Consider the following function:

$$f(x_1, x_2) = x_1^2x_2+x_1x_2^3$$

$f$ is a function of $(x_1, x_2)$. The conjugate variables $(u_1, u_2)$ to $(x_1, x_2)$ are $$u_1 = \partial f/ \partial x_1 = 2 x_1 x_2 + x_2^3$$ and $$u_2 = \partial f/ \partial x_2 = x_1^2 + 3x_1x_2^2.$$

One can construct $$g=f - u_1 x_1$$ which, to my understanding, replaces $x_1$ to its conjugate variable, $u_1$. The differential of $g$ is

$$dg = u_2 dx_2 - x_1 du_1$$ thus $g=g(u_1, x_2)$. In words, $g$ is a function of the old variable $x_2$ and the conjucate variable to $x_1$, $u_1$.

Now I fail to see this in the above example. I can compute

$$g = f - u_1x_1 = -x_1^2x_2$$ but $g$ is then still a function of $(x_1, x_2)$, not $(u_1, x_2)$. What am I missing here?

In theory, I could invert the $u_1=u_1(x_1, x_2)$ and $u_2=u_2(x_1, x_2)$ equations, then plug $x_1(u_1, u_2)$ and $x_2(u_1, u_2)$ into $f(x_1, x_2)$ to get $f(u_1, u_2)$. But that's just a simple change of variable. What's the point of the Legendre transformation concept then?

Also, in this particular example, inverting the $u(x)$ equations doesn't seem obvious. How would one get the expression for $g(u_1, x_2)$?

  • 1
    $\begingroup$ Not an answer, but the inverting isn't very hard in this case: just solve for $x_1 = \frac{u_1-x_2^3}{2x_2}$. Hamiltonians are usually quadratic so when you do this in that case all the inverting is just inverting linear functions as well. $\endgroup$
    – jacob1729
    Apr 27, 2019 at 10:54

1 Answer 1


I can show you the steps that you need to calculate the Legendre transformation, you have to look at the theory behind .

for a given function of two variable $f(x,y)$ we are looking for the Legendre function $\tilde{f}(u,v)$

Step I:

The determinant of matrix $A$ must be unequal zero. where $A$ is:

$$A= \left[ \begin {array}{cc} {\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,y \right) & \left( {\frac {\partial }{\partial x}}f \left( x ,y \right) \right) {\frac {\partial }{\partial y}}f \left( x,y \right) \\ \left( {\frac {\partial }{\partial x}}f \left( x,y \right) \right) {\frac {\partial }{\partial y}}f \left( x ,y \right) &{\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,y \right) \end {array} \right] $$

Step II:

with the equations:

$$u=\frac{\partial f(x,y)}{\partial x}\tag 1$$ $$v=\frac{\partial f(x,y)}{\partial y}\tag 2$$

we get from equation (1) , $x=f_x(u,y)$ and from equation (2) $y=f_y(u,x)$ both solutions must be exist

Step III:

with this solutions $f_x$ and $f_y$ we obtain the Legendre function




$$f_x(u,y)=1/2\,{\frac {u-{y}^{3}}{y}}$$ $$f_y(u,x)=1/3\,{\frac {\sqrt {3}\sqrt { \left( x \right) \left( u-{x}^{2} \right) }}{x}} $$ $$\tilde{f}(u,v)=1/2\,{\frac {u \left( u-{v}^{3} \right) }{v}}+1/3\,\sqrt {3}\sqrt {v \left( u-{v}^{2} \right) }-1/12\,{\frac { \left( u-{v}^{3} \right) ^{ 2}\sqrt {3}\sqrt {v \left( u-{v}^{2} \right) }}{{v}^{3}}}-1/18\,{ \frac { \left( u-{v}^{3} \right) \sqrt {3} \left( v \left( u-{v}^{2} \right) \right) ^{3/2}}{{v}^{4}}} $$

  • $\begingroup$ But you need to be able to get an analytic expression for $f_x(u, y)$ and $f_y(u, x)$, correct? What happens if you can't invert equations (1) and (2)? Can you still obtain your $\tilde{f}(u,v)$? $\endgroup$
    – Botond
    Apr 29, 2019 at 15:01
  • 1
    $\begingroup$ Correct . If you don’t have analytic solution for the invert functions , you don’t get close solution for the Legendre function, I don’t think that numerical solution make sense. $\endgroup$
    – Eli
    Apr 29, 2019 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.