Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Lets consider the Bose-Einstein and Fermi-Dirac statistics:

  1. Bose-Einstein statistics:

    $$\langle n_i\rangle = \frac{1}{\exp{[(\epsilon_i-\mu)/kT]} - 1}$$

  2. Fermi-Dirac statistics:

    $$\langle n_i\rangle = \frac{1}{\exp{[(\epsilon_i-\mu)/kT]} + 1}$$

How would these laws respond to Lorentz boosts (Inertial frames being considered in this case)? It could provide some insight into the behavior of these statistical laws in fast moving bodies like galaxies moving at relativistic speed. Incidentally the quantity:$(\epsilon_i-\mu)/kT$ is a dimensionless quantity. But quantities like v1/v2 and E1/E2 are also dimensionless. But these ratios do change for Lorentz boosts.

[v and E represent velocity and energy respectively]

From the point of general relativity, is it necessary to give these laws a covariant form?

share|improve this question
add comment

1 Answer

The simple formulae you wrote are only valid in the rest frame of a material or object. If you go into another frame, you have to generalize the inverse temperature $1/kT$ to a whole 4-vector whose time component is $1/kT$. The spatial components of this "generalized inverse temperature" allow the particles to have a nonzero mean momentum; the simple distributions you wrote above have no "spatial temperature" and the average momentum per particle is zero which is clearly possible in one reference frame only.

However, without a loss of generality, if we have a perfect thermal equilibrium, it is always possible to find a frame in which the momentum (or average momentum of particles) is zero and where your simple formulae hold. In general relativity, an exact thermal equilibrium also requires the energy to be conserved and well-defined which means that the background geometry has to have a global time-like Killing vector. If you think about it for a little while, you will see that it will still be possible to define the energies $\epsilon_i$ of the one-particle states, even in a curved spacetime (with a Killing vector), and the formulae above are actually valid (without any modifications) even if the space is curved.

To discuss distributions and temperature, one only needs to define the energy of a particle (a conserved quantity – it's conserved, via Noether's theorem, because of the time translational symmetry). Whether the space around is curved is irrelevant.

share|improve this answer
    
Suppose we are interested in investigating[theoretically] these statistical laws in some fast moving celestial object.In the rest frame of the object we may apply them as know them.But wrt to the earth the object with its rest frame is moving very fast.Transformation of these statistical laws is necessary in such situations.My question is intended for such investigation. –  Anamitra Palit Nov 15 '11 at 4:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.