Help with Goldstein Classical Mechanics Exercise 1.7

[lriuui0x0]

I'm trying to solve the Goldstein classical mechanics exercises 1.7. The problem is to prove:

$$\frac{\partial \dot T}{\partial \dot q} - 2\frac{\partial T}{\partial q} = Q$$

Below is my progress, and I got stuck at one of the step.

Now since we have langrange equation:

$$\frac{d}{dt} \frac{\partial T}{\partial \dot q} - \frac{\partial T}{\partial q} = Q$$

We only need to prove:

$$\frac{\partial \dot T}{\partial \dot q} = \frac{d}{dt} \frac{\partial T}{\partial \dot q} + \frac{\partial T}{\partial q}$$

Assuming $T$ is a function of $T(q, \dot q, t)$, expanding LHS:

$$
\begin{aligned}
& \frac{\partial}{\partial \dot q} \dot T \\
=& \frac{\partial}{\partial \dot q} (\frac{\partial T}{\partial t} + \frac{\partial T}{\partial q} \dot q + \frac{\partial T}{\partial \dot q} \ddot q) \\
=& \frac{\partial ^2 T}{\partial \dot q \partial t} + \frac{\partial ^2 T}{\partial \dot q \partial q} \dot q + \frac{\partial T}{\partial q}\frac{\partial \dot q}{\partial \dot q} + \frac{\partial^2 T}{\partial \dot q \partial \dot q} \ddot q + \frac{\partial T}{\partial \dot q}\frac{\partial \ddot q}{\partial \dot q} \\
=& \frac{\partial ^2 T}{\partial \dot q \partial t} + \frac{\partial ^2 T}{\partial \dot q \partial q} \dot q + \frac{\partial T}{\partial q} + \frac{\partial^2 T}{\partial \dot q \partial \dot q} \ddot q + \frac{\partial T}{\partial \dot q}\frac{\partial \ddot q}{\partial \dot q}
\end{aligned}
$$

Expanding RHS:

$$
\begin{aligned}
& \frac{d}{dt}\frac{\partial T}{\partial \dot q} + \frac{\partial T}{\partial q} \\
=& \frac{\partial^2 T}{\partial t \partial \dot q} + \frac{\partial ^2 T}{\partial q \partial \dot q} \dot q + \frac{\partial^2 T}{\partial \dot q \partial \dot q} \ddot q + \frac{\partial T}{\partial q}
\end{aligned}
$$

Since second order partial derivative commutes, cancelling terms we get:

$$
\frac{\partial T}{\partial \dot q} \frac{\partial \ddot q}{\partial \dot q} = 0
$$

However I don't know how to show this equation is true. Is it because $\frac{\partial \ddot q}{\partial \dot q} = 0$? But can't acceleration be a function of velocity?

 

[anuttarasammyak]

I think
$$\frac{\partial {\ddot{q}}}{\partial{\dot{q}} }=\frac{d}{dt}\frac{\partial{\dot{q}}}{\partial{\dot{q}}}=\frac{d}{dt}1=0$$

 

[lriuui0x0]

I think
$$\frac{\partial {\ddot{q}}}{\partial{\dot{q}} }=\frac{d}{dt}\frac{\partial{\dot{q}}}{\partial{\dot{q}}}=\frac{d}{dt}1=0$$

Can you elaborate on why

$$
\frac{\partial \ddot q}{\partial \dot q} = \frac{d}{dt} \frac{\partial \dot q}{\partial \dot q}
$$

 

[dextercioby]

A hint would be that $q^i, \dot{q}^i, \ddot{q}^i$ are functions only on time, therefore the partial differentials of (the members of) one set with respect to (members of) another set are zero.

 

[anuttarasammyak]

Can you elaborate on why

I exchange order of $\frac{\partial }{\partial \dot{q}}$ and $\frac{d}{dt}$ applying on $\dot{q}$ observing that the initial one is $\frac{\partial }{\partial \dot{q}}\frac{d}{dt}\dot{q}$.

 

[Delta2]

I think $\dot q$ and $\ddot q$ are considered to be independent variables, pretty much the same way $q,\dot q$ are considered to be independent.

But this brings up my eternal question regarding the Lagrangian and its partial derivatives: How can $q(t)$ and $\dot q(t)$ can considered to be independent, since we can write $t=q^{-1}(q(t))$ and replace that in $\dot q(t)$ which will be $\dot q(q^{-1}(q(t))$ hence $\dot q$ seems to be some sort of composite function of $q(t)$ therefore not independent.

 

[anuttarasammyak]

But this brings up my eternal question regarding the Lagrangean and its partial derivatives: How can $q(t)$ and $\dot q(t)$ can considered to be independent, since we can write $t=q^{-1}(q(t))$ and replace that in $\dot q(t)$ which will be $\dot q(q^{-1}(q(t))$ hence $\dot q$ seems to be some sort of composite function of $q(t)$ therefore not independent.

Each $\delta q(t)$ brings corresponding $\delta \dot{q}(t)$ in variation of path in the least action principle. So they are dependent. In mathematics of deriving Lagrange equation from the least action principle, I observe partial derivatives appear "as if" q and $\dot{q}$ are independent.

The independency comes from convention in Lagrangean derivatives mathematics, not from their physical nature. It is my understanding. I appreciate someone correct and teach me for better understanding.

 

[Ibix]

But this brings up my eternal question regarding the Lagrangian and its partial derivatives: How can $q(t)$ and $\dot q(t)$ can considered to be independent, since we can write $t=q^{-1}(q(t))$ and replace that in $\dot q(t)$ which will be $\dot q(q^{-1}(q(t))$ hence $\dot q$ seems to be some sort of composite function of $q(t)$ therefore not independent.

I struggled with that too. The point is that at this stage you don't know the relationship between $q$ and $\dot{q}$. Saying that they are independent is saying that at some time $t$ the particle could be at any position $q$ at any speed $\dot{q}$ so at time $t+dt$ it could also be at any $q$ with any $\dot{q}$ - it's a statement of your total ignorance (no offence meant! ). Fortunately Euler and Lagrange have provided us with a means to identify the family of relationships between $q$ and $\dot{q}$ that extremise the action.

 

[ergospherical]

To re-state, $q$, $\dot q$, etc. are independent variables (constituting the coordinate space), unless they are evaluated on a solution to the equations of motion (in which case $q = q(t)$, $\dot q = \dot q(t)$, etc.)

[Edit: just like the coordinates $x, y$ spanning the Cartesian plane are independent variables unless they are evaluated on a specified trajectory e.g. a parabola: $y = t^2, x = t$]

 

[Delta2]

To re-state, $q$, $\dot q$, etc. are independent variables (constituting the coordinate space), unless they are evaluated on a solution to the equations of motion (in which case $q = q(t)$, $\dot q = \dot q(t)$, etc.)

[Edit: just like the coordinates $x, y$ spanning the Cartesian plane are independent variables unless they are evaluated on a specified trajectory e.g. a parabola: $y = t^2, x = t$]

I think you gave me the most satisfying answer young man, how can you be so young and so good !?? ( e hehe I know you are eto, dw).

 

[lriuui0x0]

I exchange order of $\frac{\partial }{\partial \dot{q}}$ and $\frac{d}{dt}$ applying on $\dot{q}$ observing that the initial one is $\frac{\partial }{\partial \dot{q}}\frac{d}{dt}\dot{q}$.

Is it a general math theorem that total derivative and partial derivative commutes like this? I only know that the second order partial derivative commutes. Can you link some justification?

 

[lriuui0x0]

I think $\dot q$ and $\ddot q$ are considered to be independent variables, pretty much the same way $q,\dot q$ are considered to be independent.

But this brings up my eternal question regarding the Lagrangian and its partial derivatives: How can $q(t)$ and $\dot q(t)$ can considered to be independent, since we can write $t=q^{-1}(q(t))$ and replace that in $\dot q(t)$ which will be $\dot q(q^{-1}(q(t))$ hence $\dot q$ seems to be some sort of composite function of $q(t)$ therefore not independent.

I was wondering about this a while ago as well, what I ended up with somewhat understanding is this. Take a function $f(x, y)$, now from the perspective of this function itself surely $x, y$ are independent. This doesn't prevent you to plug in $y$ to be $g(x)$ though. So you can either view $f$ as a function of two variables $x$ and $y$, or you can view $f$ as a composite function of one variable $x$. It's up to you if you want to "look inside" and analyse the function as a composite function or not. You can surely choose to analyse the function only in its own form. Treating the function on its own is the more general version of taking special combination of arguments such that they correlate with each other. Therefore the analysis on the function on its own will hold when the arguments are correlating with each other as well.

This link also might be helpful. https://math.stackexchange.com/ques...iables-where-some-are-dependent-on-each-other

 

[anuttarasammyak]

Is it a general math theorem that total derivative and partial derivative commutes like this?

I had no concern about commutability. I would appreciate it to have some counterexamples.

 

[sysprog]

But can't acceleration be a function of velocity?

Velocity is the first derivative of position with respect to time; if the directionality is not complicated (not more than a single vector), then acceleration is simply the second such derivative (the third is jerk, the fourth is jounce -- some like to call 'jounce', 'snap', so that they can jokingly (rice krispies ad joke) call the fifth and sixth 'crackle' and 'pop'.

 

[vanhees71]

I think every student of analytical mechanics stumbles over this issue, and it's partially the fault of the physicists "efficient" mathematical notation.

In the context of d'Alembert's principle and the Lagrangian form of the action principle one has to distinguish between partial derivatives noted with the symbol $\partial$ and total time-derivatives.

The partial derivatives apply to functions $f(q,\dot{q},\ddot{q},\ldots,t)$ and in this context the $q$, $\dot{q}$, $\ddot{q}$ etc. are to be interpreted as independent variables. The dots just distinguish these different variables. Thus an expression like
$$\frac{\partial f}{\partial \dot{q}}$$
means that you take the derivative wrt. to $\dot{q}$ considering all other variables of the function $f$ as constant, which is the usual definition of a partial derivatives.

When you take the total time derivative of $f$ you consider $q$ a trajectory parametrized with time and you consider $\dot{q}$, $\ddot{q}$ etc. as the corresponding time derivatives, i.e.,
$$\dot{q}=\frac{\mathrm{d} q}{\mathrm{d} t}, \quad \ddot{q}=\frac{\mathrm{d}^2 q}{\mathrm{d} t^2}, \quad \text{etc.}$$
From this you get, e.g., a new function by taking the total time derivative of $f$:
$$g(q,\dot{q},\ddot{q},\ldots,t)=\frac{\mathrm{d}}{\mathrm{d} t} f(q,\dot{q},\ddot{q},\ldots,t)=\dot{q} \frac{\partial}{\partial q} f + \ddot{q} \frac{\partial f}{\partial \dot{q}}+\cdots+\frac{\partial f}{\partial t}.$$
Again the $q$, $\dot{q}$,... are to be taken as indepenent variables of $g$.

Now we can address the question in #1. First we note that the kinetic energy is of the form
$$T=\frac{m}{2} g_{jk}(q) \dot{q}_j \dot{q}_k$$
with Einstein's summation convention (summing tacitly over indices that appear in pairs in an equation).

Then the usual form of d'Alembert's equations of motion read
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial T}{\partial \dot{q}_j}-\frac{\partial T}{\partial q_j}=Q_j.$$
Here the partial derivatives as well as the total time derivative have to be understood as explained above.

So we have (because we can assume $g_{jk}=g_{kj}$)
$$\frac{\partial T}{\partial \dot{q}_j} = m g_{jk}(q) \dot{q}_k$$
and thus
$$\frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial T}{\partial \dot{q}_j}=m g_{jk}(q) \ddot{q}_k + m \frac{\partial g_{jk}(q)}{\partial q_l} \dot{q}_k \dot{q}_l$$
and
$$\frac{\partial T}{\partial q_j}=\frac{m}{2} \frac{\partial g_{kl}}{\partial q_j} \dot{q}_{k} \dot{q}_l$$
On the other hand we have
$$\dot{T}:=\frac{\mathrm{d} T}{\mathrm{d} t}=m g_{jk}(q) \dot{q}_j \ddot{q}_k + \frac{m}{2} \frac{\partial g_{jk}}{\partial q_l} \dot{q}_j \dot{q}_k \dot{q}_l.$$
From this we get
$$\frac{\partial \dot{T}}{\partial \dot{q}_j}=m g_{jk} \ddot{q}_k+m \frac{\partial g_{jk}}{\partial q_l} \dot{q}_k \dot{q}_l + \frac{m}{2} \frac{\partial g_{kl}}{\partial q_j} \dot{q}_k \dot{q}_l.$$
Then we find
$$Q_j=\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial T}{\partial \dot{q}_j}-\frac{\partial T}{\partial q_j}=m g_{jk}(q) \ddot{q}_k + m \frac{\partial g_{jk}(q)}{\partial q_l} \dot{q}_k \dot{q}_l-\frac{m}{2} \frac{\partial g_{kl}}{\partial q_j} \dot{q}_{k} \dot{q}_l= \frac{\partial \dot{T}}{\partial \dot{q}_j}-2 \frac{\partial T}{\partial q_j}.$$