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Let

$$L:{\mathbb R}^n\times {\mathbb R}^n\times {\mathbb R}\to {\mathbb R}$$

denote the Lagrangian (it should be differentiable) of a classical system with $n$ spatial coordinates. In the action

$$S[q]=\int_{t_1}^{t_2} L(q(t),q'(t),t)\, \mathrm{d}t,$$

the first $n$ slots are evaluates at a path $q:{\mathbb R}\to {\mathbb R}^n$, the second $n$ at $q'$ and the last one is for possible explicit time dependencies.

Are generalized momenta defined as functions of the generalized coordinates, i.e.

$$p_j=\frac{\partial L(x^1,\dots,x^n,v^1,\dots,v^n,t)}{\partial v^j},$$

or as associated with a curve $q:{\mathbb R}\to {\mathbb R}^n$, i.e.

$$p_j=\left.\frac{\partial L(q^1,\dots,q^n,v^1,\dots,v^n,t)}{\partial v^j}\right\rvert_{\large{q=q(t),\ v=q'(t)}},$$?

In the latter case it's a function of time only, and $q$ is buried somewhere within it.

A question that is codependent with this might be: What is the type of the total derivative and the term $\frac{\partial \mathcal{L}}{\partial q_i}$ in differential equation in $q:{\mathbb R}\to {\mathbb R}^n$, which is usually expressed as

$$\frac{\mathrm{d}}{\mathrm{d}t} \dfrac{\partial \mathcal{L}}{\partial {\dot q_i}} - \frac{\partial \mathcal{L}}{\partial q_i} = 0.$$

ACuriousMind
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Nikolaj-K
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    For classical systems in finite dimensions, $q,\dot{q},p$ are thought as maps from $\mathbb{R}$ to $\mathbb{R}^n$. Nevertheles, it is often convenient to keep track of the explicit dependence, e.g. in $q$ and $\dot{q}$ of $p$. – yuggib Mar 28 '15 at 11:12
  • @yuggib: It seems to me that equations stated like ${\dot q}={q,H}$ and ${\dot p}={p,H}$ are completely odd, no matter which of the two perspectives you take. – Nikolaj-K Mar 28 '15 at 14:24
  • They are not odd in my opinion, just more geometric: the Poisson brackets underline the symplectic structure of systems in classical mechanics. – yuggib Mar 28 '15 at 14:51
  • @yuggib: My point was that the Poisson bracket is defined as derivatives w.r.t. q and p, and on the rigth side we take such derivatives of q and p. On the left, on the other hand, we need to speak of trajectories. That's why they are odd, without pointing out where we substitute what. – Nikolaj-K Mar 28 '15 at 16:18

1 Answers1

2

I) Many of OP's questions on how the Lagrangian formalism works is already addressed in e.g. this Phys.SE post and links therein. For instance the question about the total time derivative in the EL equations is discussed in my answer.

II) In this answer, we would like to explain mathematically the various definitions in the Lagrangian formalism (of classical mechanics). Let us for simplicity assume that there is no explicit time dependence nor higher time derivatives, i.e. leave these generalizations as an exercise.

Let the manifold $M$ be the configuration/position space. Let there be given a Lagrangian function

$$\tag{1} E~:=~TM~\ni~z:=~(q,v)~ \stackrel{L}{\mapsto}~L(q,v)~\in~ \mathbb{R}.$$

The differential is

$$\tag{2} TE~\stackrel{L_{\ast}\equiv\frac{\partial L}{\partial z}}{\longrightarrow} ~T\mathbb{R}.$$

The corresponding Lagrangian momentum function

$$\tag{3} E~\ni~(q,v)~\stackrel{(\pi_M, p)}{\mapsto}~(q,p(q,v))~\in~T^{\ast}M$$

is given in coordinates as$^1$

$$\tag{4} p~=~\frac{\partial L}{\partial v}.$$

Similarly we have

$$\tag{5} E~\ni~(q,v)~\stackrel{(\pi_M, \frac{\partial L}{\partial q})}{\mapsto}~(q,\frac{\partial L(q,v)}{\partial q})~\in~T^{\ast}M.$$

In eqs. (3) and (5) we have used suitable canonical identifications between the cotangent bundles $T^{\ast}E$ and $T^{\ast}M$.

III) Let

$$\tag{6} I~:=~[t_i,t_f] ~\stackrel{\gamma}{\longrightarrow}~M$$

be a position path/curve. Let

$$\tag{7} I ~\ni~t~ \stackrel{\tilde{\gamma}}{\mapsto}~(\gamma(t),\dot{\gamma}(t))~\in~ E$$

be the corresponding lift to the tangent bundle.

IV) The corresponding pullback

$$\tag{8} I ~\ni~t~\mapsto~(\tilde{\gamma}^{\ast} L)(t)~:=~L\circ\tilde{\gamma}(t)~\in~ \mathbb{R} $$

and

$$\tag{9} I ~\ni~t~\mapsto~(\tilde{\gamma}^{\ast} p)(t)~:=~p\circ \tilde{\gamma}(t) $$

are often in the physics literature also called the Lagrangian and the momentum (of/along the path), respectively.

V) The action functional is

$$\tag{10} C^1(I)~\ni~\gamma~\stackrel{S}{\mapsto} \int_I \!dt ~\tilde{\gamma}^{\ast} L ~\in ~\mathbb{R}. $$

VI) The Euler-Lagrange equations read

$$\tag{11} \frac{d}{dt} p\circ\tilde{\gamma} (t)~=~\frac{\partial L}{\partial q}\circ\tilde{\gamma} (t).$$

--

$^1$ In this answer, we only discuss the Lagrangian formalism. There is a corresponding Hamiltonian formalism, which we do not consider for simplicity. In particular, the Lagrangian momentum function (4) should not be confused with the Hamiltonian momentum, which is an independent variable.

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