M=∫∫∫ρdV
Z =∫∫∫(f+φ)ρdV
while ρ(x,y,z) is density,M is mass,T is temperature
f(1/ρ,T) is helmholtz free energy per unit mass
φ(x,y,z) is potential energy per unit mass
it's said that when Δz=0 the system is in equilibrium
mass doesn't change ΔM=0
then how to solve these two equations
so as to make the statement that
f+φ+ρ(∂f/∂ρ)=constant , independent of x,y,z ?

Question

M=∫∫∫ρdV
Z =∫∫∫(f+φ)ρdV
while ρ(x,y,z) is density,M is mass,T is temperature
f(1/ρ,T) is helmholtz free energy per unit mass
φ(x,y,z) is potential energy per unit mass
it's said that when Δz=0 the system is in equilibrium
mass doesn't change ΔM=0
then how to solve these two equations
so as to make the statement that 
f+φ+ρ(∂f/∂ρ)=constant , independent of x,y,z ?

kainui · Answer

How far are you able to get on your own? It looks like we need to use the chain rule in differentiating f with respect to ρ.

kainui · Answer

Maybe we can also use integration by parts and split Z into two integrals to get it to look like M? Some interesting things to play around with.

anonymous · Answer

Δ∫∫∫(f+φ)ρdV
=∫∫∫[Δ(f+φ)ρ+(f+φ)Δρ]dV
=∫∫∫[ρΔf+ρΔφ+(f+φ)Δρ]dV
=∫∫∫{ρ[(∂f/∂ρ)Δρ+(∂f/∂T)ΔT]+ρΔφ+(f+φ)Δρ}dV
=∫∫∫{[f+φ+ρ(∂f/∂ρ)]Δρ+ρΔφ+ρ(∂f/∂T)ΔT}dV
then i don't know how to get any further

anonymous · Answer

Where, $\Delta=\nabla^2$?

anonymous · Answer

Never mind, that just looks awkward, what do you mean, then, by $\Delta M=0$?

anonymous · Answer

it's calculus of variation.To my knowledge,i think it's about time.
the mass doesn't change from time to time i think

anonymous · Answer

That's what I thought, if you want to make it so, make sure to not use capital delta $\delta M$ is more appropriate. I'll be right back, see what I can do.

anonymous · Answer

I must say, I'm currently stumped.

It seems like a continuity equation, but I'm not sure what to do with the variations, since it's not a functional we're referring to.

anonymous · Answer

I mean, we could simply go with the assumption that the system is in equilibrium when, given some small $|\delta \rho|>0$ we have that $\delta z=0$. Then it's quite simple.

anonymous · Answer

well after reading books i find every similar problem it goes that way
[f+φ+ρ(∂f/∂ρ)]Δρ
then f+φ+ρ(∂f/∂ρ) is constant
[g+φ]Δρ
then g+φ is constant 
what i can't figure out is that when it's with Δρ then it's ''coefficient'' must be constant

anonymous · Answer

does it get something to do with  ΔM=∫∫∫ ΔρdV=0 ?

anonymous · Answer

That's what I'm thinking. Perhaps integration by parts on the first integral; but that gives an odd result.

anonymous · Answer

What book is this, anyways? It'd be nice to know a bit more about the context.

anonymous · Answer

well that book is in chinese and it's a bit old

anonymous · Answer

@dan815

anonymous · Answer

Again, if we do go by the first definition (which I what I think it means) that for some variation $\delta \rho \ne 0$, we have that $\delta z=0$, then the statement is fairly clear. I mean, this is the only thing that makes sense, because any other variation yields unwanted terms, so we'd need stricter conditions to clean them up.

anonymous · Answer

but what about other terms like ρΔφ+ρ(∂f/∂T)ΔT ?

anonymous · Answer

All are zero because each is independent of $\rho$.

anonymous · Answer

why Δφ is zero? i'm wondering if  Δ means the variation from time to time

anonymous · Answer

In this case, it just means a given variation. $\delta \phi$ is zero since it is completely independent of density (only dependent on position), and $\delta T$ is the same.