Rate of change of vectors in circular motion

Question

I am reading An Introduction to Mechanics by Kleppner and Kolenkow. It said when the rate of change in a vector is always perpendicular to itself, then the vector's magnitude doesn't change; only its direction changes.

I'm having a contradiction with my thought process. Here's how:

In limit as $\Delta t \to 0$, if vector's change is perpendicular to itself, then there would be a infinitesimal change in magnitude of vector (visualize triangle law of vector addition). Since a finite time interval is made up of infinitely many such infinitesimal intervals, then wouldn't this infinitesimal change in magnitude would add up to a finite change in a finite time interval? Wouldn't this mean that the radius of a uniform circular motion changes in time?

Answer 1

You are multiplying an infinitesimal by infinity and expecting the result to be meaningful when it is actually undefined. This is leading you astray.

There are various ways to see that the magnitude of the velocity is unchanged. My favourite is to write the magnitude as the dot product:

$$ |v|^2 = \mathbf v \cdot \mathbf v $$

If we differentiate this wrt time we get:

$$ 2 |v| \frac{d|v|}{dt} = 2 \mathbf v \cdot \mathbf a $$

If $\mathbf a$ is normal to $\mathbf v$ then the dot product on the right hand side is zero. Since $|v| \ne 0$ this means:

$$ \frac{d|v|}{dt} = 0 $$

Answer 2

Usually, when we talk in physics, all our 'basis' vectors are unit length. If we have a set of three orthonormal basis vectors of form $\{ e_1, e_2 , e_3\}$ then a vector in can be written as:

$$ V= \sum_{i=0}^3 V^i e_i$$

Now suppose you were to differentiate this vector with time, let's say this was a velocity vector.

$$ \frac{dV}{dt} = \sum_{i=0}^3 \frac{dV^i}{dt} e_i + V^i \frac{de_i}{dt}$$

So, consider the $ \frac{de_i}{dt}$ term, clearly since it's unit length by definition, it's length can not increase. This means that the only way that it can change with time is rotations.

These type of transformations are useful when you have rotating coordinate systems

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See this answer and linked post to question:

Answer 3

if vector's change is perpendicular to itself, then there would be a infinitesimal change in magnitude of vector

This is actually false. Denote the magnitudes of the vector and its rate of change by $r$ and $v$ respectively. Since the change is perpendicular, the new magnitude after a time $\mathrm{d}t$ is $$\sqrt{r^2 + v^2 \mathrm{d}t^2}.$$ We can apply the binomial expansion to obtain $$r + \frac{v^2}{2r} \mathrm{d}t^2 + \ldots.$$

From this, we can see that the highest term of the change is $\mathrm{d}t^2$. It does not contain any term in $\mathrm{d}t$. Therefore, there is no first-order change in $r$ when the change is perpendicular to the original vector, i.e. in the limit, $\mathrm{d}r/\mathrm{d}t = 0$. So the magnitude does not change in any finite amount of time.