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My current understanding of the topic does not allow me to answer this question.

Also, by 'object' I mean a theoretical massless macroscopic object (if we assume such an object is possible, and if such an assumption is absurd, then let the object be a group of massless particles).

Since at a speed of $c$, time freezes in the system of reference of this object, any motion ought to be instantaneous from its perspective. But a displacement that takes no time is only possible if the velocity allowing it has no limit. Does that make this velocity essentially infinite in that system of reference (and not just equal to $c$)?

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  • $\begingroup$ Remember that there is one more parameter: time, yes, velocity, yes, but also distance. When moving at velocity $c$, you move with $c$ seen from anyones eyes - not with a different velocity seen from some specific point or frame. But time, as you say, and also distance/length becomes distorted until it all fits. $\endgroup$
    – Steeven
    Commented Apr 28, 2017 at 7:46
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    $\begingroup$ A photon (or other massless body) doesn't have a rest frame. Also see Would time freeze if you could travel at the speed of light?. $\endgroup$
    – PM 2Ring
    Commented Apr 28, 2017 at 7:48

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What you are probably forgetting is that both distance and time intervals between events seen by relatively moving inertial observers are affected by the Lorentz transformation.

It is true that if observer $B$ moves at speed $v$ relative to $A$, and calculates the "rate" given by distance travelled, as measured in frame $A$ divided by the time elapsed for that travel as measured in frame $B$, you will get a quantity with the dimensions of velocity that approaches infinity as $v\to c$. However, this quantity is not a useful velocity because it mixes quantities measured in two different inertial frames.

Let $B$ travel a speed $v$ relative to $A$ such that $B$ runs through an $A$-measured distance $v$ in unit time. There might be two pins, stationary with respect to $A$, a distance $v$ apart, and $A$ can verify that it takes unit time for $B$ to run the distance between them using clocks at the pins that have been properly synchronized.

Now look it is from $B$'s frame. The pins are moving by and the time taken to run between them is $1/\gamma(v)=\sqrt{1-\frac{v^2}{c^2}}$, which you already understand. However, from $B$'s perspective, the distance between the pins , also shrunken by the same factor is $v/\sqrt{1-\frac{v^2}{c^2}}$, so that the two effects cancel out and $B$ concludes that $A$ is moving relative to $B$ at velocity $-v$.

This result is known as relativistic reciprocity. It can be derived from basic symmetry assumptions, or it is often in first relativity courses taken as a postulate whence to derive the Lorentz transformation.

Another velocity-related quantity that may help your intuition and which genuinely does approach infinity as $v\to c$ is the rapidity. Rapidity works like the following.

  1. You begin in frame $A$ and boost so that you now move at unit velocity relative to $A$ to reach frame $B_1$. You can choose any velocity at all as your unit.

  2. Now you do the same again, i.e. make the same unit relative velocity boost from frame $B_1$ in the same direction as before.

  3. Keep doing the same thing, so that at step number $n$, you boost from frame $B_{n-1}$ to frame $B_n$m in the same direction as all the other boosts.

  4. Otherwise put, frame $B_n$ results from $n$ successive applications of unity velocity co-linear boosts.

  5. Then the rapidity of this relative motion is proportional to the number of boosts. That is, we can define the rapidity of the motion of $B_n$ relative to $A$ as $k\,n$, where $k$ is some scaling constant that we are free to define. Let's just set $k=1$ for now (although the convention usually sets it to $1/c$).

  6. It should also be clear that one can have a fraction of a unit velocity boost. A rapidity of $1/2$ results from the application to frame $A$ of a boost whose square (i.e. two successive applications thereof) boosts from $A$ to $B_1$.

Now, it is no more difficult in theory (see practical problems below) to make the $n^{th}$ boost than it is to make the first one. By Galileo's principle, this must be true: your physical situation, ability to accelerate and so forth cannot change whatever your relative velocity to any other observer (including those comoving with your motion at any former time). So it is clear that we can have a relative motion of any finite rapidity.

The universal signalling speed limit $c$ works as follow. The time dilation in each frame $B_n$ relative to frame $A=B_0$ increases with $n$, so that, from $A$'s standpoint, each successive jump adds less speed so that $B_n$'s speed, relative to the initial frame $A$, asymptotes to $c$. This is the mechanism that prevents greater than $c$ relative motion between observers: from the boosted observer's point of view, there's no barrier ever to your accelerating indefinitely.

No finite sequence of finite boosts can induce a relative speed of $c$, and $c$ corresponds to infinite rapidity.


Practical Problems with Large Rapidities

I spoke of practical problems. You're almost certainly going to vaporize yourself eventually if you try to move with speeds of an appreciable fraction of $c$ relative to the intergalactic matter and gas in any region. See Randal Munroe's "Relativistic Baseball" Whatif.

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    $\begingroup$ This was very helpful, and you’re right, I was overlooking the distortion of space ensuing in the direction of motion. At this point it should have been apparent that the only reason an infinite velocity sounded plausible was because indeed two frames of reference were being used to conclude this absurd velocity (rather, from B’s perspective, B is covering no distance in no time, whereas A moves at c). Although, all is still not clear to me at this stage, it already makes more sense. The remaining gaps should gradually close as I dig deeper into the topic. Thanks. $\endgroup$
    – Stephen
    Commented Apr 28, 2017 at 14:52
  • $\begingroup$ (the Practical Problem was interesting and somewhat funny.) $\endgroup$
    – Stephen
    Commented Apr 28, 2017 at 14:52
  • $\begingroup$ @StephenSafee Yes, Randal Munroe comes up with some gems. Usually very well researched and beautifully written, too. See also his xkcd comics as well. $\endgroup$ Commented Apr 28, 2017 at 15:02
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I would answer "yes".

Suppose we have constructed a spaceship which can travel with speed of light. It's not easy because the mass of this ship must be 0. But I think it's ok for thought experiment. There are some massless particles after all. No idea how to construct some complex system of massless particles. Because we can not travel on this ship (we are not massless!) the ship itself would need to "observe" what's going on around it and record what it sees so that later we will read the journal. I do not see any reasons why this MUST be impossible.

(Note, that to construct the ship which travels faster than light IS impossible even in the thought experiment.)

So, let's suppose we have such a ship. We program it to travel to a galaxy far far away and then back to us.

Then we read ship's journal. "Start! Ooops, I'm already at destination. Ok, go back now. Hehe, I have returned! Looks like they forgot about me already, but in some old archives they found a record that several billion years ago they did send me out to visit the galaxy far far away!"

And if the speed of ship is not exactly $c$ but a little bit less, the situation will be approximately the same. The ship's journal would tell us that the journey took, let's say, 1 second. Or 0.01 second. Or whatever small amount you want.

UPDATE.

Even when you travel with a speed close to $c$ you would not see any outer object moving faster than $c$.

When you travel with a speed of light... Well, even in a thought experiment there is no "when" - you get to any destination instantly.

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  • $\begingroup$ Why is the answer "yes", when you say that the speed is still $c$ or a bit less than $c$? The question is, if velocity goes towards infinity. $\endgroup$
    – Steeven
    Commented Apr 28, 2017 at 7:49
  • $\begingroup$ @Steeven Because journey to an arbitrary distant place takes zero time from the point of view of the ship. $\endgroup$
    – lesnik
    Commented Apr 28, 2017 at 7:53
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    $\begingroup$ But distance is also flattened so that zero distance can easily be travelled in the zero time. That argument doesn't say that velocity is infinite. $\endgroup$
    – Steeven
    Commented Apr 28, 2017 at 7:55
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    $\begingroup$ You must have close to speed of light $c$, since that will flatten the distance so that you can get there in an almost-zero time without speeding up further. $\endgroup$
    – Steeven
    Commented Apr 28, 2017 at 8:23
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    $\begingroup$ You can't Lorentz-transform $c$ to a speed less than $c$. But anyway, as Rod's answer shows, if your speed is $v$ relative to some inertial frame both times & distances that you measure in that frame contract at the same rate, so you will measure objects that are at rest in that frame to have a speed of $-v$ in your frame, no matter how close your speed is to $c$ in that frame. $\endgroup$
    – PM 2Ring
    Commented Apr 28, 2017 at 9:02

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