I'm currently reading a nice introductory book (german, could be translated as "Physics with a pencil"). The author works a lot with differential calculus and antiderivatives (integrals will be used later). I'm stuck at a kind of mathematical question inmidst the easier physics:

So he takes the equation of force:

$m \ddot{\vec{r}} = \vec{F}$

and dot-multiplies it with the speed $\dot{\vec{r}}$:

$m \dot{\vec{r}} \cdot \ddot{\vec{r}} = \dot{\vec{r}} \cdot \vec{F}$

but I do not understand how he gets to the next step:

$\delta_t \left(\frac{1}{2}m\dot{\vec{r}}^2\right) = \frac{d\vec{r} \cdot \vec{F}}{dt}\ \ \ $ (with $\frac{mv^2}{2} = E_{kin}$)

because if I resolve the last equation again I find something different than what we began with, because the derivation of the left part leaves me no single $\dot{\vec{r}}$.

$\frac{2}{2} m \ddot{\vec{r}} = \dot{\vec{r}} \cdot \vec{F}$

Maybe I misinterpret the application $\delta_t$ to both $\vec{r}$ or the $\frac{d\vec{x}}{dr}$-notation not applying to $\vec{F}$ here? (The author states before that he use the partial differentiation symbol $\delta_t$ in the same way as the total $\frac{d\vec{x}}{dr}$ because the partial states exactly what is to be derived.)

  • $\begingroup$ I wrote an answer below. One additional comment. Be careful in distinguishing vectors from scalars. You write $\vec K$, but kinetic energy isn't a vector. That might be just a typographical error, but this is one area where it pays to be scrupulous with your notation. $\endgroup$ – Ted Bunn Mar 30 '11 at 18:47
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    $\begingroup$ Trivial language point: I changed "kinetical" to "kinetic" in the title. I don't know if "kinetical" exists as an English word, but it's never used in this context. $\endgroup$ – Ted Bunn Mar 30 '11 at 18:48
  • $\begingroup$ Ted Bunn has the answer. Also, note that it's not a partial derivative. The partial derivative of that expression with respect to time is simply zero, because time is not expressly used in it. It's a total derivative with respect to time. Using the two symbols interchangeably doesn't make sense. $\endgroup$ – Mark Eichenlaub Mar 30 '11 at 18:48
  • $\begingroup$ @Mark That is a bone you would have to pick with the author (Hermann Schulz, Hannover). He claims that a clearly defined $\delta_x$ is (in this case) clear enough to be used instead of the total $\frac{d}{dx}$ because you always have to understand what you derive there and not only apply symbols. It is known that $\vec{r}$ depends on time. $\endgroup$ – Leonidas Mar 30 '11 at 19:15
  • $\begingroup$ @Ted Sorry, the $\vec{K}$ was just a leftover from me translating the german Kraft to Force. $\endgroup$ – Leonidas Mar 30 '11 at 19:16

I think you're just making a calculus error. $$ {d\over dt}\left({1\over 2}m\dot{\bf r}^2\right)={1\over 2}m{d(\dot{\bf r}\cdot\dot{\bf r})\over dt}={1\over 2}m(\ddot{\bf r}\cdot\dot{\bf r}+\dot{\bf r}\cdot\ddot{\bf r})=m\dot{\bf r}\cdot\ddot{\bf r}. $$ You can tell that something has gone wrong in your calculation, because the units of $(d/dt)({1\over 2}m\dot{\bf r}^2)$ aren't the same as those of $m\ddot{\bf r}$. Also, the first one is a scalar, and the second is a vector.

By the way, I wrote vectors in boldface rather than with arrows. Both notations are fine. I don't like having multiple things (arrows and dots) piled on top of symbols, but that's just my taste.

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  • $\begingroup$ Oh, ok. Paying more attention to the distinction of deriving vectors in the future :) $\endgroup$ – Leonidas Mar 30 '11 at 18:58

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