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  Obtaining the Maxwell-Boltzmann distribution from the string-theoretical thermal path integral?

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Can the Maxwell-Boltzmann distribution be obtained as the zero-mode part of the string-theoretical thermal path integral? And if this is indeed the case, can somebody explain this in some mathematical detaills to me?

asked Jun 15, 2014 in Theoretical Physics by Dilaton (6,240 points) [ no revision ]

If this question is nonsense I blame @LubošMotl for it, as it was him who made a corresponding comment at the beginning of this post (directly above the paragraph titled "Field theory"). So he better come here and explain this ... ;-)!

The statement Lubos is making is that the dynamics of the low-energy theory is what Maxwell is describing with Maxwell equations and thermodynamics, and that if string theory is correct, as Motl assumes, the theories Maxwell gave us are a description of the dynamics of nearly massless part of the string spectrum.

1 Answer

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There is no string thermal path-integral, gravity is incompatible with a uniform constant energy density. Constant energy density, constant temperature, leads to gravitational collapse when the volume is too big. This is why you can't have a consistent string thermal equilibrium state.

One consequence of this is that it is not possible to speak about thermal analogs of T-duality clearly in flat space-time, as people sometimes try to do, using imaginary time periodic path integrals. The presence of a thermal energy density curves space, and constant energy density is inconsistent.

The closest analog of a pure thermal state in a gravitational theory is a deSitter space, because semiclassically this space has a constant temperature, and a uniform cosmological constant. This type of state is not easy to incorporate into string theory, there is no consensus on how to do it.

answered Jun 15, 2014 by Ron Maimon (7,730 points) [ revision history ]

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