# Lateral momentum of light

1. Aug 11, 2011

### JVNY

Can light have lateral momentum, and if so as a particle or wave or both? If a person on a train throws a ball up, then from the train's frame of reference the ball goes directly up then down (ignoring air resistance). From the ground's frame of reference, the ball goes at an angle upward and forward, then downward and forward. This is because relative to the ground, the ball has forward (lateral) momentum.

Next, consider a light clock oriented perpendicular to the forward direction of a rocket traveling close to c. From the rocket's frame of reference, the light in the clock goes directly up then down (after reflecting off the mirror at the top of the clock). From the ground's frame of reference, the light goes just like the ball, upward and forward, then downward and forward.

Since we should treat relativity events as "what you see is what you get" (rather than trusting our instincts), does this mean that the light in the clock has lateral momentum, just like the ball? Can light have this lateral momentum as a particle? As a wave?

2. Aug 11, 2011

### ghwellsjr

If the person on the train left the ball sitting on a table, would you say that momentum was the reason it was moving forward in the ground's frame of reference?
They aren't the same because the ball is accelerating and doesn't travel in a straight line when viewed from the ground whereas the light is not accelerating. It's not a good comparison.
Actually, in Special Relativity, it's the other way around: we trust our "instincts" to define what we can't "see".

Light does have momentum, but not in the way you are understanding it.

3. Aug 11, 2011

### bcrowell

Staff Emeritus
You seem to be confusing two different things. In the light clock example, the light ray's momentum is parallel to its direction of propagation, but has a component transverse to the direction of motion of the clock.

A light ray's momentum is always parallel to its direction of propagation. This is true regardless of specific models of light such as ray, particle, or wave.

You may be getting confused by visualizing a light wave as if the vibration of the wave moves from side to side. It doesn't. What points sideways is a field vector.

4. Aug 11, 2011

### JVNY

I actually think of light in this case as a particle, because it is easier to analogize to the ball. The question might reduce to: in which direction does the light propagate from the frame of reference of the ground? It appears to propagate at a diagonal from the ground's reference, but perpendicular from the rocket's reference. It is your "transverse" feature that I am wondering about.

Change the example to light emitted perpendicularly from the rocket (no light clock to get in the way). If the light hits something in the same frame of reference as the ground, will the light affect the object the same as would a light shone at the object from a fixed position on the ground at the same angle that a ground observer sees the rocket's light propagating?

5. Aug 11, 2011

### Mentz114

6. Aug 11, 2011

### JVNY

This assumes that light propagates a certain way, but there are many types of light that propagate in different ways, and the light clock must work the same for all of them (a diagonal from the ground observer's reference).

Consider instead a laser, as discussed earlier in the thread you refer to. If the rocket has a laser light clock, the result is the same as I describe earlier. The tightly focused beam will go straight up then down from the rocket's frame of reference, but at an angle from the ground reference. There is no wave propagation as in your movie, just a linear propagation.

Imagine the laser being a tube of certain length bolted onto the rocket, perpendicularly. The ground observer would not consider the laser light to be coming out of the laser at an angle; the light could come out only parallel to the laser body, which is perpendicular to the rocket. Something else causes the diagonal propagation. The cause seems to be the same as for the ball. The diagonal comes from combining the two vectors, the forward vector of the rocket's travel and the perpendicular vector of the light's emission.

7. Aug 11, 2011

### Mentz114

The explanation in terms of momentum is true, of course, but it's the kind of explanation that doesn't explain much. We can't carry the billiard ball analogy too far for massless propagation.

There's probably a way to model this with optical wave packets or something similar that gives a more satisfactory explanation.

8. Aug 11, 2011

### Bill_K

No, there is only one type of light and it all propagates only one way.
Laser light propagates as waves too.
Yes he would. The effect is called aberration. During the time the light takes to get from the bottom of the tube to the top, he sees the tube move sideways. He must also see the light beam move sideways, or it would not escape the tube.

JVNY, the fact you're missing is that a propagation vector is different from a position vector. You're assuming that they're the same. Under a Lorentz transformation a propagation vector does not transform the same way. Your "bolted perpendicular tube" is a spacelike vector. It has only one component, say the y-component. In four dimensions its components written out are (0, 0, y, 0). Under a Lorentz boost in the x-direction, it remains exactly the same: (0, 0, y, 0).

But the propagation vector for light is a null vector. In addition to the space components it also has a time component. Written out: (k, 0, k, 0). That's because it represents something that is moving. Under a Lorentz boost in the x-direction it will become (γk, γvk/c, k, 0). Note that the propagation vector now has both an x-component and a y-component. That means it has changed spatial direction. In the new reference frame the light propagates at an angle.

9. Aug 11, 2011

### JVNY

Although the explanation is well beyond me, I think it gets to the same place as bcrowell -- the light actually propagates at an angle, as you put it, or with a transverse component as he puts it. This is what I was wondering about. So to return to an earlier question in the string: if the light strikes an object in the ground's frame of reference, would it affect the object the same way as a light shone from the ground at the same angle that the ground observer sees the light propagating from the rocket? Thanks.

10. Aug 11, 2011

### ghwellsjr

Symmetrical situations produce equal results, so:

If the rocket is firing the light perpendicular to its direction of travel (assumed to be parallel to the earth's surface) so that it hits something on the ground at an angle, then to get the same result on the spacecraft, the earth should fire an identical light directly up so that it hits an object on the moving spacecraft with the same angle. If you aim the firing source in a different angle, the result will not be the same.

11. Aug 11, 2011

### JVNY

I was suggesting that the target is fixed with respect to the ground (say a sign), and light fired from the rocket strikes the target at the diagonal angle (say 45 degrees, because the rocket is traveling forward and emits the light downward toward the ground). Separately, a light shone from the ground upward at the same angle (45 degrees) strikes the sign. Ignoring gravitational effects, does the light from the rocket have the same effect when it strikes the sign as the light from the ground? I expect yes, given that you say that symmetrical situations produce symmetrical results. Thanks in advance.

12. Aug 13, 2011

### JVNY

To restate and elaborate:

What fundamental law of physics causes light to propagate diagonally from an inertially moving emitter?

Consider a basic light clock with an emitter attached perpendicularly to a rocket. The light clock emits a narrow burst of laser light which travels to a mirror, reflects, and then returns to the emitter. In the rocket's frame of reference, the light travels at all times perpendicularly to the rocket and parallel to the emitter. A ground observer would see and agree that this is the case in the rocket's frame of reference. It is consistent with how light propagates from an emitter on the ground, and the laws of physics are the same in both inertial systems.

However, the ground observer would also see that in his frame of reference the light in the clock propagates diagonally, at an angle between the rocket and the emitter. What causes the light to propagate in this different direction?

One possible answer is momentum. A ball emitted from the rocket at a constant velocity would act the same as the light, moving perpendicularly in the rocket's frame of reference, but diagonally in the ground's frame of reference. The combination of the two momentum vectors (forward vector of rocket's travel and perpendicular vector of emission) determines the ball's vector in the ground's frame of reference. This is a light as particle explanation.

There might also be a light as wave explanation. If an "ether" surrounded the rocket, then the "ether" would sweep the propagating wave along with it, causing the wave to move forward as it moves away from the rocket, generating the diagonal propagation. The wave theory of light no longer requires an "ether," so presumably there a wave explanation of the diagonal that does not rely on "ether."

There may be a more complicated explanation. Consider an accelerating rocket with a light emitter. As the rocket accelerates, each new pulse of light from the emitter propagates at a more acute angle to the rocket in the ground observer's frame of reference (90 degrees when the rocket is at rest, then 89 degrees at a certain velocity, then 88 degrees at a higher velocity, etc.). This is like the path of light entering the pinhole of Einstein's accelerating elevator (although in the opposite direction -- light in the elevator appears to bend farther away from the direction the elevator is moving as it accelerates, but light from the emitter propagates closer to the direction the rocket travels as it accelerates). Does the rocket affect space time? Gravity causes space time to curve; acceleration is equivalent to gravity; therefore acceleration causes space time to curve? So the rocket is moving along space time that it had curved in the course of its acceleration, which causes the emitted light to propagate at a diagonal?

13. Aug 13, 2011

### Staff: Mentor

Maxwell's equations. Remember that Maxwell's equations are invariant under the Lorentz transform. So if you have a solution to Maxwell's equations in one frame then it is also a solution to Maxwell's equations when Lorentz transformed into another frame.

14. Aug 13, 2011

### ghwellsjr

Here's what I said:
This is not the same as your situation so I'm not saying that the light from the rocket has the same effect on the sign as when a light from the ground is shining from the same angle. For one thing, the light from the rocket would only be momentary while from the ground it would be constant. Doesn't this show that the two situations are not symmetrical?

15. Aug 13, 2011

### ghwellsjr

In my first post to you I said:
You are reading more into relativity than is there.

Consider the light clock on the rocket. Turn off the light. Think about a point midway between the two mirrors. There's nothing there, right? Ask yourself how that point "propapgates" when viewed from the ground reference. Do you explain this as momentum?

16. Aug 13, 2011

### JVNY

I was thinking about the laser emitting a pulse of light that you can follow, kind of like the small laser dot in the blue background in the picture in this link: http://www.symmetrymagazine.org/breaking/2009/04/21/worlds-first-hard-x-ray-laser-switches-on/

Instead you could leave the laser on, so it continues to produce a beam, without any reflector. Say the rocket is traveling 100 feet off the ground. The laser is pointed 90 degrees off the side of the rocket. There is a 100 foot tower with a laser on it (the "laser tower" below). There is another 100 foot tower with a target on it between the rocket and the laser tower, at a distance from them (the "target tower" below). As the rocket travels it fires its laser, at 90 degrees to the rocket. However, from a ground observer's frame of reference the laser beam travels at an angle and strikes the target. The laser on the other tower is angled so that when it fires, its laser beam goes at the same angle and strikes the target. Here is the view that someone has on the ground, looking up:

Rocket >
\
\
\
X Target tower
/
/
/
Laser tower

17. Aug 13, 2011

### JVNY

Sorry, the diagram did not post as I had done it. I'll try again. Ignore the periods that I have to add to keep everything aligned.

Rocket >
........\
..........\
............\
..............\
................X Target tower
............../
............/
........../
......../
Laser tower