# A Chain rule - legendre transformation

1. Dec 26, 2016

### sgh1324

let

df=∂f/∂x dx+∂f/∂y dy and ∂f/∂x=p,∂f/∂y=q

So we get

df=p dx+q dy

d(f−qy)=p dx−y dq

and now, define g.

g=f−q y
dg = p dx - y dq

and then I faced problem.

∂g/∂x=p←←←←←←←←←←←←←←← book said like this because we can see g is a function of x and p so that chain rule makes ∂g/∂x=p

but I wrote directly ∂g/∂x so i can get a result,

∂g/∂x=∂f/∂x−y ∂q/∂x−q ∂y∂x=p−y ∂q∂x - 0(y is independent of x)

I know that variation y is independent of x, but I'm not sure that q is also the independent function of x. what if the func f is xy??? if I set like that,

g=f−qy=xy−xy=0

it's not same with p!!!

mathematical methods in the physical sciences, Mary L. Boas 3rd edition page231
http://www.utdallas.edu/~pervin/ENGR3300/Boaz.pdf

2. Dec 26, 2016

### Useful nucleus

In taking partial derivatives , it is recommended to explicitly show which variable is maintained constant. So you want to calculate (∂g/∂x)|q. There are two ways. The easiest one, is to look at the differential of g and set dq=0 since you are holding q constant. This way you get (∂g/∂x)|q = p. The second is to use your method. Then,

(∂g/∂x)|q=(∂f/∂x)|q - q (∂y/∂x)|q --------- (1)

notice that (∂f/∂x)|q ≠ p. Correct is (∂f/∂x)|y = p.

It is possible to manipulate expression(1) above to reach to the same conclusion that we obtain easily from the differential which is (∂g/∂x)|q = p.

Also note that there no need to assume that y is independent of x. In fact in the context of applying Legendre Transform to Thermodynamics, the natural variables do depend on each other. For example U=(S,V,N) where U is the internal energy, and we have a dependence of S on N and V.