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Before I start, I want to say that this is not a duplicate of "Is it possible for information to be transmitted faster than light by using a rigid pole?", Since point A is not a real object, it is possible for A to exceed the speed of light. Notice that I have already taken the bending effect into account and decided that if the widths of the bars are small enough, then it's safe to treat the scissor as if it weren't bending. In other words, when they are thin, then the motion will obey the equation in a small region around point O. Please read to the end and have a look on my analysis in the last part before you explain.


Suppose we have a "scissor", which is composed by two bars with a same width of length "l". Bring them closer to each other, during this process, we have the equation(#):$$v=-\frac l4\frac{cos\frac{\theta}2}{sin^2\frac{\theta}2}\omega$$, where v is the velocity of A, \theta is the angle between BC and BA as shown in the picture.It's also an arbitrary function of time t, or $\theta=\theta(t)$. $\omega$ is the angular velocity of $\theta$, thus $\omega=\frac d{dt}\theta(t)$.Here is the derivation:
1. Draw a line vertical to L from A, intersecting with L at C.

  1. Since $sin\theta=\frac{AC}{AB}$, $$AB=\frac{AC}{sin{\theta}}$$.
  2. Since $cos\frac{\theta}2=\frac{AO}{AB}$, $$AO=AB*cos\frac{{{\theta}}}2$$
  3. from 2 and 3 we have: $$AO=\frac{cos\frac{\theta}2}{sin{\theta}}*AC$$
  4. Since $AC=l$, the equation in 4 becomes: $$AO=\frac{cos\frac{\theta}2}{sin\theta}*l$$or$$AO=\frac l{2sin\frac {\theta}2}$$
  5. Differentiate each side with respect to t, and denote $\frac d{dt}AO$ as "v", getting the equation(#): $$v=-\frac l4\frac{cos\frac{\theta}2}{sin^2\frac{\theta}2}\omega$$
    Again, we just look at a small area around point O where our equation is obeyed. Now, at any time $t_0$, $\theta$ and $\omega$ are independent of each other, thus $\omega$ is a free variable. At t= 0, let's set the value of $\omega$ really large and constant, so that the instantaneous speed of A is larger than even the speed of light. Later, since the factor $\frac{cos \frac{\theta}2}{sin^2 \frac {\theta}2}$ in equation(#) is quite large when $\theta$ is close to 0, and since large $\omega$ leads to small $\theta$ during a short period of time, the velocity of A will be increased by this factor. Thus overall, $\omega$ is set to be a large constant, l is a constant, the only changing factor is increasing. Therefore, the velocity of A will increase. Meanwhile, we emit a beam of light. A is in a superluminal motion. And A will arrive at a detector first, causing the bars to touch the detector, and sending the command to the machine. Then the light beam comes later, thus being detected later. Notice that here again, we place the detector close enough to point O, so that our equation is valid.

Remark: Someone may argue that we are not sure if A could attain a speed higher to that of light. So suppose at t=0, the force has already affected the area around point O, and is now traveling further.Meanwhile, ${\theta}_0=\frac{\pi}3$ A is speeded up to 0.999c, then since our formula is valid at t=0 and later period of time, let's apply it and set $l=0.001m, c=3*10^8m/s,{\theta}_0=\frac{\pi}3$. Then plug them in, we get $$\omega=-3.461*10^{11}rad/s $$. Now we want to know its velocity at $t=1*10^{-12}s$, and we get $v=6.89062*10^8m/s$, faster than light.


According to relativity, this shall violate casualty, and therefore could never happen. With which kind of mechanism is this paradox solved? I have taken the bending effect into account, and decided that as long as the widths of the bars are small enough, then it's safe to treat the scissor as if it weren't bending.

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marked as duplicate by Brandon Enright, BMS, Valter Moretti, Kyle Kanos, Prahar Apr 4 '14 at 16:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @BrandonEnright These are different if you think about it. In that case, the effect of the force travels at a finite speed which is lower than the speed of light, in my case, the force still travels at a finite speed, but still, the geometric point A could exceed the speed of light in the initial period of time if l is large but not too large. $\endgroup$ – user43796 Apr 4 '14 at 5:01
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    $\begingroup$ See also math.ucr.edu/home/baez/physics/Relativity/SR/scissors.html $\endgroup$ – Brandon Enright Apr 4 '14 at 5:08
  • $\begingroup$ @BMS Think about it, the geometric point A could exceed the speed of light since it's not a real object $\endgroup$ – user43796 Apr 4 '14 at 5:09
  • $\begingroup$ The link provided here doesn't help. Note that if the time interval is extremely small, "A" still could exceed the speed of light. It will sure then slow down though $\endgroup$ – user43796 Apr 4 '14 at 5:13
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    $\begingroup$ Your assumption that the scissors are not bending can not be right. If the width is smaller, it is even easier to bend, as far as my understanding goes. $\endgroup$ – Bernhard Apr 4 '14 at 7:12
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Can we pass the information faster than light?

No, as a matter of principle. Because the quickest "passing of information" (i.e. the signal front) is what's called "light" (or "light signals being exchanged") for the purposes of kinematics according to (S)RT.

Referring then to specifics of the proposed setup:

Suppose we have a "scissor", which is composed by two bars [...] Bring them closer to each other [...]

Either this process of "being brought closer to each other" is conducted by each bar segment independently, according to some schedule which had been agreed upon in advance. Then the speed at which "the intersection point $A$" thereby propagates, being a variant of phase speed, may indeed exceed the speed of light. But then evidently there's no information being passed (along with $A$ propagating), because the entire process occured precisely as expected based on the prior agreement.

Or otherwise: the bars (scissor blades) won't close rapidly enough (or generally: each of their segments wouldn't be stopped from following the pre-ordained schedule rapidly enough) to actually pass information from fulcrum $O$ "along the bars" to any particular segment (such as $A$) faster than the speed of light.

EDIT:
I'd like to try to express the final paragraph above ("Or otherwise: [...]") in a perhaps more compelling way:

Or otherwise: the bars (scissor blades) start and continue to close, element by element, in response to information transmitted from the fulcrum $O$ (either "in direct response", and/or "in response to preceding blade elements"). If so, the speed at which this signal is transmitted is by definition less, or at most equal, to the speed of light; and by the given setup that's also the speed at which the "geometric object $A$" propagates. Accordingly, in this case, the motion of the bars (scissor blades) would have been not "perfectly rigid", but just "as rigid (or sloppy)" as they happened to respond, in this trial.

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  • $\begingroup$ Point A could exceed the speed of light since it's a geometric object, but then what if I place a detector close enough to where A starts to move? After analysing mathematically, I found out than the time when A exceeds the speed of light is in the beginning, if we place a stick there, and so that A could touch it, then will the information be passed faster than light? $\endgroup$ – user43796 Apr 4 '14 at 5:23
  • $\begingroup$ user43796: "Point A could exceed the speed of light since it's a geometric object" -- Yes; similar to a "(wave) phase". Therefore its speed is phase speed. Another example is the "(virtual) motion" of a laser pointer dot over a screen. "but then what if I place a detector close enough to where A starts to move?" -- Yes, $A$ can and will arrive at (and pass by) detectors. But: thereby there's no information/signal being passed from fulcrum $O$ to any of those detectors (or likewise from one detector to another). $\endgroup$ – user12262 Apr 4 '14 at 5:41
  • $\begingroup$ If one side there is a machine generating a beam of light, the other side is a detector which detects the light when the bar doesn't block the generator, and place them in the right place where the light beam is lose to where A starts. Wherever A is, the light will be blocked so that it can't travel to the other side. So, At first A travels faster than light, thus block the light beam quickly. Won't the information be transported faster than light then? "thereby there's no information/signal being passed from fulcrum O to any of those detectors" Can you explain that in this case? $\endgroup$ – user43796 Apr 4 '14 at 5:43
  • $\begingroup$ user43796: "[...] Wherever A is, the light will be blocked" -- Allright, I think I got that setup: just e.g. a series (or "bank") of cameras with the scissor blades making a shutter. They're watching some movie. The shutter will close. (According to your setup, i.e. "sequentially rapidly progressing"; or, just as another consideration: the shutter might instead close in front of individual cameras in some arbitrary order.) Now: Is there information being passed from one camera to the other?? How should one camera know what "the final scene" was, that some other camera saw before being shut? $\endgroup$ – user12262 Apr 4 '14 at 6:20
  • $\begingroup$ No, not a series of cameras. What I meant is, imagine there is a machine generating light beam to a photon detector on the other side. Wherever point A goes, you can't shoot a beam of light to a detector on the other side. And place the equipment close to where A starts. a photon will start to fly toward the photon detector from where A starts when A starts to move. Within $t_0$, A block the beam, while the photon, which won't get there until t,t>$t_0$, hasn't got there yet. Point A and photon travel for a same distance, the photon gets there later. $\endgroup$ – user43796 Apr 4 '14 at 7:06

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