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I have a question on kinematics.

Say the path traced by a particle is given by a Koch curve or Koch snowflake.

enter image description here

Now consider the particle starts from some arbitrary point $A$ on the curve and continues moving with some acceleration. It moves a finite distance on the curve and reaches another point $B$ which is different from $A$ and the particle has not crossed the same point twice.

So there is a net finite displacement covered in a finite time. Hence the particle has a finite average velocity.

But the curve is not differentiable at any point, by definition of the curve. So the particle has no instantaneous velocity at all points of the path taken.

QUESTION: Can a particle have no instantaneous velocity at all points of the path taken but still a finite average velocity?

Is this possible? Can anyone explain this?

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  • $\begingroup$ How do you get a trajectory that is not differentiable in the first place!? Trajectories are solutions to the equations of motion, which are differential equations, hence physical trajectories are always differentiable. $\endgroup$ – ACuriousMind Nov 24 '15 at 17:21
  • $\begingroup$ @ACuriousMind Edited my question. Is it okay? And for the downvoter, what is the reason for downvoting my question? Please tell me. $\endgroup$ – SchrodingersCat Nov 24 '15 at 17:25
  • $\begingroup$ This question (v2) seems to be more about mathematical idealization than actual physics. Example: The Weierstrass function is a continuous function that is differentiable nowhere, or in phys-speak: A 1D path without instantaneous velocity anywhere, but with a well-defined finite average velocity between any two given times. $\endgroup$ – Qmechanic Nov 24 '15 at 17:27
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    $\begingroup$ @ACuriousMind I don't think so. The question is about the representation of physical systems (particle motion) by a particular kind of mathematical model (nondifferentiable paths), which makes it a physics question as far as I'm concerned. Even if the representation being asked about isn't valid, that doesn't stop it from being a physics question. $\endgroup$ – David Z Nov 24 '15 at 17:39
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    $\begingroup$ @ACuriousMind You are forgetting about Brownian motion, which is nowhere differentiable and conceivable in classical mechanics, and the fact that generally solutions to differential equations do not have to be differentiable, not even once, as in elastic collision problems. Of course, those things are idealizations, but no more so than material points and differentiable paths. $\endgroup$ – Conifold Nov 24 '15 at 21:49
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No, it's not possible, because one of the underlying assumptions of kinematics is that all paths are at least twice differentiable. Before you complain about this requirement, remember that physics is about building models that can be used to describe and predict measurements. Measurements always have some amount of uncertainty, and even if you suppose that it is possible for a particle to travel along a nondifferentiable path $x(t)$, it is still always possible to construct a twice-differentiable path that matches $x(t)$ to any desired level of precision. That twice-differentiable path is what you use for the model.

Even beyond that, make sure not to mix up "no instantaneous velocity" with "zero instantaneous velocity". Usually we use these terms interchangeably in physics, but we have the luxury of doing so because (we normally assume) paths are always differentiable and thus there is not really any such thing as, literally, having no instantaneous velocity. If you want to work with nondifferentiable paths, then you have to be more careful. It's conceivable that in such a model, a particle could have a perfectly well defined average velocity between any two points in time and yet never have an instantaneous velocity. This is still fine (if useless) because no physical process actually measures instantaneous velocity. The closest you get is an exceedingly short-time average, e.g. over roughly a period of oscillation of an EM wave when using the Doppler effect.

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    $\begingroup$ First you say that "No, it's not possible" and then you write that " If you want to work with nondifferentiable paths, then you have to be more careful. It's conceivable that in such a model, a particle could have a perfectly well defined average velocity between any two points in time and yet never have an instantaneous velocity. ". Looks contradictory, doesn't it? $\endgroup$ – SchrodingersCat Nov 24 '15 at 17:17
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    $\begingroup$ @Aniket: He first says that your situation is not possible in the standard models and then makes an effort to think about a generalized model where it could make sense. There is no contradiction. $\endgroup$ – ACuriousMind Nov 24 '15 at 17:25
  • $\begingroup$ @Aniket No, not really. The first paragraph is saying that it's not possible (for a particle to have no instantaneous velocity) because one of the underlying assumptions of kinematics forbids it. The second paragraph is about the hypothetical scenario where that assumption is dropped and you are working in some alternate system of kinematics that does allow nondifferentiable paths. (ninja edit: I see ACuriousMind already said this as well) $\endgroup$ – David Z Nov 24 '15 at 17:26
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    $\begingroup$ @Aniket Kinematics makes no such assumptions, almost every path of Brownian motion is continuous and nowhere differentiable, so paths of Brownian particles will have average but no instantaneous velocities. math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/McKnight.pdf This is speaking in terms of mathematical models. And speaking physically there is no more such a thing as instantaneous velocity, or twice differentiable paths for that matter, as there is such a thing as no instantaneous velocity, and non-differentiable paths. These are all mathematical idealizations. $\endgroup$ – Conifold Nov 24 '15 at 21:39
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    $\begingroup$ Thanks @Conifold for this valuable information. I suppose this is the answer to my question. $\endgroup$ – SchrodingersCat Nov 25 '15 at 10:19
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The length along any segment of the Koch snowflake is infinite. It has finite area but infinite perimeter. So, for a particle to move from one place on the snowflake to another it would have to travel an infinite distance. This is why differentiability is important.

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    $\begingroup$ It is worth pointing out that generic trajectories of (mathematical) Brownian motion are of exactly this type, they are nowhere differentiable and therefore of "infinite length" between any two points. eventuallyalmosteverywhere.wordpress.com/2012/02/18/… Nonetheless Brownian motion is perfectly fine as a mathematical model of a physical process.This is why differentiability is just an idealization that is convenient in some contexts, and not in others. $\endgroup$ – Conifold Nov 24 '15 at 22:27
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This is an excellent question. It is pedantic to whine about non differentiability ,however there is in fact a point to be made on that topic . We seem to be conflating the derivative dy/dx with the time derivative dx/dt(x is the position here). One may have a differentiable path but still the instantaneous velocity can remain undefined. Whether the converse is also false I do not know. First of all I am not mathematically adept enough to obtain an expression for the koch snowflake or any non differentiable continuous function for that matter. But we do know a simple continuous non differentiable path . Suppose particle A moves with velocity u and suddenly changes direction. At this point the particle certainly has infinite acceleration. But what about the velocity? You seem to think it is undefined.

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  • $\begingroup$ The velocity at that point is a null vector if you consider the highest point of its trajectory in vertical motion or free fall. $\endgroup$ – SchrodingersCat Nov 25 '15 at 10:18
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It is possible if the instantaneous velocity is regularized in appropriate mathematical sense. Take $ \sqrt{x} $ for example. It has no derivative at the origin, yet in any finite interval about 0 $$ \frac{\Delta \sqrt{x}}{\Delta x} $$ is finite.

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