One of hamilton's equations is $(\frac{\partial H}{\partial q} )_t = -(\frac{\partial p}{\partial t}) _q$. But isn't it $\frac{\partial L}{\partial q} = \frac{dp}{dt}$? If H = L(i.e. V = 0), what happens? Am I confusing which variable is fixed on the partial derivation?
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2$\begingroup$ If $V=0$ and the kinetic energy doesn't explicitly depend on $q$, then both of your equations will turn out to be zero $\endgroup$– CourageFeb 16, 2016 at 17:07
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$\begingroup$ The double pendulum, which contains cos(q1-q2) in its kinetic lagrangian, is a counterexample of that statement. $\endgroup$– L-CFeb 16, 2016 at 18:20
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If $H=L$, then $$ L=H=\sum_{a}\frac{\partial L}{\partial \dot{q}^{a}}\dot{q}^{a}-L $$ so that $$ L=\frac{1}{2}\sum_{a}\frac{\partial L}{\partial \dot{q}^{a}}\dot{q}^{a} $$ from which follows $$ H=L=\sum_{a}\ A_{a}(q)\ (\dot{q}^{a})^{2} $$ You can check by yourself that you get the same equations from both the Lagrangian and the Hamiltonian formalism: $$ \frac{d}{dt}\ A_{a}(q)\ \dot{q}^{a}=\frac{1}{2}\sum_{b}\frac{\partial A_{b}}{\partial q^{a}}\ (\dot{q}^{b})^{2} $$ where $$ p_{a}=2\ A_{a}(q)\ \dot{q}^{a} $$
answered Feb 16, 2016 at 17:28