0

$f(x,y)=x^3y^2$ the goal is the Legendre-transformed function: $g(x,u)=uy-f(x,y)$ where $u=\frac{∂ f}{∂ x}$ and $v=u=\frac{∂ f}{∂ y}$ where g(x,u) isn't explicitly dependent on y.

I derived $u=x^3y^2$. I heard that one now needs to reshape the equation to $y=\sqrt{(\frac{u}{3x^2})}$, but after that step I am not sure which variables are variables in the function $g(x,u)=uy-f(x,y)$ and which variables are "constants" and I am confused, why I need to reshape everything into the form y=... I thought I could just plug this into the equation and that's it.

I would be grateful for any advice!

Eletie
  • 3,213

1 Answers1

1

The idea is that if $f$ is considered a Lagrangian, you see it as $f(x,v)=x^3 v^2$. Now the momentum can be calculated as $p=\frac{\partial f}{\partial v}=2x^3 v$. Then the Hamiltonian $g= pv - f(x,v)$ expressed as a function of $(x,p)$ by replacing $v$ as a function of $(x,p)$ would be $p \frac{p}{2x^3} - x^3 (\frac{p}{2x^3})^2$.

C Tong
  • 338
  • Thank you very much for answering! –  Dec 20 '20 at 17:42
  • Is it then correct, that I get g(x,u)=$u\sqrt{\frac{u}{3x^2}}-\frac{xu}{3}$? –  Dec 20 '20 at 17:43
  • If you are using g=$uy-f$, you must use $u=\frac{\partial f}{\partial y}$ instead of $u=\frac{\partial f}{\partial x}$. – C Tong Dec 20 '20 at 19:50
  • I am confused as to why you patially derived f in respect to v and not in respect to y because in my lecture notes it says that $=\frac{∂}{∂}$ when we defined the Legendre transform –  Dec 20 '20 at 20:35
  • I was using $\partial f/\partial y$ but you were not doing that in your original post. $u=\frac{\partial x^3y^2}{\partial y}=2x^3y$, but you concluded $u=x^3 y^2$. – C Tong Dec 20 '20 at 21:20