Observational Effects of "FTL" spotlight from laser pointer?

[russ_watters]

The problem is that there is nothing of any physical interest here.

Why not? You can see it. I find that interesting.

We've had similar threads about sonic booms.

 

[russ_watters]

I think it's the part I'm most interested in too.

Having thought about it a little more, I think what the person sees is:

Start: Origin point only.

Next: When the beam sweeps past him he instantly sees a solid line back to the origin point, that retracts to the origin point, before it and the origin point disappear.

Simultaneously, he sees the point continuing past him at some speed.

...still thinking about what those speeds are.

 

[russ_watters]

The problem is that there is nothing of any physical interest here.

The definitions being used for those in this thread should be obvious: they are the definitions that imply that "something" cannot "move" faster than light. Yes, there are other definitions that could be adopted for those words, but they are irrelevant to the discussion in this thread.

Guys!

This debate about "something" or "nothing" is not helpful. The OP did not introduce these words. The OP wants to know about what the observer sees. Period. Arguing about whether what the observer sees is "something" or "nothing" is not relevant or helpful.

 

[Ibix]

Simultaneously, he sees the point continuing past him at some speed.

...still thinking about what those speeds are.

Mathematically, you just need to note that if you see a point a distance $d$ away from you illuminated at time $t$ then it was actually illuminated at $t_r=t-d/c$. Then you write down the "equation of motion" of the laser spot - i.e. an expression for $x(t_r)$, $y(t_r)$ and $z(t_r)$ - and hence write down $d(t_r)$. That gives you an expression for $t(t_r)$, which may be multivalued - in which case you'll find you see the spot in multiple places.

 

[AndromedaRXJ]

Having thought about it a little more, I think what the person sees is:

Start: Origin point only.

Next: When the beam sweeps past him he instantly sees a solid line back to the origin point, that retracts to the origin point, before it and the origin point disappear.

Simultaneously, he sees the point continuing past him at some speed.

...still thinking about what those speeds are.

I know I asked this already, but so this isn't equivalent to what a tachyon would do? Or rather, suppose a tachyon went the exact same speed as the dot. Would the observation of its motion be the same?

 

[PeroK]

I know I asked this already, but so this isn't equivalent to what a tachyon would do? Or rather, suppose a tachyon went the exact same speed as the dot. Would the observation of its motion be the same?

There is no mystery about a tachyon if you consider a single reference frame. It's simple kinematics. Speed is distance/time.

The problem comes when you consider the motion in another reference frame, using the Lorentz transformation. Then you have, for example, the Tachyon moving in the opposite direction in that reference frame. And, if you use tachyons to transmit messages you can create causality problems, by getting a response to a message before you have sent it.

 

[Ibix]

I know I asked this already, but so this isn't equivalent to what a tachyon would do? Or rather, suppose a tachyon went the exact same speed as the dot. Would the observation of its motion be the same?

Depends how you're observing its motion. If you cover your plane with tachyon detectors, each one of which flashes a little light when it spots a tachyon, then yes.

But why do that? If tachyons are real, why not have your tachyon detectors signal to you with tachyons? The implications of that kind of thinking are true paradoxes in relativity (not like the twin paradox et al) such as the tachyonic anti-telephone (google is your friend), which is a less bloodthirsty implementation of shoot-your-own-grandad. That leads us to believe that tachyons don't exist.

 

[PeterDonis]

Then you have, for example, the Tachyon moving in the opposite direction in that reference frame.

This can happen with ordinary particles moving on timelike worldlines.

 

[PeroK]

This can happen with ordinary particles moving on timelike worldlines.

Yes, of course. I meant that the Tachyon could start from a source A and reach a target B in one frame, but in another frame it would be at the target B before it's at the source A.

 

[TeethWhitener]

Set up a series of reference frames all at rest with each other, at a linear distance $a$ from one another. The origin is our observer, and each of the other reference frames can be labeled by their distance from the origin: $ka, k\in \mathbb{Z}$. Let's say a light beam sweeps over the surface at a local speed of $bc, b>1$. This is (in the limit where $a\to 0$) roughly equivalent to a system where a light pulse is emitted from $-ka$ at $t=0$ in the observer's frame, a second pulse is emitted from $-(k-1)a$ at $t=\frac{a}{bc}$ in the observer's frame...an $n$th pulse is emitted from $(n-k-1)a$ at $t=\frac{na}{bc}$ in the observer's frame.

The observer will see the pulse at the origin (the $(k+1)$th pulse) first, at a local time of $t=\frac{(k+1)a}{bc}$. At this time, the $k$th pulse will have traveled a distance of $\frac{a}{b} < a$, and the prior pulses (that is, the $n<k+1$ pulses) will have traveled a distance of $\frac{(k-n)a}{b} < (k-n)a$. The observer will thus see these pulses in reverse order, with the $n$th pulse arriving at $t_{n<k+1} = \frac{na -(n-k-1)ab}{bc}$, or $\frac{a(n-k-1)(1-b)}{bc}$ after the first pulse (this quantity is positive since $n-k-1$ and $1-b$ are both negative for $n<k+1$).

For the $n>k+1$ pulses, all of the above formulas remain valid, but with the sign switched: $t_{n>k+1} = \frac{na +(n-k-1)ab}{bc}$, or $\frac{a(n-k-1)(1+b)}{bc}$ after the first pulse. Thus, the observer sees the $(k+1)$th pulse first, then we can compare, e.g., the $k$th and $(k+2)$th pulses to determine the behavior of the recession of the pulses. The $k$th pulse arrives at $\frac{a(b-1)}{bc}$ after the $(k+1)$th pulse, and the $(k+2)$th pulse arrives at $\frac{a(b+1)}{bc}$ after the $(k+1)$th pulse. Extrapolating linearly, the $(k+1+p)$th pulses arrive at $p\frac{a(b\pm 1)}{bc}$ after the initially observed pulse, with the plus holding for $p>0$ and the minus for $p<0$. So the "backward receding" pulses ($n<k+1$) will appear to recede faster than the corresponding "forward receding" pulses by a factor of $\frac{b-1}{b+1}$ (that is, the backward receding pulses appear to the observer to take $\frac{b-1}{b+1}$ as long to complete as an equal number of corresponding forward receding pulses).

 

[TeethWhitener]

The really interesting thing about this problem is that the closer $b$ is to $c$, the faster the backward receding pulses appear in relation to the forward receding pulses. Intuitively, this is because the light has time to "pile" up as the speed of the sweep gets closer to $c$. Or conversely, the limit where $b\to\infty$ is just the limit where all pulses are synchronized at $t = \frac{(k+1)a}{bc}$ in the rest frame of the observer.

 

[Shane Kennedy]

So from what I understand, it's possible to have a very powerful laser pointer where you point at an arbitrarily large and far away surface and make the spotlight appear to move at faster than light. E.g. you can be holding the laser in your hand and you simply flick your wrist.

My question is, what does an observer on that surface see? Will the observer see the spotlight arrive at one location before it leaves its initial location? Can there be any observer that will see this? Does the image of the spotlight continuously appearing on the surface, simply count as a continuous series of events which can be seen in any order since they're not causally connected?

The spot doesn't move FTL. When the orientation of the pointer changes, light still has to travel from the pointer to the target, along the new path.

 

[RandyD123]

The spot of light might appear to be moving across the surface of the target but it's not. At all times light only moves from the laser to the target, no light moves across the surface of the target.

I think this is the key to the OP's question. If I understand this correctly I think this is what would happen. Let use the moon's north and south pole. If a person is standing on each pole and a laser is point at person A (north pole) from earth, it will take x amount of seconds for the laser light to reach him. Once it does and the person on Earth "flicks" his wrist to move the laser to the south pole, it will still take x amount of seconds for that beam to travel across the moons surface to get to the south pole. Meaning that even though the person on Earth has the laser pointing at the south pole it is still going to take x amount of seconds to get there. Even if you had people all across the moons surface, no one will see the beam until it gets there and the person on Earth can't see the reflected light until it gets back. We could use two planet as examples, it doesn't change anything. If I point a laser at Mercury and then quickly spin around and point my laser at Pluto, I still have not defeated the speed of light. If there were a string of people holding mirrors between Mercury and Pluto, once I have sent my laser beam, I could be back in the house eating dinner before the event is over for me (waiting for the reflected light to come back). The whole key is in the reflection. And no one along the way can possibly see the beam until it gets to them, at the speed of light!

 

[RandyD123]

So from what I understand, it's possible to have a very powerful laser pointer where you point at an arbitrarily large and far away surface and make the spotlight appear to move at faster than light. E.g. you can be holding the laser in your hand and you simply flick your wrist.

My question is, what does an observer on that surface see? Will the observer see the spotlight arrive at one location before it leaves its initial location? Can there be any observer that will see this? Does the image of the spotlight continuously appearing on the surface, simply count as a continuous series of events which can be seen in any order since they're not causally connected?

Just because you "flick" your wrist, really has no bearing on anything. You've moved your wrist from point A to point B, but that doesn't mean the light has got from point A to point B FTL. It doesn't matter from what side you are watching from.

 

[votingmachine]

I skimmed this thread so it may be in one of the longer answers.

It would look like a line of light. If you had a flashlight that emitted a line of light and flashed it at that distant target, the person there would see a momentary line of light appear. If you did the laser trick, there would be illuminated surface in essentially a line that appears at WAY faster than the optics in the eye and brain are handled. You would see a line of light appear and disappear in the smallest fraction of time that you could perceive.

Say that the light scattering was strong enough to stimulate the eye for about 15 meters in both directions. You described the interval of that 30 meters being illuminated as FASTER than the interval it would take for a photon to move that 30 meters. 30meters/c is an imperceptible time interval. So the light that strikes your eye is essentially hitting simultaneously.

If you had an array of sensors and a really good clock, you could measure that the light hit earlier at different points. on the path being illuminated.

If you turn on a light in a dark room, the light appears to light up all the room at once. Even though really, the distant parts of the room are lit later than the near parts. We simply don't process that time interval. If you set up a series of mirrors in the room, and a set of sensors, you could find that the light goes from the bulb to the near parts of the room before it gets to the far parts and back.

The laser offers a narrow beam, and the location where it scatters from a surface appears to move as our eye sees that location. But we are not seeing movement of a source except indirectly.

 

[jbriggs444]

The spot doesn't move FTL.

The illuminated spot can move FTL. Just like the point of intersection of the blades in a super-luminal scissors. What cannot move FTL is the information that the person holding the laser pointer had flicked his wrist.

Suppose that you are standing on a uniform flat plane. The laser pointer is far above and the beam is sweeping past you from left to right at some large multiple of the speed of light. You are six feet tall.

[One foot is approximately one light-nanosecond]
T=-6 ns: You look up and see the laser directly
T=0 ns minus a fraction: Illuminated spot on the ground passes a point eight feet to your left.
T=0 ns: Illuminated spot on the ground passes your feet.
T=0 ns plus a fraction: Illuminated spot on the ground passes a point eight feet to your right.
T=+6 ns: You see the illuminated spot at your feet.
T=+10 ns minus a fraction: You see the illuminated spot ten feet to your left.
T=+10 ns plus a fraction: You see the illuminated spot ten feet to your right.

As @russ_watters suggested in #24, you would expect to "see" a spot approximately at your feet separating and moving away in both directions.

Edit: fixed the dimensions to make for nice 3-4-5 triangles.

 

[phinds]

Just because you "flick" your wrist, really has no bearing on anything. You've moved your wrist from point A to point B, but that doesn't mean the light has got from point A to point B FTL. It doesn't matter from what side you are watching from.

You are pointing the light at a surface say 10 light seconds away from you and slightly to your left. You turn the light on. The light won't get there for 10 seconds. As soon as you turn the light on, you flick your wrist so that within 1 second, you are now pointing the light at a point that is 5 light seconds away from the first location and it's slightly to your right. That light won't get there for 10 seconds during which time the light to the first location is 9 seconds away from arriving.

10 seconds after you turned the light on, the beam arrives at the first location and 1 second later it arrives at the second location. The locations are 5 light seconds apart so the beam has "moved" across the surface at 5c.

 

[russ_watters]

Just because you "flick" your wrist, really has no bearing on anything. You've moved your wrist from point A to point B, but that doesn't mean the light has got from point A to point B FTL. It doesn't matter from what side you are watching from.

Others have answered most of this, but...

A small, but key clarification: light does not travel from A to B, only the illuminated spot travels from A to B. Since it is not a real object, it is not constrained by the speed of light.

Geometry introduces some weird effects (such as the unzipping line), but because the spot isn't real, it can re-locate at any speed...if you're willing to wait for the start.

 

[RandyD123]

Others have answered most of this, but...

A small, but key clarification: light does not travel from A to B, only the illuminated spot travels from A to B. Since it is not a real object, it is not constrained by the speed of light.

Geometry introduces some weird effects (such as the unzipping line), but because the spot isn't real, it can re-locate at any speed...if you're willing to wait for the start.

When you shine the laser at an object in the sky, those light photons are real. And if you sweep the sky with the laser on, those photons are real as well. All photons will get there when they are supposed to, but none will travel FTL. Is that not correct?

 

[russ_watters]

When you shine the laser at an object in the sky, those light photons are real. And if you sweep the sky with the laser on, those photons are real as well. All photons will get there when they are supposed to, but none will travel FTL. Is that not correct?

Yes, it is correct. So, you ok with the rest now...?

 

[phinds]

When you shine the laser at an object in the sky, those light photons are real. And if you sweep the sky with the laser on, those photons are real as well. All photons will get there when they are supposed to, but none will travel FTL. Is that not correct?

Do you think my reply in post #47 implies otherwise?