1
$\begingroup$

I wasn't really sure how to phrase the title. Suppose I have a purely time-dependent function $x(t)$, and I want to know its time derivative $\dot{x}:=\dfrac{dx}{dt}$. Then I ask how the time derivative of $x$ changes as $x$ changes. So I'm considering $\dfrac{d\dot{x}}{dx}$. Does this make sense? Because in all applied courses I've taken you're told it's not "proper maths" but you can basically divide through by a differential if it appears both in the numerator and denominator of a fraction. So $\dfrac{d\dot{x}}{dx}=\dfrac{d}{dx} \dfrac{dx}{dt}=\dfrac{d}{dt}.$ This doesn't really make sense to me - how can we have taken a derivative and arrived at a differential operator?

The motivation for this question is that I'm doing an applied course where I'm calculating the Euler-Lagrange equations for some Lagrangian, and the Lagrangian is explicitly a function of $\dot{x}$, and the E-L equations involve taking the partial $x$ derivative of the Lagrangian. I understand that in taking the partial derivative with respect to $x$, you hold all other variables constant. Does $\dot{x}$ count as a separate variable? I'm pretty sure that when I did an introductory dynamics course a year or two ago, I was told that the derivative $\dfrac{\partial{\dot{x}}}{\partial{x}}$ is zero, but I don't think it was ever explained.

It's occurred to me more recently that you only ever build upon what is taught previous courses, and that there's not really very many opportunities to go over old concepts that you should already know, so it might be a stupid question but if I don't ask I'm stuck with the misunderstanding. Thanks for any help!

$\endgroup$
  • 1
    $\begingroup$ In Euler-Lagrange setting $\partial \dot x/ \partial x $ is zero because they are functions of $t$. It does not mean that $d \dot x/ d x$ is zero.It is $\ddot x / \dot x$. $\endgroup$ – Alexander Vigodner Aug 14 '14 at 19:24
2
$\begingroup$

I dont' think it is necessary redefining $\dot x= v$ and introduce any EulerLagrage equation here. It seams for me that you want to estimate a sensitivity of "velocity" with respect to "coordinate". Basically you need a ratio of "acelleration" $\ddot x$ and "velocity" $\dot x$. And indeed $$ \frac {d (dx/dt)} {dx}=\frac {d (dx/dt)} {dt (dx/dt)} =\frac{\ddot x}{\dot x} $$ This is how I understand the problem. I hope it might help you.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That is correct. But, when using E-L equations (what @Lammey is really doing in his course), you must consider $\dot{x}$ as another variable. $\endgroup$ – cjferes Aug 14 '14 at 17:54
  • $\begingroup$ OK, no problem replace $\dot x =v$ and $\ddot x= \dot v$. What is the problem? $\endgroup$ – Alexander Vigodner Aug 14 '14 at 19:21
  • $\begingroup$ That's not the issue. He is trying to use E-L equations (that's the motivation for his problem). So, he thinks he needs to compute $\frac{d\dot{x}}{dx}$, but he doesn't because in E-L equations, $x$ and $\dot{x}$ are treated as distinct variables -not related. That's why I recommended using $v=\dot{x}$, to avoid unnecessary confusions. $\endgroup$ – cjferes Aug 14 '14 at 19:51
  • $\begingroup$ There is no any problem in Euler equation to write $dv/dq$. It is not zero. Partial derivatives ($t$ is constant) are zero. Full derivative through time $t$ are not. $v=v(t)$, $q=q(t)$, $dv/dq= \dot v /\dot q$. $\endgroup$ – Alexander Vigodner Aug 14 '14 at 20:09
  • $\begingroup$ From Wikipedia (en.wikipedia.org/wiki/Euler_LaGrange), Section "Basic method" under "Classical Mechanics": " Compute $\frac{\partial L}{\partial \dot{q}}$ and from it, $\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}}$. It is important that $\dot{q}$ be treated as a complete variable in its own right, and not as a derivative." There's nothing wrong with the full derivative through time. It's just that E-L equations needs $\dot{x}$ to be treated as a variable, not as a derivative. $\endgroup$ – cjferes Aug 14 '14 at 20:21
3
$\begingroup$

For the Euler-Lagrange equations, you have to consider $\dot{x}$ as a separate variable. If you're dealing with typical mechanics problems, it's easier to denote $\dot{x}=v$ and avoid confusion.

I don't know if that helps you with understanding the procedure.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.