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If momentum is, $$p = \frac{\partial \mathcal{L}}{\partial \dot{q}}$$ and force is, $$ F = \frac{dp}{dt}$$

and by Euler-Langrange equations,

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

then why isn't force,

$$ F = \frac{\partial \mathcal{L}}{\partial q} $$

Am I getting my definition of momentum wrong? Or am I wrong about force being the time derivative of momentum?

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    $\begingroup$ Sometimes the definition of conjugate momenta in classical mechanics is not the usual kind which is used in $F=ma=dp/dt$, so perhaps that's the problem? $F = \frac{\partial \mathcal{L}}{\partial q}$ is true in rectilinear coordinates with a single particle moving around in some potential surface. $\endgroup$ – DumpsterDoofus Feb 19 '14 at 5:02
  • $\begingroup$ @Nikhil What makes you think it's not? $\endgroup$ – David Z Feb 19 '14 at 5:10
  • $\begingroup$ What happens when you consider Lagrange's equations in Cartesian coordinates? These notes should answer your question and like a good set of notes should leave you with new questions. isites.harvard.edu/fs/docs/icb.topic1288789.files/… $\endgroup$ – sunspots Feb 19 '14 at 5:12
  • $\begingroup$ The quantity you call force above is actually known as "generalized force" in the context of lagrangian mechanics $\endgroup$ – Manishearth Feb 21 '14 at 1:19
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General remarks.

The momentum you define in the first equation, namely \begin{align} p = \frac{\partial L}{\partial \dot q} \end{align} is not necessarily the same momentum that appears in Newton's Second Law. This momentum is called the canonical momentum conjugate to $q$, and it can be quite different from the momentum you're used to (the one appearing in the Second Law) depending on which generalized coordinates $q$ one chooses to parameterize the configuration space of the system.

Example: Free particle on the plane.

Consider, a particle moving freely in the plane. We can choose polar coordinates $(\rho, \phi)$ on the plane with which to parameterize the plane, and in these coordinates, the Lagrangian reads \begin{align} L(\rho, \phi) = \frac{1}{2}m(\dot \rho^2 + \rho^2\dot\phi^2), \end{align} in which case, the canonical momentum corresponding to $\phi$ reads \begin{align} p_\phi = \frac{\partial L}{\partial \dot \phi} = m\rho^2\dot\phi. \end{align} What is this quantity? Is it the momentum appearing in the Second Law? Certaintly not; the units are not right. So then what should be make of this? Well, recall that the angular momentum of such a particles is \begin{align} \mathbf L = \mathbf x\times \mathbf p \end{align} where $\mathbf p = m\dot{\mathbf x}$ is the momentum you're used to from the Second Law. If we write this in polar coordinates, then we obtain \begin{align} \mathbf L = \rho\hat{\boldsymbol\rho} \times m(\dot\rho\hat{\boldsymbol \rho} + \rho\dot\phi\hat{\boldsymbol\phi}) = m\rho^2\dot\phi\hat{\mathbf z} = p_\phi\hat {\mathbf z}, \end{align} so we find that $p_\phi$ is the angular momentum of the particle! Notice, however, that if we were to write the same Lagrangian in cartesian coordinates $(x,y)$, then we would find that the conjugate canonical momenta are \begin{align} p_x = m\dot x, \qquad p_y = m\dot y \end{align} which are the components of the momentum that appears in the Second Law. In other words, the canonical momentum and momentum that appear in the Second Law are equal only if one chooses to write the Lagrangian in cartesian coordinates.

Addendum. cartesian coordinates

Consider a system of a single particle moving in three dimensions under the influence of a conservative, velocity-independent force $\mathbf F$, namely a system for which there exists a function $V = V(\mathbf x)$, the potential energy, such that \begin{align} \mathbf F = -\nabla V \tag{$\star$} \end{align} The Lagrangian for such a system is given by \begin{align} L(\mathbf x, \dot{\mathbf x}) = \frac{1}{2} m\dot{\mathbf x}^2 - V(\mathbf x), \end{align} from which it follows that \begin{align} \frac{\partial L}{\partial x^i} = -\frac{\partial V}{\partial x^i}, \end{align} but by $(\star)$ the expression on the right hand side is simply the $i$th component $F_i$ of the force, so we do indeed have \begin{align} \frac{\partial L}{\partial x^i} = F_i \end{align} in this case. This can be generalized to any number of particles in a straightforward manner. In fact, it is discussed in the notes to which airwoz linked in his comment.

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  • $\begingroup$ Are you saying that in Cartesian coordinates, $F = \frac{\partial \mathcal{L}}{\partial r}$ is actually true? $\endgroup$ – XYZT Feb 19 '14 at 6:47
  • $\begingroup$ @NikhilMahajan Under appropriate circumstances, yes. See the addendum. $\endgroup$ – joshphysics Feb 19 '14 at 7:02
  • $\begingroup$ I think you're missing a starred equation here, but otherwise great answer. $\endgroup$ – Kyle Kanos Feb 19 '14 at 13:31

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