Can We Travel Faster Than Light?

[aaryan0077]

I know that relativity doesn't permits a motion at speed of light or fast, as the Lorentz factor becomes infinite and we get absurd results, but what about wormholes or something, say wrap drive and what about http://en.wikipedia.org/wiki/Alcubierre_drive" and
tachyons.
Also I've heard that in quantum chromodynamics that electrons normally travel at speed greater than c, ans some less than c, so that average speed is c.
So does this all means that we can go faster than light, if so than what will happen to casualty?

 

[Hermit]

I know that relativity doesn't permits a motion at speed of light or fast, as the Lorentz factor becomes infinite and we get absurd results, but what about wormholes or something, say wrap drive and what about http://en.wikipedia.org/wiki/Alcubierre_drive" and
tachyons.
Also I've heard that in quantum chromodynamics that electrons normally travel at speed greater than c, ans some less than c, so that average speed is c.
So does this all means that we can go faster than light, if so than what will happen to casualty?

Special relativity explains that an information cannot travel at speed greater than c, because the Lorentz factor would become a complex number ; but there are situations where "something" goes faster than c. For example, imagine a laser pointed forward a planet ; then, you turn the laser forward another planet ; this action during one second. If there is a distance of one light-year between the two planets, the spot will have a speed of one light-year per second.

 

[HallsofIvy]

"The spot" not being an actual physical entity, of course.

 

[Dale]

but what about wormholes or something

The whole point about wormholes etc. is that you always have a timelike worldline (ie v<c always). Wormholes are about taking a shortcut, not about going fast.

if so than what will happen to casualty?

You can have any two of the three: relativity, causality, and FTL. Right now it looks like relativity and causality, not FTL.

 

[vin300]

Special relativity explains that an information cannot travel at speed greater than c, because the Lorentz factor would become a complex number ; but there are situations where "something" goes faster than c. For example, imagine a laser pointed forward a planet ; then, you turn the laser forward another planet ; this action during one second. If there is a distance of one light-year between the two planets, the spot will have a speed of one light-year per second.

No, light will reach the spot only at c so the speed of the spot will prove to be c The light would be like a string rotated in the air.

 

[Doc Al]

No, light will reach the spot only at c so the speed of the spot will prove to be c The light would be like a string rotated in the air.

No, there's nothing preventing a "spot" from moving faster than c. As has been pointed out, the spot is not a physical entity, so there's no issue with relativity.

 

[diazona]

No, light will reach the spot only at c so the speed of the spot will prove to be c The light would be like a string rotated in the air.

The spot will appear to travel at 1 ly/s, since at one time the laser is shining on one planet, and 1 second later the laser is shining on the other planet. But all this will only be seen on the planets some time after the person holding the laser moves it, because of the finite speed of light.

 

[vin300]

No, the spot does not move at>c, because you do not see it to be moving at>c as light takes longer to reach the spot and reflect back

 

[Doc Al]

No, the spot does not move at>c, because you do not see it to be moving at>c as light takes longer to reach the spot and reflect back

The travel time of the light to the surface is irrelevant.

 

[vin300]

The point here is that he said the spot moves at >c and all agreed.That is what is clearly wrong to me you cannot say the spot has moved unless you see it

 

[Doc Al]

The point here is that he said the spot moves at >c and all agreed.That is what is clearly wrong to me you cannot say the spot has moved unless you see it

You may have to wait for the light to reach you, but you'll deduce that the "spot" moved at a speed greater than c. See: http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html#3

 

[sylas]

The point here is that he said the spot moves at >c and all agreed.That is what is clearly wrong to me you cannot say the spot has moved unless you see it

Doc Al is correct. The time it takes for the light to get from the spot to you is irrelevant.

For example, consider a laser beam swept across the face of the moon; assume that it is bright enough that you can see it; and that it takes 1 microsecond to sweep from one side to the other. The dot moves faster than c.

From Earth, you see the dot sweeping across the face of the moon a bit over a second afterwards. But you still see the dot sweeping over the face of the moon in a microsecond.

If you are closer, or further away, you'll change the time between when you see the spot moving and when it actually swept across the moon; but that makes no difference to how fast the spot is moving or how fast you see it moving.

Cheers -- sylas

PS. For pedants in the audience; yeah, I know there will be tiny differences due to different angles at one side of the sweep or the other. It's still seen to be moving at almost the same velocity it actually did sweep across the surface, and that speed is much greater than lightspeed.

 

[vin300]

The scissor Gedanken exp is wrong The theory essentially says no mass can move at speeds greater than c Mass elements farther than approximately 3*10^7 away are not free to move in(neglect curvature)
Also, there is shadow only if there is light, so the speed of shadow is restricted to be equal to the speed of light

 

[sylas]

The scissor Gedanken exp is wrong The theory essentially says no mass can move at speeds greater than c Mass elements farther than approximately 3*10^7 away are not free to move in(neglect curvature)

"scissor"?

I'm guessing you mean a case where you close scissors sufficiently fast and the cutting edge where the two arms meet moves faster than light.

This edge is not a "mass element". The sweeping dot is not a "mass element". There is no violation of relativity or the limit of light speed on transmission of information or particles.

Added in edit: Found it. You are referring to The Superluminal Scissors. It is indeed precisely as I guessed, and it is not "wrong". In the specific example given, the contact point moves faster than the speed of light, but there is no matter moving faster than light speed.

 

[vin300]

This edge is not a "mass element".

Yes?Then what is it?

The sweeping dot is not a "mass element"..

It is not, but if you think of my answer you will find it not to be greater than c

 

[vin300]

. But you still see the dot sweeping over the face of the moon in a microsecond.

No no no
Look at doc's reply#11

 

[sylas]

Yes?Then what is it?

The point in space where the two blades meet of course. The particles involved in that point keep changing as the scissors close.

It is not, but if you think of my answer you will find it not to be greater than c

Your answer is flatly wrong. This isn't hard. Abstract points like the meeting point of two scissor blades or a dot of light sweeping over the face of the moon easily go much faster than c. There's no particles moving with the dot, or the cutting point on scissors.

I have no idea why you don't follow this. It isn't hard.

 

[vin300]

The point in space where the two blades meet of course. The particles involved in that point keep changing as the scissors close.



Your answer is flatly wrong. This isn't hard. Abstract points like the meeting point of two scissor blades or a dot of light sweeping over the face of the moon easily go much faster than c. There's no particles moving with the dot, or the cutting point on scissors.

I have no idea why you don't follow this. It isn't hard.

OK I hadn't read it properly
The explanation is right, but it takes much much longer to close the scissor as written there,it is written that it takes longer for the tips to come close, and note that I had initially written the light travels as a string rotated in the air, analogous to bending of blades here and from the numeric mention in the previous post the contact point moves at c towards the tip, not greater than c

 

[vin300]

I'll explain. In the gedanken expt, the scissor is closed in 1/10 of a sec. If the edges were not to bend, the scissor had to be of length c/10 meters the farthest mass element has velocity c.If in 1/10 sec the edges overlap, the velocity of the farthest element and the contact point are both same=c.Even if they were bent, the velocity of the farthest element(at 1 ly meters,original lt)is c and the velocity of cp is c
The blades are said to be bent because it takes apparently longer time for the farther elements to come close
The scissors flex at c/10 meters

 

[vin300]

Abstract points like the meeting point of two scissor blades or a dot of light sweeping over the face of the moon easily go much faster than c. There's no particles moving with the dot, or the cutting point on scissors.

.

There's no particles moving with the dot, but that does not mean you can see the universe rolling around you at superluminal speeds.You perceive it as moving slower than supposed, because it takes time for light from such large distances to reach you

 

[vin300]

Shadows do not travel faster than light as mentioned in post#13,
If you still say contact point can move faster than light, it is mentioned in the explanation of gedanken exp that it takes longer for the scissor edges to come close completely(it would be possible to say how much if the angle were mentioned) so the contact point moving faster than light would surpass the tip of the scisor

 

[matheinste]

Hello all,

Is the following reasoning valid?

The events of the laser beam stopping its illumination of planet A and the event of the laser beam first illuminating planet B are events in spacetime. Events in spacetime do not move and so the question of how fast an event travels to another event is meaningless. It is of course true that, if the spatial separation of the planets is large enough (this is presumably intended by the questioner) it would require a superluminal signal to leave the first event, beam stopping illuminating planet A, and be present at the second event, beam first hitting planet B, in the frame in which they are all assumed to be at rest with respect to each other.

Matheinste.

 

[DrGreg]

vin300

Consider this. You are on Earth and you aim a laser beam at a target that is 2 light-seconds wide and one light-minute distant. Your laser is aimed with the help of a telescopic sight with crosshairs.

Up until 12:00:00 the laser is aimed at the left edge of the target. You can see both the crosshairs and the laser spot on the left of the target.

Between 12:00:00 and 12:00:01, you change the aim to the right edge of the target. From 12:00:01, you see the crosshairs on the right edge of the target, but you still see the spot on the left edge, because it takes a minute for the light to reach the target.

The light aimed at the left edge continues to reach the target until 12:01:00. The light aimed at the right edge reaches the target from 12:01:01 onwards. This is because each photon of light travels in a straight line in whatever direction you aim it, and takes exactly one minute to get there.

You, on Earth, still have to wait another minute for the reflected light to get back to earth. Up until 12:02:00 you see the laser spot on the left side of the target. From 12:02:01 you see the laser spot on the right side of the target.

Between 12:02:00 and 12:02:01, you see the spot move 2 light-seconds in one second. And you know the spot actually moved between 12:01:00 and 12:01:01. Whichever way you look at it, the spot moved 2 light-seconds in one second, which is twice the speed of light.

If you still think that's impossible, which step in the above argument do you think is wrong?

(All times and distances measured relative to Earth. Gravitational effects ignored.)

 

[russ_watters]

Shadows do not travel faster than light as mentioned in post#13,

If you can make a point of light travel at faster than C (and you can), you can make a shadow travel faster than C. The principle is exactly the same.

If you still say contact point can move faster than light, it is mentioned in the explanation of gedanken exp that it takes 1 ly for the scissor edges to come close completely so the contact point moving faster than light would surpass the tip of the scisor

The speed of the contact point is not constant, it accelerates as the scissors close.

 

[Wangf]

i think this is explained somewhere else. the individual photon does not travel > c.

The illusion that the laser travels faster than c is because once you move the laser to another spot, it is different photons that hits the new spot.

 

[Wangf]

Shadows, moving laser lights do not travel faster than c.

I am just a financial guy, but still know moving laser spots do not travel faster than c. it is clearly explained elsewhere just like i said above.

Moving the laser on moon, or other planets, is like shooting a bullet to north and then turn 180 degree shooting another bullet to south, you cannot say the bullet travels faster than c, because they are different bullets..

same things, they are different photons once you move your laser, the same with shadows, they are different shadows cause by blockage of different photons.

 

[vin300]

vin300

Consider this. You are on Earth and you aim a laser beam at a target that is 2 light-seconds wide and one light-minute distant. Your laser is aimed with the help of a telescopic sight with crosshairs.

Up until 12:00:00 the laser is aimed at the left edge of the target. You can see both the crosshairs and the laser spot on the left of the target.

Between 12:00:00 and 12:00:01, you change the aim to the right edge of the target. From 12:00:01, you see the crosshairs on the right edge of the target, but you still see the spot on the left edge, because it takes a minute for the light to reach the target.

The light aimed at the left edge continues to reach the target until 12:01:00. The light aimed at the right edge reaches the target from 12:01:01 onwards. This is because each photon of light travels in a straight line in whatever direction you aim it, and takes exactly one minute to get there.

You, on Earth, still have to wait another minute for the reflected light to get back to earth. Up until 12:02:00 you see the laser spot on the left side of the target. From 12:02:01 you see the laser spot on the right side of the target.

Between 12:02:00 and 12:02:01, you see the spot move 2 light-seconds in one second. And you know the spot actually moved between 12:01:00 and 12:01:01. Whichever way you look at it, the spot moved 2 light-seconds in one second, which is twice the speed of light.

If you still think that's impossible, which step in the above argument do you think is wrong?

(All times and distances measured relative to Earth. Gravitational effects ignored.)

The boldened part and the further is wrong in your explanation.
The light aimed at the right edge reaches the target from 12:01:01 onwards.
That was your statement, it will come back only at 12:03:01 it takes two light minutes for the light to reach there and come back.
The 1lm and 1ls hypotenuse is the wavefront made of last set of photons reaching there at 12:01:00 and the first sets of photons ejected in the course of transition, the last of this being at the observer at 12:01:01 (since the length of the hypotenuse increases with the transit time, the time for the spot to transfer depends on this time,obvious)This hypotenuse of light photons reflect and reach back such that you see the original spot till 12:02:00.The final spot lags behind the original spot one minute and a second
Note:my para in gigan's post has been deleted due to errors and unnecessary exp

 

[Gigan]

Consider the right triangle in your eg.three sides of 1 ls, 1lm, 1lm.The last set of photons reaches the observer at 12:04:1/2 the laser is tilted at 12:02:1/2 half a sec after the spot is seen.The 1lm hypotenuse is the wavefront made of last set of photons reaching there after 1lm and the first sets of photons ejected in the course of transition, the last of this being at the observer.This hypotenuse of light photons reflect and reach back such that you see the original spot till 12:04:1/2 and the final spot two lm after that event at 12:06:1/2 so you interpret it has taken light 2lm to cover the distance of 1ls
The time is registered lesser for longer distances of the spots and lesser distance from sorce to surface

I've been following this thread for a while and feel I can chime in now.

You would see the second spot at 12:04:1.5 not 12:06:1/2

Your aim was at the first object until 12:02:1/2 then in the course of 1 second your aim is switched to the second object.

So Photons are on their way towards object 1, and then 1 second later, photons are on their way to object 2.

Since they are both going the same distance at the same velocity, the first photon from the second object would have to be seen 1 sec after the last photon from the first object.If it appeared at 12:06:1/2 then you're saying it takes 4 minutes for light to travel 2 light minutes

 

[vin300]

. The speed of the contact point is not constant, it accelerates as the scissors close.

Yes, maybe proportional to cot(angle) or cosec(angle) is it?
approaches infinity as they become parallel

 

[vin300]

The problem with Dr.Greg's explanation is that he thought light from the new point reaches 1sec after the last view of the original point so it moved at a superluminal speed.
That isn't what is happening When the original point is last viewed the light has just reached the new point(1 sec lag) and has to travel equal distance back
That proves it does not move at superluminal speeds

 

[sylas]

The problem with Dr.Greg's explanation is that he thought light from the new point reaches 1sec after the last view of the original point so it moved at a superluminal speed.
That isn't what is happening When the original point is last viewed the light has just reached the new point and has to travel equal distance back
That proves it does not move at superluminal speeds

That's just false. You continue to view a dot on the original left hand side for 59 seconds after the dot has actually reached the new point on the right.

The example involves light traveling a distance of one light-minute to the screen, and one light-minute back. Since it goes at the speed of light, you see a dot two minutes after you shine the beam to initiate a dot. Hence you see the dot on the left, and then a second later on the right, but all one minute after the dot sweeps across the screen.

Therefore you see the dot sweeping across the screen between 12:02:00 and 12:02:01.

The dot actually sweeps across the screen between 12:01:00 and 12:01:01.

Put numbers on it to check. You say that "When the original point is last viewed the light has just reached the new point". That's silly. You've been given a precise description of where the laser is aimed. What is that time you think that the original point is "last viewed" by the observer? The answer there is 12:02:00.

What is the time that you think the light actually reaches the the right hand side? The answer there is 12:01:01.

You've got a complete description of the experiment. Put numbers on your answers.

Cheers -- sylas

 

[vin300]

You continue to view a dot on the original left hand side for 59 seconds after the dot has actually reached the new point on the right.

No it takes one second more than the period you see the dot on the left side for it to actually reach the right.
If this is the basic misconception, then for the new set of photons to reach there, it takes only one second while the distance is one light minute.What about what you said in this para below?
Look at the third quote, It should be clear then

involves light traveling a distance of one light-minute to the screen, and one light-minute back. Since it goes at the speed of light, you see a dot two minutes after you shine the beam to initiate a dot.

Right

Hence you see the dot on the left, and then a second later on the right, .

No you don't You see the dot on the left for two minutes after you deviate the laser then you see it sweeping for another one minute, if you realize the diagonal formation for one minute after you deviate the laser. The last photon that hits the original point followed by lagging new photons forms a diagonal of length one light minute and one light sec which reflect back in a pattern such that you see the whole one minute and one sec span between the original and final spot.Is it fine now?
It may be clear this way- youshoot the photons in different directions- such that every new photon is diagonally behind the previous by a particular distance(which sum up to one light minute and one light second)or a particular period (which sum up to one minute and a sec)

Therefore you see the dot sweeping across the screen between 12:02:00 and 12:02:01.

The dot actually sweeps across the screen between 12:01:00 and 12:01:01.

Put numbers on it to check. You say that "When the original point is last viewed the light has just reached the new point". That's silly. You've been given a precise description of where the laser is aimed. What is that time you think that the original point is "last viewed" by the observer? The answer there is 12:02:00.

What is the time that you think the light actually reaches the the right hand side? The answer there is 12:01:01.

All that can be resolved

 

[Doc Al]

vin300: You seem to be quite stuck on this one. DrGreg's example is perfectly correct, as is sylas's commentary. Did you read the link I gave earlier?

No you don't You see the dot on the left for two minutes after you deviate the laser then you see it sweeping for another one minute, if you realize the diagonal formation for one minute after you deviate the laser. The last photon that hits the original point followed by lagging new photons forms a diagonal of length one light minute and one light sec which reflect back in a pattern such that you see the whole one minute and one sec span between the original and final spot.Is it fine now?

No, not fine. The spot takes 1 second to sweep across. (Of course you "see" the 1 second sweep two minutes after you first fired the laser.)

All that can be resolved

Nothing to resolve.

 

[sylas]

Vin... do you understand the experiment described? You seem to be the only person not getting this. Make sure you read and understand the account and indicate the times at which events occur.

There is a screen, one light minute away. It is two light seconds wide. You have a laser aimed at the left side of the screen. OK?

At 12:00:00 you swing the laser to point towards the right side of the screen, taking one second to do so.

Hence, at 12:00:01 the laser is aimed at right side of the screen.

One minute after this, the photons which left the laser between 12:00:00 and 12:00:01 arrive at the screen.

• Photons leaving at 12:00:00 arrive at the left side of the screen, at 12:01:00.
• Photons leaving at 12:00:00.383 arrive at 12:01:00.383..., a point 0.766 light seconds from the left side and 1.234 light-seconds from the right side.
• Photons leaving at 12:00:01 arrive at the right side of the screen, at 12:01:01.

There is a a dot of light (not a particle) which moves from the left to the right over the time span 12:01:00 to 12:01:01 --- one second.

To SEE the dot, the light has to get back to your eye again... which takes another minute.

Hence at 12:02:00 you see the last view of the dot on the left side of the screen. At 12:02:01 you see the dot on the right side of the screen. Between 12:02:00 and 12:02:01 (one second) you see the dot sweeping over the screen... one minute after it actual did sweep over the screen.

No it takes one second more than the period you see the dot on the left side for it to actually reach the right.

That's not really coherent... the "period" we see the dot on the left side is indefinite. The experiment says that the laser has been trained on the left side, so you've been watching it there for some time.

You see the dot leave the left side two minutes after you cease pointing at the left side. You see the dot arrive at the right side two minutes after you start pointing at the right side. Therefore you see the dot leaving the left side one second before you see it arriving at the right side. You see it traversing the screen for that one second.

Put times on it. The correct answer is that you see the dot leave the left side 12:02:00 and you see it arrive at the right side at 12:02:01.

No you don't You see the dot on the left for two minutes after you deviate the laser then you see it sweeping for another one minute, if you realize the diagonal formation for one minute after you deviate the laser. The last photon that hits the original point followed by lagging new photons forms a diagonal of length one light minute and one light sec which reflect back in a pattern such that you see the whole one minute and one sec span between the original and final spot.Is it fine now?

No. You are still completely wrong.

You deviate the laser at 12:00:00. At 12:00:01 is it now pointing to the right, because the example involves turning the laser over a duration of one second.

The photons which leave the laser at 12:00:01 are directed to the right side of the screen. They arrive at the screen at 12:01:01, making a dot. You see the dot, on the right, at 12:02:01.

You are correct that you see the for on the left for two more minutes, up until 12:02:00.

Note that you see it on the right one second after seeing it on the left. Not one minute later. It only takes two minutes for light to get to the right side of the screen and back; NOT three minutes.

It may be clear this way- youshoot the photons in different directions- such that every new photon is diagonally behind the previous by a particular distance(which sum up to one light minute and one light second)or a particular period (which sum up to one minute and a sec)

All that can be resolved

Everyone so far has resolved it except you. OF COURSE you shoot different photons in different directions, but because they are different direction, what the heck are you summing them for?

The line of photons approaching the screen are in a line that is almost, but not quite, parallel to the screen, and they are all moving almost perpendicular to that line.

Think. At 12:00:00 to 12:00:01 you twist the laser.

Ten seconds later at 12:00:11, the photon headed for the left is 11 light seconds from the laser, and the photon headed for the right is 10 light seconds from the laser. They are spead out over a span of about one third of a light second, from the leftmost to the right most, as they all continue to advance at c towards the screen.

Cheers -- sylas

 

[vin300]

OK I got it my stupid mistake