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Given $$ L=L(t, x_i, \dot x_i) $$ as a function of generalized coordinates/velocity, and $$ p_i:=\frac{\partial L}{\partial \dot x_i}, $$ how can we calculate $$\frac{\partial L}{\partial p_i}?$$

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    $\begingroup$ Presumably by thinking of $\dot{x}$ as a function of $p$. Is this from a textbook? How did you encounter this? $\endgroup$ – Qmechanic Sep 14 at 15:43
  • $\begingroup$ It's not from a textbook (someone edit and add "homework" I kept it). I need that in general inside the formalism of Lagrangian mechanics. How having $\dot{x}(p)$ helps? $\endgroup$ – chkone Sep 14 at 15:59
  • $\begingroup$ $\dfrac{\partial L}{\partial p_i}=0$ by the same reasoning that $\dfrac{\partial H}{\partial \dot{x}_i}=0$, see A mathematically illogical argument in the derivation of Hamilton's equation in Goldstein. $\endgroup$ – Frobenius Sep 14 at 20:34
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I guess the exact answer depends on the context where you found this.

But mathematically, this could be done with the chain rule: $$ \frac{\partial L}{\partial p_i} = \frac{\partial L}{\partial \dot{x}_i} \frac{\partial \dot{x}_i}{\partial p_i}. $$

Then, as suggested in the comment, you'd have to work out $\dot{x}_i(p_i)$ in order to compute $\partial \dot{x}_i/\partial p_i$.

A possible application of this is comuting the Hamiltonian $H(x_i, p_i)$ from the Lagrangian $L(x_i, \dot{x}_i)$.
In (classical) electromagnetism you know that the conjugate momentum $p_i$ ( $= \partial L/\partial \dot{x}_i$ ) $= \gamma m_0 \dot{x}_i$, i.e. $p_i(\dot{x}_i)$.

You can just invert this by writing $\dot{x}_i = p_i/(\gamma m_0)$, i.e. $\dot{x}_i(p_i)$.
This allows you to find the Hamiltonian $H = p_i \dot{x}_i - L = \gamma m_0 c^2$.

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    $\begingroup$ This is a situation where the partial derivatives need to specify which variables are being held constant while the differentiation is done—because there are more variables than underlying degrees of freedom. $\endgroup$ – Buzz Sep 15 at 0:20
  • $\begingroup$ @Buzz Good point. Should I treat $\dot{x}_i$ and $p_i$ on equal footing? That is, whatever is held constant with respect to one is also held constant with respect to the other? $\endgroup$ – SuperCiocia Sep 15 at 0:39
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@SuperCiocia: The context is what you mention. Expressing Hamilton Equations only related to Lagrangian, generalized coordinated/velocity, it is not applied to a specific dynamical system, it's general.

Particulary this one $\frac{\partial H}{\partial p_i}=\dot{x_i}$

If my math are correct we have: $$ \begin{align} \frac{\partial H}{\partial p_i} & = \frac{\partial \sum{\dot{x_k}p_k} - L}{\partial p_i} = \dot{x_i} \\ & = \sum{\frac{\partial \dot{x_k}p_k}{\partial p_i}} - \frac{\partial L}{\partial p_i} \\ & = \sum{\dot{x_k}} - \frac{\partial L}{\partial p_i} (??? NotSure) \\ & = \dot{x_i} - \frac{\partial L}{\partial p_i} (Or ??? NotSure) \\ \end{align} $$

What should give us: $$ \frac{\partial L}{\partial p_i}=\sum{\dot{x_k}}-\dot{x_i} $$ Or $$ \frac{\partial L}{\partial p_i}=0 $$ As mentionned by @Frobenius

If it's correct didn't help to express $\frac{\partial H}{\partial p_i}=\dot{x_i}$ only in function of term $(t, x_i, \dot{x_i}, L)$.

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