# Differentiate the Lagrangian wrt. momentum?

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 $$\frac{\partial L}{\partial p_i}?$$

• Presumably by thinking of $\dot{x}$ as a function of $p$. Is this from a textbook? How did you encounter this? – Qmechanic Sep 14 at 15:43
• 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? – chkone Sep 14 at 15:59
• $\dfrac{\partial L}{\partial p_i}=0$ 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. – Frobenius Sep 14 at 20:34

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 $$H = p_i \dot{x}_i - L = \gamma m_0 c^2$$.

• 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. – Buzz Sep 15 at 0:20
• @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? – SuperCiocia Sep 15 at 0:39

@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 $$(t, x_i, \dot{x_i}, L)$$.