# How does light slide sideways?

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1. Mar 19, 2015

### Rising Eagle

The laser ranging of the moon doesn't make sense. The moon has a special mirror on it that doesn't reflect incoming light along the same angle as the angle of incidence only on the opposite side of the perpendicular, but instead reflects light back exactly on the same path as the light's approach. A telescope is aimed from earth directly to the moon's reflector. A laser is aimed through the telescope, the laser beam hits the reflector and bounces back exactly to the same telescope for detection and time of flight measurement.

Consider the fact that the moon and earth are in parallel motion in space at 370 kilometers per second. The problem is that in the 2.5 seconds trip of the laser beam to the moon, the moon has moved and the light should miss the reflector completely. Even worse, in the 5 seconds of the round trip of the laser beam, the earth has moved quite a bit through space and the light should miss the telescope which has moved along with it.

The only way for this to work is if light slides sideways to parallel the motion of the moon and earth which keeps the light beam always at the same angle of approach to the reflector, no matter what the path of the laser beam through space actually is. This sliding simulates the conditions as if the earth and moon were both hanging motionless in space when the measurement is done. Arguing that the telescope is aimed forward to intersect the moon in the same way a quarterback aims for his receiver where he thinks he will be when the football arrives doesn't work because the reflector still returns the light back to where the telescope was when the laser beam was first sent, and not to where the telescope has moved to.

Does light move sideways? By what mechanism? SR doesn't really explain it. Do its postulates even predict it? How else could laser ranging be made to work?

2. Mar 19, 2015

### Staff: Mentor

Motion is frame-dependent, and changing frames cannot affect any physics. So you can just analyze the experiment in a frame that is also moving at 370 km/sec in the same direction as the earth and the moon. Then the only motion involved is the residual relative motion of the earth and the moon.

(Btw, this 370 km/sec figure is arbitrary anyway. There is no absolute motion, so I could just as easily say the earth and moon are moving at 99.99999% of the speed of light relative to a cosmic ray passing by. That doesn't change the physics at all.)

No, all that means is that the laser should point at where the moon is going to be in 1.25 seconds, instead of where it is right now. (Btw, it's 1.25 seconds one way and 2.5 seconds round trip, not 2.5 seconds one way and 5 seconds round trip.) The light doesn't have to slide sideways; it just has to be aimed right. It's no different than shooting at a moving target: you have to lead the target in order to hit it.

No, all that means is that the beam has to be wide enough to encompass the relative motion of earth and moon during the time of flight of the light. In fact, as the article linked below states, the beam is 1.8 km wide when it reaches the moon, and the reflected beam expands similarly on its way back to earth.

http://physics.ucsd.edu/~tmurphy/apollo/basics.html

Last edited: Mar 19, 2015
3. Mar 19, 2015

### FactChecker

In any non-accelerating inertial reference frame, the behavior of light observed in that reference frame is as though the reference frame is stationary. That is the basis of Special Relativity. There is no "ether" in space that would make the light beam you are talking about fall behind that straight line from the Earth to the moon.

4. Mar 19, 2015

### Rising Eagle

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My bad about the time of flight. I remembered it wrong. As to your point about leading, it can't work. Both the moon and the earth are each others' target and the automated reflector on the moon does not lead the earth when it returns the light beam. I mention that in the paragraph of my original post.

5. Mar 19, 2015

### DaveC426913

You're still missing the point. Earth and Moon are stationary. They are not moving at all. (You are always at rest in your own frame of reference). It is the rest of the universe that is moving.

As Peer says:
Consider: if you throw a tennis ball at a fellow passenger aboard a plane moving at 500mph, does the tennis ball have to "slide sideways" at 500mph?

6. Mar 19, 2015

### Staff: Mentor

It does if both the earth and the moon are motionless, since no "leading" is required then. They aren't exactly motionless with respect to each other, so there is a small effect due to that relative motion during the time of flight of the light; but as I noted in my post, the spreading of the beam compensates for that--the beam is not an infinitely thin line, so it doesn't all travel in one single direction and is not all reflected in one single direction.

The real question is, given that my analysis is correct in a frame in which the center of mass of the earth-moon system is motionless (so the only motion is the relative motion of earth and moon in the center of mass frame), how would SR analyze the experiment in a frame in which that center of mass is moving at 370 km/sec? I'll leave you to ponder that one, but here's a hint: Lorentz transformations affect angles as well as lengths and times.

7. Mar 19, 2015

### Staff: Mentor

Not completely; they do have some relative motion even in the center of mass frame of the Earth-Moon system. But it's a lot smaller than 370 km/sec, and the spreading of the beam during flight is more than enough to compensate for the relative motion during flight.

8. Mar 19, 2015

### Rising Eagle

Ok, so light is sliding sideways. Is there a mechanism for the sideways slide or just supposition? Is the light coming from the laser connected to the laser and therefore sliding sideways with the laser independent of the target? Also, is the reflected light from the target sliding sideways with the reflector independent of the laser?

9. Mar 19, 2015

### DaveC426913

Don't want to confuse him with details.

10. Mar 19, 2015

### DaveC426913

11. Mar 19, 2015

### Staff: Mentor

That's not what he said.

Look up Lorentz transformations and do the math; start with a frame in which the Earth and Moon are at rest (at least to a good enough approximation), and then transform into a frame in which they are moving at 370 km/sec, and see what you get. It won't be a description of light "sliding sideways".

12. Mar 19, 2015

### Rising Eagle

Forgive my confusion, but consider a second earth-moon pair. Say pair one is moving east toward an origin at nonrelativistic speed with earth-moon oriented north south. Say pair two is moving west toward an origin with the same orientation. Each pair shoots laser beams between the earth and moon and back again. What will the earth and moon in one pair see when observing the other pair and vice verse? Wouldn't each pair see themselves as stationary and the other as in motion with a laser beam moving sideways at the same rate as the earth and moon that are exchanging them? It tells me that there is some kind of "momentum" the light photons have to carry on the same speed of the sliding motion as the launching platform. Another observer sitting high above the plane with the two pairs will see two pairs moving towards each other and light beams within each pair moving sideways. Clearly I don't fully comprehend the nuances of relativity, because I don't see the mechanism. The lasers are not pointed at an angle, yet the light travels at an angle. The angle of attack is along the line of sight between the earth and moon in each case and not in the direction of propagation. Please explain. Is the scenario flawed?

BTW, I never mentioned absolute reference frame in this thread. That was another thread to describe a scenario with no second observer. Please don't assume that is what I am assuming here. I asked how is it that light slides sideways and what mechanism. So far in this thread it was said that SR treats the frame as if it is stationary. Ok, In the two pair scenario what is happening? Please disregard relative motion between earth and moon and their individual rotations; those motions are not relevant to my original post.

13. Mar 19, 2015

### Staff: Mentor

If the two observers that are stationary with respect to each other see the beam moving straight and the two observers that are moving relative to them see the beam - and the two observers - moving "sideways", is the beam "really" moving "sideways" or not?

Have you read about the "light clock" thought experiment? You're describing it:
http://www.emc2-explained.info/The-Light-Clock/#.VQuaQI7F-So

Last edited: Mar 20, 2015
14. Mar 20, 2015

### Rising Eagle

I say yes. The motion of the emitter is impressed on the photons. How, I don't know. Consider the observer above the plane looking down.

No. I just now made it up to focus attention on my real question. I'll look over your thought experiment. Thx for the link.

15. Mar 20, 2015

### A.T.

What one has to consider is not the relative motion of the centers, but of the surfaces. But even without the relative motion, with both tidally locked, their common rest frame is rotating, so you have Coriolis effects diverting the beam.

Last edited: Mar 20, 2015
16. Mar 20, 2015

### jartsa

How about this: Inside the Laser device sideways sliding light waves are amplified as they reflect back and forth between two moving mirrors.

17. Mar 20, 2015

### bahamagreen

The moon is orbiting 1.023 km/s - mean orbital velocity with respect to the earth-moon barycenter
The moon is 1.25 light seconds away

In the 1.25 seconds the beam travels from earth to moon:
the moon advances 1.28km in orbit
the earth's surface at Apache Point Observatory rotates 0.5km

In the 1.25 seconds the beam returns from moon to earth:
the moon advances another 1.28km in orbit
the earth's surface at Apache Point Observatory rotates another 0.5km

Using line of sight aiming with 1.25 second latency:
center of beam arrives at moon 1.28km off target reflector
target reflector is 0.38km outside beam radius

I doubt the reflectors are even visible to make a line of sight aim prior to engaging the beam, but with the beam on and the locations of the reflectors known, hunting and sighting the return beam is confirmation that the correct leading is established.

18. Mar 20, 2015

### FactChecker

Only real experiments finally convinced scientists. In the late 1800's, scientists thought that light traveled through a stationary "ether" at a certain speed. They tried to experimentally detect the effect on light of the motion of the Earth through the ether. That never worked. They could never detect an effect of non-accelerating motion on the speed of light. The most famous experiment was the Michelson-Morley experiment.

19. Mar 20, 2015

### Staff: Mentor

You said before that you weren't looking for absolute motion. It is inadvertent, but I think you are. The only motion that is relevant here is that the source and receiver are stationary with respect to each other. An infinite number of other possible observers might say the light is sliding sideways, but none of them need agree on what direction it is "actually" traveling. So when you say "the motion of the emitter" you are inadvertently suggesting it has an absolute state of motion. Different observers do not have to agree on its direction of motion.

Hecht, if you add another observer moving in the opposite direction, both might say the beam is moving sideways, but one will say it is moving to the left while the other says it is moving to the right! Which way is it "really" moving?

It is also worth noting that the "beam" is not a solid object. It isn't required to even be a line, much less travel parallel to itself. If you spin in place while holding a laser pointer (or water hose) the beam will be a spiral. That is in fact the shape seen in the lunar ranging experiment (see above about the coriolis effect).

Last edited: Mar 20, 2015
20. Mar 20, 2015

### DaveC426913

What if you were right? What if remote observer B expects to see light sliding sideways to catch up to a receding Moon?

OK, now what if there were a third observer C, moving East at twice the speed that E/M is moving? From their reference frame, they actually see E/M moving away from them to the West. So, for them, the light beam obviously must slide West to catch up.

How can the light beam slide East and West at the same time?

Last edited: Mar 20, 2015
21. Mar 20, 2015

### Rising Eagle

I agree my thinking might be leading me to inadvertently acknowledge an absolute motion, but by analysis, not by assumption. I observe and I ask. The idea that any observer sees a consistent picture as all others providing that all observers apply a transformation to correct for their points of view is a powerful one. But only applies when the transformed results for all observers agrees with the "bigger (objective?) picture." There is an implied symmetry or invariance going on.

Consider just one earth-moon pair. let the earth be the observer fixed at the origin and the moon be in eastern linear motion along x toward the y axis at some fixed y. At time 0, earth shoots a photon aimed to meet up with the moon's reflector at a certain place and time. The photon hits the reflector and is returned precisely to the nonmoving earth exactly retracing the forward path on its journey back to earth while the moon continues moving. At the time of the reception back on earth the earth and moon are both aligned on the y axis. Let the initial position of the moon and its starting point at time 0 be chosen to assure this alignment. The distances the moon travels before and after it receives the photon are equal.

Now, reverse the observers. The earth is in motion in the other direction. The moon is at the origin and the earth moves west toward the y axis at some fixed -y. The initial relative positions of earth and moon are the same as before and the earth's velocity is equal and opposite. At time 0, the earth shoots a photon at the moon with the exact same aim and timing. The earth is in motion and so the photon slides sideways adding to the angular aim which assures the photon meets the moon's reflector at the same place and time. Remember, same dynamics, different observer. The earth continues to move and by the time the photon returns, the earth is at the y axis and where the photon was reflected to.

The earth is responsible for the aim (and therefore the return angle) which is identical in the two points of view. The moon is responsible for returning the photon back along the same path as the photon came in on. The only way to make these two stories consistent is by having the reflected photon in the first part slide sideways in accord with the moon's velocity. In fact, each part of the story needs light to slide sideways to make it consistent with the dynamics of the other. What is this mechanism? When a photon is launched directly at a target, the velocity of the photon seen by the target and the emitter is not related to the velocity of the emitter. But off angle there is a motion of photon that depends on the motion of the emitter. How is this possible? It is as if the emitter is the primary observer and the target is somehow an observer of a different sort.

22. Mar 20, 2015

### Staff: Mentor

No: the whole point of Special Relativity is that there isn't a "bigger objective picture". Transformations enable the infinite number of observers to come to an agreement about what happened, but the choice of which frame to view the events from is totally arbitrary. I prefer typically viewing events from the local frame that the event is happening in, but that's just a personal preference. For this, that's the emitter's frame.
I'm not sure what else to tell you, but this is getting circular. There doesn't have to be a mechanism for something that is basically an optical illusion. Again, with spraying water from a hose: if you wave the hose back and forth, the water looks like it is zig-zagging, but it isn't: it is always traveling straight away from where it was emitted, in the direction it was emitted. So is the light.

Again, I don't like the term "slide sideways" here. It really doesn't make sense. A photon is essentially a point particle and it has a speed and direction of motion. How does inserting the concept of "slide sideways" impact that? It seems to me that it doesn't unless you consider the illusion that the beam of photons is a solid object that should be moving along its axis. Please be explicit: Let's say on an xy axis, the photon is moving up, along the Y-axis at C. Add "slide sideways" to it. Now what is it doing?
Both (all) observers are always stationary with respect to the beam of light. That's the other postulate of SR.

Again, maybe you need to read that light clock thought experiment more. For the light clock, the local observer is bouncing a photon back and forth between two mirrors, always tracing the same path, over and over again. For a remote observer who is moving with respect to the local observer, the photon appears to trace a zig-zag.
As I said, I prefer to view things from the rest frame of the emitter. If you want to label that frame the "primary" one, that works for me.

23. Mar 20, 2015

### DaveC426913

What??

So Earth is at (0,0) and the Moon is moving along the x-axis toward the Y-axis. It's going to collide with the Earth.
Or is the Moon at y=1 sliding past Earth?

Either way, this is not the co-moving Earth-Moon system you started with.

24. Mar 20, 2015

### DaveC426913

I think this is key to RE's question. And it will likely result in a lot more information being returned from searches about it.

25. Mar 20, 2015

### DrGreg

Maybe this diagram helps

Here's something (it could be a pulse of light, it could be a ball) bouncing up and down in a train.

An observer in the train (top) infers the thing is moving vertically up and down.

An observer on the ground (bottom) infers the thing is "sliding sideways" in a zig-zag path.

This is valid in both relativistic and Newtonian mechanics.