Faster than light

[John Connellan]

I just want to make sure I understand why a beam of light or shadow from a beam of light cannot move faster than light in the direction perpendicular to the direction of the photons.

You all know the idea that if you point a laser to the moon and wiggle your hand, that a spot on the moon will appear to travel faster than light - the same with shadows.

It is my understanding that, because the beam of light leaves the laser pointer at velocity c, as you change the angle of the pointer, the beam sort of lags behind as shown in the diagram below.

http://ie.mg1.mail.yahoo.com/ya/download?fid=Draft&mid=1_1892_AGLPjkQAAVIRSWixEwJzUUJLfQY&pid=2&tnef=&YY=1231597858984&newid=1&clean=0&inline=1

So at a later time, although you are pointing the laser at a certain angle, the spot appearing on the moon will be at a different angle along the path L. Also, no matter how fast you change the laser angle, the spot on the moon cannot travel faster along the path L than c, and instead, the lag time between the laser and the spot grows larger.

is this a sound argument?

[cosmictraveler]

is this a sound argument?

No! It is a "light" argument. :p

[John Connellan]

No! It is a "light" argument. :p

LOL, trust you Cosmic to pick up on that one :D

[Vern]

So at a later time, although you are pointing the laser at a certain angle, the spot appearing on the moon will be at a different angle along the path L. Also, no matter how fast you change the laser angle, the spot on the moon cannot travel faster along the path L than c, and instead, the lag time between the laser and the spot grows larger.

There's no problem with illusions such as this. Many exist that seem to move faster than light. It is even possible that stars moving toward us may approach at angles such that their measured speed seems faster than light. You can work it out with arithmetic; it has been awhile since I saw it though and I don't have an example.

[John Connellan]

There's no problem with illusions such as this. Many exist that seem to move faster than light. It is even possible that stars moving toward us may approach at angles such that their measured speed seems faster than light. You can work it out with arithmetic; it has been awhile since I saw it though and I don't have an example.

Yeah, but from the above, I can't see how the spot can even appear to move faster than light because of the limit on the speed of the photons emmitted from the laser pointer in the first place

[Boris2]

http://www.phy.duke.edu/research/photon/qelectron/proj/infv/fast_tut.ptml

[Fraggle Rocker]

Is this a sound argument?What you see "moving" is not the source of the photons, and it's not even the photons themselves. It's simply a series of images that are not comprised of the same elements. To use today's terminology, it's "virtual matter," not "real matter."

If you shine your flashlight on Venus and then you turn around and shine it on Mars, sure it looks like the beam of light has moved a few hundred million miles in a fraction of a second, i.e., faster than c. But none of the individual photons are moving that fast. Each beam consists of a different batch of photons. So relativity has not been violated.

[BenTheMan]

What you see "moving" is not the source of the photons, and it's not even the photons themselves. It's simply a series of images that are not comprised of the same elements. To use today's terminology, it's "virtual matter," not "real matter."

I don't know if this is quite right, I'll have to think about it. My first reaction is to say that this is incorrect, and it is real photons, not virtual photons.

John---remember that the issue is whether or not you can send a signal faster than light, using the tricks you talked about.

Another cute example of the type you listed is to imagine an infinitely long pair of scissors closing. Eventually, the place where the two edges meets moves faster than light.

[John Connellan]

If you shine your flashlight on Venus and then you turn around and shine it on Mars, sure it looks like the beam of light has moved a few hundred million miles in a fraction of a second, i.e., faster than c.

No, what I'm saying is that you cannot shine a spot of light on venus and then on mars faster than the time it would take light to travel from venus to Mars. Therefore no information travels faster than c

[John Connellan]

John---remember that the issue is whether or not you can send a signal faster than light, using the tricks you talked about.

Like I said, I blieve that these tricks cannot send a signal faster than light and I believe this one in particular does not even appear as if it is sending a signal faster than c

Another cute example of the type you listed is to imagine an infinitely long pair of scissors closing. Eventually, the place where the two edges meets moves faster than light.

The scissors (like the very long pole thought experiment) cannot move faster than the speed of sound in that material. The more rigid the scissors the faster you can move it. However, a very long scissors will simply appear to sway at the speed of sound in the material

OK, gotta go. I will finish this conversation tonight I hope :)

[Cellar_Door]

http://img211.imageshack.us/img211/6929/factcheckinghd5.jpg

[camilus]

http://img211.imageshack.us/img211/6929/factcheckinghd5.jpg


LOL hahahahahaha

[John Connellan]

CNN eh? :rolleyes:

[Neddy Bate]

What is all this about scissors? Let us perform an experiment with these scissors.

If we construct scissors from two laser beams, the intersection of the two beams (lines), can be considered a mathematical point, which moves along an axis. Can it move faster than c?

Surely, the lines are always curves?

[swivel]

Surely, the lines are always curves?

Precisely. Laser beams are not rigid, they are created by emission. So the analogy with scissors won't work.

If the scissors could be moved, despite their infinite nature, there would have to be an infinite length of scissor which is approaching the other side faster than the speed of light. In fact, the scissors wouldn't need to be infinitely long to exceed 'c'. If you compute the starting angle and the length of time you choose to "snip" the scissors in, you can calculate how long the scissors need to be to have the tips just breaking the speed of 'c', either relative to one another or relative to a fixed plane, prior to closure.

Edit: My math skills are extremely rusty, but if we place the tips of the scissors at a distance of 300,000,000 meters from one another and we are able to close the scissors in one second from an angle of 30 degrees, the arms of the scissors would "only" need to be 579,555,495 meters long each. The tips will be traveling, relative to each other, a little bit over 'c' when they meet.

Of course, these calculations are for left-handed scissors, so they wouldn't cut shit even if you could move them that fast.

[John Connellan]

Can it move faster than c?

No. Because as Swivel points out, light is not "rigid" and as you move the beam of photons, the beam will change direction like a type of swaying wave. This results in even the intersection never exceeding c.

[przyk]

No. Because as Swivel points out, light is not "rigid" and as you move the beam of photons, the beam will change direction like a type of swaying wave. This results in even the intersection never exceeding c.
Sure it will. If you go back to imagining a laser pointing at the moon, remember that every the part of the beam will be delayed by the same second or so. Wiggle your hand while pointing the laser, and the image of the laser beam will move around the moon's surface in the same pattern, just delayed by a little over one second (and magnified).

[swivel]

Sure it will. If you go back to imagining a laser pointing at the moon, remember that every the part of the beam will be delayed by the same second or so. Wiggle your hand while pointing the laser, and the image of the laser beam will move around the moon's surface in the same pattern, just delayed by a little over one second (and magnified).

When the moon is being illuminated by the laser, each "bit" of the arc is made up of a photon that has left the laser, bounced off the moon, and returned to your eye all on its own. What the overall "pattern" of a bunch of pricks of light, all moving at 'c' are doing is an illusion.

Imagine standing on the moon. You can see the red laser light flying back and forth as someone on Earth flicks their wrist. Could it appear that this red pattern was "moving" faster than the speed of light? Sure, but in reality it is just a bunch of discrete points creating an animation. It would be like watching a bunch of distant stars twinkle in a pattern and calling that an object moving faster than the speed of light.

[John Connellan]

Sure it will. If you go back to imagining a laser pointing at the moon, remember that every the part of the beam will be delayed by the same second or so. Wiggle your hand while pointing the laser, and the image of the laser beam will move around the moon's surface in the same pattern, just delayed by a little over one second (and magnified).

No, it will be delayed yes, but you will also find that at some point, the faster you wiggle your finger, the spot on the surface of the moon will not "appear" to be travelling any faster than c.

Think about it. You are shining light at one angle. After a few seconds, it gets to the moon. You have a spot. Change the angle almost instantaneosly and the spot will not travel over to the new angle at the same speed. It will be limited by c in the direction it is traveling.

[LogicTech]

The argument is all wrong przyk, as you are trying to assume an absolute frame of reference. There is no preferred frame of reference; the speed of light will always appear to be c in every frame of reference.

[prometheus]

There is no preferred frame of reference; the speed of light will always appear to be c in every frame of reference.

This is true, but the semantics are somewhat misleading. The speed of light does not appear to be c in every frame, it actually is c in every frame. From this all of the myriad consequences of relativity fall out. :)

[Montec]

Hello all

Actually the round trip time for light will always measure out to be "c" in all frames.

:)

[przyk]

When the moon is being illuminated by the laser, each "bit" of the arc is made up of a photon that has left the laser, bounced off the moon, and returned to your eye all on its own. What the overall "pattern" of a bunch of pricks of light, all moving at 'c' are doing is an illusion.
I know.
No, it will be delayed yes, but you will also find that at some point, the faster you wiggle your finger, the spot on the surface of the moon will not "appear" to be travelling any faster than c.
Let's say for argument's sake that the moon is exactly one light second away (since the exact delay won't change anything). If at t = 0 seconds you're pointing your laser at one side of the moon, you're guaranteed a spot on that side of the moon at t = 1 s. If you're quick and at t = 0.01 s (we can use Jupiter instead if this is too fast for you) you're pointing your laser at the opposite side of the moon, then the photons it shoots off will guarantee a spot on that opposite side of the moon (since there are no photon-photon interactions, it doesn't matter what the laser was doing earlier) at t = 1.01 s. If your laser was always turned on and pointing at somewhere on the moon, then the spot travelled across at least over 5400 km of lunar surface (half the moon's circumference) in 1/100th of a second, or at least 540,000 km/s. For comparison, the speed of light is just under 300,000 km/s.
Think about it. You are shining light at one angle. After a few seconds, it gets to the moon. You have a spot. Change the angle almost instantaneosly and the spot will not travel over to the new angle at the same speed. It will be limited by c in the direction it is traveling.
Swivel's reply applies to you here: the spot isn't a material object. It's different photons reflecting off the moon's surface at different times essentially independently of one another, so there's no reason for the spot to be constrained by the speed of light.
The argument is all wrong przyk, as you are trying to assume an absolute frame of reference.
No I'm not.
the speed of light will always appear to be c in every frame of reference.
I've known this much about relativity since I was thirteen, but thanks anyway.