Paradox within the twin paradox

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  • #1
cshum00
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1) So let's say we have twin A and twin B.
2) Twin A stays on Earth.
3) Twin B goes on a trip nearly at the speed of light.
4) Twin A sees twin's B clock moving slower.
5) Twin B sees twin's A clock moving slower.
6) When twin B returns, twin A is older.

#6 Implies that twin B's clock has indeed moved slower. But how can it be since #4 and #5 says that each twin see the other twin's clock moving slower?
 

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  • #2
JesseM
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1) So let's say we have twin A and twin B.
2) Twin A stays on Earth.
3) Twin B goes on a trip nearly at the speed of light.
4) Twin A sees twin's B clock moving slower.
5) Twin B sees twin's A clock moving slower.
6) When twin B returns, twin A is older.

#6 Implies that twin B's clock has indeed moved slower. But how can it be since #4 and #5 says that each twin see the other twin's clock moving slower?
That's not a paradox within the twin paradox, that is the twin paradox (i.e. the seeming paradox is the argument that each should be able to say the other is moving and therefore aging slower, but only one can actually be younger when they reunite). The resolution is that the time dilation formula which says moving clocks run slower than stationary ones only works in inertial frames of reference, but in order for the two twins to move apart and then later reunite one of them has to accelerate to turn around, a non-inertial form of motion (and he'll be able to tell he was moving non-inertially because he'll feel G-forces during the acceleration, which can be measured by an accelerometer). For a more extensive discussion of the twin paradox, this page is pretty good.
 
  • #4
teachmemore
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What if the twins where born on two separate planets in the galaxy at the same time (by a supernatural being). Then one twin decided to travel to see the other twin many light years across the galaxy. He flew at close to the speed of light across the galaxy until he finally approached the planet of the other twin. Before he decelerates to land at the planet, has he not aged more than his twin?
 
  • #5
teachmemore
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Oh wait. duh. Lets say the twin is born in travel. No acceleration either.
 
  • #6
teachmemore
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Can not this problem be greatly simplified?

There are two stop clocks/watches (counters from 0 to infinity) in space. 1 clock is moving toward the other at near c. Both clocks are simultaneously reset to 0. When the moving clock passes the stationary clock, it has ticked fewer times.

No acceleration. Just two counters in space.

Surely, there is a way to resolve this paradox without bringing acceleration into it.
 
  • #7
ghwellsjr
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What if the twins where born on two separate planets in the galaxy at the same time (by a supernatural being). Then one twin decided to travel to see the other twin many light years across the galaxy. He flew at close to the speed of light across the galaxy until he finally approached the planet of the other twin. Before he decelerates to land at the planet, has he not aged more than his twin?

I guess if a supernatural being is involved, then that supernatural being can choose to make the twins to be born at the same time, but unless that supernatural being decides to communicate that fact to us beings bound by Mother Nature and which reference frame that supernatural being prefers, then we won't have any way of knowing that they were born at the same time in a particular reference frame.

But we can image that two twins widely separated in space were born at the same time in a particular reference frame and then see what happens when one or both of them travel in that same reference frame, and then we can transform the whole situation into another reference frame and see what happens in that other reference frame.

You haven't specified any reference frame for your example but let's suppose it is the one in which the twins are both stationary when they were born. Now one of them travels as you describe and meets his twin. The traveling twin will be younger, not older, than the twin that remained at rest in the reference frame. You got it backwards. Don't be concerned about accelerations, they don't significantly change the analysis.

But this is not the whole truth because, as I pointed out in my referenced link, if we analyze the situation from another reference frame we could conclude that the other twin ended up younger or from another reference frame they end up the same age. It is not possible to make an absolute statement about the ages of the twins in your scenario, except that the aging rate changes while the one twin accelerates.
 
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  • #8
ghwellsjr
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Oh wait. duh. Lets say the twin is born in travel. No acceleration either.

As I said in the previous post, the acceleration won't significantly change the conclusions that we come to.

But now you have made the problem of not specifying a reference frame more obvious because we have each twin at rest in a different reference frame. In the "traveling" twin's reference frame, the "stationary" twin will age at a slower rate and in the "stationary" twin's reference frame, the "traveling" twin will age at a slower rate. You also have to be concerned about how you define them to be born at the same time since they are no longer in the same reference frame.
 
  • #9
teachmemore
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I guess if a supernatural being is involved, then that supernatural being can choose to make the twins to be born at the same time, but unless that supernatural being decides to communicate that fact to us beings bound by Mother Nature and which reference frame that supernatural being prefers, then we won't have any way of knowing that they were born at the same time in a particular reference frame.

But we can image that two twins widely separated in space were born at the same time in a particular reference frame and then see what happens when we one or both of them travel in that same reference frame, and then we can transform the whole situation into another reference frame and see what happens in that other reference frame.

You haven't specified any reference frame for your example but let's suppose it is the one in which the twins are both stationary when they were born. Now one of them travels as you describe and meets his twin. The traveling twin will be younger, not older, than the twin that remained at rest in the reference frame. You got it backwards. Don't be concerned about accelerations, they don't significantly change the analysis.

But this is not the whole truth because, as I pointed out in my referenced link, if we analyze the situation from another reference frame we could conclude that the other twin ended up younger or from another reference frame they end up the same age. It is not possible to make an absolute statement about the ages of the twins in your scenario, except that the aging rate changes while the one twin accelerates.

Ya. That was a typo. And please read my successive posts. Neither twin accelerates.

Ok. Sorry, the reference frame is that of the stationary twin/clock. When the one twin/clock passes within close range, the stationary twin takes a peak at his clock and sees that it has not ticked as many times as his own.
 
  • #10
teachmemore
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Ah, but both twins take a peak at one another's clock.

The twin/clock that is passing near c, takes a peak at the stationary one and sees that it has ticked more times than his own.
 
  • #11
teachmemore
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As I said in the previous post, the acceleration won't significantly change the conclusions that we come to.

But now you have made the problem of not specifying a reference frame more obvious because we have each twin at rest in a different reference frame. In the "traveling" twin's reference frame, the "stationary" twin will age at a slower rate and in the "stationary" twin's reference frame, the "traveling" twin will age at a slower rate. You also have to be concerned about how you define them to be born at the same time since they are no longer in the same reference frame.

That is the paradox. I don't see how you are reconciling it. When they see each other, as they pass at close range, their observations do not correspond with that which they have perceived for the trip. That is the paradox.
 
  • #12
teachmemore
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Lets make the scenario even more clear. The clocks show the number of ticks in symbolic notation.

So that when the moving twin passes the stationary twin, from his reference frame, he sees that the number of ticks on the clock in the stationary reference frame is a much larger number than his own.

On the other hand, the stationary twin, from his reference frame, looks at the clock in the moving reference frame and sees a number that is smaller than the number on his clock.
 
  • #13
ghwellsjr
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Can not this problem be greatly simplified?

There are two stop clocks/watches (counters from 0 to infinity) in space. 1 clock is moving toward the other at near c. Both clocks are simultaneously reset to 0. When the moving clock passes the stationary clock, it has ticked fewer times.

No acceleration. Just two counters in space.

Surely, there is a way to resolve this paradox without bringing acceleration into it.

As I said in the previous post, you haven't specified a reference frame in which to describe your scenario so we will get different answers just by choosing different reference frames. You need to pay attention to relativity of simultaneity.
 
  • #14
teachmemore
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As you pointed out, acceleration just confuses the problem and has no significance in resolving the paradox. That is why I removed it.
 
  • #15
teachmemore
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As I said in the previous post, you haven't specified a reference frame in which to describe your scenario so we will get different answers just by choosing different reference frames. You need to pay attention to relativity of simultaneity.

Now you are just beating around the bush. The reference frames are clearly implied and have been explicitly stated in following posts.
 
  • #16
HallsofIvy
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If you have two separate beings each moving relative to other, each will see that others clock as ticking slower and will see the other person as aging more slowly. There is no paradox there. That's just "relativity".

The only time a "paradox" occurs is if you insist that the two people, or clocks, start out stationary relative to one another and then, later, after the motion of one at high speed relative to the other, they are again stationary relative to one another. Since both see the other as aging more slowly how can one be younger than the other? Of course, to do that, have them starionary relative to one another at one time, moving rapidly relative to one another, then stationary again, requires acceleration.
 
  • #17
teachmemore
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I will state the reference frames again for you.

Frame 1: The reference frame of the twin/clock moving near c.
Frame 2: The stationary frame of the second twin/clock.

Take the reading from both reference frames of the clock in the other reference frame and compare them to the clock in their own reference frames.
 
  • #18
ghwellsjr
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The "resolution" of the paradox is that we cannot say anything about times on clocks that move around, only clocks that are stationary with respect to each other (in other words, stationary in the same reference frame), because when we transform moving clocks from one reference frame to another, we can get different answers, depending on the reference frames we choose. If you want to compare in an absolute sense, the times displayed on two clocks, they have to start in the same location, move around any way you want, and then come back together in the same location, then the times on the two clocks, and thus their histories of aging rates, will give the same answer no matter what reference frame you analyze the situation in.
 
  • #19
teachmemore
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If you have two separate beings each moving relative to other, each will see that others clock as ticking slower and will see the other person as aging more slowly. There is no paradox there. That's just "relativity".

The only time a "paradox" occurs is if you insist that the two people, or clocks, start out stationary relative to one another and then, later, after the motion of one at high speed relative to the other, they are again stationary relative to one another. Since both see the other as aging more slowly how can one be younger than the other? Of course, to do that, have them starionary relative to one another at one time, moving rapidly relative to one another, then stationary again, requires acceleration.

I understand that already. You have not said anything new that clears up the paradox presented here.

Maybe there is a problem with what I am proposing.

I am proposing that as these twins/clocks pass one another - they are very close, but still moving in alternate reference frames - they take a peek at one another's clock and see a discrepancy. If this is not possible, please explain to me why it is not possible that they should be able to see one another's clocks while they are moving in different reference frames.

Thank you.
 
  • #20
teachmemore
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The "resolution" of the paradox is that we cannot say anything about times on clocks that move around, only clocks that are stationary with respect to each other (in other words, stationary in the same reference frame), because when we transform moving clocks from one reference frame to another, we can get different answers, depending on the reference frames we choose. If you want to compare in an absolute sense, the times displayed on two clocks, they have to start in the same location, move around any way you want, and then come back together in the same location, then the times on the two clocks, and thus their histories of aging rates, will give the same answer no matter what reference frame you analyze the situation in.

Ah! Thank you ghwellsjr! Now we are getting somewhere. I will think about this for a while and come back with more questions if there is something about this explanation that doesn't make sense to me.
 
  • #21
ghwellsjr
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I will state the reference frames again for you.

Frame 1: The reference frame of the twin/clock moving near c.
Frame 2: The stationary frame of the second twin/clock.

Take the reading from both reference frames of the clock in the other reference frame and compare them to the clock in their own reference frames.

I hope you understand that you have to specify both twins in a single reference frame and then you can do it over again by specifying both twins in a different reference frame and you can't cheat, you have to use the Lorentz Transform and you have to pay attention to the issue of simultaneousness. Otherwise, you will be specifying two different scenarios that seem like they are the same scenario and then when you get two different final answers, you will think that there is a paradox when in reality, you started out with two different situations.

It is a lot easier to start with the twins together at the same location at the same age because that scenario does not require any difficult transformation when you analyze it from different reference frames and there are no simultaneity issues to contend with. That is what I did in my referenced link in post #3. Then it is easy to see why the different reference frames show different age differences between the twins half way through the Twin Paradox and why the different reference frames show the same age difference when the twins come together at the end. Once you understand that situation, you can split the problem apart like you are trying to do and see why it is the specification of the problem that results in the different solutions you are getting.
 
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  • #22
teachmemore
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I hope you understand that you have to specify both twins in a single reference frame and then you can do it over again by specifying both twins in a different reference frame and you can't cheat, you have to use the Lorentz Transform and you have to pay attention to the issue of simultaneous. Otherwise, you will be specifying two different scenarios that seem like they are the same scenario and then when you get two different final answers, you will think that there is a paradox when in reality, you started out with two different situations.

It is a lot easier to start with the twins together at the same location at the same age because that scenario does not require any difficult transformation when you analyze it from different reference frames and there are no simultaneity issues to contend with. That is what I did in my referenced link in post #3. Then it is easy to see why the different reference frames show different age differences between the twins half way through the Twin Paradox and why the different reference frames show the same age difference when the twins come together at the end. Once you understand that situation, you can split the problem apart like you are trying to do and see why it is the specification of the problem that results in the different solutions you are getting.

OK. So what you are telling me is that it is impossible to synchronize two clocks that are in different frames of reference?

Please confirm. Thank you.
 
  • #23
teachmemore
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I hope you understand that you have to specify both twins in a single reference frame and then you can do it over again by specifying both twins in a different reference frame and you can't cheat, you have to use the Lorentz Transform and you have to pay attention to the issue of simultaneous. Otherwise, you will be specifying two different scenarios that seem like they are the same scenario and then when you get two different final answers, you will think that there is a paradox when in reality, you started out with two different situations.

It is a lot easier to start with the twins together at the same location at the same age because that scenario does not require any difficult transformation when you analyze it from different reference frames and there are no simultaneity issues to contend with. That is what I did in my referenced link in post #3. Then it is easy to see why the different reference frames show different age differences between the twins half way through the Twin Paradox and why the different reference frames show the same age difference when the twins come together at the end. Once you understand that situation, you can split the problem apart like you are trying to do and see why it is the specification of the problem that results in the different solutions you are getting.

Ah yes. I totally understand what you mean when you talk about cheating though. I will think about the actual exchanges of information between these reference frames in more detail.
 
  • #24
ghwellsjr
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OK. So what you are telling me is that it is impossible to synchronize two clocks that are in different frames of reference?

Please confirm. Thank you.

There are two issues with your question. First off, the term "synchronize two clocks" means to end up with two clocks that continue to maintain the same time on both of them. This can only happen if they are at rest with each other and they never accelerate, in other words, they continue to be at rest in a particular reference frame. We can define them to be synchronized in our thought experiments or we can go through the thought process to bring them into synchronization in our thought experiments or we can go through the actual steps to synchronize real clocks in real experiments. But you have to understand that whether we merely define them to be synchronized in our thought experiment, or we go through the steps to bring them into synchronization in our thought or real experiments, we are still defining them to be synchronized. This is what Einstein's second postulate is all about. It defines the one way speed of light to be the same in all directions in a particular reference frame.

The second issue is concerning clocks that are in motion relative to each other. Now, of course, they can no longer maintain their sychronization because they will be running at different rates but we can set them to the same time when they are in the same location. Really, we set one of them to the time on the other one at the moment they are colocated.
 
  • #25
Grep
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OK. So what you are telling me is that it is impossible to synchronize two clocks that are in different frames of reference?

Please confirm. Thank you.
Makes it pretty hard to synchronize anything when people in two different frames of reference disagree about what was, or wasn't, simultaneous.
 
  • #26
cshum00
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That's not a paradox within the twin paradox, that is the twin paradox (i.e. the seeming paradox is the argument that each should be able to say the other is moving and therefore aging slower, but only one can actually be younger when they reunite). The resolution is that the time dilation formula which says moving clocks run slower than stationary ones only works in inertial frames of reference, but in order for the two twins to move apart and then later reunite one of them has to accelerate to turn around, a non-inertial form of motion (and he'll be able to tell he was moving non-inertially because he'll feel G-forces during the acceleration, which can be measured by an accelerometer). For a more extensive discussion of the twin paradox, this page is pretty good.

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.

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?

I guess there is a missing piece in my basics to tie the right knowledge together.
 
  • #27
teachmemore
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OK. So the thought experiment is revised. I have removed any forms of cheating. Everything takes place within the confines of reality/special relativity. Hopefully this is more well defined and clear.

We have two counters that can read and compare one another's count via an encrypted light pulse.

The two counters are in two different frames of reference. One is moving near c. The other is stationary. The moving reference frame is moving directly toward the non-moving reference frame.

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. (there are other ways to accomplish this, but this coincidence is physically possible and simple) (knowledge of this coincidence by either counter is not required for the thought experiment to be paradoxical)

The two light pulses, carrying information to each clock of the reset, reach each clock. At this point in time, from each of their own respective reference frames, they "see" the other clock reset itself, while in their own reference frame, they have already begun counting.

They continue to send out light pulses to one another at every count. The counter moving at a speed near c continues on toward the second counter.

When these two reference frames reach each other, they should be able to instantly compare counts at the instant they touch. At this very instant, the moving counter should have a smaller count than the stationary reference frame and these consecutive counts should have been confirmed by all data transmitted between the two counters during the expedition, so that all exchanges of information are consistent with the final result.

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; because the moving counter read the stationary counter as being at 0 when it read its own counter at a value greater than 0. By the time these two counters touched and instantaneously exchanged the data of their current count, the moving counter would realize that it is now behind the stationary counter. Even though they are still in two different, but touching reference frames, this exchange of data is still made possible by special relativity, and their should be no discrepancy in this data exachange. To make it more clear why this is all possible, lets use some figures. Lets say that they are not touching but are 1 light second apart, yet their counters have been ticking at 1 minute intervals. By the time they are 1 light second apart, they should be able to transmit one another's count between reference frames in a single second and the moving counter will be able to determine that its count is behind that of the stationary counter - all within the laws of special relativity.

Please let me know if anything about this thought experiment is unclear.

Basically, for the counter in the moving reference frame to be able to ever receive a light pulse from the other reference frame that tells it its own count is behind, it would have to receive the information of each concecutive count prior to it. These two counters should be able to approach one another close enough in order to exchange information showing the moving counter that its count is behind the stationary counter, without changing reference frames, which means that the moving counter is perceiving the counter in the stationary reference frame as counting faster than itself.

This, to me, appears completely, mathematically provable. Please show any errors or assumptions I may have made.

I think some would claim that this defies special relativity. But I don't believe that it does.
 
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  • #28
teachmemore
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Makes it pretty hard to synchronize anything when people in two different frames of reference disagree about what was, or wasn't, simultaneous.

My new revised thought experiment clears that up. The paradox is independent of any agreement between the two reference frames. The paradox comes about from only what is perceived by each reference frame from the moment the experiment is set up.
 
  • #29
teachmemore
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There are two issues with your question. First off, the term "synchronize two clocks" means to end up with two clocks that continue to maintain the same time on both of them. This can only happen if they are at rest with each other and they never accelerate, in other words, they continue to be at rest in a particular reference frame. We can define them to be synchronized in our thought experiments or we can go through the thought process to bring them into synchronization in our thought experiments or we can go through the actual steps to synchronize real clocks in real experiments. But you have to understand that whether we merely define them to be synchronized in our thought experiment, or we go through the steps to bring them into synchronization in our thought or real experiments, we are still defining them to be synchronized. This is what Einstein's second postulate is all about. It defines the one way speed of light to be the same in all directions in a particular reference frame.

The second issue is concerning clocks that are in motion relative to each other. Now, of course, they can no longer maintain their sychronization because they will be running at different rates but we can set them to the same time when they are in the same location. Really, we set one of them to the time on the other one at the moment they are colocated.

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.
 
  • #30
teachmemore
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There are two issues with your question. First off, the term "synchronize two clocks" means to end up with two clocks that continue to maintain the same time on both of them. This can only happen if they are at rest with each other and they never accelerate, in other words, they continue to be at rest in a particular reference frame. We can define them to be synchronized in our thought experiments or we can go through the thought process to bring them into synchronization in our thought experiments or we can go through the actual steps to synchronize real clocks in real experiments. But you have to understand that whether we merely define them to be synchronized in our thought experiment, or we go through the steps to bring them into synchronization in our thought or real experiments, we are still defining them to be synchronized. This is what Einstein's second postulate is all about. It defines the one way speed of light to be the same in all directions in a particular reference frame.

The second issue is concerning clocks that are in motion relative to each other. Now, of course, they can no longer maintain their sychronization because they will be running at different rates but we can set them to the same time when they are in the same location. Really, we set one of them to the time on the other one at the moment they are colocated.

Your second point regarding the synchronization of two clocks could be used to synchronize two clocks at a given point in time that are in different reference frames and different locations.

Of course, this synchronization has to be done relative to a reference frame, so it would be done relative to the stationary reference frame.

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.
 
  • #31
ghwellsjr
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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.
 
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  • #32
JesseM
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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.
 
  • #33
teachmemore
77
0
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 travelling 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.
 
  • #34
teachmemore
77
0
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.
 
  • #35
teachmemore
77
0
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.
 

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