Derivation of Entropy Eq - For Smart People

[qbslug]

hello smart people
I came across a weird equation for entropy and do not know how it was derived.
It looks like this

S = N(dS/dN)[V,E] + V(dS/dV)[N,E] + E(dS/dE)[N,V]

note that these technically partial derivatives

So how do you derive this equation. I am assuming you start with the first law of thermodynamics but I am confused how they got this result.

 

[Count Iblis]

Actually, this follows from the fact that S is an extensive quantity: it scales with system size if you keep the intensive vsriables (temperature, pressure etc.) constant. The internal energy, volume and number of particles are, of course, also extensive variables.

Let's see what happens if we specify a system by the variables E, V, and N, and then increase the system size by a factor lambda. The entropy must increase by lambda too:

S(\lambda E, \lambda V,\lambda N) = \lambda S(E, V, N)

Suppose we put:

\lambda= 1 +\epsilon

and expand both sides in powers of epsilon. Then equating the coefficient of epsilon on both sides gives you the desired equation. This is Euler's theorem for homogeneous functions (a special case of it).

You can then use the fundamental thermodynamic relation:

dE = T dS - P dV + \mu dN

To express the partial derivatives in terms of the temperature, pressure and chamical potential.

 

[qbslug]

Thanks for your input. I t makes more sense.
But what do you mean by expand in powers of epsilon and why do you need to replace lambda with 1+epsilon?
Can't you just use euler's theorem of homogeneous functions and be done.
I wasn't aware of euler's theorem so no wonder I was lost.

 

[Count Iblis]

That's how you derive Euler's theorem. We have:

S\left(\lambda E, \lambda V,\lambda N\right) = \lambda S(E, V, N)

Putting lambda = 1 + epsilon and expanding to first order in epsilon gives for the left hand side:

S\left[(1+\epsilon) E, (1+\epsilon) V,(1+\epsilon) N\right] =<br /> S\left(E, V,N\right) + \epsilon\left[E\left(\frac{\partial S}{\partial E}\right)_{V,N} + V\left(\frac{\partial S}{\partial V}\right)_{E,N}+N\left(\frac{\partial S}{\partial N}\right)_{E,V}\right]<br />


On the right hand side you have:

S(E, V, N) + \epsilon S(E, V, N)

If you equate the coefficient of epsilon on both sides you get:

S(E, V, N) =E\left(\frac{\partial S}{\partial E}\right)_{V,N} + V\left(\frac{\partial S}{\partial V}\right)_{E,N}+N\left(\frac{\partial S}{\partial N}\right)_{E,V}