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Is possible to achieve an infinite speed in real life?

I know beforehand that relativity theory forbids anything from travel faster than speedlight, so please stay with reading the full question first before answer - is more a conceptual/mathematical issue

For generality I will work with dimensionless units.

Imagine a point-particle system which position in time is described by the equation: $$x(t)=\frac{t}{2}\log(t^2)$$ and the particle starts moving at time $t_0=-0.2$ with initial speed and the experiments ends measuring its position at time $t_f=0.2$ following the equation $x(t)$. position_vs_time

From the plot it look like something quite feasible to be done, but if I focus on their derivative I will note that there is a point in space where it maximum speed will rise to infinity without any discontinuity on the position of the particle: $$\frac{d}{dt}x(t) = 1+\frac{\log(t^2)}{2}$$

Since the maximum speed diverges $\lim\limits_{t\to 0}\|\dot{x}(t)\|_\infty \to \infty$:

  1. Is this system attainable under current interpretation of physics laws?
  2. Where I am messing up with the concept of speed and its divergence?
  3. If this is an achievable system, Do you know a mechanic system that behaves like this?

Motivation (Added later)

Since I am receiving some downvotes, I would like to made an explanation of why this question is interesting:

In this video I found by accident on youtube, Terence Tao, one of the most important nowadays mathematician, is reviewing finite-time blow up solutions to differential equations. There, he talk about the possibility of having this solutions which reaches infinity in finite time for the Navier-Stokes equation, which is indeed a widely known physical system.

Thinking in the speed limits rise in Relativity due to Einstein, I was wondering if this talk is just a mathematical curiosity which cannot be happening in real life, but then I saw this video about the Euler´s disk toy, where, for the wobbling rate the solutions indeed shows to have a finite time blow up (it just a frequency ratio so inertia restrictions indeed are not applying).

On this question, also I figure out that sometimes a finite time blow up could be thought as the reciprocal of a solution of finite duration to a differential equation (meaning here, it achieves zero by it own dynamics and stays there forever after).

So if some people are researching this kind of things, I was thinking that maybe something that achieving the blow up in their speed profile could also be possible (at least is possible in the math).... this is why I keep unitless/dimensionless the differential equations, since maybe a model where inertia is not involved could fit the presented situation.

Another example, the point particle could be the displacement of a shadow, as the example shown on this video.

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    $\begingroup$ Note infinite speed implies infinite kinetic energy which s clearly unphysical. $\endgroup$ Apr 4 at 22:09
  • $\begingroup$ @StephenG-HelpUkraine You are right, but it only applies if inertia is directly involved, as example, the wobbling ration of the Euler's Disk toy indeed shows a finite-time blow up behavior (see here)... this is why I keep dimensionless the function, to keep it open to every possible model. $\endgroup$
    – Joako
    Apr 4 at 22:29
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    $\begingroup$ 1. You can't just write down any function $x(t)$ and call it a "physical system". The trajectories $x(t)$ of a physical system are derived as solutions to its equations of motion, not postulated a priori. 2. If your question is just whether a function of the type $t\mathrm{log}(t^2)$ occurs anywhere in physics, your question is much too broad/unfocused to be on-topic here. 3. It is completely unclear what you mean by infinite speed being "achievable" when you simultaneously seem to say what we can ignore relativity. $\endgroup$
    – ACuriousMind
    Apr 4 at 23:11
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    $\begingroup$ It is not appropriate to modify a question in a way that invalidates responses you have already received to the original question. I hav reverted the inappropriate edit $\endgroup$
    – Dale
    Apr 5 at 0:19
  • $\begingroup$ @Dale I did it because it was closed and the comment was to made the question more specific. So if I can´t change it without violating the site standards I will delete it and ask the proper version again. Thank for taking the time to edit it. $\endgroup$
    – Joako
    Apr 5 at 2:04

2 Answers 2

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Is this system attainable under current interpretation of physics laws?

No, it is not.

Even if we backed off from current physical laws and considered only the physical laws known prior to Einstein, it still would be unattainable. Not only does the first derivative of the position diverge at $t=0$ but the second derivative also diverges. This means that the force required to make such a profile is infinite. We do not know of any attainable way to produce an infinite force.

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  • $\begingroup$ Thanks for answering. I think your answer apply only if inertia is involved, and could be there some experiments where it is not involved. please check the added motivation. $\endgroup$
    – Joako
    Apr 4 at 22:38
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I know beforehand that relativity theory forbids anything from travel faster than the speed of light [...]

Doesn't this answer your question?

  1. No, because no massive particle can travel faster than the speed of light.
  2. Just because you write an expression $x(t) = \ldots$ doesn't mean that it's a physically realizable trajectory for a particle.
  3. It isn't.

From the plot it look like something quite feasible to be done [...]

Then the plot is deceiving, specifically around $t=0$.

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  • $\begingroup$ Thanks for answering. I think your answer apply only if inertia is involved, and could be there some experiments where it is not involved. please check the added motivation. $\endgroup$
    – Joako
    Apr 4 at 22:39
  • $\begingroup$ @Joako Sure - If you're not talking about the motion of a particle (e.g. if you're talking about something like the tip of a shadow) then the relativistic "speed limit" does not necessarily apply. However, you are explicitly talking about the motion of a point particle, so it does. $\endgroup$
    – J. Murray
    Apr 4 at 22:43
  • $\begingroup$ Thanks for your comment. Actually I don´t know how to explain it better, like a "point in space" sound like is a fixed point, maybe you cold do a suggestion, like for the geometric point for where scissors close (I am not native English speaker, so apologies in advance). $\endgroup$
    – Joako
    Apr 4 at 22:54

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