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

1. Oct 18, 2014

### 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.

2. Oct 18, 2014

### 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.

3. Oct 18, 2014

### 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?

4. Oct 18, 2014

### ghwellsjr

Yes, and they would see it for one second.
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?

5. Oct 18, 2014

### Staff: Mentor

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.

6. Oct 18, 2014

### Simon Bridge

(1)
A 186000ml long beam travelling through space.

(2)
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)
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)
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?

7. Oct 19, 2014

### 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

8. Oct 19, 2014

### 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 :)

9. Oct 21, 2014

### 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?

10. Oct 22, 2014

### PAllen

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).

11. Oct 22, 2014

### 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.

12. Oct 22, 2014

### A.T.

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

13. Oct 22, 2014

### PAllen

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.]

Last edited: Oct 22, 2014
14. Oct 22, 2014

### A.T.

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

15. Oct 22, 2014

### PAllen

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.

16. Oct 22, 2014

### A.T.

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

Last edited: Oct 22, 2014
17. Oct 22, 2014

### Simon Bridge

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.

18. Oct 23, 2014

### 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.

19. Oct 23, 2014

### Simon Bridge

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.

20. Oct 23, 2014

### PAllen

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. ]

Last edited: Oct 23, 2014