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I am solving the same problem as the person in this link was solving. But I am having a different kind of doubt, i.e related to the derivative of radius.

then differentiating $\dfrac{dE}{dt} = \dfrac{dE}{dr} \dfrac{dr}{dt} = \dfrac{1}{2}\dfrac{ke^2}{r^2}\dot{r}$ and setting equal to the Larmor formula gives $\dfrac{1}{2}\dfrac{ke^2}{r^2}\dot{r} = -\dfrac{2}{3}\dfrac{e^2a^2}{c^3} = -\dfrac{2}{3}\dfrac{e^2k^2e^4}{c^3m^2r^4}$ or $$r^2dr = -\dfrac{4}{3}\dfrac{ke^4}{c^3m^2}dt$$

Here, I am unable to figure out why the time derivative of radius (magnitude of position vector) is not equal to the magnitude of tangential velocity of the electron. I think it is indeed wrong to take it as tangential speed because that eventually gives an absurd equation which is not possible physically. But still I'm not getting why $\dot{r} \neq v$. Any help would be greatly appreciated.

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    $\begingroup$ For a comprehensive account of circular motion check Klepnner and Kolenkow's chapter 1 $\endgroup$ Jul 26 at 20:31
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    $\begingroup$ @GedankenExperimentalist Ok Thanks. Will check Klepnner. $\endgroup$
    – Albedo
    Jul 27 at 6:52

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You can easily think of counterexamples for when the rate of change of the magnitude of the position vector is not equal to velocity.

e.g. uniform circular motion. The magnitude of the position vector is unchanging, yet the velocity is obviously non-zero.

Basically, if the position vector and velocity vectors are not parallel, then $\dot r=\text d|\mathbf r|/\text dt$ will not be equal to $v=|\text d\mathbf r/\text dt|$.If you do want to relate them, for 2D motion we have

$$\frac{\text d|\mathbf r|}{\text dt}=\frac{\text d}{\text dt}\sqrt{x(t)^2+y(t)^2}=\frac{x(t)\cdot\dot x(t)+y(t)\cdot\dot y(t)}{\sqrt{x(t)^2+y(t)^2}}=\mathbf v\cdot \hat r$$

i.e., the rate of change of the magnitude of the position vector is determined by the component of the velocity along the position vector. This should make sense, as you can only change the magnitude of a vector by adding vector components parallel to it.

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  • $\begingroup$ Thanks a lot to reiterate this concept. My profs have repeatedly told not to confuse them, but still I did :( I wish I could accept both the answer. $\endgroup$
    – Albedo
    Jul 27 at 6:51
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    $\begingroup$ @Albedo No worries! The answers are basically the same, Jensen's is more concise though. $\endgroup$ Jul 27 at 14:34
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The time derivative of $r$ is not $v$, because this does not take into account motion purely in the $\hat \theta$, $\hat \phi$ direction.

Imagine circular motion:

$$\frac{dr}{dt}= 0$$

yet there is clearly a velocity.

The correct relation is: $$\frac{d}{dt} (r \hat r) = \vec{v}$$

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    $\begingroup$ $\hat r$ is a function of time. $\endgroup$ Jul 26 at 19:12

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