Not a scientist, just wondering -- Shining a laser pointer into space

[tetra]

Thought Experiment 1:
I shoot a laser into space for one second. Just before I let off the trigger, the face of the beam is ~186,000 miles away. What happens to the beam when I let off? Does it instantly cease to exist? Does the tail compress into the stopped head and go poof :)? Does that 186,000 mile long beam keep going until it hits something? Or, something completely different?

Thought Experiment 2:
I devise a perfectly mirrored containment vessel such that when I shine a laser into the container, all the bounced beams intersect at the exact same point inside. What happens at that point? In other words, how do the beams interact, or do they at all, or is any interaction maybe related to and dependent on the frequency of the beam?

Thanks in advance for teaching me.

 

[Matterwave]

1) Light is not a continuous object like many of the things we are familiar with. There will be a 186,000 mile long "beam" that will move in space, and eventually spread out (even lasers diffract a little) to encompass a wider and wider area. If you were from far away, you would see a 1 second light pulse, pass by you (in other words, you would see the laser light be a pulse that is 1 second long).

2) Light will interfere with each other, certainly; however, when you have a lot of light bouncing around in there, the interference will tend to average out and you will be left with what is essentially just a cavity full of light.

 

[tetra]

1.) if the laser was aimed at an observer 186,000 away, what would they see? if a one second pulse, would they see it one second after firing it? in other words, after I stopped firing it.

2.) assume all beams converge through a single point. aside from any interference that may occur, would there be an additive effect of the power of the beam at that focal point?

 

[ghwellsjr]

1.) if the laser was aimed at an observer 186,000 away, what would they see? if a one second pulse, would they see it one second after firing it? in other words, after I stopped firing it.

Yes, and they would see it for one second.

2.) assume all beams converge through a single point. aside from any interference that may occur, would there be an additive effect of the power of the beam at that focal point?

Your perfectly mirrored containment vessel must have a hole in it in order to shine the laser beam inside, correct? What's to keep the light from also coming out of that same hole? If your vessel were 1/2 light second in size, then I suppose you could shine the laser inside for one second and just when the first reflected part of the beam reached the hole you could shut it and keep all the light bouncing around between two points, in which case you wouldn't really need a vessel, just two mirrors, correct?

 

[Nugatory]

1) Assuming that the observer and the laser are at rest relative to one another (if they aren't, the problem is underspecified and we need more information)... The observer will see a one-second pulse starting one second after the laser was turned on. If the laser was turned on at noon and turned off at at 12:00:01 to generate the one-second pulse, the observer would be illuminated by laser light starting at 12:00:01 and ending at 12:00:02.

2) yes. A more prosaic example would be the way that a magnifying glass can be used to focus sunlight on a point - all the rays of light passing through the glass come together at one point, and the intensity adds. On a sunny day you can burn holes in a piece of paper with a three-inch magnifying glass that way.

So far, none of this has anything to do with relativity - it's pure classical physics.

 

[Simon Bridge]

(1)

I shoot a laser into space for one second. Just before I let off the trigger, the face of the beam is ~186,000 miles away. What happens to the beam when I let off? Does it instantly cease to exist? Does the tail compress into the stopped head and go poof :)? Does that 186,000 mile long beam keep going until it hits something? Or, something completely different?

A 186000ml long beam traveling through space.

(2)

I devise a perfectly mirrored containment vessel such that when I shine a laser into the container, all the bounced beams intersect at the exact same point inside. What happens at that point? In other words, how do the beams interact, or do they at all, or is any interaction maybe related to and dependent on the frequency of the beam?

Light does interfere with other light and the details tend to depend on the cavity.
Two equal intensity laser beams trained on the same spot can be expected to produce about twice the intensity there than either one alone.
Lots of random reflections would quickly produce a cavity full of light, with the details of the light depending on the shape of the cavity and the wavelength of the light.
However - you are imagining a lot of mirrors arranged so that there are a lot of beams crossing the same place right?
What happens there depends on the details of the arrangement of the mirrors - if you are careful you can, in principle, get from zero to number-of-crossing-beams-squared times the single-beam intensity due to interference. Usually nobody is that careful and you end up with an average Nx intensity... much like when you switch on several light bulbs.

(2.1)

if the laser was aimed at an observer 186,000 away, what would they see? if a one second pulse, would they see it one second after firing it? in other words, after I stopped firing it.

Using a normal stopwatch - the observer would get a one second flash of light, and the pulse would start one second after you started firing or exactly when you stopped firing. If they waved at you when they received the pulse, you'd see the wave 1 second after you finished firing. You've experienced the same thing with sound waves.

(3.1)

assume all beams converge through a single point. aside from any interference that may occur, would there be an additive effect of the power of the beam at that focal point?

The "interference" is the additive effect of one beam of light on another.
That is how light adds up.

Where is all this coming from?

 

[tetra]

Where is all this coming from?[/QUOTE]

just thinking about lasers and how light behaves. thank you all for your thoughtful and well articulated answers. sorry it was in the wrong room though...

thanks

 

[Simon Bridge]

Random musings hit all of us from time-to-time.
If I knew the context I may be able to help you better ... but enjoy anyway.
Cheers :)

 

[Thecla]

Simon , You say it is a 186,000 mile long light beam traveling through space.
How can that be?
The whole beam is moving at the speed of light, so wouldn't that beam Lorentz contract to zero length?

 

[PAllen]

Simon , You say it is a 186,000 mile long light beam traveling through space.
How can that be?
The whole beam is moving at the speed of light, so wouldn't that beam Lorentz contract to zero length?

The beam creator and beam receiver are not in relative motion, so they see the beam the same length. The beam was generated as 1 light second long in their mutual rest frame, so anyone at rest with respect to the creator will see the beam this long. You 'see' how long a beam is by absorbing a small part of it and seeing how long you get signal for. If you detect the beam for 1 second, you know it was 1 light second long > 186,000 miles.

Length contraction applies between observers in relative motion. Note, there is no such thing as a frame for light - see our FAQ on this at the top of this forum.

The way a differently moving observer would see such a beam of light is governed not by length contraction but by Doppler. Someone moving toward the emitter would see the beam shorter by the Doppler factor, and someone moving away would see it longer by the Doppler factor (because the beam has the same number of wave crests for every observer).

 

[Thecla]

I was thinking of looking at the beam sideways. The way you describe it the beam enters your eye directly,straight on.
How can you look at it sideways?
Suppose the beam is traveling through a fog and we can see the beam the way we see a searchlight beam rotating on top of a tall building on a foggy night.This second observer can see the light scattered at 90 degrees toward him perpendicular to the beam. So a fraction of the photons are scattered toward the second observer and the rest of the photons will travel straight on to the intended observer
What the second observer would see if he is a few hundred thousand mile away would be a 186,000 mile long beam moving through the fog at the speed of light.

 

[A.T.]

What the second observer would see if he is a few hundred thousand mile away would be a 186,000 mile long beam moving through the fog at the speed of light.

The visual size of the beam would continuously change. It would be longer than 186,000 miles when approaching, and less when receding.

See Figure 9 here:
http://www.spacetimetravel.org/tompkins/node4.html

 

[PAllen]

The visual size of the beam would continuously change. It would be longer than 186,000 miles when approaching, and less when receding.

See Figure 9 here:
http://www.spacetimetravel.org/tompkins/node4.html

Does that apply here? I think no[edit: yes]. That is all about relating rest frame description of 'something' to appearance when moving rapidly relative to observer. Here, there is no such thing as rest frame description, and both emitter and receiver and fog have no relative motion.

[Edit: Ok, I see the argument applies irrespective whether there is any rest length.]

 

[A.T.]

That is all about relating rest frame description of 'something' to appearance...

No, it's about relating the observers frame description to appearance.

 

[PAllen]

No, it's about relating the observers frame description to appearance.

Yes, I cross posted a correction. Of course, if you factor out light delays, you conclude the beam is 1 light second long at all times.

 

[A.T.]

Yes, I cross posted a correction. Of course, if you factor out light delays, you conclude the beam is 1 light second long at all times.

That's correct, but Thecla seems to talk about the visual appearance in post #11.

 

[Simon Bridge]

I was thinking of looking at the beam sideways.
How can you look at it sideways?

Simpler than the "fog" perhaps: only 1D geometry and does not rely on "visual appearance".
You set up an array of detectors (for simplicity: all stationary with respect to each other) and record which ones go off.
Any observer stationary wrt the detector array gets the proper length of the beam as measured by that array.

Talking about what someone sees with their eyes is a little trickier ... you have to factor all kinds of other stuff in.

 

[Thecla]

After some thought I realize that a 186,000 mile beam of light is not a" thing" it is an aggregate of light particles moving in the same direction consisting of mostly empty space. It can't appear to be contracted.
Consider this analogy:
You have 187 rocket ships sitting on a very long runway. Each rocket is 1000 miles directly behind the other. They all take off at the same time and all travel at 1000 miles per hour. To an observer on the ground it would appear to be a caravan 186,000 miles long. If they all were programed to accelerate
to 99.9% of the speed of light at the same time,the caravan, moving at nearly the speed of light, would still be 186,000 miles long to the observer on earth. The rockets themselves would contract to an earthly observer, but the space separating the rockets would not contract.

 

[Simon Bridge]

After some thought I realize that a 186,000 mile beam of light is not a" thing" it is an aggregate of light particles moving in the same direction consisting of mostly empty space. It can't appear to be contracted.

That's not why.

If you put a row of stones 1 mile long along the ground, anyone moving with respect to the ground will measure less than a mile of stones. The distance between the stones measures shorter, as well as the width of the stones.

It is the same with any length - it's a geometric effect: this line being an accumulation of stones does not matter.

The trick with keeping these things straight is to make the reference frame explicit.

With the line of spaceships example, you forgot to specify which reference frame the separation is measured in.

So you have space-craft separated by 10000miles in the frame of some observer A.
That line of spacecraft accelerate to close to the speed of light (btw: length contraction happens at all speeds)
OK - so when they accelerate, are they maintaining their separation in their own reference frame or in A's reference frame?
The spacecraft have their own reference frame S.
If they maintain their separation in A's frame, what separation is measured in S's frame?
If they maintain their separation in S's frame, then what happens to the separation in A's frame?

The situation with the light is that a train of light-waves (SR is not QM) are leaving the observer A with a particular wavelength in A's reference frame.
x wavelengths leave so the length of the train is $x\lambda$ in A's frame. Observer B, moving at some speed wrt A, will see a shorter train of waves ... since the speed is the same, and the number of wavelengths is the same, then the wavelength must be shorter.

 

[PAllen]

The situation with the light is that a train of light-waves (SR is not QM) are leaving the observer A with a particular wavelength in A's reference frame.
x wavelengths leave so the length of the train is $x\lambda$ in A's frame. Observer B, moving at some speed wrt A, will see a shorter train of waves ... since the speed is the same, and the number of wavelengths is the same, then the wavelength must be shorter.

With light, the determining factor is Doppler. Unlike with length contraction, direction of relative motion matters. If B is moving away from A, the wavelength will increase; towards it will decrease; and neither factor will be $\gamma$ of the relative motion. With length contraction, the length of sequence of canon balls launched by A will be considered less by B for either direction of (colinear) motion, and the factor will be $\gamma$ .

[edit: Actually a sequence of launched canon balls is not a simple length contraction case either. If B is moving away from A, this sequence will get larger; if B is moving towards A, smaller. These factors will be the ratio of gamma for the canon ball speed (v, relative to A) and gamma for the relativistic addition/subraction of u and v, where u is B's speed relative to A. ]

 

[Simon Bridge]

Yeah - everything depends on who is doing the measuring - which is why all the earlier stuff about timing: the idea was to isolate the particular physics under investigation. The doppler shift is often taught as including a length contraction ... so a sort-of combination of effects but IRL I don't think it makes sense to separate them out.

But I like the "sequence of balls" thing. The key - as always with this stuff - is to get really explicit about what you are talking about. Really really spell it out using the language of relativity ... like talk about reference frames etc.

So: let's see if I can produce a better exercise...

A sends n balls per time t_A at speed u_A towards B.
Thus A deduces that the spatial separation of the balls is $s_A=u_At_A/n$

B counts n balls passing in time t_B at speed u_B
Thus B deduces the separation of the balls to be $s_B=u_Bt_b/n$

(B and A have relative velocity v. Should be able to express v in terms of the other velocities.)

A and B get together later and compare notes - what do they find?
Will probably help to separate the cases that B and A approaching each other from B and separating.

 

[PAllen]

Well, still a little problem...

What if v = u_A, i.e. u_B=0 ? s_B should be maximum in this case not 0! (! is surprise not factorial ;)).

 

[Simon Bridge]

You'd be saying that B counts n balls passing in an infinite amount of time doing speed zero ... how does that work out?

But I'd like to see OP try some of the math here ... it is hard to do relativity by analogy and intuition, and the math is really what OP needs to get used to in order to gain understanding here.

 

[PAllen]

You'd be saying that B counts n balls passing in an infinite amount of time doing speed zero ... how does that work out?

It doesn't. The set up is not quite right. I'm writing up the correct set up and answer. You can try too.

 

[PAllen]

I like to write this in terms of u of balls and v observer B. Then, without loss of generality, take n=1, and time between balls as 1. Then the equations for the two balls per A are:

x= ut and x = ut - u, with separation of u.

Then, just Lorentz transform these with a boost of v, and compute the delta in terms of x'. I'll give the answer for checking, using Simon's notation for the two distances:

s_B/s_A = (√(1-v²/c²)) / (1 - uv/c²)

which is seen to give γ(v) when u=v as expected.

 

[Simon Bridge]

It doesn't. The set up is not quite right.

I suspect we may be using a different definition for "work" and "quite right"...

I set it up so both A and B time the same number of balls, and use that as their "way to calculate the separation" ... so they have each agreed on a definition of "separation" in those terms. They are both doing the calculation entirely inside their own reference frames.

The exercise that I had in mind writing was to use special relativity to predict the outcome of the two measurements.
You may have another exercise in mind? Or it may be that I need to be more precise about some of the description?

You seem to be saying that $u_Bt_B$ dosn't converge?
Well, A and B are making naive measurements and comparing them.
The next step is to improve on the concept, should the naive idea fail to hold up under scrutiny.
It's a gambit.

The fact you saw it right away testifies to your experience - but students often need to learn the hard way.
Of course, that was off the top of my head - if you can come up with a better one then I'm all ears :)

I'm writing up the correct set up and answer.

... I love the way considering someone's question can set off these explorations...

You can try too.

Maybe later ... :)

Then, just Lorentz transform these with a boost of v, and compute the delta in terms of x'. I'll give the answer for checking, using Simon's notation for the two distances:

... now try to figure how to explain that at OPs level.

But I'd have started the same way - put $u=u_A$ and then expressed $u_B$ in terms of v and u.
Fire just two balls ... etc. the key is usually to realize that B changes position in A's frame.

 

[PAllen]

I like to write this in terms of u of balls and v observer B. Then, without loss of generality, take n=1, and time between balls as 1. Then the equations for the two balls per A are:

x= ut and x = ut - u, with separation of u.

Then, just Lorentz transform these with a boost of v, and compute the delta in terms of x'. I'll give the answer for checking, using Simon's notation for the two distances:

s_B/s_A = (√(1-v²/c²)) / (1 - uv/c²)

which is seen to give γ(v) when u=v as expected.

Also, of note, if you take u=c (crests of light, pulses of light), then you get the relativistic Doppler formula, as might be expected:

√((1+v/c)/(1-v/c))

 

[PAllen]

I suspect we may be using a different definition for "work" and "quite right"...

I set it up so both A and B time the same number of balls, and use that as their "way to calculate the separation" ... so they have each agreed on a definition of "separation" in those terms. They are both doing the calculation entirely inside their own reference frames.

The exercise that I had in mind writing was to use special relativity to predict the outcome of the two measurements.
You may have another exercise in mind? Or it may be that I need to be more precise about some of the description?

You seem to be saying that $u_Bt_B$ dosn't converge?
Well, A and B are making naive measurements and comparing them.
The next step is to improve on the concept, should the naive idea fail to hold up under scrutiny.
It's a gambit.

Yes, my wording was too strong, and the discontinuity is removable, which can salvage this method. Just for curiosity, I show doing it this way as well.

It is still easiest to work using u (velocity of balls per A), and v (velocity of B per A), as will be seen. Of course, when you want it:

u_B = (u-v)/(1-uv/c²) = velocity of balls per B

Then the tricky part is getting t_B. For all cases except when u=v, you can look at all quantities in A frame, and note that the separation/closing speed between B and the balls is simply (u-v), where v ∈ (-c,c). Being a closing speed, not a relative speed, there is no problem that this can result in a value > c. Then the time for B to get from one ball to another is, per A is simply:

u/(u-v)

[Of course, you see the problem, which we ignore for the moment, with the case of u=v.] Then, the time A sees elapse on B's clock for this period is:

(1/γ(v))* u/(u-v), purely by time dilation.

Then the distance per B becomes (given formula for u_B above):

u_B *[(1/γ(v))* u/(u-v) ] = u(√(1-v²/c²)) / (1 - uv/c2)

exactly as before, with the problematic u-v being trivially removable in the limit.