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$f(x,y)=x^3y^2$ 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
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1 Answers1

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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
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