5
$\begingroup$

I read an idea in mathematics about periodic functions that a constant function is also a periodic function with an undefinable period.

So , suppose a body is at rest (for an observer). This means that it's position doesn't change with time for that observer. So everytime the observer sees the body, it is at its initial position.

So can we say that the body is in a periodic motion with an undefinable period ?

$\endgroup$
1
  • 1
    $\begingroup$ I have no idea that what should I add as tags. So please edit or suggest one if someone finds a suitable tag. $\endgroup$ – Ankit May 16 at 14:01
8
$\begingroup$

A function $f(t)$ is said to be periodic with period $T$ if the following holds for all $t$:

$$ f(t)=f(t+T). $$

As such, the constant function is clearly periodic and every real number is a period. However, it is common to define 'the period' of a function to be the smallest real number that is a period. For example the function:

$$f(t)=\sin 2t $$ is periodic with period $2\pi$, but it is also periodic with period $\pi$. As such whilst you might strictly say "$2\pi$ is a period" if you asked for 'the period', you'd find the answer "$\pi$".

But since every number is the period of the constant function, and there is no smallest positive real number, we arrive at a contradiction and there is thus no single period. This is the source of the tension that "the constant function is periodic, but its period is undefined".


However, this is really a purely terminological issue. In any given case, if you have a theorem relating to period functions (eg maybe you know that in a periodic potential, the solutions to the time-independent Schroedinger equation are governed by Bloch's thoerem) then the constant case is probably a much easier special case that you can solve independently. You will probably find that the constant function does satisfy most theorems about periodic functions but you may need to be careful on occasion.

NB As User SolomonSlow points out, a better terminology is to distinguish between "a period" and "the fundamental period". However it appears that the OP is specifically quoting a text where the word "fundamental" does not appear, which I would say is somewhat common.

$\endgroup$
2
  • 1
    $\begingroup$ There's a name for what you're trying to get at when you say "the period..." You are talking about the fundamental period of the function. The fundamental period of a constant function is not defined. But, like you say, we can analyze a constant function as if it was periodic with period P for any P that we care to choose. $\endgroup$ – Solomon Slow May 16 at 16:11
  • 1
    $\begingroup$ @SolomonSlow agreed that that terminology is clearer but I've definitely seen both used and the OP is clearly quoting a text that doesn't include the word 'fundamental' when they write "a constant function is also a periodic function with an undefinable period." $\endgroup$ – jacob1729 May 16 at 22:26
6
$\begingroup$

You can say what you like, depending on how you define your terms. For example, you can say that a body at rest is "in motion" with speed $v=0$, but in most circumstances this would not be a helpful use of terminology. The phrase "in motion" is ordinarily used to explicitly indicate that the body's state of motion is other than "at rest". That means its motion is such that its location at a given moment is not the same as its location at at least some other times.

In mathematics, to say that a function is periodic is to say something purely about that function. One would say, for example, that if, for all $x$, $$ f(x + L) = f(x) $$ then the function $f(x)$ is periodic, and if $L$ is the smallest number for which this is true, then $L$ is the period. So this leads to the idea of a flat function being "periodic" in this sense.

However when we talk about periodic motion we ordinarily mean both that the body is moving and that the motion is periodic. So if $y(t)$ is the position of the body at time $t$, then for periodic motion we require both that for all $y$, $$ y(t + T) = y(t) $$ and that there is some $\tau$ such that $$ y(t + \tau) \ne y(t). $$

$\endgroup$
3
  • 1
    $\begingroup$ It may be worth adding that if defining 'the period' of a function (with definite article) to be the smallest such positive $L$, then a constant function indeed has no well defined period. $\endgroup$ – jacob1729 May 16 at 15:30
  • $\begingroup$ @jacob1729 Thanks! ; I made an edit $\endgroup$ – Andrew Steane May 16 at 16:31
  • $\begingroup$ @AndrewSteane thanks for your answer. 🙂 $\endgroup$ – Ankit May 17 at 7:48
3
$\begingroup$

I don't know what a mathematician would say, but as a physicist I would not see any value in describing a body at rest as being in periodic motion with an undefinable period. Motion implies displacement over time- if the body is never displaced then it is never in motion. You might say that a body at rest is in motion at zero speed, but I cannot see why that would be of any practical help.

$\endgroup$
0
$\begingroup$

As a specific example of periodic motion, an object oscillating under harmonic motion will satisfy the differential equation (assuming a one-dimensional problem)

$$\frac{d^2x}{dt^2}+Ax=0.$$

A solution of this could be in the form

$$x=a\sin(\omega t),$$

where $A=a\omega^2$. We could define a stationary object to be in the limit where the frequency or amplitude vanishes i.e. $\omega\rightarrow0$ or $a\rightarrow0$. In this sense, the stationary object could have a well-defined period with a vanishing amplitude or a well-defined amplitude with an infinitely long period.

$\endgroup$
2
  • 1
    $\begingroup$ This is incorrect, periodic motion with period $T$ is anything satisfying $x(t+T)=x(t)$ for all $t$. Whilst you can Fourier decompose these into sums of expressions satisfying equations like yours, the sum as a whole will not since the value of $A$ will be different for each. $\endgroup$ – jacob1729 May 16 at 15:27
  • 1
    $\begingroup$ @jacob1729 Yes, this would be the specific case of harmonic motion? $\endgroup$ – jamie1989 May 16 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.