#### khashayar

Hey Guys
I'm kinda new to relativity. I've worked realy hard to understand the idea but there are still some thought going in my head trying to convince me that it doesn't make sense or I don't understand it. I was wondering if you can help me.
So here is my first problem regarding special relativity. Sorry that it's long but i had to explain the thought experiment.

In the twin paradox the problem is solved by invoving accelaration needed for the traveler to come back and arguing that the traveler is not in one inertia frame of reference at all time. Now let’s consider the following scenario. Suppose we have three identical stopwatches, and a car that can travel at speeds close to speed of light. One of the stop watches is placed at the start point of the track and another at the finish line. The third stop watch is installed on the car. Let’s assume that the road is perfectly straight and flat and that the stop watches are designed such that as soon as they meet they turn each other on if they were off and off if they were on. Also let all the stopwatches to be initially off.
The driver starts at some point far from the start point accelerating and reaches a speed close to that of light before he reaches the start line. As soon as he passes the start line both stopwatches start counting. When he reaches the finish line the stopwatch at the finish line starts counting and the stopwatch on the car stops. Driver slows down and brings the stop watch back to us. Meanwhile we have picked the other two stop watches from the start line and finish line. We stop both of them at the same time. Subtracting the time on the first stopwatch from the second one we have the time that took the car to travel from start to finish line from our point of view. The stopwatch on the car on the other hand should have the time measured from the driver’s point of view for completing the drive.
Since the stop watch installed on the car was not working while the driver was accelerating or slowing down it could not be affected by acceleration, and also we assume that the other two stop watches are brought together at really low speeds with almost zero acceleration. Also there was no communication between the driver and the observer involved to obtain the results.
Now the question is, “Which of the values is greater? The one measured by the stopwatch on the car, or the one obtained from subtracting the values shown by the other two stopwatches.” Suppose that we say since the car was in motion with respect to the observer the value shown by the stopwatch on the car is smaller, but then the driver can claim that he was not in motion and it was the observer and everything else which were moving at that speed in the oposit direction.

I used Lorentz_Einstein trasformations to find the time needed for the trip from two different points of view, but my results were crazy. They were telling me no matter what the time frrom observers poit of view is longer than the time recorded by the car. Did I make a mistake somewhere? Or is it simply because the action of stop watches being turned on and off happenes in the same space coordinate with respect to the car frame of reference?
Thanks

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The car should indeed have a longer time. The reason? As you said, the stopwatches affect each other. Well, since they cannot send an instaneous signal to each other, they must communicate at the most at light speed. Since the car is going very close to light speed as it passed the first stopwatch (or vice versa from the car's point of view) It will take time for the signal to catch up and start the starting stopwatch, and again when it passes the final one it will take time for the signal to affect the car stopwatch to turn it off. So thus the car signal will indeed be longer as it should.

In fact, this is no paradox that I can see, unless I am somehow missing the main point of it and visualizing the thought experiment wrong. So perhaps others can verify this?

#### khashayar

Brad_Ad23 I do understand your explanation. However the thought experiment is designed in such a way that the distance between the car's stop watch and the other two stop watches as they pass each other is almost zero so the signal travel between the two passing each other takes almost zero time. I try to think of it as the stop watch on the car hitting the other two stop watches start/stop buttons gently as it passes by them.

Put that way, it would change my explination.

Since that is the case, it falls back to the same explination as the Twin's paradox. Because the car accelerates, the symmetry will be broken, regardless of whether the clock is on or not, it is time itself that slows. Still, I can't seem to find a way to show this. It could be some form of system entanglement that is occuring.

I better wait and see if any other people come up with explinations.

#### Hurkyl

Staff Emeritus
Gold Member
The stopwatch in the car will measure less time.

From the drivers POV:

Case 1:

The car deccelerates before the two stopwatches are brought together.

One of the peculiar things about accelerations is that the rate of time observed on distant clocks varies by how far away the clock is and in which direction. In particular, during the deceleration, he observes time running much faster for the start stopwatch than the finish stopwatch, fast enough to make up for the slowdown due to time dilation.

Case 2:

The car deccelerates after the two stopwatches are brought together.

I think the key is that when velocities are near the speed of light, slight differences can lead to large percentile differences in time dilation... but I don't have the time atm to sit down and draw up the space-time diagrams to see for sure what's going on.

#### Janus

Staff Emeritus
Gold Member
Case 3:

The stop watches at the start line and finish line are stopped simultaneously before they are moved together, and before the ship deccelerates. (This can be done by just allowing for the light signal propagation delay time between them.)

Then you come up against the "Failure of Simultaneity", Where, by the two clock's frame, they stop at the same instant, but from the frame of the ship's clock, one continues to run after the other stops, accumulating the proper time difference.

#### LURCH

So could this be accurately described by saying that bringing the car to a stop in order to compare watches should be viewed as an acceleration which brings the car into a common frame of refference with the other two watches? That is the key event, right; the car changing from one inertial frame of reference to another? If the car kept moving, the driver would always measure the clocks on the track to be slow, and the observer on the track would always measure the clock in the car to be slow?

#### TonyG

The car's stopwatch will read less time.

In the stationary frame (the two external stopwatches), the time measured will be L/v, where L is the distance to the finish line and v is the car's velocity.

In the moving (car's) frame, the time measured will be L'/v, where L' is the (Lorentz-contracted) distance to the finish line.

Since L'<L, the car's watch will read less time. Specifically,
dt = (L-L')/v, where L' = L *(1-(v/c)^2)^1/2.

As far as the argument about inertial frames, the car does change reference frames, breaking the symmetry -just like in the "twin paradox" case.

thanks alot guys

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