Lagrange multiplier for bose- Einstein stats

[siresmith]

Hi,

Why is -BEi used instead of +BEi as the lagrange multiplier for indistinguishable particles? How is it justified?

I've been reading a book about statistical mechanics and it introduces lagrange multipliers first for distinguishable particles- it has ln(ni) + a + BEi = 0.

(where a is alpha, B is beta, Ei is the energy)

The thing is, for the indistinguishable case, it has -BEi instead of BEi, so that it looks like ln () + a -BEi = 0. it goes on to show how this becomes the Boltzmann distribution is the number of levels,g, is much greater than the number of particles, n. But it seems that this only works if you have -BE rather than +BE.

So it looks like some sort of ad hoc measure to make things work. don't get it, is there a reason that its - instead of +?

thanks

 

[azntoon]

.The reason for the use of -BEi instead of +BEi as the lagrange multiplier for indistinguishable particles is because a negative sign is required to ensure that the probability of each state is positive. This is due to the fact that the energy levels of indistinguishable particles tend to be strongly degenerate; that is, they have more than one particle in the same state. Therefore, if the lagrange multiplier was positive, then the probability of each state would be zero, which is not desirable. Therefore, the negative sign is used to ensure that the probability of each state is positive.

 

[denian]

for your question. The use of -BEi instead of +BEi as the Lagrange multiplier for Bose-Einstein statistics is justified by the nature of indistinguishable particles. In classical mechanics, particles are considered to be distinct and can be identified by their individual properties, such as position and momentum. However, in quantum mechanics, particles are described by wave functions and cannot be uniquely identified. This means that the energy levels of indistinguishable particles are not independent and are correlated, resulting in a negative sign for the Lagrange multiplier.

To understand this further, let's consider a system of N identical particles in a specific energy level Ei. According to Bose-Einstein statistics, the probability of finding a particle in this energy level is given by:

pi = 1 / (exp[(Ei - μ)/kT] - 1)

where μ is the chemical potential, k is the Boltzmann constant, and T is the temperature. Now, if we take the logarithm of both sides and use the definition of the Lagrange multiplier, we get:

ln(pi) = -BEi + a

where a is a constant. This shows that the Lagrange multiplier for indistinguishable particles is indeed -BEi, as you have observed. This negative sign is necessary to account for the indistinguishability of particles and is not just an ad hoc measure.

Furthermore, the use of -BEi as the Lagrange multiplier leads to the correct Boltzmann distribution in the limit of g >> n, as you have mentioned. This is because in this limit, the term -BEi becomes much smaller compared to ln(ni) and can be neglected, resulting in the familiar form of the Boltzmann distribution.

In conclusion, the use of -BEi as the Lagrange multiplier for Bose-Einstein statistics is justified by the nature of indistinguishable particles and is necessary to correctly describe their behavior. I hope this helps clarify any confusion you may have had.