In chapter 1 problem 11 of Goldstein I'm asked to show that Lagrange's equations: $$\frac d {dt} \biggr(\frac {\partial T} {\partial \dot {q_j}} \biggr) - \frac {\partial T} {\partial {q_j}}=Q_j$$ can be written as:

$$\frac {\partial \dot T} {\partial \dot {q_j}} - 2\frac {\partial T} {\partial {q_j}}=Q_j.$$

All the solutions I've found start with the following 2 lines:

$$\dot T = \sum_i \frac {\partial T} {\partial q_i} \dot {q_i} +\sum_i \frac {\partial T} {\partial \dot {q_i}} \ddot {q_i} +\frac {\partial T} {\partial t}$$

$$\frac {\partial \dot T} {\partial \dot {q_j}} = \sum_i \frac {\partial T} {\partial q_i \partial \dot {q_j}} \dot {q_i} +\sum_i \frac {\partial T} {\partial \dot {q_i} \partial \dot {q_j}} \ddot {q_i} +\frac {\partial T} {\partial t} +\frac {\partial T} {\partial q_j}=\frac d {dt} \biggr(\frac {\partial T} {\partial \dot {q_j}} \biggr)+\frac {\partial T} {\partial q_j}$$ and move on from here.

My problem is that the $\ddot q_i$ are not being held constant when taking the partial derivative, they are functions of $\textbf q,\dot {\textbf {q}},t$, so the second line should include another term: $$\sum_i \frac {\partial T} {\partial \dot {q_i}} \frac {\partial \ddot {q_i}} {\partial \dot q_j},$$ which messes up the derivation. What am I missing here?


1 Answer 1


Short answer

The short answer is that $\ddot{q}_i$ does not depend on $\dot{q}_j$.

Longer answer, with details

Let's repeat the computation, explicitly writing the independent variables the functions depends on. The kinetic energy $T$ is a function of the generalized coordinates $q_i$, their time derivative $\dot{q}_i$ and time $t$. Generalized coordinates and their derivative are functions of time $q_i(t)$, $\dot{q}_i(t)$. The kinetic energy can be written as

$T(\dot{q}_i(t), q_i(t), t)$.

Before computing time derivative of $T$, let's clearly state that the partial derivative w.r.t. $\dot{q}_j$ is a function of $\dot{q}_i(t), q_i(t), t$ as well,

$\dfrac{\partial T}{\partial \dot{q}_k} (\dot{q}_i(t), q_i(t), t)$,

so that its time derivative reads

$ \dfrac{d}{dt} \left(\dfrac{\partial T}{\partial \dot{q}_k}(\dot{q}_i(t), q_i(t), t) \right)= \left[\dfrac{\partial}{\partial t} + \ddot{q_j}(t) \dfrac{\partial}{\partial \dot{q}_j} + \dot{q_j}(t) \dfrac{\partial}{\partial q_j} \right] \dfrac{\partial T}{\partial \dot{q}_k}(\dot{q}_i(t), q_i(t), t) \qquad \qquad (1)$.

Now, we can evaluate the time derivative of the kinetic energy

$\dfrac{d T}{dt}(\ddot{q}_i(t), \dot{q}_i(t), q_i(t), t) = \dfrac{\partial T}{\partial t}(\dot{q}_i(t), q_i(t), t) + \ddot{q_j}(t) \dfrac{\partial T}{\partial \dot{q}_j} (\dot{q}_i(t), q_i(t), t) + \dot{q_j}(t) \dfrac{\partial T}{\partial q_j}(\dot{q}_i(t), q_i(t), t) $,

so that its derivative w.r.t. $\dot{q}_j$ reads (without explicitly writing the independent variables of the functions, for brevity),

$\dfrac{\partial}{\partial \dot{q}_k}\dfrac{d T}{dt} = \underbrace{\dfrac{\partial}{\partial \dot{q}_k}\dfrac{\partial T}{\partial t} + \ddot{q_j}(t) \dfrac{\partial}{\partial \dot{q}_k}\dfrac{\partial T}{\partial \dot{q}_j} + \dot{q_j}(t) \dfrac{\partial}{\partial \dot{q}_k}\dfrac{\partial T}{\partial q_j}}_{\frac{d}{dt}\frac{\partial T}{\partial \dot{q}_k}} + \dfrac{\partial T}{\partial q_k} \qquad \qquad (2)$.

Finally, using (2) in Lagrange equation, you get the desired equation

$\dfrac{\partial}{\partial \dot{q}_k} \dfrac{d T}{d t} - 2 \dfrac{\partial T}{\partial q_k} = Q_k$.


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