- #1
MathematicalPhysicist
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The problem is to show that for a fermionic gas the entropy is given by:
[tex]\sigma=-\int d\epsilon D(\epsilon )[f(\epsilon )log(f(\epsilon )-(1-f(\epsilon )log(1-f(\epsilon )][/tex] where D(epsilon) is the derivative operator wrt epsilon, and f(epsilon) is fermi-dirac distribution function.
Now what I think is that I only need to show that the entropy equals minus the integrand, but I'm not sure where did the minus come from.
I mean the entropy is defined as logarithm of the number of possible states, the function that counts this number is: (f^f)*((1-f)^(1-f))
cause f counts the number of possible states there are below the chemical potential and 1-f above it, and we take a power of themselves because there sum equals the number of states of the system.
but I don't where did the minus sign come from, can you help me on this?
thanks in advance.
[tex]\sigma=-\int d\epsilon D(\epsilon )[f(\epsilon )log(f(\epsilon )-(1-f(\epsilon )log(1-f(\epsilon )][/tex] where D(epsilon) is the derivative operator wrt epsilon, and f(epsilon) is fermi-dirac distribution function.
Now what I think is that I only need to show that the entropy equals minus the integrand, but I'm not sure where did the minus come from.
I mean the entropy is defined as logarithm of the number of possible states, the function that counts this number is: (f^f)*((1-f)^(1-f))
cause f counts the number of possible states there are below the chemical potential and 1-f above it, and we take a power of themselves because there sum equals the number of states of the system.
but I don't where did the minus sign come from, can you help me on this?
thanks in advance.