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In the Introduction to Mechanics, Kleppner and Kolenkow, the authors make a statement about vectors, specifically:

If the instantaneous change in a vector is always perpendicular to the vector, the vector must rotate.

This follows them establishing that, if $\vec{\Delta A}$ is perpendicular to $\vec{A}$, A would now be $\sqrt{A^2 + (\Delta A)^2}$. And so as $\Delta A \rightarrow 0$, A would become $\sqrt{A^2 + dA^2}$ after the change, which is simply $A$ and there would hence be no change in magnitude.

My problem arises when they establish that is $\vec{dA}$ is always perpendicular to $\vec{A}$, can I not establish that is perpendicular for all given instants of time? And if that is true, then for every instantaneous change in time, $dt$ there is some change in $A$, so supposedly, as we move along every instant in time, $\vec{dA}$ acts upon $A$, and so:

at $t_0$, $A=A_0$,

at $t_0+dt$, $A=\sqrt{A_0^2 + dA^2}$,

at $t_0+2dt$, $A=\sqrt{\sqrt{A_0^2 + dA^2}^2 + dA^2}$= $\sqrt{A_0^2 + 2\cdot{}dA^2}$. . .

and after some finite time interval,

$t_0+n\cdot{}dt$, $A$= $\sqrt{A_0^2 + n\cdot{}dA^2}$, where $n \rightarrow \infty$

So supposedly, since the square of a differential is not exactly $0$, as cited in this question: Why is it considered that $(\mathrm d x)^2=0$? there should be some observable change in the magnitude of the vector, which according to the text, and most other resources, experiences no change whatsoever in its magnitude. So well, my question would be, why exactly, does the magnitude not change?

Note: I am not extremely well versed in mathematics, so it is relatively straightforward that I may make wrong assumptions which are rather obvious to some, so please bear with me, I only really wish to understand a doubt which I have no other way to.

This was originally posted on the physics exchange site but the moderator removed it, citing: "not suitable for this site." And so, I have come here, asking a question which perhaps more than a little physics inclined.

Adyansh Mishra
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  • The way you are thinking about this, as time passes the acceleration ceases to be perpendicular to the direction of motion, and velocity will increase. But if the acceleration is always perpendicular, velocity will not change. Furthermore, if we parametrize with respect to $s$ (arc length or distance traveled) this derivative of motion will always be perpendicular to the direction of travel. – user317176 Aug 05 '23 at 00:12
  • I have, as far as I am aware, never once assumed that acceleration ceases to be perpendicular to the direction of the motion – Adyansh Mishra Aug 05 '23 at 00:33

2 Answers2

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Suppose we have a vector $\vec{v}(t)$ that is always perpendicular to its rate of change. This gives us the differential equation

$$\vec{v}\cdot \vec{v'} = 0 \implies \vec{v}\cdot\vec{v} = |\vec{v}|^2 = C$$

so the magnitude can never change. A vector in the physics sense is only a magnitude and a direction, so if $\vec{v'}$ is nonzero that means only the direction is changing i.e. a rotation.

Ninad Munshi
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  • Could you perhaps explain it relative to my assumptions so that I could understand where I went wrong? – Adyansh Mishra Aug 05 '23 at 00:31
  • @AdyanshMishra no, given the forum. This is a math stackexchange and I can only go so far with heuristics that don't make any sense to me. I have to stick to a mathematically proper explanation. A lot of reasoning with differentials can be waived away with "it doesn't work this time because differentials aren't rigorous, but I promise it will work next time." That's not to say it doesn't have its value but as a refresher tool for experts, not a teaching tool for first time learners. – Ninad Munshi Aug 05 '23 at 00:33
  • That is a pity, but well there doesn't seem to be a better alternative, although as I told you, I am not very well versed in mathematics, I am a high school student, so the bit about you going from the dot product being 0 to the constant property to velocity didn't make much sense to me, could you perhaps lay it down a bit more obviously for me? – Adyansh Mishra Aug 05 '23 at 04:37
  • @AdyanshMishra go backwards and take the derivative of $\vec{v}\cdot\vec{v} = C$. What do you get? – Ninad Munshi Aug 05 '23 at 04:42
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The setup as described is indeed somewhat woolly. The idea is that $\Delta s$, the total length traveled, is constant over the discussion of $dA=|\vec{dA}|$ becoming infinitesimal, so that $n\cdot dA=\Delta s=const.$

Thus indeed $n\cdot (dA)^2=\Delta s\cdot dA =\frac{\Delta s}{n}$ is likewise infinitesmal, or converges to zero as the subdivision becomes ever finer, $n\to\infty$. The distance to the center is, in the limit curve, unchanged, the motion is on a circle.

Lutz Lehmann
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  • I don't really understand how $n\cdot{dA}=\Delta s$, do you mind explaining that? – Adyansh Mishra Aug 05 '23 at 09:11
  • If the $\vec{\Delta A_k}$ are macroscopic, your calculation shows an outward spiral. The aim however is to make the updates $\vec{dA_k}$ microscopic, but for some fixed macroscopic distance of the composite motion. Thus the fixing of the path length. So that $dA$ is infinitesimal, $n$ near infinite and reciprocal to $dA$. The other answer goes more in the systematic insight that you apply the Euler method to $\dot A=RA$ with $R^2=-I$ denoting the 90° rotation. From that side, the Euler method is order 1, its error proportional to the step length $dA=|A|,dt$. – Lutz Lehmann Aug 05 '23 at 13:23