Obviously this is impossible in relativity; however, if we ignore relativity and use only Newtonian mechanics, is this possible? How (or why not)?

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    $\begingroup$ The question as posed is trivial. If the force as a function of time F(t) integrates to infinity, then of course the answer is yes. $\endgroup$ – Ben Crowell Aug 5 '11 at 2:45
  • $\begingroup$ @Ben Crowell, what is infinite speed? It's nonsense... $\endgroup$ – AndyK Aug 5 '11 at 6:20
  • $\begingroup$ Vector quantities like velocity etc. cannot be infinite. This is meaningless. You should reformulate your question (the title). $\endgroup$ – AndyK Aug 5 '11 at 16:56
  • $\begingroup$ @ANKU: It's not meaningless - please see BebopButUnsteady's answer $\endgroup$ – BlueRaja - Danny Pflughoeft Aug 5 '11 at 17:23
  • $\begingroup$ @ANKU One comment about one topic is enough. $\endgroup$ – user68 Aug 5 '11 at 20:20

The answer is yes in some unintersting senses: Take two gravitational attracting point particles and set them at rest. They will attract each other and their velocity will go to $\infty$ in finite time. Note this doesn't contradict conservation of energy since the gravitational potential energy is proportional to $-1/r$. This isn't so interesting since it's just telling you that things under gravity collide. But its technically important in dealing with the problem of gravitationally attracting bodies.

Now a more intersting question: Is there a situation where the speed of a particle goes to infinity without it just being a collision of two bodies?

Suprisingly, the answer to this question is yes, even in a very natural setting. The great example is given Xia in 1995 (Z. Xia, “The Existence of Noncollision Singularities in Newtonian Systems,” Annals Math. 135, 411-468, 1992). His example is five bodies gravitational interacting. With the right initial conditions one of the bodies can be made to oscillate faster with the frequency and amplitude going to infinity in finite amount of time.


Here is an image. The four masses $M$ are paired into two binary systems which rotate in opposite directions. The little mass $m$ oscillates up and down faster. It's behavior becomes singular in finite time.

Five body setup

  • $\begingroup$ Amazing! Just the answer I was looking for. Here is a link to the paper mentioned, and here is an article trying to explain it in laymen's terms. Could you perhaps add some more information about the "five bodies interacting?" $\endgroup$ – BlueRaja - Danny Pflughoeft Aug 4 '11 at 18:26
  • $\begingroup$ We can tell the same for atracting point charges. As the distance r decreases the force between them increases. Taking limit $r\to 0$ you get $F\to \infty$, therefore $u\to \infty$. But this means that the speed aproaches the infinity, not that the velocity is infinite. Take a look at definition of the limit. $\endgroup$ – AndyK Aug 4 '11 at 19:07
  • $\begingroup$ I'm having trouble understanding your diagram. I'm fine with the fact that $m$ is a comparatively small mass, but where does it go when it oscillates to infinite speeds? Does it get out of the gravity well eventually? And how do the 2 binary systems maintain the distance from each other? Is it supposed to be implied that they're rotating, or that they're oscillating as well? $\endgroup$ – Alan Rominger Aug 4 '11 at 20:32
  • $\begingroup$ @Zassounotsukushi Check page 8 of the paper. $\endgroup$ – mmc Aug 4 '11 at 22:51
  • $\begingroup$ @Zassounotsukushi: Imagine the left-half of the position-vs-time graph 1/t, with t < 0. No matter what speed you give me, I can find a time t < 0 where the object is moving at an even greater speed; thus at the limit, the object's speed is infinite. Xia found a case where, under Newton's laws, this happens (though the graph behaves a bit more like (1/t) sin(1/t)). At t=0, there's what we call a singularity; I'm not really sure how to answer the question of where the object is, or what it looks like, at or after t=0, partly because singularities don't make physical sense. $\endgroup$ – BlueRaja - Danny Pflughoeft Aug 17 '11 at 16:52

If by "ignoring relativity", you mean ignoring the fact that nothing can move faster than the speed of light, then the answer is still no.

Since kinetic energy is proportional to the square of the speed, infinite speed would mean infinite energy, which you cannot provide, whatever the amount of time you are considering.

  • $\begingroup$ I'm sure the quest to accelerate to infinity will be easier due to the numerous physically implausible inventions that are possible in this world without relativity :-P $\endgroup$ – Alan Rominger Aug 4 '11 at 18:52
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    $\begingroup$ Others have provided simple counterexamples to your statement. $\endgroup$ – Ben Crowell Aug 5 '11 at 2:36

Lab Reference Frame

Let's start assuming a force is exerted on the object from equipment "lab" reference frame, meaning that there is negligible recoil on the lab frame from accelerating the object. Not only is it the case that $F=ma$, but the power delivered to the object grows as $P=v F$. Take a device like the large hadron collider, and just completely wave off the technical difficulty of applying force to an object of progressively increasing velocity, and let's not the following.

$$v' = a = \frac{F}{m} = \frac{P}{v m}$$

The requirement is that $v \rightarrow \infty$ in a finite time. You know, just for fun let's use an actual functional form.

$$v(t) = -\frac{1}{t}$$

For t from $-\infty$ to $t=0$.

$$P = m v' v = -\frac{m}{t^3}$$

So the power delivered must increase at the hilariously fast rate of $1/t^3$ as $t$ goes to zero, not that it matters because we all knew this would result in requiring infinite energy, but this just shows exactly how intangible it is.


In the case of a rocket the propellant has to be hauled along with the payload, but the tradeoff is that you then don't have the $P=v F$ proportionality, since the propellant itself is still moving after being ejected. The equation of motion for a rocket that starts with mass $M$ and ends with mass $m$, ejects propellant with speed $v_e$ is as follows. I'll add in the approximation for $M \gg m$, because obviously that must be the case since we're talking about going to infinity.

$$v = v_e \ln{ \frac{M}{m} } $$

This is certainly interesting. It is interesting to note that the speed the rocket can reach is proportional to the speed the propellant is ejected at, which is also not limited by the speed of light. The energy required to expel that propellant also isn't a problem (haha) because it is not the case that $E=mc^2$ in this world. But it would be skirting the problem to say that the propellant is ejected at infinite speed, so we still seek a way for the above expression to limit to infinity with $v_e \neq \infty$. But we would also really like $\frac{M}{m} \neq \infty$, because that would require either an infinitely large starting mass or an infinitely small ending mass. Neither of these options are appealing, unless atoms also don't exist in this world, allowing us to use fractal math to claim that an infinitely small chuck of something went to infinite speed.

To the extent that we don't make these absurd assumptions, your request is impossible, even given the already absurd assumption of allowable superluminal speed.

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    $\begingroup$ "Let's start assuming a large reference frame acts on the object." What does this mean? A frame of reference is not an object that exerts forces on other objects. $\endgroup$ – Ben Crowell Aug 5 '11 at 2:37
  • $\begingroup$ @Ben Edited the answer for clarity. $\endgroup$ – Alan Rominger Aug 5 '11 at 2:48
  • $\begingroup$ Will somebody PLEASE explain the downvotes, because as far as I can tell this is a useful and thoughtful answer. If there is something that you know that I don't, of course I want to hear it. $\endgroup$ – Alan Rominger Aug 5 '11 at 12:49
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    $\begingroup$ Downvoted because it's too limited in imagination --- as the top comment makes clear. There seems to be a great deal of "classical" work that people are in the dark about. You've done a great job of explaining a particular situation, but clearly failed to account for the full general case. $\endgroup$ – genneth Aug 6 '11 at 16:01
  • $\begingroup$ @genneth That's fair. I would just add that the case of two passing charges fully satisfies the requirement of delivering the rate of power specified, that being the $1/t^3$ form I gave. It's interesting to put things in these terms, because it shows that no answer can really be "right". Since it's a made-up world, quantum mechanics could prohibit the flyby that leads to infinite velocity... or it could not. $\endgroup$ – Alan Rominger Aug 6 '11 at 22:45

It sounds like the answer is yes for particles with mass but zero physical extent. I assume if the particles have finite extent (or we bring in quantum mechanics), then the answer is no, as it would be if we brought in relativity.


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