As one key argument in Introductoryt QM Class, we've been taught to use a Lagrangian and Hamiltonian generalized description of a dynamic systems, which follows the Euler-Lagrange or Hamilton equation of motion. However, I stumbled upon a doubt using the total time derivative inside the firts EL equation.

Consider this problem:

An object with mass $m$ is connected to a one-dimensional spring, with elastic constant $k$ and relaxed lenght $l_0$. Derive the system's generalized equation of motion using Lagrange's formalism.

At first, I know that inside this conservative system, the lagrangian is written as follows: $$ \mathcal{L}(x,\dot{x})=E_{k}(\dot{x})-E_{p}(x)=\frac{1}{2}m\dot{x}^2-kl_0x-\frac{1}{2}kx^2 $$ so far, so good. Now, using the first Euler-Lagrange equation and taking the correct derivatives, this expressions are obtained: $$ \frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{x}}\right)=\frac{\partial\mathcal{L}}{\partial x}\quad\to\quad m\frac{d\dot{x}}{dt}=-k(x+l_0) $$ now I'm supposed to treat $d\dot{x}/dt=\ddot{x}$, but I just can't because the operator $d/dt$ is a total derivative with respect to time for the function $\dot{x}$, which depends on $x$ because it's its derivative, so I thought it had to be expressed like this: $$ \frac{d\square}{dt}=\frac{\partial\square}{\partial t}+\dot{\mathbf{q}}\cdot\nabla_{\mathbf{q}}\square\quad\to\quad\frac{d\dot{x}}{dt}=\frac{\partial\dot{x}}{\partial t}+\dot{x}\frac{\partial\dot{x}}{\partial x}\neq\frac{\partial\dot{x}}{\partial t} $$ but I know there is a catch in my reasoning, but it doesn't seem trivial to me. I also learned from this source that treating $\partial\dot{x}/\partial x\neq0$ is the right thing to do to demonstrate the second Hamilton relation.

Remark: Beware of the following incorrect derivation of the $\partial H/\partial q = -\partial L/\partial q$: from the definition $H=p\dot{q}-L$, we have $\partial H/\partial q=\partial(p\dot{q})/\partial q-\partial L/\partial q$. Since neither $p$ nor $\dot{q}$ depends on $q$, the first term on the right-hand side is zero, and we arrive at the desired result, $\partial H/\partial q = -\partial L/\partial q$.

The error in this reasoning is that $\dot{q}$ does depend on $q$, because it is understood that when writing down $H$, $\dot{q}$ must be eliminated and written in terms of $q$ and $p$. So the first term on the right-hand side is in fact not equal to zero.

The final question is: how can I treat the total time differential inside the first EL equation? Why and where is my reasoning wrong?


1 Answer 1


Your reasoning is wrong in that it doesn't take into account the space your variables and equations belong to. This is not in any ways your fault: I have NEVER seen a thorough explanation of Lagrangian and Hamiltonian dynamics in an introduction to quantum mechanics. Here is how it works.

Lagrangian and Hamiltonian dynamics (at least in the context of particle dynamics, i.e. not of field dynamics) are dual theories, in the sense that they are equivalent theories that are defined on dual spaces (in the sense of vector spaces). The Lagrangian is a function given on a set known as a tangent bundle. The Hamiltonian is a function given on a set known as a cotangent bundle, which, in a sense that will be clear soon, is the dual space to the tangent bundle.

Let's say you have a configuration space $X$. This can be, for example, the 3-dimensional space of the positions of a particle in physical space, or the 3N-dimensional space of the positions of N particles in physical space. The tangent bundle associated to $X$ is denoted with $TX$ and consists of the space of vectors attached to each and every point of $X$. There is a rigorous definition of $TX$, but for the purpose of answering to your question, let us define $TX$ as $$ TX=\coprod_{x\in X}T_{x}X $$ where $T_{x}X$ is the vector space which contains the vectors attached to $x\in X$, and $\coprod$ is the disjoint union of its argument for $x\in X$, i.e. $$ TX\ni(x^{i},\dot{x}^{i}) $$ where we denote with $\dot{x}^{i}$ the $i$-th component of some vector. The Lagrangian $\mathscr{L}$ is a real function on $TX$, while the action is a functional on the space of curves on $X$: $$ S[x(t)]=\int_{a}^{b}\mathscr{L}(x^{i}(t),\dot{x}^{i}(t))\ dt $$ where $x:\Bbb{R}\to X$ is a curve and the $\dot{x}^{i}$'s are given by $dx^{i}/dt$. The action is minimized when the Euler-Lagrange equations $$ \frac{\partial\mathscr{L}}{\partial x^{i}}(x(t),\dot{x}(t))-\frac{d}{dt}\frac{\partial\mathscr{L}}{\partial \dot{x}^{i}}(x(t),\dot{x}(t))=0 $$ are satisfied. These equations, though, are equations on $TX$, with an additional $\Bbb{R}$-dependence (in a rigorous formulation, $TX$ should really be replaced with $T(\Bbb{R}\times X)$). In $TX$, $x^{i}$ and $\dot{x}^{i}$ are independent variables, so indeed here there is no derivative $\partial \dot{x}^{i}/\partial x^{j}$.

The cotangent bundle associated to $X$ is denoted with $T^{*}X$ and defined by

$$ T^{*}X=\coprod_{x\in X}T^{*}_{x}X $$ where $T^{*}_{x}X$ is the dual space of $T_{x}X$. We have $$ T^{*}X\ni(x^{i},p_{i}) $$ where the $p_{i}$'s are the components of the element in $T_{x}^{*}X$. The Hamiltonian $\mathscr{H}$ is a function on $T^{*}X$. To go from Lagrangian to Hamiltonian dynamics, we define the so called Legendre transform associated to $\mathscr{L}$. This is the function $$ \Bbb{F}\mathscr{L}:TX\to T^{*}X $$ defined by $$ \Bbb{F}\mathscr{L}(q^{i},\dot{q}^{i})=\left(q^{i},\frac{\partial\mathscr{L}}{\partial \dot{q}^{i}}(q^{i},\dot{q}^{i})\right) $$ Let's suppose that $\Bbb{F}\mathscr{L}$ is invertible. Then we have a function $\Bbb{F}\mathscr{L}^{-1}:T^{*}X\to TX$ and we can define the Hamiltonian $\mathscr{H}$ as $$ \mathscr{H}:T^{*}X\to \Bbb{R},\qquad\quad \mathscr{H}(q^{i},p_{i})=\min_{\dot{q}}\left\{p_{i} \dot{q}^{i}-\mathscr{L}\left(q^{i},\dot{q}^{i}\right)\right\} $$ It's easily seen that $$ 0=\frac{\partial}{\partial \dot{q}^{i}}\ \left\{p_{i} \dot{q}^{i}-\mathscr{L}(q^{i},\dot{q}^{i})\right\}=p_{i}-\frac{\partial \mathscr{L}}{\partial \dot{q}^{i}}\left(q^{i},\dot{q}^{i}\right) $$ i.e. iff $(q^{i},p_{i})=\Bbb{F}\mathscr{L}(q^{i},\dot{q}^{i})$ (so that in turn $(q^{i},\dot{q}^{i})=\Bbb{F}\mathscr{L}^{-1}(q^{i},p_{i})$). Again, $q^{i}$ and $p_{i}$ are independent variables in $T^{*}X$. Moreover, as the $\dot{q}^{i}$'s in $\mathscr{H}$ are held fixed by the minimum condition, $$ d\mathscr{H}=\dot{q}^{i}dp_{i}-\frac{\partial\mathscr{L}}{\partial q^{i}}dq^{i} $$ and of course $$ d\mathscr{H}=\frac{\partial\mathscr{H}}{\partial q^{i}}dq^{i}+\frac{\partial\mathscr{H}}{\partial p^{i}}dp^{i} $$ both with the correct relation between $(q^{i},\dot{q}^{i})$ and $(q^{i},p_{i})$, so that $$ \dot{q}^{i}(t)=\frac{\partial\mathscr{H}}{\partial p^{i}}(q^{i}(t),p_{i}(t)),\qquad\quad \dot{p}_{i}(t)=\frac{d}{dt}\frac{\partial\mathscr{L}}{\partial \dot{q}^{i}}\left(\Bbb{F}\mathscr{L}^{-1}(q^{i}(t),p_{i}(t))\right)=\frac{\partial\mathscr{L}}{\partial q^{i}}\left(\Bbb{F}\mathscr{L}^{-1}(q^{i}(t),p_{i}(t))\right)=-\frac{\partial\mathscr{H}}{\partial q^{i}}\left(q^{i}(t),p_{i}(t)\right) $$ These are equations on $T^{*}X$ and, as such, there is no derivative $\partial p_{i}/\partial x^{j}$. Again, in order to be able to derivate with respect to $t$, $T^{*}X$ should really be extended to $T^{*}(\Bbb{R}\times X)$.

The problem with the switch between Lagrangian and Hamiltonian dynamics is that it must be formulated in VERY rigorous terms. The formulation I gave above is not by any means the only one, and it doesn't take care of many aspects which in more complicated theories are not by any means trivial. In each of the spaces $TX$ and $T^{*}X$, the $q^{i}$'s are independent of the $\dot{q}^{i}$'s and the $p_{i}$'s, but the $\dot{q}^{i}$'s in the action are not independent of the $q^{i}$'s, and the $(q^{i},p^{i})$'s are made to depend (ONLY in the switch between the two formulations, not in the dynamical equations) upon the $(q^{i},\dot{q}^{i})$'s by the Legendre transform.

  • 1
    $\begingroup$ Wonderfully thorough answer, although I think a footnote that a non-invertibility of the Legrendre transform leads to constraints on the Hamiltonian and gauge symmetries on the Lagrangian side couldn't hurt. $\endgroup$
    – ACuriousMind
    Mar 17, 2016 at 0:16
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    $\begingroup$ Yes, it wouldn't. I guessed that the OP isn't familiar with constraints, so I left it out. Do you think that it is a good idea to add the note, for future readers? $\endgroup$ Mar 17, 2016 at 0:19
  • $\begingroup$ @GiorgioComitini Molte Grazie! Excellent answer indeed, I forgot to mention that I'm at an undergraduate level, but I believe that I've understood the catch in my reasoning. Feel free to add the note for future readers, it'll be surely obscure to me, but I've already had my clarification. $\endgroup$
    – TheVal
    Mar 17, 2016 at 15:24
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    $\begingroup$ @Andrea. I am an undergraduate too (in Catania), but I mainly study as a self-learner. It's always good to have a confrontation with peers; as such, if you want to, add me on facebook so that we can keep in touch. I have the sombrero galaxy as profile pic. $\endgroup$ Mar 17, 2016 at 16:02
  • $\begingroup$ @ACuriousMind. I decided not to add the note, as I believe that the answer to the question should be clear to those who know anything about constraints and gauge transformations in a geometrical setting. $\endgroup$ Mar 17, 2016 at 16:02

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