# What is the result of this laser/time dilation scenario?

1. Dec 7, 2011

### Acumen

At an origin there is an observer (A) who fires a laser at a series of beacons in a straight line spaced 186,000 miles (1 light second) apart each.

At the same time as the laser is fired, a second observer (B) starts traveling on a path perpendicular to that of the laser, stopping and returning along the same path so that they return to the origin. B travels at such a speed so that there is a 4 to 1 time dilation for the total time of his journey.

When B returns to the origin, A's clock sees 4 seconds have passed, B's clock sees 1 second passed. What is the resulting state of the beacons for both observers?

As I understand it, the result should be that:

A sees 4 beacons have been hit by the laser.
B sees 1 beacon has been hit by the laser.

How does SR explain that when B returns 4 beacons have been hit?

2. Dec 7, 2011

### ghwellsjr

They both will see that two beacons have been hit.

3. Dec 7, 2011

### Acumen

Yes that much better than 4, oops. So that makes sense to me for A, but if B sees 2 are hit, that means that the light traveled 744,000 miles (round trip 2 beacons) in 1 of B's seconds. That would mean it traveled faster than the speed of light, no?

4. Dec 7, 2011

### Janus

Staff Emeritus
No. For one thing, due to length contraction, the distance between beacons will only be 1/4 light year for B, and for the other, B would measure light as traveling at c relative to himself, it would be the beacons that would be moving, first towards him and then away.

5. Dec 7, 2011

### Acumen

I was under the impression length only contracted in the direction of motion? Since the laser is perpendicular should not the distances remain the same?

Also if B does observe 2 beacons hit due to contraction, to me it seems that even though the speed of the laser is constant, that the speed of contraction and expansion exceeded the speed of light?

The second beacon would contract from 372,000 to 93,000 and expand to 372,000 = 558,000 mile round trip contraction and expansion in 1 second.

Is this faster than light contraction allowed?

6. Dec 7, 2011

### ghwellsjr

When you measure the speed of light, you have to remain at a constant velocity during your measurement and it requires a mirror placed some constant measured distance away from you and you need a clock to measure how long it takes for the light to make a round trip, so while B is traveling away from A at right angles to the direction the laser is pointing in, he can't measure that light. He needs A to fire another laser pointing in his direction. Or you could have said there was a bright flash of light traveling in all directions. Then he can make a measurement of the round-trip speed of light as he's traveling away and then make another measurement as he's coming back.

But some time during his trip, he will be able to see the light reflecting off the first beacon (assuming it reflects in all directions) but he won't see the light reflecting off the second beacon until the moment he gets back.

7. Dec 7, 2011

### ghwellsjr

I think you overlooked the fact that B is traveling at right angles to the direction the laser is pointing and to where the beacons are located.

8. Dec 7, 2011

### Acumen

So even if he never observes the laser hitting the second beacon until the moment he returns, hasn't he now had only 1 second pass and sent information 744,000 miles? Does this not mean that even though he knows 4 seconds passed from A's perspective, that from his perspective a signal was sent faster than the speed of light?

9. Dec 7, 2011

### ghwellsjr

No, as I said, if he wants to measure the speed of light, he needs a reflector that is traveling with him. The beacons are not traveling with him.

10. Dec 7, 2011

### Staff: Mentor

B's frame is non-inertial, so it is expected for light to travel at speeds other than c in it.

Last edited: Dec 7, 2011
11. Dec 8, 2011

### Acumen

So as I understand it, non-inertial frames allow things to travel with a perceived speed faster than light, which is called proper velocity?

Why would the beacons need to be traveling with B if he only records the light when he is about to leave and as soon as he returns to the inertial frame?

12. Dec 8, 2011

### ghwellsjr

Proper velocity does not apply to light.

Special Relativity defines the progress of light to be c in any inertial frame. There is no standard way to define the characteristics of a non-inertial frame, but in general, you cannot consistently define the speed of light to be c everywhere in it for all times and in all directions like you can with an inertial frame. The reason we need to define the speed of light is because we cannot observe the progress of light like we can observe the progress of moving material objects that always travel at less than the speed of light. We use light to observe these objects but there is nothing faster than light with which to observe light so its progress will remain a mystery apart from a definition of a Frame of Reference.
I'm not sure if you're asking about observer A measuring the round trip speed of light from him to the second beacon and back to him, or defining the one-way speed of light according to an inertial frame in which he is at rest. In either case, your scenario doesn't allow him to do either one.

Last edited: Dec 8, 2011
13. Dec 8, 2011

### Staff: Mentor

Yes.

Proper velocity is not a commonly used concept. I am not familiar enough with it to talk about its behavior in non-inertial frames.