# 2 Clock Experiment

#### Prosoothus

Registered Senior Member
I've been at sciforums for over a year debating relativity, and yet it still confuses me. I have conjured up a thought experiment, and would appreciate any feedback regarding the results of the experiment according to SR. Maybe this will help me understand relativity a little better.

Here's the thought experiment (see attachment):

There are two atomic clocks: clock A and B. Both of them are off, and their times are set to 0, at location s0. The two clocks begin accelerating. They both reach a speed of .90c before they reach point s1. As a result, when they reach s1 they are both travelling at a speed of .90c and ARE NOT accelerating. When they reach point s1, the beam of laser 1 hits both of the clocks. The clocks detect the signal using photocells. As soon as the two clocks detect the laser beam, they switch on and start to tick. They both continue to travel at .90c, and continue to tick, until they reach lasers 2a and 2b. When the clocks detect the laser beams from lasers 2a and 2b, they switch off (they record the time and stop ticking). Finally, the clocks are brought together so that an observer can see their recorded time. What time will the observer see on each clock?

Note:The clocks are only ticking when they are travelling at a constant velocity (not accelerating). Also, there are no gravitational fields in the area that can influence the clocks.

Edit: I changed the distance measurements in the illustration to make calculations easier. Attachment is in next post.

Tom

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Attachment:

Answer: The time will be 1 second on both clocks.

James,

No time dilation???

From the view point of each observer moving with their respective clock (at 0.9c), they do not travel 270 000 km but
117 690 km. Thus the time elapsed of each clock is 0.436 sec.

ryans,

From the view point of each observer moving with their respective clock (at 0.9c), they do not travel 270 000 km but 117 690 km. Thus the time elapsed of each clock is 0.436 sec.

Relative to what? Clock A, clock B, or the lasers?

Relativity dictates that relative to the lasers, both clocks are moving at .90c. However, relative to clock A, clock B is moving at .99448c and relative to clock B, clock A is moving at .99448c. As a result, relative to clock A, clock B should be ticking slower, but relative to clock B, clock A should be ticking slower.

You have three frames of reference (lasers, clock A, and clock B) which relativity claims are all valid. Unfortunately, each clock can only show one time at any given moment. What is the preferred frame of reference for the clocks, and why?

Tom

The distance specified as 270 000 km was assumed to be that measured by an observer at rest with the lasers i.e. the starting and the finishing point.

You have three frames of reference (lasers, clock A, and clock B) which relativity claims are all valid. Unfortunately, each clock can only show one time at any given moment. What is the preferred frame of reference for the clocks, and why?

There are no preferred frames of reference, it's all relative.

ryans,

There are no preferred frames of reference, it's all relative.

If each clock can only show one time, then the combination of the two times on the two clocks can only show one of the three frames of reference (or perhaps a fourth). Which frame of reference will the clocks show?

Tom

Have you remebered that from clock A's perspective, the length travelled by clock B is no longer 270 000 km.

ryans,

Have you remebered that from clock A's perspective, the length travelled by clock B is no longer 270 000 km.

You're right, but why does that matter? From clock A's frame of reference, clock B was travelling at .9944c. Because of this, clock B should be ticking slower relative to clock A. Regardless of the distance that clock B is travelling relative to clock A, whether it's 1 km or 10 million kilometers, the time on clock B can't catch up to the time on clock A relative to clock A.

Tom

Yes, but you are assuming from clock A's frame of reference, clock B will pass its laser at the same time Clock A passes its own. This will only be true for the observer at the starting point, who sees both clocks travelling at 0.9c. From clock A's perspective, clock A and clock B do not pass their repective markers at the same instant. Simultanity just flew out the door my friend.

ryans,

From clock A's perspective, clock A and clock B do not pass their repective markers at the same instant. Simultanity just flew out the door my friend.

Clocks A and B do pass their markers at the same instant, even though clock A sees clock B pass its marker after clock A passes its marker. This delay is not caused by time dilation or length contraction, but is due to the finite speed of light.

I hope that your not claiming that delays caused by the finite speed of light are needed in order to preserve relativity.

Tom

all things given

in a motionless universe then each clock would have the same time measurement.
but in ours they will have diffrent times due to the motion of the universe it's self as
if speed relitive with rotation or expansion then Sa= r or e + or - and Sb= r or e + or -
example plane leaves ny at 600mph heading west plane a
leaves ny at 600mph heading east plane b which returns first assumeing no stops.

with respect

Originally posted by Prosoothus
ryans,

Clocks A and B do pass their markers at the same instant, even though clock A sees clock B pass its marker after clock A passes its marker. This delay is not caused by time dilation or length contraction, but is due to the finite speed of light.

I hope that your not claiming that delays caused by the finite speed of light are needed in order to preserve relativity.

Tom

No, this does not occur due to light signal delay, but due to time and length dilation.

Consider the situation from A's perspective. Due to length contraction, A will measure the distance between laser 1 and laser 2A as 117690 km. he will measure the same distance between lasers 1 and 2B.

At .9c, A will measure the time it takes to cross this distance as .4359 sec. Now he will also measure laser 2B's velocity as .9c relative to himself. He will also measure clock B's velocity as .994475c relative to himself, for a relative velocity difference of 0.094475c. (He will see clock B move at .094475c relative to lasers 1 and 2B. ) By A's clock this means that it will take 4.15241 sec for clock B to intersect laser 2B, after factoring out the light siganl delay. (light signal delay would add another 235380/300000 = .7846 sec)

Thus clock A will stop before clock B from A's perspective(A stopped after .4359 sec, and B after 4.15241 sec, as measured by A) . If we now apply time dilation to clock B as seen by A, we see that clock B will appear to run slow by a factor of sqrt(1-.994475c²/c²) = .10497

4.15241 * .10497 = .4358 sec. Thus clock A will see clock B read .4358 sec as it crosses laser 2b, which is the same time that clock A reads when it crossed laser 2A. What Clock B sees with respect to clock A will be exactly the same.

So when you bring both clocks together, They both agree that they should read the same time.

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Oops. Sorry. ryans and Janus58 are right.

The time displayed by the clocks is 0.436 seconds. The time displayed by a clock at rest with respect to laser 1 is 1 second.

There is no signalling involved here, Tom, so there are no effects from light delays. All effects are due to time dilation and length contraction. All observers agree on the final times displayed.

Don't tell me that after so long on this board, you're still making the same elementary mistakes as MacM. (?)

Choices

James R.,

Don't tell me that after so long on this board, you're still making the same elementary mistakes as MacM. (?)

REPLY: Actually I prefer that you call them "Choices" than "Mistakes" but I'm sure you won't agree.

MacM:

We are discussing relativity here. If you ignore the relativity of simultaneity in a relativity problem which requires it then you are making a mistake.

Oh

James R.,

I guessed you might say that.

Janus58,

Thanks for the explanation. I see now that SR is a mathematically consistent theory. Your explanation can be used to explain MacM's three clock problem as well.

However, I still don't believe that the math of relativity matches physical reality. I still have a problem with understanding how space can be contracted and not contracted at the same time. For example you state:

A will measure the distance between laser 1 and laser 2A as 117690 km. he will measure the same distance between lasers 1 and 2B.

At .9c, A will measure the time it takes to cross this distance as .4359 sec.

I may agree that clock A measures a contracted distance. However, I don't agree with your assumption that the measured distance is the actual distance that clock A must traverse. I do, however, understand that you must accept that the measured distance is the actual distance if you accept the principle of invariance of light. I still don't believe that the principle of invariance of light is valid. I think more experiments need to be done in order for me to be convinced of its validity (experiments done far away from the strong gravitational field of the Earth).

Tom