Paradox within the twin paradox

[ghwellsjr]

You have a misunderstanding of what a reference frame is. It is a co-ordinate system with time added. Think of the graphs you made in school. You drew a horizontal line and a vertical line. At the intersection you wrote the number zero to specify the origin. Then you put positive numbers off to the right and upwards and negative numbers off to the left and downwards. Even though the numbers stopped because of the limited size of your paper, you knew that they really extended all the way to infinity in all directions. Then if you wanted to describe two objects, you specified their locations in terms of the x,y co-ordinates and you could calculate their distance apart or whatever.

Now what if someone told you that they were going to make two graphs on two separate pieces of paper and put one object in one of the graphs and another object in the other graph and then they started moving the pieces of paper around trying to explain how the two objects in the two separate graphs moved in relation to one another. Wouldn't you say they were mixed up?

That's what you are doing in your thought experiment. Instead, you must start with one and only one reference frame. That's like your graph. You can say that one counter is stationary at the origin and the other one is one light year off to the left (for example) and traveling at 0.99995c toward the origin and at time zero both counters are set to zero and they send out a signal once per minute. Then we can analyze what will happen approximately one year later when the traveling counter reaches the stationary counter. They will have different counts, the traveling one will have a much lower count on it.

Just think about something. When this scenario starts, neither counter will have any knowledge of what the other counter is doing. It will take almost one year for the signals coming from the traveling counter to reach the stationary counter and then within less than half an hour, all the signals will arrive in a burst and then the counter will arrive. At a speed of 0.99995c, the traveling counter's time has slowed to 1% of normal. Since there are 525600 minutes in a year, this counter will have only sent out 5256 signals during his entire trip but they will arrive during the last 26.28 minutes (1-0.99995 or 0.00005 times the number of minutes in a year). That's a rate of 200 signals per minute as measured by the stationary clock.

On the other hand, the traveling counter will not see anything from the stationary counter until after about half a year. Then it will start seeing signals coming in two per minute (they are traveling towards him at the speed of light and he is traveling toward them at almost the speed of light, which approximately doubles the rate at which he receives the signals), except since his clock is running at 1% of normal, he will think they are coming in at 100 times that rate which is 200 per minute. And by the time he gets to the stationary clock, he will have received almost a year's worth of signals sent out at one per minute or very nearly 525600.

Now if you want to analyze the same scenario in another reference frame, the one in which the traveling counter is stationary, you have to correctly transform everything in the first frame to the second frame. You just can't say that the two counters will be zero at the start at the same time because that would be a different scenario.

 

[JesseM]

If what you say about synchronization were true, then it would be impossible to resynchronize a clock. Or for instance to resync a slave server to its master. Of course, this is semantics. But truly, mechanisms that are synchronized fall out of synchronization all the time.

You seem not to be understanding the problem, the issue is not that there is any difficulty in "synchronizing" clocks given some frame's definition of simultaneity, the problem is that different frames have different definitions of simultaneity and thus different opinions about whether any given pair of distant events (like two clocks showing the same reading) happened "at the same time" (same t-coordinate in that frame) or "different times", and the laws of physics work exactly the same in all inertial frames so there is no physical reason to think one frame's definition of simultaneity is more "correct" than any other's.

So, this part of your thought-experiment is just ill-defined:

Now, by shear coincidence these two counters reset themselves at the very same instant and send out a light pulse toward one another in the same instant.

"At the same instant" according to what frame's definition of simultaneity? If the events of the counters resetting themselves are simultaneous in one inertial frame, those same events are non-simultaneous in other inertial frames.

The relativity of simultaneity is also why we can't give any objective answer to which of two twins has aged more if they are both moving inertially (rather than one turning around so they can compare clocks at a single location)--for example, if twins Alice and Bob depart from the same location when they are both aged 30, and they move apart inertially at constant relative speed 0.6c, then in Bob's rest frame the event of his turning 40 is simultaneous with the event of Alice turning 38 so she is aging slower in his rest frame, but in Alice's rest frame the event of her turning 38 is simultaneous with the event of Bob being aged 36.4 so he is the one aging slower in her rest frame.

 

[teachmemore]

You have a misunderstanding of what a reference frame is. It is a co-ordinate system with time added. Think of the graphs you made in school. You drew a horizontal line and a vertical line. At the intersection you wrote the number zero to specify the origin. Then you put positive numbers off to the right and upwards and negative numbers off to the left and downwards. Even though the numbers stopped because of the limited size of your paper, you knew that they really extended all the way to infinity in all directions. Then if you wanted to describe two objects, you specified their locations in terms of the x,y co-ordinates and you could calculate their distance apart or whatever.

Now what if someone told you that they were going to make two graphs on two separate pieces of paper and put one object in one of the graphs and another object in the other graph and then they started moving the pieces of paper around trying to explain how the two objects in the two separate graphs moved in relation to one another. Wouldn't you say they were mixed up?

That's what you are doing in your thought experiment. Instead, you must start with one and only one reference frame. That's like your graph. You can say that one counter is stationary at the origin and the other one is one light year off to the left (for example) and traveling at 0.99995c toward the origin and at time zero both counters are set to zero and they send out a signal once per minute. Then we can analyze what will happen approximately one year later when the traveling counter reaches the stationary counter. They will have different counts, the traveling one will have a much lower count on it.

Just think about something. When this scenario starts, neither counter will have any knowledge of what the other counter is doing. It will take almost one year for the signals coming from the traveling counter to reach the stationary counter and then within less than half an hour, all the signals will arrive in a burst and then the counter will arrive. At a speed of 0.99995c, the traveling counter's time has slowed to 1% of normal. Since there are 525600 minutes in a year, this counter will have only sent out 5256 signals during his entire trip but they will arrive during the last 26.28 minutes (1-0.99995 or 0.00005 times the number of minutes in a year). That's a rate of 200 signals per minute as measured by the stationary clock.

On the other hand, the traveling counter will not see anything from the stationary counter until after about half a year. Then it will start seeing signals coming in two per minute (they are traveling towards him at the speed of light and he is traveling toward them at almost the speed of light, which approximately doubles the rate at which he receives the signals), except since his clock is running at 1% of normal, he will think they are coming in at 100 times that rate which is 200 per minute. And by the time he gets to the stationary clock, he will have received almost a years worth of signals sent out at one per minute or very nearly 525600.

Now if you want to analyze the same scenario in another reference frame, the one in which the traveling counter is stationary, you have to correctly transform everything in the first frame to the second frame. You just can't say that the two counters will be zero at the start at the same time because that would be a different scenario.

I agree. I don't see how what you say here disagrees with what I describe.

The traveling counter will perceive the stationary counter as ticking faster than it from the time it receives the first signal.

I'm sorry if I confused you with the two reference frames. I could have separated the analysis into two, one for each reference frame to make it clearer, but it wouldn't have changed anything.

 

[teachmemore]

You have a misunderstanding of what a reference frame is. It is a co-ordinate system with time added. Think of the graphs you made in school. You drew a horizontal line and a vertical line. At the intersection you wrote the number zero to specify the origin. Then you put positive numbers off to the right and upwards and negative numbers off to the left and downwards. Even though the numbers stopped because of the limited size of your paper, you knew that they really extended all the way to infinity in all directions. Then if you wanted to describe two objects, you specified their locations in terms of the x,y co-ordinates and you could calculate their distance apart or whatever.

Now what if someone told you that they were going to make two graphs on two separate pieces of paper and put one object in one of the graphs and another object in the other graph and then they started moving the pieces of paper around trying to explain how the two objects in the two separate graphs moved in relation to one another. Wouldn't you say they were mixed up?

That's what you are doing in your thought experiment. Instead, you must start with one and only one reference frame. That's like your graph. You can say that one counter is stationary at the origin and the other one is one light year off to the left (for example) and traveling at 0.99995c toward the origin and at time zero both counters are set to zero and they send out a signal once per minute. Then we can analyze what will happen approximately one year later when the traveling counter reaches the stationary counter. They will have different counts, the traveling one will have a much lower count on it.

Just think about something. When this scenario starts, neither counter will have any knowledge of what the other counter is doing. It will take almost one year for the signals coming from the traveling counter to reach the stationary counter and then within less than half an hour, all the signals will arrive in a burst and then the counter will arrive. At a speed of 0.99995c, the traveling counter's time has slowed to 1% of normal. Since there are 525600 minutes in a year, this counter will have only sent out 5256 signals during his entire trip but they will arrive during the last 26.28 minutes (1-0.99995 or 0.00005 times the number of minutes in a year). That's a rate of 200 signals per minute as measured by the stationary clock.

On the other hand, the traveling counter will not see anything from the stationary counter until after about half a year. Then it will start seeing signals coming in two per minute (they are traveling towards him at the speed of light and he is traveling toward them at almost the speed of light, which approximately doubles the rate at which he receives the signals), except since his clock is running at 1% of normal, he will think they are coming in at 100 times that rate which is 200 per minute. And by the time he gets to the stationary clock, he will have received almost a years worth of signals sent out at one per minute or very nearly 525600.

Now if you want to analyze the same scenario in another reference frame, the one in which the traveling counter is stationary, you have to correctly transform everything in the first frame to the second frame. You just can't say that the two counters will be zero at the start at the same time because that would be a different scenario.

You have a misunderstanding about my understanding. I just have to be incredibly clear and well defined in order to aid the way you interpret descriptions because you seem to be easily confused by the process of interpretation.

 

[teachmemore]

You seem not to be understanding the problem, the issue is not that there is any difficulty in "synchronizing" clocks given some frame's definition of simultaneity, the problem is that different frames have different definitions of simultaneity and thus different opinions about whether any given pair of distant events (like two clocks showing the same reading) happened "at the same time" (same t-coordinate in that frame) or "different times", and the laws of physics work exactly the same in all inertial frames so there is no physical reason to think one frame's definition of simultaneity is more "correct" than any other's.

So, this part of your thought-experiment is just ill-defined:

"At the same instant" according to what frame's definition of simultaneity? If the events of the counters resetting themselves are simultaneous in one inertial frame, those same events are non-simultaneous in other inertial frames.

The relativity of simultaneity is also why we can't give any objective answer to which of two twins has aged more if they are both moving inertially (rather than one turning around so they can compare clocks at a single location)--for example, if twins Alice and Bob depart from the same location when they are both aged 30, and they move apart inertially at constant relative speed 0.6c, then in Bob's rest frame the event of his turning 40 is simultaneous with the event of Alice turning 38 so she is aging slower in his rest frame, but in Alice's rest frame the event of her turning 38 is simultaneous with the event of Bob being aged 36.4 so he is the one aging slower in her rest frame.

I agree this was ill-defined. I understand the relativity of simultaneity and should clarify this thought experiment by defining this simultaneity with the description I provided in post #30. I expected someone might make the objection you made here.

 

[ghwellsjr]

Both counters measure exactly the same signal rate coming from the other counter, in this case 200 signals per minute. And from that, they each deduce that the other one's clock is running at 1% of their own clock rate.

You stated in your experiment:

If both counters view the other as going slow (ie. the light pulses, which represent the speed of time in the alternate reference frame from each reference frame are arriving with greater interval than the counters' own counter ticks), than this WOULD be impossible to reconcile;

So let's see what would happen if there were not the reciprocal time dilation. The traveling counter would send out 100 times more signals and the signal rate that the stationary counter would measure would be 20,000 signals per minute for a total of about 525600. And the traveling counter would receive the same number of signals but at 2 per minute. This is not reciprocal.

So hopefully, you still agree, and if that's that case, has the paradox gone away for you?

 

[teachmemore]

Let me be clear. I'm not trying to say that there is actually any paradox. I'm just trying to say that when the data transmission between the two counters is analysed under the laws of special relativity, the counters don't appear to "perceive" one another as running slower than themselves and yet, this does not seem to defy special relativity in any way.

 

[John232]

Since when from the Big Bang have two objects always traveled at a constant speed relative to each other?

 

[teachmemore]

Both counters measure exactly the same signal rate coming from the other counter, in this case 200 signals per minute. And from that, they each deduce that the other one's clock is running at 1% of their own clock rate.

You stated in your experiment:So let's see what would happen if there were not the reciprocal time dilation. The traveling counter would send out 100 times more signals and the signal rate that the stationary counter would measure would be 20,000 signals per minute for a total of about 525600. And the traveling counter would receive the same number of signals but at 2 per minute. This is not reciprocal.

So hopefully, you still agree, and if that's that case, has the paradox gone away?

OK. Maybe you have found where there is confusion. To me, the signal rate is what the clock uses to determine how fast the other clock is counting. So the clocks appear to perceive one another as counting faster than themselves, not slower.

 

[ghwellsjr]

Opps, we have simultaneity of posting (#36 and #37)

 

[teachmemore]

Opps, we have simultaneity of posting (#36 and #37)

That is impossible! I'm in a different reference frame (spaceship).

 

[pervect]

I agree this was ill-defined. I understand the relativity of simultaneity and should clarify this thought experiment by defining this simultaneity with the description I provided in post #30. I expected someone might make the objection you made here.

A special switching device in the stationary reference frame, but in a different location in space would be triggered by a passing counter, which would simultaneously reset its counter. The device would simultaneously send a light pulse to the stationary counter. The stationary counter, by knowledge of the distance between the device and itself would then set its counter to 0 and add the amount of time to its counter required to send the light signal to it from the device.

Which distance do you mean? The distance in the stationary frame? "The distance" will in general depend on the frame.

 

[teachmemore]

Which distance do you mean? The distance in the stationary frame? "The distance" will in general depend on the frame.

That is correct. The distance between the device and the counter in the stationary frame as measured from the stationary frame. Both the device and one counter are in the same frame, but at a great distance apart.

 

[ghwellsjr]

OK. Maybe you have found where there is confusion. To me, the signal rate is what the clock uses to determine how fast the other clock is counting. So the clocks appear to perceive one another as counting faster than themselves, not slower.

No, the rate at which the signals arrive is not the same as the rate they are sent out and this is caused by the relative motion. The point is that when both counters/clocks/observers are sending out pulses/signals/ticks at what they perceive to be the same rate as defined by their own clocks, then the rate at which they are received by the other one, taking into account the light travel time, their relative motion, and as measured by their own clock, results in them both measuring an identical signal rate coming from the other one.

In your scenario, you had the traveling counter approaching the stationary one which made the signal rate go up. But consider what happens when the traveler goes past the stationary one. Now the signal rate will go down from 200 per minute to 1 every 200 minutes but it still calculates to the same reciprocal time dilation factor.

 

[teachmemore]

No, the rate at which the signals arrive is not the same as the rate they are sent out and this is caused by the relative motion. The point is that when both counters/clocks/observers are sending out pulses/signals/ticks at what they perceive to be the same rate as defined by their own clocks, then the rate at which they are received by the other one, taking into account the light travel time, their relative motion, and as measured by their own clock, results in them both measuring an identical signal rate coming from the other one.

In your scenario, you had the traveling counter approaching the stationary one which made the signal rate go up. But consider what happens when the traveler goes past the stationary one. Now the signal rate will go down from 200 per minute to 1 every 200 minutes but it still calculates to the same reciprocal time dilation factor.

I did not say that the rate at which they arrived is the same as the rate at which they are sent out. I don't know where you got that from. In fact, I said the exact opposite. Yes what you say is obvious to me.

Also I need to correct something in this thought experiment that arises from my post #30 there. The stationary counter could not possibly send out its first pulse when the device is triggered.

So the thought experiment needs to be revised. When the stationary counter receives its first pulse, it calculates the elapsed time in its reference frame based on the distance from the triggered device and begins to send pulses back to the moving counter.

We can clarify this experiment in the following way:

These counters simply count at an even time interval in their own reference frame, and they also count incoming light pulses.

Once the stationary counter has done its calculation and set its counter, it sends out a number of pulses equivalent to the its starting count, and then continues to count at a 1 minute interval.

When the moving counter approaches the stationary counter, it has counted fewer times than the number of pulses it has received and the stationary counter has counted greater times than the number of pulses it received.

The question then becomes this:

If the two counters are equally relative to one another, how did one counter send more pulses than the other? Both the number of pulses and the number of counts should be equal.

I don't believe that this defies special relativity. I think there is a difference between what can be perceived by data transmission and what special relativity says about relative motion and time.

 

[John232]

In the light clock thought experiment there is no uncertainty of the speed and position of the photon. I would think that the number of counts should be equal also if the position was simultanous for the photon for each observer, but they are not certain of the exact position. They seem to know a lot about the speed of the photon too...

 

[JesseM]

Also I need to correct something in this thought experiment that arises from my post #30 there. The stationary counter could not possibly send out its first pulse when the device is triggered.

Instead of having a scheme with a "switching device" sending a signal to get both counters started, you could just have both counters programmed to calculate the current time in the stationary frame, and then have them both start sending signals at a pre-decided time-coordinate in the stationary frame.

The question then becomes this:

If the two counters are equally relative to one another, how did one counter send more pulses than the other? Both the number of pulses and the number of counts should be equal.

I don't believe that this defies special relativity. I think there is a difference between what can be perceived by data transmission and what special relativity says about relative motion and time.

Each counter sees the same ratio between (rate other counter's signals are arriving)/(rate I am sending signals). In this case, since the counters are moving towards each other each sees the other's signals coming in faster than they themselves are sending signals. However, if they each start sending signals simultaneously in the stationary frame (as would be true under my assumption above), then their view of one another is not completely symmetrical, because the delay that each counter sees between the time it started sending signals and the time it sees the first signal from the other counter will not be equal due to the relativity of simultaneity, so the stationary counter sees a greater delay and thus receives less signals in total from the moving counter than the moving counter receives from the stationary counter.

For example, suppose in the stationary frame the stationary counter is at x=0 light-years, and the moving counter is moving towards it at 0.6c, and at t=0 years the moving counter is at x=30 light-years, and both counters start sending signals at t=0, each sending signals at a rate of 1 per year according to their own clock. Then the two counters will meet after a time of 30/0.6 = 50 years in this frame. Since the moving counter sends the first signal at t=0 from 30 light-years away, the stationary counter receives the first signal at t=30, a delay of 30 years. Meanwhile, at t=18.75 years, the moving counter will be at position x=30 - 0.6*18.75 = 18.75 light-years, and naturally the first signal from the stationary counter will also be at x=18.75 light-years at time t=18.75 years, so that's when the moving counter receives the first signal. And in fact the delay is even smaller according to the moving counter's clock, since in the stationary frame it's running slow by \sqrt{1 - 0.6^2} = 0.8, so the moving counter only experiences a delay of 0.8*18.75 = 15 years between sending its own first signal and receiving the first signal from the stationary counter.

After each starts receiving signals from the other, they will each receive them at a rate of 2 per year according to their own clock, since the relativistic Doppler effect equation says f_{received} = f_{sent} \sqrt{\frac{1 + v/c}{1 - v/c}}, so with v/c = 0.6 the received frequency is greater than the sent frequency by a factor of \sqrt{\frac{1.6}{0.4}} = \sqrt{4} = 2. So if the stationary counter receives the first signal when its own clock reads 30 years, and meets the moving counter when its own clock reads 50 years, it must have received 40 signals from the moving counter in this time. Meanwhile if the moving counter receives its first signal at t=18.75 years and meets the stationary counter at t=50 years, a time-interval of 31.25 years in the stationary frame, according to its own clock the time between receiving the first signal and meeting the stationary counter must be 0.8*31.25 = 25 years, so it has received a total of 50 signals from the stationary counter.

 

[John232]

In the light clock thought experiment there is no uncertainty of the speed and position of the photon. I would think that the number of counts should be equal also if the position was simultanous for the photon for each observer, but they are not certain of the exact position. They seem to know a lot about the speed of the photon too...

Both observers would agree on the number of ticks of a certain clock but their own separate clocks would tick by different amounts since the other clock would be seen to use a longer distance of travel for the photon because of its motion while their own clock would use a shorter distance so then they would see each others time slow down relative to themselves.

 

[JesseM]

I am confused about the "moving clocks" part. For twin A on Earth, twin B is traveling is at a velocity +v. For the twin B who is traveling, twin A on Earth is moving at a velocity -v. And i thought that Special Relativity refers that no inertial frame is special; meaning twin B moving nearly at the speed of light should not be special.

Yes, no inertial frame is special. But if B turns around to return to Earth, then B has changed velocities relative to every inertial frame, there is no inertial frame that says that it's A that changed velocities rather than B. So, the situation is not symmetrical. And every inertial frame agrees that a constant-velocity path between two points elapses more time than a path with varying velocity, this is analogous to the way a straight line between two points on a 2D plane always has a shorter length than a path with varying slope (I expand on this geometrical analogy in [post=2972720]this post[/post]).

The only part that i see is different is that twin B requires acceleration +a to go somewhere and then an acceleration -a with respect to twin A to come back. Then if acceleration is the cause, how come Lorentz factor in time dilation does not include acceleration?

Again you may find it helpful to read the geometrical analogy in the post above. In terms of the analogy, the idea that time dilation depends only on velocity is analogous to the fact that a particular quantity on a 2D plane--namely, the rate that a car's odometer is increasing relative to increase in x-coordinate--can be expressed solely as a function of the slope S of the path the car is driving on (with S analogous to velocity) and not the rate the slope is changing dS/dx (analogous to acceleration), with the odometer increasing by an amount \Delta x \sqrt{1 + S^2} for each increase in x-coordinate \Delta x (analogous to the fact that the time dilation equation says a moving clock's reading will increase by an amount \Delta t \sqrt{1 - v^2/c^2} when the time-coordinate of the frame you're using increases by \Delta t). And yet it is still true that a constant-slope path between two points (a straight line) always has a shorter distance than a path with varying slope, analogous to the fact that a constant-velocity path between two events (like the twins departing from one another and reuniting) always has a greater elapsed time than a non-constant-velocity path between the same two events.

 

[Grep]

I recognize quality explaining when I see it, and this thread delivers.

 

[ghwellsjr]

I said:

Both counters measure exactly the same signal rate coming from the other counter, in this case 200 signals per minute.

Then you said:

OK. Maybe you have found where there is confusion. To me, the signal rate is what the clock uses to determine how fast the other clock is counting. So the clocks appear to perceive one another as counting faster than themselves, not slower.

And from that, I deduced that since I had just said that each counter measures the signal rate coming from the other counter at 200 signals per second (when they are each sending out 1 signal per minute), that you had concluded that they "appear to perceive one another as counting faster than themselves, not slower". And so I said:

No, the rate at which the signals arrive is not the same as the rate they are sent out and this is caused by the relative motion.

And then you said:

I did not say that the rate at which they arrived is the same as the rate at which they are sent out. I don't know where you got that from. In fact, I said the exact opposite. Yes what you say is obvious to me.

If it's obvious to you, then why did you say "appear to perceive one another as counting faster than themselves, not slower"? Do you understand that there is a reciprocal relationship between the rates at which two moving observers receive periodic signals from the other one when they are both emitting at the same rate and that this can only happen if they each perceive the other's clock as running slow? And do you understand that the reason why the rate is much higher than might be obvious is because the traveling counter is going toward the stationary counter and as soon as it passes the stationary counter, the rate suddenly drops from 200 signals per minute to 0.005 signals per minute (or one signal every 200 minutes) because they are now getting farther apart, but the same reciprocal time dilation applies? I'm only asking because these things are not obvious to me.

Also I need to correct something in this thought experiment that arises from my post #30 there. The stationary counter could not possibly send out its first pulse when the device is triggered.

So the thought experiment needs to be revised. When the stationary counter receives its first pulse, it calculates the elapsed time in its reference frame based on the distance from the triggered device and begins to send pulses back to the moving counter.

We can clarify this experiment in the following way:

These counters simply count at an even time interval in their own reference frame, and they also count incoming light pulses.

Once the stationary counter has done its calculation and set its counter, it sends out a number of pulses equivalent to the its starting count, and then continues to count at a 1 minute interval.

When the moving counter approaches the stationary counter, it has counted fewer times than the number of pulses it has received and the stationary counter has counted greater times than the number of pulses it received.

I wish you hadn't brought this up. I really don't understand what the problem is that you are trying to fix here. Look at my interpretation in post #31 of your scenario. Since you are defining what is happening in a particular reference frame, it is perfectly legitimate for you to say that both counters start at the same exact time. So let's go back to that scenario and don't worry about how the counters knew when to start counting and simultaneously start emitting their signals.

OK
The question then becomes this:

If the two counters are equally relative to one another, how did one counter send more pulses than the other? Both the number of pulses and the number of counts should be equal.
.

I thought I explained this in post #31 but let me reiterate. The traveling counter is counting at a slower rate than the stationary counter. In fact, it is counting at 1% so it only emits 1/100 of the number of signals and its counter is 1/100 of the stationary counter at the time when they meet. So the traveling counter emitted 5256 signals and the stationary counter counted 5256 signals and when they met, the traveling counter communicated to the stationary counter that it was on count 5256 which matched what the stationary counter counted. And during the same one-year interval of time, the stationary counter emitted 525600 signals and the traveling counter counted all 525600 signals and when they met, the stationary counter communicated to the traveling counter that it was on count 525600 which matched what the traveling counter counted.

Now, maybe you are asking why there are two different counter numbers involved here and that's because, as far as the traveling counter is concerned, the whole scenario took place in 3.65 days whereas for the stationary counter, it took a whole year. But they both are emitting at the same rate of 1 signal per minute (according to their own timebase) and they are both detecting (once the detection starts) at 200 signals per minute (again, according to their own timebase), which, based on the relative velocity and knowing the rate of signal emission, they can each calculate the other one to have a slow timebase of 1% of their own.

OK
I don't believe that this defies special relativity. I think there is a difference between what can be perceived by data transmission and what special relativity says about relative motion and time.

Of course it doesn't defy SR but I don't know what you mean in the next sentence--no idea at all.

So now are all the issues cleared up or do you still have unresolved questions?

 

[teachmemore]

I said:

Then you said:

And from that, I deduced that since I had just said that each counter measures the signal rate coming from the other counter at 200 signals per second (when they are each sending out 1 signal per minute), that you had concluded that they "appear to perceive one another as counting faster than themselves, not slower". And so I said:

And then you said:

If it's obvious to you, then why did you say "appear to perceive one another as counting faster than themselves, not slower"? Do you understand that there is a reciprocal relationship between the rates at which two moving observers receive periodic signals from the other one when they are both emitting at the same rate and that this can only happen if they each perceive the other's clock as running slow? And do you understand that the reason why the rate is much higher than might be obvious is because the traveling counter is going toward the stationary counter and as soon as it passes the stationary counter, the rate suddenly drops from 200 signals per minute to 0.005 signals per minute (or one signal every 200 minutes) because they are now getting farther apart, but the same reciprocal time dilation applies? I'm only asking because these things are not obvious to me.

I wish you hadn't brought this up. I really don't understand what the problem is that you are trying to fix here. Look at my interpretation in post #31 of your scenario. Since you are defining what is happening in a particular reference frame, it is perfectly legitimate for you to say that both counters start at the same exact time. So let's go back to that scenario and don't worry about how the counters knew when to start counting and simultaneously start emitting their signals.

I thought I explained this in post #31 but let me reiterate. The traveling counter is counting at a slower rate than the stationary counter. In fact, it is counting at 1% so it only emits 1/100 of the number of signals and its counter is 1/100 of the stationary counter at the time when they meet. So the traveling counter emitted 5256 signals and the stationary counter counted 5256 signals and when they met, the traveling counter communicated to the stationary counter that it was on count 5256 which matched what the stationary counter counted. And during the same one-year interval of time, the stationary counter emitted 525600 signals and the traveling counter counted all 525600 signals and when they met, the stationary counter communicated to the traveling counter that it was on count 525600 which matched what the traveling counter counted.

Now, maybe you are asking why there are two different counter numbers involved here and that's because, as far as the traveling counter is concerned, the whole scenario took place in 3.65 days whereas for the stationary counter, it took a whole year. But they both are emitting at the same rate of 1 signal per minute (according to their own timebase) and they are both detecting (once the detection starts) at 200 signals per minute (again, according to their own timebase), which, based on the relative velocity and knowing the rate of signal emission, they can each calculate the other one to have a slow timebase of 1% of their own.

Of course it doesn't defy SR but I don't know what you mean in the next sentence--no idea at all.

So now are all the issues cleared up or do you still have unresolved questions?

All agreed.

But, the issue is that the traveling counter has seen 525600 signals over a duration of 3.65 days. While the stationary counter has seen 5256 signals over the duration of a year. Which means that the stationary counter sees the traveling counter as running slower than itself, while the traveling counter sees the stationary counter as running faster than itself. You don't see a problem with this? I think most would say that both counters should see the other one as running slow. But that doesn't appear to be the case.

These signals are what the counters use to measure the rate of time in the alternate reference frame. Since both counters are approaching one another, should they not both see the other counter as counting slower than themselves? If the motion of these reference frames is only "relative", than why should time in one reference frame move slower than time in the other?

What I'm trying to say is that the fact that one counter is moving faster than the other is significant to this problem. That these two counters are moving relative to one another is not enough information, we also need to know that one is moving near c, and the other is not.

Does this make sense?

The light signals are like the ticking of a clock. These light signals are what are used to perceive the rate of time in the alternate reference frame.

 

[JesseM]

All agreed.

But, the issue is that the traveling counter has seen 525600 signals over a duration of 3.65 days. While the stationary counter has seen 5256 signals over the duration of a year. Which means that the stationary counter sees the traveling counter as running slower than itself, while the traveling counter sees the stationary counter as running faster than itself. You don't see a problem with this? I think most would say that both counters should see the other one as running slow. But that doesn't appear to be the case.

It will be the case if you look at the rate that each is seeing signals during a period when they were actually receiving signals the whole time, not the total number of signals each received during a period which might include a lot of "dead air" because it was before they had received the first signal from the other one. In this example the gamma-factor is 100, so the relative velocity must be 0.999949998749938c, which means the relativistic Doppler shift is sqrt[(1 + 0.999949998749938)/(1 - 0.999949998749938)] = 199.994999875905, so if each is sending signals at 1 per minute, then during any period when they are receiving signals from the other, they will be receiving them at a rate of about 199.995 per minute.

What I'm trying to say is that the fact that one counter is moving faster than the other is significant to this problem. That these two counters are moving relative to one another is not enough information, we also need to know that one is moving near c, and the other is not.

Does this make sense?

Definitely not, there is no objective truth about which is "moving near c" and which is "stationary", that depends on your frame of reference. Whatever is happening with the counters in the "stationary" frame, you could construct an analogous pair where the choice of when each counter started sending signals was different, in such a way that the counter which was "travelling" relative to the stationary frame would be the one to receive 5265 signals over the course of a year of its own proper time (would actually be 100 years in the stationary frame), while the counter that was at rest in the stationary frame would receive 525600 over 3.65 days in the stationary frame, and yet each counter started sending signals simultaneously in the rest frame of the traveling counter (not simultaneously in the stationary frame).

 

[teachmemore]

No, the rate at which the signals arrive is not the same as the rate they are sent out and this is caused by the relative motion. The point is that when both counters/clocks/observers are sending out pulses/signals/ticks at what they perceive to be the same rate as defined by their own clocks, then the rate at which they are received by the other one, taking into account the light travel time, their relative motion, and as measured by their own clock, results in them both measuring an identical signal rate coming from the other one.

In your scenario, you had the traveling counter approaching the stationary one which made the signal rate go up. But consider what happens when the traveler goes past the stationary one. Now the signal rate will go down from 200 per minute to 1 every 200 minutes but it still calculates to the same reciprocal time dilation factor.

Oh geeze. I am so sorry. I do apologize. What you said here was clearly not obvious to me, because it was exactly what I've been trying to put my finger on, and it is exactly what I meant when I made that confusing statement about data transmission being different than relative motion/time dilation.

Ya. My whole point by all this is that what somebody visualizes from their frame of reference doesn't necessarily correspond with the time dilation factors.

The light pulses reaching the counters are what is visualized; which is different than the relative time of these two reference frames. Which is the same for both. And each alternate reference frame is slower for both; even though what they visualize is very different.

 

[JesseM]

Oh geeze. I am so sorry. I do apologize. What you said here was clearly not obvious to me, because it was exactly what I've been trying to put my finger on, and it is exactly what I meant when I made that confusing statement about data transmission being different than relative motion/time dilation.

Ya. My whole point by all this is that what somebody visualizes from their frame of reference doesn't necessarily correspond with the time dilation factors.

The light pulses reaching the counters are what is visualized; which is different than the relative time of these two reference frames. Which is the same for both. And each alternate reference frame is slower for both; even though what they visualize is very different.

By "what they visualize" do you mean how fast they see the signals coming in? Like I said, that's given by the relativistic Doppler shift equation, and for two observers moving at constant relative velocity the visual rates are completely symmetrical (if counter #1 sees signals from counter #2 coming in at about 200 per minute, then counter #2 must also see signals from counter #1 coming in at about 200 per minute).

 

[ghwellsjr]

It's all explained, right here in post #31:

...
Just think about something. When this scenario starts, neither counter will have any knowledge of what the other counter is doing. It will take almost one year for the signals coming from the traveling counter to reach the stationary counter and then within less than half an hour, all the signals will arrive in a burst and then the counter will arrive. At a speed of 0.99995c, the traveling counter's time has slowed to 1% of normal. Since there are 525600 minutes in a year, this counter will have only sent out 5256 signals during his entire trip but they will arrive during the last 26.28 minutes (1-0.99995 or 0.00005 times the number of minutes in a year). That's a rate of 200 signals per minute as measured by the stationary clock.

On the other hand, the traveling counter will not see anything from the stationary counter until after about half a year. Then it will start seeing signals coming in two per minute (they are traveling towards him at the speed of light and he is traveling toward them at almost the speed of light, which approximately doubles the rate at which he receives the signals), except since his clock is running at 1% of normal, he will think they are coming in at 100 times that rate which is 200 per minute. And by the time he gets to the stationary clock, he will have received almost a year's worth of signals sent out at one per minute or very nearly 525600.

 

[teachmemore]

By "what they visualize" do you mean how fast they see the signals coming in? Like I said, that's given by the relativistic Doppler shift equation, and for two observers moving at constant relative velocity the visual rates are completely symmetrical (if counter #1 sees signals from counter #2 coming in at about 200 per minute, then counter #2 must also see signals from counter #1 coming in at about 200 per minute).

How is this symmetrical if there is an unequal number of pulses exchanged between these counters? At any time near the end of the voyage, the traveling counter can determine that he is moving faster than the stationary one by comparing the total pulses received with the total counts made.

 

[teachmemore]

It will be the case if you look at the rate that each is seeing signals during a period when they were actually receiving signals the whole time, not the total number of signals each received during a period which might include a lot of "dead air" because it was before they had received the first signal from the other one. In this example the gamma-factor is 100, so the relative velocity must be 0.999949998749938c, which means the relativistic Doppler shift is sqrt[(1 + 0.999949998749938)/(1 - 0.999949998749938)] = 199.994999875905, so if each is sending signals at 1 per minute, then during any period when they are receiving signals from the other, they will be receiving them at a rate of about 199.995 per minute.

Definitely not, there is no objective truth about which is "moving near c" and which is "stationary", that depends on your frame of reference. Whatever is happening with the counters in the "stationary" frame, you could construct an analogous pair where the choice of when each counter started sending signals was different, in such a way that the counter which was "travelling" relative to the stationary frame would be the one to receive 5265 signals over the course of a year of its own proper time (would actually be 100 years in the stationary frame), while the counter that was at rest in the stationary frame would receive 525600 over 3.65 days in the stationary frame, and yet each counter started sending signals simultaneously in the rest frame of the traveling counter (not simultaneously in the stationary frame).

This here is what I need to think about.

I was under the impression that the traveling counter counts more slowly simply because it is travelling, and so, knowledge about which one is traveling is crucial to determining which counter counts the fewest times.

 

[JesseM]

How is this symmetrical if there is an unequal number of pulses exchanged between these counters? At any time near the end of the voyage, the traveling counter can determine that he is moving faster than the stationary one by comparing the total pulses received with the total counts made.

Again, the rate that they each see signals arriving from the other counter is symmetrical. But if you impose the rule that they both start sending signals at the same time in the stationary frame, then the stationary counter won't see signals start to arrive (beginning with the first one sent by the traveling counter) until shortly before they meet, while the traveling counter will see signals start to arrive long before meeting the stationary counter. But this asymmetry has nothing to do with the basic laws of physics, it's just a consequence of the rule that both started sending signals simultaneously in the stationary frame; if you instead imposed the rule that both started sending signals simultaneously in the traveller's rest frame, then the reverse would be true, with the stationary counter seeing signals start to arrive long before they meet and the traveling counter not seeing signals start to arrive until shortly before they meet.

 

[JesseM]

This here is what I need to think about.

I was under the impression that the traveling counter counts more slowly simply because it is travelling

By "counts more slowly" do you just mean how fast its own clock is ticking, or are you talking about the rate it receives signals from the other counter? If you're just talking about clock rates (the rate it counts seconds), then relative to the "stationary" frame its clock does tick more slowly, but not in any absolute sense. In the traveling counter's own rest frame, its clock is ticking normally while the stationary counter's clock is ticking slowly.

and so, knowledge about which one is traveling is crucial to determining which counter counts the fewest times.

It's crucial to understand that in relativity all inertial frames are equally valid as far as the basic laws of physics are concerned (the http://nobelprize.org/educational/physics/relativity/postulates-1.html), there is no frame-independent objective truth about who is "really" moving at a greater velocity, for any pair of objects in relative motion you can choose either object's inertial rest frame to analyze the situation, and the laws of physics will work the same way in this frame as any other frame (including the law that says a clock moving at velocity v relative to the frame is running slow by \sqrt{1 - v^2/c^2} relative to the time coordinate of that frame).