Question on the reasoning behind the light clock thought experiment.

1. May 20, 2011

Inducer

I read that the speed of light is independent on the relative velocity of the source. So why is it that the beam zig-zags if there is relative motion between an observer and a light clock. This zig-zagging implies the vector addition of velocities. I know that the speed c is kept constant during this "addition".This to me doesn't make sense.Here is a link to a Wikipedia article (http://en.wikipedia.org/wiki/Time_dilation) that includes this thought experiment along with the derivation that considers it.

2. May 20, 2011

JesseM

The light's speed is independent of the source, but its velocity vector is not, since the direction does depend on the motion of the source. This is necessary in order to preserve the first postulate of SR which says the laws of physics should work exactly the same in every inertial frame, which can also be translated to mean that if you're in a sealed room moving inertially, you'll observe the same result to any experiment regardless of the room's velocity relative to any external landmark like the Earth. If a laser pointed vertically (orthogonal to the floor and ceiling) has the beam go up to a spot on the ceiling directly above the laser for one sealed room, this must be true in every such sealed room. But this means that if the sealed room is unsealed so that an observer who sees the room in motion can see what's going on inside, then in order for the beam to end up at that spot on the room's ceiling, this outside observer must see the laser beam slanted relative to the laser emitter itself (since the spot on the ceiling will have moved between the time a photon was emitted from the laser emitter and the time it reached the ceiling)

3. May 20, 2011

A.T.

The moving observer must see the light hit the mirrors, just like the stationary one. The speed of the beam is still c for him

4. May 20, 2011

Inducer

So JesseM, you are saying that the fact that the light beam hits the ceiling at that particular spot is in agreement between an observer within the room and the observer outside of the room.

5. May 20, 2011

Inducer

I also must point out that I have not yet encountered relativity in an educational curriculum, in other words it has not been taught to me.

6. May 20, 2011

JesseM

Yes, although the important issue isn't really inside vs. outside but rather velocity relative to the room and emitter. Anyway, all observers always agree about events localized to a particular point in space and time like whether a particular spot on a ceiling gets hit by a laser or not (imagine there was a bomb which could be set off by a light-sensitive diode at that spot--if different frames could disagree about whether the light hit it, they could disagree about whether the bomb exploded, whether a person near the bomb lived or died, etc.!)

7. May 20, 2011

JesseM

If you're interested in learning the basic ideas there are some links to online resources on this thread.

8. May 20, 2011

Inducer

Now I see the relevance of events, thank you.

9. May 21, 2011

Realname

I have a question somewhat related to this discussion and hoped I wouldn't have to make a new thread.

So, you have two observers holding clocks. One accelerates to nearly the speed of light away from the other, we'll name the traveler person A and the stationary one person B. To each person, the others clock is slowing down as they separate. The way I interpret this is that the light they are receiving from each other is taking an increasingly longer time to travel between them due to the increase in distance.

What would happen if person A were to abruptly stop his travel and move back towards person B at the same speed? Would time not appear to be faster than normal for the opposite person of each observer? They would be catching up through the opposite persons outbound light-waves, and time would appear to be passing at about 2x. When person A arrives back at person B's position, the apparent increase in the clocks' speeds would have compensated for the prior slowing down at his departure. Thus having both clocks be at equal times and both observers having aged equally.

Something tells me I'm wrong.

10. May 21, 2011

ghwellsjr

You are correct regarding each observer seeing the other one's clock as running slower (by the same amount) as they are separating and the light taking longer to reach each other but as soon as A turns around, he immediately sees B's clock as running faster than normal whereas B does not see A as turning around until way later at which point he will see A's clock as running faster (by the same amount) but since A sees B's clock as running slow for half the time and fast for half the time and since B sees A's clock running slow most of the time and fast for a rather short time, when they get back together, A will have seen B's clock accumulate much more time than B will have seen A's clock accumulate. And of course, when they get back together each will have viewed their own clock as accumulating exactly what the other one says his clock should have accumulated.

11. May 21, 2011

Realname

If person A is travelling nearly the speed of light, he would arrive at person B's location momentarily after the image of him turning around reaches person B, correct? So in the instant after person A sees person B turning around, the apparent speed of time that person B is observing of the traveler (A) would be extremely high.

Now if I'm interpreting this right, you mean the clocks do end up at the same time? I'm curious about this as I always hear stories explaining relativity saying that if I went on such a journey I would be much younger than my friends upon my return.

12. May 21, 2011

ghwellsjr

When A finally sees B turning around, A will see B's clock running extremely fast. But B has been seeing A's clock running at exactly the same high rate of speed for a lot longer, in fact for half the trip. A only sees B's clock running at this high rate of speed for the very last small portion of the trip.
No, the clocks do not end up with the same accumulated time on them. B's clock has accumulated much more time than A's clock. When they reunite, A's observation of his own clock's accumulated time matches B's observation of the accumulated time on A's clock and B's observation of his own clock's accumulated time matches A's observation of the accumulated time on B's clock. They both agree that B has aged much more than A and they both can watch the process in their own time.

You commented earlier:
When person A arrives back at person B's position, the apparent increase in the clocks' speeds would have compensated for the prior slowing down at his departure. Thus having both clocks be at equal times and both observers having aged equally.​
I was pointing why "the apparent increase in the clock's speeds" does not compensate for the prior slowing down. It's because this "compensation" does not last for the same ratio of time for each observer. The traveler sees it for exactly have the trip while the other one sees it for only a very short portion of the trip.

You were suggesting an explanation:
The way I interpret this is that the light they are receiving from each other is taking an increasingly longer time to travel between them due to the increase in distance.​
This is true for the first half of the trip, but for the second half of the trip it's the other way around:
The light they are receiving from each other is taking an increasingly shorter time to travel between them due to the decrease in distance.​
The only problem is that while this happens immediately on turn-around for A, it doesn't happen immediately for B because the light that A has been sending to B hasn't gotten to B yet and won't for a very long time. So B continues to see A as traveling away from him for much longer than half the trip and it is only near the end of the trip that B sees A turn around and come back to him with his clock running faster. But it cannot compensate for the slowing down during the outbound part of the trip because it isn't for the same length of time. So B sees A's clock as having accumulated much less time than his own when they reunite.

Last edited: May 21, 2011
13. May 21, 2011

Inducer

I've done some thinking about the nature of an event in different reference frames; I made some simple calculations pertaining to the following scenario. There is a room moving at speed v and a beam of light is made to go from one side of the room to another in the same direction and parallel to v. There is a stationary observer and an observer within the room and these observers correspond on their observations. The stationary observer measures how far the room moves in the time it takes the beam to reach the other side (obviously taking into account the constancy of the speed of light). Now the observer in the room tells the observer that relative to him the beam is moving at the speed of light and therefore it takes D/c seconds to reach the other side where D is the length of the room. The observer outside then calculates the distance the room moved in D/c seconds which tells the stationary observer which location that the observer within the room sees the light beam hit the wall. The stationary observer sees the beam hit the wall when the room had covered Dv/(c-v) meters but according to the observer within the room the beam would have hit the wall when the room has covered a distance Dv/c.Therefore the event doesn't occur at the same time or place for each observer and the condition required by these equations for both distances to equal is v=0. Contraction or time dilation could account for this.

14. May 21, 2011

Realname

The thought that is troubling me on this explanation is this:

Person A turns around at nearly the speed of light. Directly in front of him is the light that will show him turning around. As his trip continues he will slowly fall behind this light and the gap will fill with outbound light-waves he is sending on his journey home which are also travelling faster than him but still behind the image of his turning around. So A sees B sped up to about 2x for a long period of time. However, wouldn't the area of light in front of A show the equivalent story except extremely fast in a very short period of time for person B?

Also, if person A doesn't return and simply stays at a far distance, would he still be at the same point of time as person B (regarding actuality not appearances of the opposite side)?

The way I'm visualizing it is that when they separate everything is equal and either point of reference (person A or B) is identical. However, once one side moves back towards the other, symmetry is lost due to the light already in exchange between the two sides. So would it be only on person A's return journey that any difference in apparent elapsed time would arise?

Hopefully I explained this clearly enough. Thanks so much for your patience :)

Last edited: May 21, 2011
15. May 21, 2011

ghwellsjr

Opps, I made a mistake in this quote which I fixed. Maybe you could fix it in your quote of me to avoid confusion. I changed the A to B immediately following "hasn't gotten to...".
Yes, that appears to be an almost correct summary of what I'm saying except that I would have called the light-waves as "inbound" instead of "outbound". I use the terms "inbound" and "outbound" to refer to the two portions of the travelers trip, not to the direction the light is traveling.

But I think the point you are failing to grasp is that when B finally sees A turning around and A's clock appears to be running faster, the rate at which B sees A's clock running faster is exactly the same as the rate that A has been seeing B's clock running faster. If one sees the two rates for exactly half the time and the other sees the rates for a different percentage of the time, then the two clocks will end up with a different accumulated time on them.
We cannot say what the relative time is on the two clocks that are separated in terms of "actuality not appearances". It all depends on the reference frame in which the clocks are assigned times. In the frame in which both A and B started out at rest and ended up at rest, we can say and they both will say that A's clock will have accumulated less time on it but other frames moving with respect to both of them can say other things and there is no absolute truth to any of them.
No, if they use Einstein's method to evaluate the synchronization of their clocks, they will determine that A's clock has accumulated less time. But you must realize that Einstein's synchronization is arbitrary and doesn't mean that the A's clock has "really" accumulated less time, other methods will produce different conclusions.

As far as symmetry goes, as soon as A accelerates at the beginning of his trip, symmetry is already lost.

16. May 21, 2011

Realname

I believe I understand it now.

So the traveler (A) has alotted less time than the stationary person B because of time dilation. I'm understanding this as: because the speed of light can not be surpassed, any increase in speed through space must be compensated for with a slowing of time. The closer your speed comes to the speed of light the slower things pass.

Would this be a correct description?

Also, I'm failing to understand why the mass of an object increases when it moves closer to the speed of light?

Thanks.

17. May 21, 2011

ghwellsjr

I'm not sure I understand what you are describing but it doesn't sound right to me. I have not been talking about time dilation. I have been talking about Relativistic Doppler. There is a big difference. Relativistic Doppler has to do with what observers with relative motion between them see of each others' clock rates and, after the transients caused by any accelerations are over with, it is symmetrical, assuming that they are traveling along a single line, toward or away from each other. Each one observes the other one's clock as going faster, if traveling toward each other, or slower, if traveling away from each other. Relativistic Doppler is the net result of what you described in your first post on this thread:
"the light they are receiving from each other is taking an increasingly longer time to travel between them due to the increase in distance"​
and time dilation.

If you take out the first component (the one you described), then you will be left with Time Dilation which is always the moving clock running at a slower rate. Both Relativistic Doppler and Time Dilation are symmetrical and reciprocal effects.

So in your scenario when A accelerates away from B, after the acceleration has ended and each observer is looking at the other's clock, they will each see the other one's clock as running slower, but they will interpret that to mean that the effect has two parts and when they take out the part that has to do with the light in transit, they will be left with Time Dilation. They will each interpret the other one's clock as running slower than their own.

Now this may seem pointless to discuss but when you consider the inbound part of the trip, after they both see the other one's clock as running faster than their own, when they take out the effect of the light in transit, they will again be left with the same Time Dilation, that is, each one will interpret the other one's clock as running slower than their own (even though they see the other one as running faster than their own).

So during the entire trip, both observers will interpret the other one's clock as running slower than their own because of Time Dilation because each one can view the other one as moving with respect to themself. So this, by itself, doesn't help to understand why the traveler ends up with less accumulated time on his clock.

The key to understanding how Time Dilation provides a good explanation is to pick a reference frame. An obvious choice is the one in which they both start out at rest and end up at rest. In this frame, B is always stationary and never experiences Time Dilation while A always experiences Time Dilation during his whole trip. Now it is obvious that A will end up with less accumulated time on his clock. But you must not think that this reference frame is the only one that you can analyze the situation in. You can use any reference frame but it will be more complicated and since all reference frames end up with the same final answer, as far as I'm concerned, you can stop with the easiest frame.

18. May 21, 2011

GrayGhost

Good question. The zig zagging does require a relative motion between the 2 POVs, as Galileo pointed out. However, this has no impact on light's speed. The fact that light's speed is c per all inertial POVs, and the same mechanics hold equally good in any inertial frame, requires only that each POV measure space and time differently. That's what allows it all to make sense, although most are skeptical for quite some time until they come to grips with it.

GrayGhost