# Momentum in Lagrangian mechanics

In the context of translation symmetry for lagrangian mechanics i was given this statement:

For a mechanical system $$\frac{∂L}{∂\dot{q}_i}=p_i$$ is the momentum.

I have no idea where this comes from.

To supplement the previous answers, consider the Lagrangian for a particle in a 1D potential $$V(q)$$ with a speed $$v = \dot{q}$$ and mass $$m$$: $$L = \frac{1}{2} m\dot{q}^2 - V(q)$$. Then the generalized momentum is: $$p = \partial L/ \partial \dot{q} = m\dot{q}$$. This matches the expression for momentum of a classical particle $$mv$$.

Lagrangian mechanics, though, is much more general as pointed out by the other answers. It allows one to study other systems with other generalized momentum associated with translational invariance.

Calling $$\frac{\partial L}{\partial \dot q}$$ plainly "momentum" is a little awkward, especially for people first learning about Lagrangian mechanics. More accurately it is called "generalized momentum" just as $$q$$ are the generalized coordinates.

The reason for the term "generalized momentum" is that you know what momentum is for systems of point bodies in cartesian coordinates $$x$$ (Newtonian mechanics). Then you transition to Lagrangian mechanics and notice that what you have always recognized as cartesian momentum since Newton, can be represented as $$\frac{\partial L}{\partial \dot x}$$. Then you notice that this is a conserved quantity if space is homogeneous (i.e. if there are no potential energy terms contained in $$L$$ that are fixed in space, or in other words, that explicitly depend on $$x$$). Finally you notice that Lagrangian mechanics can be applied in any generalized coordinates, especially if it can be expressed as a variational principle (Hamilton's principle). If the generalized coordinate space happens to be homogeneous as well (e.g. for a freely rotating joint with the angle as generalized coordinate), then you can conclude that what you have called "generalized momentum" (e.g. the angular momentum of the body that is constrained by the revolute joint) is also conserved.

You could also give it a name very different from "generalized momentum", but then you would miss the analogy of being conserved under appropriate conditions (homogeneity of the associated generalized coordinate space).

It's rather a generalized momentum. The generalized momenta of a mechanical system are defined in a different way to conventional linear and angular momentum.

Consider a holonomic mechanical system with Lagrangian $$\mathcal{L}=\mathcal{L}(q,\dot{q},t)$$. Then the scalar quantity $$p_j$$ defined by $$p_j=\frac{\partial \mathcal{L}}{\partial \dot{q}_j}$$ is called the generalized momentum corresponding to the coordinate $$q_j$$ It's also called the momentum conjugate to $$q_j$$

Edit: The above is defining property of momentum. All you can do is to make this significant: As I pointed out it's not linear or angular momentum but plays a much more border role in classical mechanics which can be seen from Noether's theorem, I'm stating it here.

Let $$\mathcal{S}$$ be a holonomic mechanical system with $$\mathcal{L}(q,\dot{q},t)$$ and let $$\{\mathfrak{M}^\lambda\}$$ be a one-parameter famility of mappings that have the action $$q\rightarrow q^\lambda$$

where $$q^\lambda=q$$ when $$\lambda=0$$. If the mappings $$\{\mathfrak{M}^\lambda\}$$ leave $$\mathcal{L}$$ invariant in the sense that $$\mathcal{L}(q^\lambda,\dot{q}^\lambda,t)=\mathcal{L}(q,\dot{q},t)$$ for all $$\lambda$$, then the quantity $$\sum_{j=1}^np_j\left[\frac{\partial q^\lambda_j}{\partial \lambda }\right]_{\lambda=0}$$ is conserved in any motion $$\mathcal{S}$$.

Note that the conserved quantity is not generally one of the momenta but linear combination of all of them with coefficient depending on generalized coordiante.

• It is the momentum, assuming that you correctly set up the Lagrangian. It comes from Noether's theorem. en.m.wikipedia.org/wiki/Noether%27s_theorem Apr 11, 2021 at 16:02
• I'll still say that's the defining equation of generalized momenta. Apr 11, 2021 at 16:34
• Your statement coincides with mine. Given a Lagrangian the Noether theorem tells you what the momentum is. Apr 11, 2021 at 16:36
• I added a little explanation to account noether's theorem Apr 12, 2021 at 9:15