数学上：一组独立变量到另一组独立变量的变换

物理上：Lagrange力学中的广义坐标和广义速度变换为Hamilton力学中的该功能一坐标和广义动量；热力学函数$U,H,F,G$之间的变换.

$u(x)=\frac d{dx}f,\to x(u)$:

have $g(u)=x(u)u-f(x(u)),\frac{dg}{du}=\frac{dx}{du}\cdot u+x(u)-\frac {df}{dx}\cdot\frac{dx}{du}=x(u)$

then we get $x(u)$ (exachange the roles of $x$ and $u$) from a Legendre transformation[$f$&$g$ are Legandre transforms of each other.]

$x(u)=\frac{d}{du}g$

勒让德变换可以写成以下形式：从$F(\boldsymbol u,\boldsymbol w)$变换到$F^*(\boldsymbol v,\boldsymbol w)$，其中$\boldsymbol u,\boldsymbol v$是变换对，$\boldsymbol \omega$是不参与变换的变量。勒让德变换满足下式：

$$ F^*(\boldsymbol v,\boldsymbol w)+F(\boldsymbol u,\boldsymbol w)=\boldsymbol u\cdot \boldsymbol v,\text{where}\ \boldsymbol v=\nabla_{\boldsymbol u}F(\boldsymbol u,\boldsymbol w) $$

And the inverse formula, but the symmetric relation.

$$ F^(\boldsymbol v,\boldsymbol w)+F(\boldsymbol u,\boldsymbol w)=\boldsymbol u\cdot \boldsymbol v,\text{where}\ \boldsymbol u=\nabla_{\boldsymbol v}F^(\boldsymbol v,\boldsymbol w) $$

Futher more, the varible $\boldsymbol \omega$ don’t join the transformation.

$$ \nabla_{\boldsymbol w}F^*(\boldsymbol v,\boldsymbol w)+\nabla_{\boldsymbol w}F(\boldsymbol u,\boldsymbol w)=0 $$

Lagrangian to Hamiltonian

https://web.physics.wustl.edu/alford/physics/Legendre_introduction.pdf

8.2: Legendre Transformation between Lagrangian and Hamiltonian mechanics

$L(q,\dot q,t)$

E-L方程：

$$ \frac{\partial L}{\partial q_i}=\frac d{dt}\frac{\partial L}{\partial\dot{q}_i} $$

$$ \begin{aligned} \frac{dL}{dt} & = \frac{\partial L}{\partial q_i}\dot{q}^i+\frac{\partial L}{\partial\dot{q}_i}\ddot{q}^i \\&=\frac d{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right)\dot q^i+\frac{\partial L}{\partial\dot{q}_i}\ddot{q}^i \\&=\frac d{dt}\left (\frac{\partial L}{\partial\dot{q}_i}\dot{q}^i \right) \end{aligned} $$

$$ \begin{aligned}\frac{dL}{dt}&=\underline{\sum_j\dot{q}_j\frac{d}{dt}\frac{\partial L}{\partial\dot{q}_j}} -\sum_j\dot{q}j\left[Q_j^{EXC}+\sum{k=1}^m\lambda_k\frac{\partial g_k}{\partial\dot{q}_j}(\mathbf{q},t)\right]+\underline{\sum_j\frac{\partial L}{\partial\dot{q}_j}\ddot{q}_j} +\frac{\partial L}{\partial t}\\&=\sum_j\frac{d}{dt}\left(\dot{q}_j\frac{\partial L}{\partial\dot{q}_j}\right)-\sum_j\dot{q}j\left[Q_j^{EXC}+\sum{k=1}^m\lambda_k\frac{\partial g_k}{\partial q_j}(\mathbf{q},t)\right]+\frac{\partial L}{\partial t}\end{aligned} $$

根据勒让德变换

$$ p=\frac{\partial L(q,\dot q,t)}{\partial \dot q} $$

$$ H(p,q,t)+L(q,\dot q,t)=p\dot q $$

the symmetric relation: