Lets consider the Bose-Einstein and Fermi-Dirac statistics: Bose-Einstein statis
ID: 1324256 • Letter: L
Question
Lets consider the Bose-Einstein and Fermi-Dirac statistics:
Bose-Einstein statistics:
Fermi-Dirac statistics:
[v and E represent velocity and energy respectively]
From the point of general relativity, is it necessary to give these laws a covariant form?
Lets consider the Bose-Einstein and Fermi-Dirac statistics: Bose-Einstein statistics: (epsiloni-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? = frac{1}{exp{[(epsiloni-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: = frac{1}{exp{[(epsiloni-mu)/kT]} - 1} Fermi-Dirac statistics:Explanation / 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 ?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
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