Light Puzzle

[Dancing David]

[/I]
Yes but - “Length contraction, according to Hendrik Lorentz, is the physical phenomenon of a decrease in length”.

What increases the length of the distances between the light beams to keep the beam and distance lengths the same when you are travelling in the same direction as the light beams?

Say the high way is painted at 20 yards alternating 20 yards pained and twenty yards unpainted.

You travel over it at speed X and measure the intervals of passage, you will get rate and length of pulse as R,L (rate=time from start to start IE start paint to start paint, start unpainted to start unpinted) (length equals time from start to stop, paint to paint or unpainted to unpainted)

Now if you travel at 2X what will the R,L be?

 

[CurtC]

It’s a basic postulate of the theory of relativity that the speed of light is constant and independent of the motion of the observer.

Yes.

A second long beam of light is therefore a second long to all observers regardless of time or place or motion.

No.

Imagine that there are one billion waves of this light in the pulse. If you're moving towards the sender, you will perceive that the pulse is briefer, but you still count the same million waves. Therefore the wavelength has shortened, IOW the frequency has increased, IOW the light is more to the blue end of the spectrum that you had observed before.

 

[Dilb]

You are stationary compared to a laser 10 light seconds away, which emits a brief pulse of light every second. Immediately after receiving the first pulse, you (in a ship) instantaneously accelerate to 0.5c, and move toward the laser.

From the laser POV: it takes 20 seconds for the ship to reach the laser, during which the laser emits 20 pulses. There are 10 pulses already in transit, so the ship receives 30 pulses in 20 seconds. The ship is time dilated, and so counts only 17.3 seconds.

From the ship POV: the laser has changed by length contraction to be only 8.66 light seconds away, so it takes 17.3 seconds for the laser to reach the ship. During this time, the laser is time-dilated, so it emits 15 pulses.

What are we missing? The time-coordinate of the laser has also changed: the interval
[latex]s^2 = -ct^2 + x^2[/latex]
between the laser emitting the 11th pulse, and the laser reaching the ship, is constant, but x and t change. In particular, this means the laser emitted those 11 pulses 20 laser-seconds ago, but 23 ship-seconds ago (if we pretend the ship was moving at 0.5c back then). The laser is time-dilated, so in those 23 seconds it emits 20 pulses. The ship receives 30 pulses total, exactly in agreement.

This illustrates the unusual result: when the ship accelerates to 0.5c, the laser changes from having already emitted 11 pulses, to having already emitted 16 pulses. This is exactly the sort of thing meant by saying "simultaneity is relative". The laser thinks emitting the 11th pulse and the ship receiving the 1st pulse happen at the same time, while the ship disagrees.

 

[MattusMaximus]

It’s a basic postulate of the theory of relativity that the speed of light is constant and independent of the motion of the observer. A second long beam of light is therefore a second long to all observers regardless of time or place or motion.

As others have mentioned, this is incorrect. You are confusing the speed of the laser light with the observed frequency (i.e. ticking of the laser "clock"), and they aren't the same thing. Make certain you understand the difference before proceeding.

 

[ynot]

Thanks for the replies.

Even though the light beams are fired from the laser they are never stationary relative to it because light always moves relative to everything at c (they don‘t accelerate to c). Is this correct?
　
If a one laser second light beam was fired at a “stationary” mirror it would measure one laser second when it returned. If the laser moved toward the mirror at half c immediately after firing the beam however would the beam measure one laser2 second when it returned to the laser?

 

[rjh01]

Thanks for the replies.

a. Even though the light beams are fired from the laser they are never stationary relative to it because light always moves relative to everything at c (they don‘t accelerate to c). Is this correct?
　
b. If a one laser second light beam was fired at a “stationary” mirror it would measure one laser second when it returned. If the laser moved toward the mirror at half c immediately after firing the beam however would the beam measure one laser2 second when it returned to the laser?

[Quote altered slightly to make it easier to answer]

a. The speed of light (in a vacuum) is always c in all references. Light does not accelerate. So the answer is yes

b. If the laser fired for one second when it returns the beam will be one second long. However as the mirror is moving towards the laser then the beam would have picked up some energy from the mirror and so will be at a higher frequency.

I am not an expert so I could be wrong.

 

[Uncayimmy]

Can we simplify this problem as follows?

Situation #1: Sender and Receiver are stationary relative to each other. Sender turns on his light for one second and turns it off. Receiver's device detects the light as having been on for one second.

Situation #2: Receiver is moving toward sender at 0.5C. Sender turns on his light for one second according to his clock. What does my device read as the duration of the pulse?

 

[ArmillarySphere]

#2 would be 0.87 seconds.

 

[ynot]

[Quote altered slightly to make it easier to answer]

a. The speed of light (in a vacuum) is always c in all references. Light does not accelerate. So the answer is yes

b. If the laser fired for one second when it returns the beam will be one second long. However as the mirror is moving towards the laser then the beam would have picked up some energy from the mirror and so will be at a higher frequency.

I am not an expert so I could be wrong.

Regardless whether you define that the mirror is moving toward the laser or laser toward the light the light beam always moves relative to both at c. So what creates this extra energy?

 

[Dilb]

The energy comes from the kinetic energy of the mirror. The light has more energy because it's a property of light that higher frequency photons have more energy than lower frequency photons.

 

[nathan]

Regardless whether you define that the mirror is moving toward the laser or laser toward the light the light beam always moves relative to both at c. So what creates this extra energy?

The first thing you need to realize is that the total energy you measure in a system is a property of your frame of reference. Total energy is kind of arbitrary -- energy's always measured WRT some baseline energy level. All inertial observers will agree that the total energy of a closed system does not change. But they way they think that energy is partitioned between parts of that system may vary. Non-inertial observes may observe energy is not conserved.

To make things simple, I'm going to make the mirror a freely moving inertial object.

In the laser's inertial reference frame (the laser is stationary in that frame), the energy increase in the beam comes from a retardation of the mirror's motion. Note there's no change in the total energy of this system, in this reference frame (as expected).

Now, note that in this set up, the mirror experiences an acceleration, and therefore it is not an inertial observer. We can consider two different inertial reference frames -- that of the mirror before reflection and that of the mirror after reflection.

Let's start with the before-reflection reference frame. The light beam comes from one direction and then bounces off to return back, and the mirror experiences a recoil. After the collision the mirror has some kinetic energy, so the reflected beam must have a lower energy than before -- it will be red shifted. The laser has the same energy as before, as its velocity is unchanged wrt the before-reflection reference frame.

In the after-reflection reference frame, the mirror comes from one direction and the light beam from the other. They meet, the mirror stops and the light beam returns from whence it came. In this case, the mirror loses energy, so the outgoing beam must have a higher energy -- it will be blue shifted.

How can the beam be both blue and red shifted? Well the reference frames we observe these phenomena in are moving relative to each other, and that accounts for it. Those observing red shift are moving in the direction of the reflected beam faster than those observing blue shift are moving in that direction.

 

[ynot]

Let’s say the mirror is moving toward or away from the laser at a speed that means one laser second equals 2 mirror seconds. The laser fires a 1 laser second long beam of light that the mirror receives as being 2 mirror seconds long. But when the mirror reflects the light beam back to the laser wouldn’t it receive the beam as still being 1 laser second long?

Given the emitting and receiving of the light beam are different events that occur at different times does either event have any effect on the other given light always travels at c for both emitter and receiver? Rather than the beam being 1 laser second long when emitted and 2 receiver seconds long when received, isn’t it always just 1 light second long regardless? The beam is I light second long when emitted at c and 1 light second long when received at c.

 

[ynot]

The energy comes from the kinetic energy of the mirror. The light has more energy because it's a property of light that higher frequency photons have more energy than lower frequency photons.

Isn’t the kinetic energy of the mirror relative to the light constant because it’s governed by c?

 

[nathan]

Isn’t the kinetic emery of the mirror relative to the light constant because it’s governed by c?

This is a meaningless statement. You measure kinetic energy in a reference frame. If that reference frame is inertial, then the you'll observe conservation of energy.

A reference frame moving at c is kind of meaningless. How would it measure the velocity of anything?

 

[Dilb]

No, why would you think that?

 

[nathan]

Let’s say the mirror is moving toward or away from the laser at a speed that means one laser second equals 2 mirror seconds. The laser fires a 1 laser second long beam of light that the mirror receives as being 2 mirror seconds long. But when the mirror reflects the light beam back to the laser wouldn’t it receive the beam as still being 1 laser second long?

ETA: No. The timing of the returning pulse as observed by the laser is less than 1 second, because the trailing edge of the pulse is reflected when the mirror has moved closer to the laser than when the leading edge was reflected. So the trailing edge travels less distance (in the reference frame of the laser), and therefore takes less time to come back. Ergo the pulse must be shortened.

Given the emitting and receiving of the light beam are different events that occur at different times does either event have any effect on the other given light always travels at c for both emitter and receiver?

I'm having trouble understanding your question. The emission of the light beam is a necessary (but not sufficient) cause of the effect of receiving the reflected beam. The receiving of the reflected beam happens after the emission of the emitted beam, and therefore cannot affect it (including effecting it).

Rather than the beam being 1 laser second long when emitted and 2 receiver seconds long when received, isn’t it always just 1 light second long regardless? The beam is I light second long when emitted at c and 1 light second long when received at c.

No. You keep asking this question. The observed interval of time between two events (the start of the pulse and the end of the pulse) is affected by the (relative) velocities of inertial observers. That the two events involve a beam of light is orthogonal to that.

 

[Towlie]

A laser in space fires one second long beams of light at one second intervals.

At this point we've already established that you really mean two second intervals - one second on and one second off. 1 + 1 = 2.

You are 10 light seconds away from the laser and you receive the beams of light 10 seconds after they were emitted and measure them to be one second long with one (actually two) second intervals. You then travel toward the laser at half c and continue to measure the light beams and intervals to be one second long because light travels relative to you at c regardless.

No.

First of all, only at speeds very close to light speed do relativistic effects begin to take effect. At only half the speed of light relativity doesn't even enter into it to any significant extent. For your scenario, forget about Einstein.

Second, when approaching the source at half the speed of light, you actually would time the intervals between pulses at one second each, and their duration would be 1/2 second each. That is, the frequency of pulses would double and their periods would be cut in half.

It takes you twenty of your seconds to reach the laser

Yes.

but as your time has been time dilated...

No. You're not going nearly fast enough for that. The only influence that would be significant here would be the Doppler Effect.

 

[nathan]

First of all, only at speeds very close to light speed do relativistic effects begin to take effect. At only half the speed of light relativity doesn't even enter into it to any significant extent. For your scenario, forget about Einstein.

At .5c SR will affect your results over Newtonian mechanics by a factor of sqrt(.75), which is about .87 i.e 13%. So, yes, I'd say you need to account for SR (which ynot is trying to understand, I think).

For time dilation to be a factor of 2, the relative speed would need to be .5 = sqrt (1 - r^2) -> .25 = 1 - r^2 -> r = sqrt (.75) -> .87c.

 

[Uncayimmy]

#2 would be 0.87 seconds.

To me that seems to go a long way towards answering the question in the OP. If my device reads it a 0.87 seconds on, then it will read it as 0.87 seconds off, correct?

The pulses of light could be considered a tick-tock of a clock, right? If so, then what the moving Receiver is observing is that the Sender's clock has slowed down.

Now, here's where the mind-**** happens for me. While stationary relative to Sender, Receiver puts a rod behind him that is one light-second long. That means they both agree it is 186,000 miles long from A (Receiver's ass) to B (end of the rod). Correct?

So, while still stationary relative to one another, both will agree that the laser will hit points A and B one second apart. Correct? If we imagine the laser beam looking like a Star Trek phaser, they will be the same length.

When Receiver is moving at 0.5c relative to Sender, then Sender will observe it hitting points A and B 0.87 seconds apart. Thus, the length of the rod has shortened due to length contraction. Sender still measures light moving at c.

What about the Receiver? The rod is stationary relative to Receiver, so he still measures it as 186,000 miles long. It will still take one second for it to go from point A to point B. Only now when it reaches point B, the "tail end" of the laser beam will have passed point A 0.13 seconds ago as observed by Receiver.

Am I doing this right?

 

[nathan]

To me that seems to go a long way towards answering the question in the OP. If my device reads it a 0.87 seconds on, then it will read it as 0.87 seconds off, correct?

The pulses of light could be considered a tick-tock of a clock, right? If so, then what the moving Receiver is observing is that the Sender's clock has slowed down.

Now, here's where the mind-**** happens for me. While stationary relative to Sender, Receiver puts a rod behind him that is one light-second long. That means they both agree it is 186,000 miles long from A (Receiver's ass) to B (end of the rod). Correct?

So, while still stationary relative to one another, both will agree that the laser will hit points A and B one second apart. Correct? If we imagine the laser beam looking like a Star Trek phaser, they will be the same length.

When Receiver is moving at 0.5c relative to Sender, then Sender will observe it hitting points A and B 0.87 seconds apart. Thus, the length of the rod has shortened due to length contraction. Sender still measures light moving at c.

All ok so far.

What about the Receiver? The rod is stationary relative to Receiver, so he still measures it as 186,000 miles long. It will still take one second for it to go from point A to point B. Only now when it reaches point B, the "tail end" of the laser beam will have passed point A 0.13 seconds ago as observed by Receiver.

What do you mean .13 seconds *ago*? That statement is implicitly assuming a Galilean universal time coordinate. One observer will consider it takes the light pulse .87 seconds to traverse the rod, the other will consider it takes 1 second. (but as you say, they both measure the same c, because the former observer also thinks the rod is shorter).

You can't directly compare times in different inertial frames as you're trying to do -- because you get paradoxes. For instance if we take your '.13 seconds ago' at face value, it implies time travel.

Part of the confusion going on here is I think using 'length' to mean 'interval of time', and to think that the constancy of 'c' in all inertial frames implies some kind of universal measure of distance using light beams. It doesn't.