4
$\begingroup$

In Classical Mechanics, the Lagrangian is a function of a generalized coordinate $q$, the corresponding generalized velocity $\dot{q}$, and time $t$.

For the Lagrangian, $$\dfrac{d}{dt}\frac{\partial L}{\partial \dot{q}} \neq \frac{\partial \dot{L}}{\partial \dot{q}},$$ that is, the two derivatives do not commute. However, what if $L$ were a function only of $q$ and $t$? Then would the total derivative in $t$ and the partial derivative in $q$ always commute?

Also, for $L(q, t)$, is $\frac{\partial \dot{L}}{\partial \dot{q}}$ always equal to $\frac{\partial L}{\partial q}$?

$\endgroup$

2 Answers 2

2
$\begingroup$

Take into account that, for a function $f$ of $q,\dot{q},t,$ the total derivative is given by $$ \frac{d}{dt}=\sum_{h=1}^n\left(\dot{q}_h\frac{\partial}{\partial{q_h}}+\ddot{q}_h\frac{\partial}{\partial{\dot{q}_h}}\right)+\frac{\partial}{\partial{t}} $$ If $f$ only depends on $q,t$ then \begin{align} \require{cancel} \frac{\partial}{\partial{q_k}}\frac{df}{dt} &=\frac{\partial}{\partial{q_k}}\left[\sum_{h=1}^n\left(\dot{q}_h\frac{\partial{f}}{\partial{q_h}}+\cancel{\ddot{q}_h\frac{\partial{f}}{\partial{\dot{q}_h}}}\right)+\frac{\partial{f}}{\partial{t}}\right]=\\ &=\sum_{h=1}^n\left(\dot{q}_h\frac{\partial}{\partial{q_h}}\frac{\partial{f}}{\partial{q_k}}+\right)+\frac{\partial}{\partial{t}}\frac{\partial{f}}{\partial{q_k}}=\frac{d}{dt}\frac{\partial{f}}{\partial{q_k}} \end{align} Note that the two derivatives commute also if $f$ depends on $\dot{q}.$
Also, for $\partial\dot{L}/\partial\dot{q}$ and $\partial{L}/\partial{q}$ when $L$ does not depend on $\dot{q},$ while the former is zero, the latter in general is not.

$\endgroup$
2
  • $\begingroup$ How come $\frac{\partial}{\partial q_k}(\dot{q_h}\frac{\partial f}{\partial q_h})=\dot{q}_h\frac{\partial}{\partial q_h}\frac{\partial f}{\partial q_k}$? I can see commuting partials being useful eventually, but how do we know that $\dot{q}_h$ does not depend on $q_k$? $\endgroup$
    – Shady Puck
    Oct 12, 2023 at 4:26
  • 1
    $\begingroup$ @ShadyPuck they are independent variable, like in $y'' = f(x,y,y')$ you can obtain by derivation $y''' = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}y' +\frac{\partial f}{\partial y'}y''$ $\endgroup$ Oct 12, 2023 at 21:59
0
$\begingroup$

In general a Lagrangian $L$ that depends on functions $\phi:M\to N$ and their $k$ derivatives, i.e. a function $\newcommand{\R}{\mathbb R}$ \begin{equation} L:J^k(M,N) \to \R \end{equation} can be written in local coordinates as \begin{align} L(j^k_x\phi) &= L(x,\phi_{i}(x),\phi_{i,\mu}(x),\phi_{i,\mu\nu}(x),\dots,\phi_{i,\mu_1\dots\mu_k}(x)) \\ &= L(x^{\mu},\phi_{i,(\alpha)}(x)). \end{align} where the indices after the commas represent partial derivatives and $(\alpha)$ represents a list of indices of size $\geq 0$. The total derivative along the $\mu$-th coordinate in $M$ is the operator \begin{equation} \frac{dL}{dx^{\mu}} = \frac{\partial L}{\partial x^{\mu}} + \frac{\partial L}{\partial \phi_{i}}\phi_{i,\mu} + \frac{\partial L}{\partial \phi_{i,\nu}}\phi_{i,\nu\mu} + \dots + \frac{\partial L}{\partial \phi_{i,\nu_1\dots\nu_k}}\phi_{i,\nu_1\dots\nu_k\mu} .\end{equation} we abreviate this as \begin{equation} \frac{dL}{dx^{\mu}} = \frac{\partial L}{\partial x^{\mu}} + \frac{\partial L}{\partial \phi_{i,(\alpha)}}\phi_{i,(\alpha)\mu} \end{equation} where the repeated indices imply sum. Then for any list of indices $(\alpha)$ we get \begin{align} \frac{\partial}{\partial\phi_{i,(\alpha)}}\frac{dL}{dx^{\mu}} &= \frac{\partial}{\partial\phi_{i,(\alpha)}} \left( \frac{\partial L}{\partial x^{\mu}} + \frac{\partial L}{\partial \phi_{j,(\beta)}} \phi_{j,(\beta)\mu} \right) \\ &= \frac{\partial^{2} L}{\partial\phi_{i,(\alpha)}\partial x^{\mu}} + \frac{\partial^{2} L}{\partial\phi_{i,(\alpha)}\partial \phi_{j,(\beta)}} \phi_{j,(\beta)\mu} + \frac{\partial L}{\partial \phi_{j,(\beta)}} \frac{\partial \phi_{j,(\beta)\mu}}{\partial\phi_{i,(\alpha)}} \\ &= \left( \frac{\partial^{2} L}{\partial x^{\mu}\partial\phi_{i,(\alpha)}} + \frac{\partial^{2} L}{\partial \phi_{j,(\beta)}\partial\phi_{i,(\alpha)}} \phi_{j,(\beta)\mu} \right) + \frac{\partial L}{\partial \phi_{j,(\beta)}} \delta^{i}_j\delta^{(\alpha)}_{(\beta)\mu} \\ &= \frac{d}{dx^{\mu}}\frac{\partial L}{\partial\phi_{i,(\alpha)}} + \frac{\partial L}{\partial \phi_{i,(\beta)}} \delta^{(\alpha)}_{(\beta)\mu} .\end{align} So the operators $\frac{d}{dx^{\mu}}$ and $\frac{\partial}{\partial\phi_{i,(\alpha)}}$ don't quite commute, but almost. The difference is \begin{equation} \frac{\partial }{\partial \phi_{i,(\alpha)}}\frac{dL}{dx^{\mu}} - \frac{d}{dx^{\mu}}\frac{\partial L}{\partial\phi_{i,(\alpha)}} = \frac{\partial L}{\partial \phi_{i,(\beta)}} \delta^{(\alpha)}_{(\beta)\mu} .\end{equation} This makes clear that the operators fail to commute exactly when $(\alpha)=(\beta)\mu$ for some list of indices $(\beta)$ such that $L$ depends on $\phi_{i,(\beta)}$. I.e., whenever the variable $\phi_{i,(\alpha)}$ is the $\mu$-th derivative of a variable $\phi_{i,(\beta)}$ upon which $L$ depends.

For example, for $M=\R$ and $N=\R^{3}$, take \begin{equation} L(t,\phi_{i}(t),\dot\phi_{i}(t)) = \frac{m}{2}\|\dot\phi(t)\|^{2} - V(\phi_3(t)) .\end{equation} Then since the $\dot\phi_i$ appear in $L$, the derivatives \begin{equation} \frac{\partial }{\partial\ddot\phi_i}\frac{dL}{dt} \qquad \text{and} \qquad \frac{d}{dt}\frac{\partial L}{\partial\ddot\phi_i} \end{equation} won't match up. Also, since $L$ depends on $\phi_3$, the derivatives \begin{equation} \frac{\partial }{\partial\dot\phi_3}\frac{dL}{dt} \qquad \text{and} \qquad \frac{d}{dt}\frac{\partial L}{\partial\dot\phi_3} \end{equation} won't agree, but all the other derivatives will commute.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .