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Suppose we define a unit vector $\vec r$ along radial direction for a particle in uniform circular motion at an angular frequency $\omega$. Then we can write:

$$\vec r = \cos(\omega t)\hat i + \sin(\omega t)\hat j$$

And the modulus of this vector is one. If we differentiate this to get the velocity we get:

$$\vec v = \omega(-\sin(\omega t)\hat i + \cos(\omega t)\hat j)$$

And the modulus of the velocity is $\omega$. Differentiate again to get the acceleration and we get:

$$\vec a = \omega^2(-\cos(\omega t)\hat i - \sin(\omega t)\hat j)$$

And the modulus of the velocity is $\omega^2$. In general we find:

$$ \left|\frac{d^n\vec r}{dt^n} \right| = \omega^n $$

This feels weird, because for any $\omega > 1$ the modulus of each successive derivative keeps increasing without limit as we increase $n$, and yet that's what the calculation tells us. Is there some way to understand intuitively what it means for the successive derivatives to increase all the way to infinity for $n \to \infty$?

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  • $\begingroup$ I can't quite follow your math. But be careful to distinguish $r = |\vec{r}|$ and $\vec{r}$ . I'm not sure you made that confusion, but check yourself to make sure you didn't. (to avoid this kind of thing I never use $r = |\vec{r}|$, rather I always use $|\vec{r}|$. It's more cumbersome to write, but makes things clear. It also provides a notation advantage in that letters are usually used for real numbers, but the magnitude is a positive-definite real. But that's just me.) $\endgroup$
    – garyp
    Oct 12 '20 at 16:15
  • $\begingroup$ the rate of change of successive derivatives of position vector is infinite None of these rates of change is infinite. What you wrote isn’t what you meant to write, I think. $\endgroup$
    – G. Smith
    Oct 12 '20 at 16:15
  • $\begingroup$ Switching the order of the $\hat i$ and $\hat j$ terms midway through is confusing. I originally thought that you had differentiated incorrectly. It is conventional to always put $\hat i$ first. $\endgroup$
    – G. Smith
    Oct 12 '20 at 16:23
  • $\begingroup$ Your edit made it more confusing. You can make subscripts with _. Unit vectors should have a hat. Vectors like velocity and acceleration should have an arrow. $\endgroup$
    – G. Smith
    Oct 12 '20 at 16:29
  • $\begingroup$ So sorry sir can you again put that edit again please $\endgroup$ Oct 12 '20 at 16:34
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I think there is a teachable moment here, in that I think we can get some insight through a different way to frame your question.

The first thing to note is that $\omega$ has dimensions of $[{\rm time}]^{-1}$ (in SI units, ${\rm s}^{-1}$). Therefore, it does not make sense to compare $\omega$ to a dimensionless number, like $1$, so saying $\omega>1$ or $\omega<1$ is not a well-formed statement. The numerical value of $\omega$ can be larger than, less than, or equal to one, depending on the unit system chosen.

In fact, for uniform circular motion, we can always choose units where $\omega=1 /({\rm unit\ of\ time})$. This amounts to saying that we choose our unit of time so that the particle makes one revolution of the circle in one unit of time -- or in more physical terms, we use the particle itself as our clock (think of the particle like a second hand on a stopwatch).

In these units, all of the derivatives take on the same value, since $\omega=1$ in our units. So this should indicate to you that nothing is actually "blowing up" physically. What you are seeing is a kind of self-similarity, where the motion and all of its derivatives is constantly changing.

The apparent growth you were finding at larger and larger values of $n$ is an artifact of the units you use to describe the system. Let's consider the first derivative. The x component of the velocity has to change from $v_x = 2\pi R \omega $ to 0 over the span of one quarter of an orbit around the circle. The time it takes for a quarter cycle is $T_{1/4}=1/(4\omega)$, since $1/\omega$ is by definition the time for one cycle. Therefore the acceleration over this period needs to be $a_x=-2\pi R \omega / (1/4\omega) = - 8\pi R \omega^2$. There are two powers of $\omega$ in this expression: one coming from the original x component of the velocity which needed to be lost over a quarter-cycle, and another power accounting for the unit of time over which the deceleration took place. This explains why there is a factor of $\omega^2$ in the acceleration (2nd derivative), but note it does not imply that there is any divergence. If we chose to measure the motion of the particle in different units, the numerical value of the acceleration could be larger or smaller or stay the same as the velocity.

A similar argument works for every successive derivative of motion.

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  • $\begingroup$ It looks like your $\omega$ is smaller than the one in the question by a factor of $2\pi$. $\endgroup$ Dec 13 '20 at 13:27
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you start with $\vec v_1$

$$\vec v_1= \left[ \begin {array}{c} \rho\,\cos \left( \varphi _{{0}} \right) \\ \rho\,\sin \left( \varphi _{{0}} \right) \end {array} \right] $$ the first time derivative is :

$$\vec v_2=\vec {\dot{v}}_1=\omega\,S(\frac \pi2)\,\vec v_1$$ with $$S(\varphi)=\left[ \begin {array}{cc} \cos \left( \varphi \right) &-\sin \left( \varphi \right) \\ \sin \left( \varphi \right) & \cos \left( \varphi \right) \end {array} \right] $$

$$\vec v_2=\left[ \begin {array}{c} -\omega\,\rho\,\sin \left( \varphi _{{0}} \right) \\ \omega\,\rho\,\cos \left( \varphi _{{0}} \right) \end {array} \right] $$

where $\vec{v}_2 \perp\vec{v}_1$

the second time derivative is :

$$\vec v_3=\vec {\ddot{v}}_1=\omega\,S(\frac \pi2)\,\vec v_2= \left[ \begin {array}{c} -{\omega}^{2}\rho\,\cos \left( \varphi _{{0} } \right) \\ -{\omega}^{2}\rho\,\sin \left( \varphi _{{0}} \right) \end {array} \right] $$

where $\vec{v}_3 \perp\vec{v}_2$

$$\vec v_n= {\frac{d^n}{dt^n}}\vec{v}_1=\omega\,S(\frac \pi2)\,\vec v_{n-1}$$ where $\vec{v}_n \perp\vec{v}_{n-1}$

conclusion : a new derivative vector is obtain by rotate the pervious derivative vector with rotation angle $\pi/2$ and multiplying with $\omega$

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Is there some way to understand intuitively what it means for the successive derivatives to increase all the way to infinity

Similar phenomenon goes on for derivatives of position in a linear movement. Fourth derivative of position is snap, and such movement is described by equation : $$ {\vec {r}}={\vec {r}}_{0}+{\vec {v}}_{0}t+{\frac {1}{2}}{\vec {a}}_{0}t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}t^{3}+{\frac {1}{24}}{\vec {s}}t^{4} $$

So $\vec r = f(t^1,t^2,t^3,t^4)$ is polynomial equation of degree 4.

Fifth derivative of position is crackle, and such movement is described by :

$$ {\vec {r}}={\vec {r}}_{0}+{\vec {v}}_{0}\,t+{\frac {1}{2}}{\vec {a}}_{0}\,t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}\,t^{3}+{\frac {1}{24}}{\vec {s}}_{0}\,t^{4}+{\frac {1}{120}}{\vec {c}}\,t^{5} $$

Position vector is described by 5-th degree polynomial equation, $\vec r = f(t^1,t^2,t^3,t^4,t^5)$

Last, but not the end,- Sixth derivative of position is pounce, with movement equation :

$$ {\vec {r}}={\vec {r}}_{0}+{\vec {v}}_{0}\,t+{\frac {1}{2}}{\vec {a}}_{0}\,t^{2}+{\frac {1}{6}}{\vec {\jmath }}_{0}\,t^{3}+{\frac {1}{24}}{\vec {s}}_{0}\,t^{4}+{\frac {1}{120}}{\vec {c}}_{0}\,t^{5}+{\frac {1}{720}}{\vec {p}}\,t^{6} $$

This is 6-th degree polynomial. In general it seems that position vector can be described by n-th degree polynomial with respect to time : $$\vec r = \vec r(t^1,t^2,\cdots,t^n)$$

Or as expressed as series :

$$ \vec {r}={\vec {r}}_{0}+\sum_{k=0}^{n} \frac{1}{(k+1)!} \vec x_k\,t^{k+1} $$

As about Physical interpretation of increasing polynomial degree in monotonic way,- hard to say. Probably that there's no technical limits of what can change. Higher order derivatives indicates that something in a system is changing at a rate a lower-order derivative is not able to cope with.

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  • $\begingroup$ How can something change that high $\endgroup$ Oct 13 '20 at 9:43
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    $\begingroup$ The third equation you made converges since (k+1)! Is greater than t^k+1 after certain extent $\endgroup$ Oct 13 '20 at 9:46
  • $\begingroup$ But it is different from the situation in the question $\endgroup$ Oct 13 '20 at 9:46
  • $\begingroup$ I think the higher degree of polynomial,- the less it appears in practice, so it can be that probability of noticing higher-order derivatives in nature gets smaller and smaller as n approaches infinity. However it can be useful in some areas, like universe expansion in increasing rate. In some scenarios, universe may be expanding at such high rates that inclusion of some sort of higher order derivates as "jolt" could be needed. $\endgroup$ Oct 13 '20 at 9:58
  • $\begingroup$ But it is different from the situation in the question I think this is not much different than your question. Linear acceleration maps to angular acceleration, linear jolt maps to "angular jolt" $$j = \frac{d\alpha}{dt}$$ and so on... $\endgroup$ Oct 13 '20 at 11:38

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