Legendre Transformation Can someone explain how the Legendre transformation for multivariable functions work? I have been trying to find Legendre transformation for a function like $$F(x, y) = x^{2}+y^{2}+xy.$$ But I don't know how to approach it. And also does Legendre transform in multivariable functions have any physical significance like Fourier transforms.
 A: Sure - there are a number of definitions of the Legendre transformation, but the one used most commonly in physics is as follows:
For each variable of your function (i.e. $x$ and $y$ for you) we define a "conjugate variable" (often called a "conjugate momentum"). Let's call these $\{p, q, \ldots\}$ (i.e you need $p$ and $q$ that are conjugate to $x$ and $y$). These variables are defined by $p = \frac{\partial F}{\partial x},\,\, q = \frac{\partial F}{\partial y},\,\, \ldots$, so are simple partial derivatives of your original function.
Then we construct a new function, the Legendre transformation, 
$$F^{\star}(p, q, \ldots) = px + qy + \ldots - F(x, y, \ldots) \bigg|_{p = \frac{\partial F}{\partial x},\,\, q = \frac{\partial F}{\partial y},\,\, \ldots}$$
What does the notation with the vertical line mean? Well you have a set of $N$ equations relating the $N$ conjugate variables to the $N$ original variables and you must solve this system of equations to express $x$, $y$, $\ldots$ in terms of the conjugate variables $p$, $q$ etc. That means that at the end of your calculation you will have a function of two variables $p$ and $q$ because you will have eliminated $x$ and $y$ in favour of the conjugate variables. 
Here are the steps you must complete:


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*$p = \frac{\partial F}{\partial x} = $?

*$q = \frac{\partial F}{\partial y} = $?

*Invert these equations to write $x = x(p, q)$ and $y = y(p, q)$ in terms of $p$ and $q$. 

*Form $F^{\star}(p,q) = p x(p, q) + q y(p, q)  - F\big(x(p, q), y(p, q)\big)$. Then $F^{\star}$ is your Legendre transform.


Two other interpretations of the Legendre transform are (restrict to one variable for simplicity)


*

*The derivative of the Legendre transform $F^{\star \prime}(p)$ is the inverse function of the derivative of the original function: $F^{\star \prime} = F^{\prime \, -1}$

*For convex functions the Legendre transform is the function that maximises $px - F(x)$ over x: $F^{\star}(p) = \max_{x} \{ px - F(x) \}$. You can see how this leads to the (implicit) equation $F^{\prime}(x) = p$. 


As for applications in physics....


*

*In quantum mechanics / quantum field theory, the conjugate momentum to a variable in a Lagrangian, $\mathcal{L}(\phi, \dot{\phi})$ is defined to be $\pi = \frac{\mathcal{L}}{\partial \dot{\phi}}$. One goes to the Hamiltonian description of the problem by taking the Legendre transformation:
$$\mathcal{H}(\phi, \pi) = \pi \dot{\phi} - \mathcal{L}(\phi, \dot{\phi}) \big|_{\pi = \frac{\mathcal{L}}{\partial \dot{\phi}}}$$ that eliminates $\dot{\phi}$ from the problem. 

*In thermodynamics, starting from the internal energy (say), $U(S, V)$, we can get the Helmholtz free energy as (negative!) the Legendre transform with respect to entropy, $S$. Now since $dU = T dS - p dV$, it is clear that the conjugate variable (don't confuse $p$ with the pressure here) to $S$ is $p = \frac{\partial U}{\partial S} = T$ and so negative the Laplace transform would be 
$$F(T, V) = -\left[p S - U(S, V) \right] = -\left[ T S - U(S, V) \right] = U - TS$$
with the requirement that, for any particular gas, we express its entropy, $S$, in terms of temperature and volume, $V$, so that $F$ becomes a function of $T$ and $V$ (this is easy once we have an explicit expression for $U(S, V)$).
