suggesting that a moving body appears "relatively" warmer.

Later on Landsberg (Nature 213 (1966) 571 and 214 (1967) 903) argued that the thermodynamic quantities that are statistical in nature, such as temperature, entropy and internal energy, should not be expected to change for an observer who sees the center of mass of the system moving uniformly. This approach, leads to the conclusion that some thermodynamic relationships such as the second law are not covariant and results in the transformation rule: $$T' = T$$

So far it seems there isn't a general consensus on which is the appropriate transformation, but I may be not aware of some "breakthrough" experiment on the topic.

Main reference:

M.Khaleghy, F.Qassemi. Relativistic Temperature Transformation Revisited, One hundred years after Relativity Theory (2005). arXiv:physics/0506214.

• I think worth noting is that from practical point of view, an observer moving fast (not even near to relativistic speeds, just orders of 1 Mach) through a gaseous medium will "experience" the temperature as significantly higher, as particles collide at increased kinetic energy. Aviation recognizes a bunch of temperature quantities related to relative speed, as dictated by practical engineering necessity. So, from practical points of view (like thermal durability of materials), the rise of temperature at relativistic velocities would be massive. – SF. Oct 1 '18 at 15:46

One thing to note is observing something's temperature and thermodynamic notions of temperature aren't exactly the same thing. This is in line with @Mattia 's answer. If a star is receding form you then it will appear cooler because its radiation has been red-shifted. Does this mean that there can be a net flow of heat from us to the star (provided it's moving fast enough)? In the rest frame of the star, our radiation is red-shifted, so this would lead to a paradox.

On the other hand, for accelerating observers there is what's known as Unruh radiation, very much analogous to Hawking radiation. An accelerated observer appears to be radiating energy as though it has been heated, and in its own frame, observes the vacuum to have a thermal spectrum. Since there is acceleration, there is no requirement of thermal equilibrium.

The answer to this long standing question has been given by Landsberg. But it seems this answers was overlooked by many (including myself, see my wrong answer here).

There is no universal relativistic temperature transformation of the form $$T' = T(v)$$ .

Why? Let's look at the example of a moving black body. The black body spectrum of a moving black body shows a frequency shift due to the relativistic doppler effect. The doppler effect however depends on the angle $$\alpha$$ between observer and the source. This leads effectively to an angle dependent temperature for a moving black body:

$$T'(\alpha, v) = \frac{T \sqrt{1-\frac{v^2}{c^2}}}{1 - \frac{v}{c} \cos \alpha }$$

So an observer moving in a heat reservoir cannot detect an isotropic blackbody spectrum and hence cannot find a parameter which can be identified as temperature.

This is an important effect in astronomy. For example the cosmic microwave background shows a temperature anisotropy due to the movement of the earth relative to the background, a fact which has been explicitly calculated in the 60s, e.g.

But as Landsberg also notes they basically just rediscovered what Pauli had already published in his famous article/book about the black body radiation in a moving frame of reference:

• "So an observer moving in a heat reservoir cannot detect an isotropic blackbody spectrum" correct me if I'm wrong, but if heat has a well-defined net flow, regardless of what spectrum you perceive, there would be a definition of temperature? – 友人A Oct 14 '19 at 15:41
• When heat flows from warm to cold objects there is a temperature difference $\Delta T$. You cannot define a unique temperature just from the heat flow alone. – asmaier Oct 14 '19 at 16:50
• "there is a temperature difference ΔT" with which we could define a total ordering on some property of objects. Such a property (with suitable scaling) is temperature, as formulated in classical thermodynamics. – 友人A Oct 14 '19 at 18:16

Cubero et al. 2007: Thermal equilibrium and statistical thermometers in special relativity (http://arxiv.org/abs/0705.3328) came to the conclusion

that 'temperature' can be statistically defined and measured in an observer frame independent way.

With fully relativistic 1D molecular dynamics simulations they verified that the temperature definition given by Landsberg Nature 214 (1967) 903) defines a Lorentz invariant gas thermometer on a purely microscopic basis.

Loooking at a mole of an ideal gas you can deduce that if there ARE any consistent transformations of the thermodynamic state variables the transformation of the product $$k·T$$ is given by $$k'·T' = k·T/\gamma$$.

Planck (and others) opted for $$k' = k$$ (but his proof for this 'begs the question' !). There are very good arguments for $$T' = T$$ and hence $$k' = k/\gamma$$ . The main theorems of thermodynamics are form invariant.

$$R = k·N_{A} = P_{0}·V_{0}/T_{0}$$ can only be invariant if temperatures transform in the same way as volumes do, that is by multiplying by the root. All the details and references are to be found in