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I'm currently studying Lagrangian mechanics, and in the process, I've met the following equations in a couple of proofs. $$ \frac{\partial \mathcal{L}}{\partial q_i} = \dot p_i $$

$$ \frac{\partial \mathcal{L}}{\partial \dot q_i} = p_i. $$ Where do they come from? and do they hold indefinitely? Or only under certain conditions? Are there any general ones?

Qmechanic
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2 Answers2

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The relation below is simply the definition of the generalized momentum. It requires no proof, as is true by definition:

$$ p_i = \dfrac{\partial \mathcal{L}}{\partial \dot{q_i}} $$

Now according to Lagrangian Mechanics, we aim to find the equation of motion $q(t)$ which minimizes the action $S = \int \mathcal{L}(q, \dot{q}, t) \cdot dt$; note that we are fixing the starting points and ending points of the trajectory to specific points (i.e. what trajectory $q(t)$ which starts at $(q_1, t_1)$ and ends at $(q_2, t_2)$ minimizes $S[q]$). Turns out action is minimized if and only if this equation is satisfied:

$$ \dfrac{\partial \mathcal{L}}{\partial q} = \dfrac{d}{dt} \dfrac{\partial \mathcal{L}}{\partial \dot{q}} $$

$$ \therefore \dfrac{\partial \mathcal{L}}{\partial q_i} = \dot{p}_i $$

So the latter equation is a ramification of the axiom which states $S[q]$ needs to be minimized. The derivation of this equation is proven in many physics or functional analysis textbooks.

Edit: Your textbook should prove this itself, but you can also look into theorem 6.1 of this file https://scholar.harvard.edu/files/david-morin/files/cmchap6.pdf

Soroush khoubyarian
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The intuition behind the definition: $$ p_i = \frac{\partial L}{\partial \dot{q}_i} $$ is that for a single massive particle under some position-dependant potential energy, $V(\mathbf{q})$ and taking as generalized coordinates the cartesian coordinates (let's make it 1D and $q = x$ for simplicity): $$ L = T - V = m\frac{\dot{q}^2}{2} - V $$ So really: $$ \frac{\partial L}{\partial \dot{q}} = m \dot{q} = p $$ it coincides with linear momentum in this case. But really, it's a definition that generalizes this notion even when that's not equal to linear momentum. The second equation comes from the Lagrange equation itself: $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \dot{p} = \frac{\partial L}{\partial q} $$

Petrini
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    Alright, but when it comes to the second equation, the way it was derived would mean that it only holds for conservative systems, wouldn't it? –  May 04 '22 at 10:27
  • Yes, pretty much. Under non-conservative systems, you could use $\frac{d}{dt} \frac{\partial L}{\partial \dot{q} }- \frac{\partial L}{\partial q} = Q$ where $Q$ is a generalized force. Check Diracology's answer on this post: https://physics.stackexchange.com/a/342300/271807 – Petrini May 04 '22 at 11:16