What Causes Time Dilation in the Twin Paradox?

[Mohammed_I]

I can't fully understand why a person who makes a journey into space in a high-speed rocket will return home to find his age less than an identical twin who stayed on Earth. It makes since for the twin who stayed on earth, but for the twin who traveled into space, he sees himself at rest and sees Earth moving at constant velocity. so he should feel that time on Earth is moving slower than his time, this way he will find his twin younger than him not older. Please tell me what I am missing to help me understand this thought experiment.

Thank you.

 

[gendou2]

The trick is that one twin (the one in the rocket ship) undergoes increased acceleration (much more than 9.8 m/s), while the other does not.

http://en.wikipedia.org/wiki/Twin_paradox

 

[Fredrik]

That only proves that the two twins aren't in identical situations. It implies that something can be different, but it doesn't explain why something is different.

The part that's hard to get is that there's no way for the astronaut twin to turn his ship around without having his brother's age "jump ahead" by a large amount. You need to understand simultaneity in SR before you can understand that.

Check out http://web.comhem.se/~u87325397/Twins.PNG .

I'm calling the twin on Earth "A" and the twin in the rocket "B".
Blue lines: Events that are simultaneous in the rocket's frame when it's moving away from Earth.
Red lines: Events that are simultaneous in the rocket's frame when it's moving back towards Earth.
Cyan (light blue) lines: Events that are simultaneous in Earth's frame.
Dotted lines: World lines of light rays.
Vertical line in the upper half: The world line of the position (in Earth's frame) where the rocket turns around.
Green curves in the lower half: Curves of constant -t^2+x^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket left Earth.
Green curves in the upper half: Curves of constant -(t-20)^2+(x-16)^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket turned around.

 

[Mohammed_I]

Thank you very much Fredrik, this helped me a lot. Thank you too Gendou

 

[neopolitan]

Check out http://web.comhem.se/~u87325397/Twins.PNG .

I wrote a longer post, but it got lost.

Fundamentally, I don't buy the discontinuity at the turn around point. "A" appears to age 7 or so years on "B"'s outward leg, and around 33 years on the return leg.

You can think about it with "A" sending out constant updates by radio signal, "I have aged 1 day" - once a day, every day and working out when and where these messages and "B" will be collocated. You should be able to see that "B" will receive 7 years worth of messages on the way out and 33 years worth on the way home. There will be a few messages during the turn around period, but not 25.6 years worth. (How many depends entirely on the rates of acceleration and thus how quickly "B" turns around.)

cheers,

neopolitan

 

[Dale]

You can think about it with "A" sending out constant updates by radio signal, "I have aged 1 day" - once a day, every day and working out when and where these messages and "B" will be collocated. You should be able to see that "B" will receive 7 years worth of messages on the way out and 33 years worth on the way home. There will be a few messages during the turn around period, but not 25.6 years worth. (How many depends entirely on the rates of acceleration and thus how quickly "B" turns around.)

You are correct, what you are describing is essentially the relativistic Doppler effect for a signal frequency of 12 uHz, but presumably you would want to correct for the light travel time. So you would get an "I have aged 1 day" signal, note how far you are from Earth at that time (in your frame), and calculate when the signal was sent to determine when he had aged. You wouldn't assume that he had aged when you received the signal, but rather when he sent the signal.

 

[neopolitan]

You are correct, what you are describing is essentially the relativistic Doppler effect for a signal frequency of 12 uHz, but presumably you would want to correct for the light travel time. So you would get an "I have aged 1 day" signal, note how far you are from Earth at that time (in your frame), and calculate when the signal was sent to determine when he had aged. You wouldn't assume that he had aged when you received the signal, but rather when he sent the signal.

Yep, that is why I only talk about receiving the signals. But the signals are just reification of the _information_ about the aging of "A" according to "B". It is the only valid way of thinking about it, of putting labels on a chart like the one that Fredrik put together, imho.

cheers,

neopolitan

 

[granpa]

Yep, that is why I only talk about receiving the signals. But the signals are just reification of the _information_ about the aging of "A" according to "B". It is the only valid way of thinking about it, of putting labels on a chart like the one that Fredrik put together, imho.

cheers,

neopolitan

its absolutely true that there is a jump at the turn around point. in fact if he accelerates decelerates reaccelerates again and again then his calculation of the twins age will jump back and forth again and again (the twin will appear to move backward in time). now I don't buy that this is a 'real' effect but it certainly happens. you forget that to calculate the twins age he must take into account the speed of light. but he always measures the speed of light to be c relative to himself in spite of the fact that he just changed velocity. obviously that much have a drastic effect on his calculation.

whether such relativistic effects are 'real' or not is another discussion entirely

 

[Fredrik]

Fundamentally, I don't buy the discontinuity at the turn around point. "A" appears to age 7 or so years on "B"'s outward leg, and around 33 years on the return leg.

It's 7.2 during the outbound trip, 7.2 during the return trip, and 25.6 during an instantaneous turnaround. Nothing important changes if you make the turnaround phase last longer.

You can think about it with "A" sending out constant updates by radio signal, "I have aged 1 day" - once a day, every day and working out when and where these messages and "B" will be collocated. You should be able to see that "B" will receive 7 years worth of messages on the way out and 33 years worth on the way home. There will be a few messages during the turn around period, but not 25.6 years worth. (How many depends entirely on the rates of acceleration and thus how quickly "B" turns around.)

Yep, that is why I only talk about receiving the signals. But the signals are just reification of the _information_ about the aging of "A" according to "B". It is the only valid way of thinking about it, of putting labels on a chart like the one that Fredrik put together, imho.

You might as well say that the definition of simultaneity in SR that's accepted by everyone isn't valid, because that's what I'm using.

 

[neopolitan]

You might as well say that the definition of simultaneity in SR that's accepted by everyone isn't valid, because that's what I'm using.

But are you using precisely what is "accepted by everyone"? And if you are using that, is that definition of simultaneity supposed to be a helpful device or a proclamation on how things actually are?

Ponder this ... in your example, "B" travels for 12 years. "A" ages 25.6 years in the turnaround period. How does the universe know to age "A" 25.6 years, and not 51.2 years or 102.4 years (which apply if "B" plans to travel for 24 or 48 years?

Suppose, just suppose, that "B" travels to the turnaround point at a velocity of 0.8c, the heads off at 0.8c on the way back, but half-way home suffers a breakdown and has to limp the rest of the way at 0.001c. We'd have to assume that A and B are long lived, of course, but your figures will no longer work out. The instantaneous aging you posit at the turnaround point is an implication that the universe somehow sees the future and applies the right instantaneous aging to "A" (based on "B"'s future speeds).

I don't think that is part of the standard understanding of simultaneity.

cheers,

neopolitan

 

[neopolitan]

its absolutely true that there is a jump at the turn around point. in fact if he accelerates decelerates reaccelerates again and again then his calculation of the twins age will jump back and forth again and again (the twin will appear to move backward in time). now I don't buy that this is a 'real' effect but it certainly happens. you forget that to calculate the twins age he must take into account the speed of light. but he always measures the speed of light to be c relative to himself in spite of the fact that he just changed velocity. obviously that much have a drastic effect on his calculation.

whether such relativistic effects are 'real' or not is another discussion entirely

Try thinking about it with "B" calculating "A"'s age purely on the basis of the number of reports of one day's aging. You will see that the discontinuities just don't happen.

I will recant of course, if you can show how 25.6 years worth of daily reports will be received by "B" instantaneously, or over a relatively short period, if "B" chooses to decelerate and accelerate back homewards. (I would then also like to see how the other discontinuities work, for example if "B" decelerates on the way home and "A" moves backward in time - are the daily reports rescinded and reissued? and how?)

cheers,

neopolitan

 

[Fredrik]

But are you using precisely what is "accepted by everyone"? And if you are using that, is that definition of simultaneity supposed to be a helpful device or a proclamation on how things actually are?

That definition of simultaneity tells you how to associate an inertial frame with a physical observer moving with constant velocity. There's nothing controversial about it.

Any coordinate system is a "helpful device" because you can use it to describe "how things actually are". The problem is of course that if you use a coordinate system to describe how things actually are, the description may sound very different from a description made using another coordinate system. (Take the pole-and-barn paradox for example: In the pole's frame, there's a time when the whole pole is inside the barn. In the barn's frame, there isn't). Because of these apparent contradictions, I can't encourage the use of expressions like "how things actually are". It's better to talk about e.g. "how things are in this particular frame".

Ponder this ... in your example, "B" travels for 12 years. "A" ages 25.6 years in the turnaround period. How does the universe know to age "A" 25.6 years, and not 51.2 years or 102.4 years (which apply if "B" plans to travel for 24 or 48 years?

The universe isn't doing anything, so it doesn't have to know anything. The reason it's 25.6 and not any other number is that B is 16 light-years from Earth and that the slope of a simultaneity line is always v when the slope of the world line is 1/v. You can verify for yourself that with v=0.8c, that works out to 25.6 years.

Suppose, just suppose, that "B" travels to the turnaround point at a velocity of 0.8c, the heads off at 0.8c on the way back, but half-way home suffers a breakdown and has to limp the rest of the way at 0.001c. We'd have to assume that A and B are long lived, of course, but your figures will no longer work out. The instantaneous aging you posit at the turnaround point is an implication that the universe somehow sees the future and applies the right instantaneous aging to "A" (based on "B"'s future speeds).

If B slows down to 0.001c on the way back, he will be in a frame where his "now" is simultaneous with an earlier event on Earth, so A will be younger than "before". That doesn't mean that the universe has "seen" or "done" something. It only means that B is using a different coordinate system.

I don't see the problem.

 

[neopolitan]

I don't see the problem.

Maybe you have the right idea about how simultaneity in SR is applied, but I really doubt it after long discussions elsewhere.

To bring in simultaneity, if you must have it, consider "A" as a simple transmitter, pulsing out "A has aged x days" repeatedly (where x(m) = x(m-1)+1). Consider "B" as a similar transmitter with a similar message, "B has aged z days" (similar values of z).

Using the relevant equations, "A" may consider the sending of one message to "B" and "B"'s sending of another message to be simultaneous, whereas "B" would not agree.

Once "B" reaches the turn around point, there will a number of signals from "A" still on the way. Since the speed was 0.8c, for a period of 12 years, then in _classical_ terms there would be about 9.6 years worth of signals inbound (and last signal from "A" should be "A has aged 876 days"). In SR terms there will be 12.8 years worth of signals still on the way and the last message will be "A has aged 2629 days". (I am using your figure of 7.2 years btw, I haven't checked its validity. In any event, the actual figures are not important, it is the whole concept that matters).

"B" then turns around and goes back through the incoming barrage of signals. Over a period of 12 years ship time, in _classical_ terms, "B" would experience 12 years of signals, plus the 9.6 years worth which were inbound at the turnaround point.

In SR terms, "B" will travel through 32.8 years worth of signals before getting home and asking what the hell is wrong with "A"'s transmission schedule.

What won't happen is the sudden delivery of 25.6 years worth of signals at the turn around point or the double delivery of signals (which would be the implication of "B" ending up in 'a frame where his "now" is simultaneous with an earlier event on Earth').

The only way you could organise double delivery of signals is by traveling faster than the speed of light (and it would involved some doubling back too, unless "A" started sending out pulses before "B" left Earth).

Your mathematics might be spot on, Fredrik, but the application of them is questionable.

cheers,

neopolitan

 

[Jonathan Scott]

What won't happen is the sudden delivery of 25.6 years worth of signals at the turn around point or the double delivery of signals (which would be the implication of "B" ending up in 'a frame where his "now" is simultaneous with an earlier event on Earth').

A stream of signals arriving does not define simultaneity. You also have to calculate how far they traveled and hence when they originated. It is that calculation which has an abrupt discontinuity at the turn-around point.

 

[neopolitan]

A stream of signals arriving does not define simultaneity. You also have to calculate how far they traveled and hence when they originated. It is that calculation which has an abrupt discontinuity at the turn-around point.

But you agree that "B" will receive a rational stream of signals from "A"?

Additionally, why exactly is there an abrupt discontinuity at the turn around? When pointed in one direction and getting "A has aged x days", B will calculate that A is a certain distance away (with the assumption that the message traveled c). Once turned around and getting "A has aged x+1 days" and being in approximately the same position, why does B make a calculation that A must have aged 25.6 years over the past day?

My thinking is that B should make a calculation like this:

To the best of my knowledge we (A and B) have been separating at a constant 0.8c for a period of 12 years, so that our current separation is 9.6 light years. Any signal from A will have taken 9.6 years to get to my current position. Therefore, A is currently 9.6 years older than as indicated in the latest message I have received.

Righto, I will turn around now. Done.

Ok, now I am pretty much in the same spot as I was yesterday, with a separation of 9.6 light years from A. It is a day or so later. Again, any signal from A will have taken 9.6 years to get here, and A is currently 9.6 years older than indicated in the latest message

No discontinuity.

Is there any other sensible way that B should calculate how great the separation between the two of them is?

cheers,

neopolitan

PS Thinking about it some more ...

"A" will appear to have been sending out signals at a greater rate than "B" might have otherwise suspected, so that by the turnaround point "B" will already have received 2.4/0.6 years of signals. If "B" is clever, then an assumption will be made to the effect that "A" has and will continue to transmit signals at the same rate, then "B" should expect that there are 9.6/0.6 years worth of signals on the way at the turnaround point.

9.6/0.6 + 2.4/0.6 = 16 + 4 = 20 years

On the way back "B" will expect to intercept a total of 36 years worth of messages.

Still no discontinuity :)

 

[Jonathan Scott]

But you agree that "B" will receive a rational stream of signals from "A"?

Additionally, why exactly is there an abrupt discontinuity at the turn around? When pointed in one direction and getting "A has aged x days", B will calculate that A is a certain distance away (with the assumption that the message traveled c). Once turned around and getting "A has aged x+1 days" and being in approximately the same position, why does B make a calculation that A must have aged 25.6 years over the past day?

The change of B's frame of reference at the turn around makes all the difference to the time and space separation.

You can illustrate the ordinary 2D rotation equivalent by using "forwards" and "sideways" instead of "time" and "space" when moving in a plane. If two objects start from the same spot and move off in different directions at an acute angle relative to one another, then they are both moving less in each others' "forward" direction than the other. However, if one then turns a corner towards where the other one is going, the other one is suddenly much further "forward".

 

[granpa]

why exactly is there an abrupt discontinuity at the turn around?

I already told you the reason why in post 8. he has to take the speed of light into account. the speed of light remains the same even when he changes velocity. can't you see that thet wall change his calculation?

 

[Fredrik]

Neopolitan, you need to make an effort to understand simultaneity. Suppose that a light signal is emitted in the positive x direction at (t₁,0), gets reflected at some point along the x-axis and returns at (t₂,0), then the reflection event must be simultaneous with (t₁+(t₂-t₁)/2,0). That's how simultaneity works. Pick t₁=-T and t₂=T for simplicity. The emission event is (-T,0) and the return event is (T,0). The reflection event must be simultaneous with (0,0), and since the speed of light is 1, the spatial coordinate must be T, so the reflection event is (0,T).

Now draw the world line of someone who's moving with speed v in the positive x direction, and imagine that this person is doing the same thing we just did to find out which events are simultaneous. Note that the slope (\Delta t/\Delta x) of the world line is 1/v. You will find that to this person, two events are simultaneous if and only if they are on a line with slope v.

That's what I used to draw the diagram. Before the rocket turns around, B considers events on a blue line simultaneous. After the turnaround, he considers events on a red line simultaneous.

 

[Fredrik]

To bring in simultaneity, if you must have it, consider "A" as a simple transmitter, pulsing out "A has aged x days" repeatedly (where x(m) = x(m-1)+1). Consider "B" as a similar transmitter with a similar message, "B has aged z days" (similar values of z).

This isn't the way to "bring in simultaneity". See my previous post.

Once "B" reaches the turn around point, there will a number of signals from "A" still on the way. Since the speed was 0.8c, for a period of 12 years, then in _classical_ terms there would be about 9.6 years worth of signals inbound (and last signal from "A" should be "A has aged 876 days").

You need to specify a frame if you're going to refer to a certain time. Since you're saying "12 years" instead of "20 years", I assume that we're considering B's frame, and since you're talking about things that happened before the turnaround, I assume that we're talking about the first of B's frames, the one that's moving away from Earth.

The 9.6 and 876 figures are wrong. When (in B's frame) the rocket turns around, only 7.2 years have passed on Earth. The last message that arrived was sent by A when he had aged 4.0 years. (The turnaround event is (20,16) in A's frame, so a light signal that arrives at that event must have been emitted at (4,0)).

In any event, the actual figures are not important, it is the whole concept that matters.

I don't think it matters. It doesn't have anything to do with simultaneity.

"B" then turns around and goes back through the incoming barrage of signals. Over a period of 12 years ship time, in _classical_ terms, "B" would experience 12 years of signals, plus the 9.6 years worth which were inbound at the turnaround point.

He's going to experience 36 years of signals, since he only received 4 years of signals before he turned around. (I see now that you realized that yourself).

What won't happen is the sudden delivery of 25.6 years worth of signals at the turn around point or the double delivery of signals (which would be the implication of "B" ending up in 'a frame where his "now" is simultaneous with an earlier event on Earth').

The only way you could organise double delivery of signals is by traveling faster than the speed of light (and it would involved some doubling back too, unless "A" started sending out pulses before "B" left Earth).

Huh. You lost me here. First of all, he's only ending up in 'a frame where his "now" is simultaneous with an earlier event on Earth' if he slows down on the way back (which is the scenario you were considering at the time). The turnaround puts him in a frame where his "now" is simultaneous with a much later time on Earth. 25.6 years later to be exact.

I have no idea why you're talking about double delivery of signals and stuff like that.

Additionally, why exactly is there an abrupt discontinuity at the turn around? When pointed in one direction and getting "A has aged x days", B will calculate that A is a certain distance away (with the assumption that the message traveled c). Once turned around and getting "A has aged x+1 days" and being in approximately the same position, why does B make a calculation that A must have aged 25.6 years over the past day?

Simultaneity. (There's obviously no discontinuity in the incoming messages, but the simultaneity lines get tilted the other way when the rocket turns around).

 

[neopolitan]

The change of B's frame of reference at the turn around makes all the difference to the time and space separation.

You can illustrate the ordinary 2D rotation equivalent by using "forwards" and "sideways" instead of "time" and "space" when moving in a plane. If two objects start from the same spot and move off in different directions at an acute angle relative to one another, then they are both moving less in each others' "forward" direction than the other. However, if one then turns a corner towards where the other one is going, the other one is suddenly much further "forward".

"B" may have to a new basis of determining, in his new frame, which distant events about which he receives information in the future are simultaneous with which local events (which may be in his past), but he won't reassess events which were simultaneous with events in his previous frame. All such events are in the absolute past and cannot be resurrected.

I do not think that a range of spatially colocated events at "A" (not colocated in time) can be simultaneous for "B", ever. That would violate causality. (For example, "B" cannot receive two events which one which is dependant on the other with a delay. Imagine an egg dropping to the floor under the influence of gravity, "B" cannot receive the event "egg whole" and "egg broken" together since there is a time component to the fall between.)

Your "turning the corner" simile only works if "B" changes speed, which he doesn't really ... except during the actual turnaround. "B" changes velocity, speed is the same, only the direction changes.

cheers,

neopolitan

 

[matheinste]

Hello neopolitan.

Surely we are talking about events in B's frame(s) being simultaneous with events in A's frame and not about colocated events in space being simultaneous.I don't see how anything happening in the twins scenario violates causality. Don't forget that B is only in one frame at a time and there is no suggestion of events in one of these frames being simultaneous (for B)with events in the other. A's world of course carries on as normal and even to B it is seen as a continuous forward time progression albeit very much speeded up at turnaround.

Matheinste.

 

[neopolitan]

I can't fully understand why a person who makes a journey into space in a high-speed rocket will return home to find his age less than an identical twin who stayed on Earth. It makes since for the twin who stayed on earth, but for the twin who traveled into space, he sees himself at rest and sees Earth moving at constant velocity. so he should feel that time on Earth is moving slower than his time, this way he will find his twin younger than him not older. Please tell me what I am missing to help me understand this thought experiment.

Thank you.

I wonder if anyone else is still thinking about the OP's original question?

I have kept it in mind and attempted not to get tangled inextricably in simultaneity issues.

My point, in my first post in this topic is that there is no real discontinuity.

If there is a discontinuity it is based on the conceit that there is no change to "B"'s frame. However, in the real universe, "B" will undergo observable accelerations and will know that velocity changes have occured.

If "B" ignores that, and only if "B" ignores that, then the purely mathematical discontinuity will arise. That discontinuity will be balanced by an equal and opposite mathematical discontinuity from "A"'s point of view (as calculated by "B").

In reality, there is no discontinuity, as can be understood by considering the flows of information possible between "A" and "B".

Fredrik, I'll highlight something for your edification. Please observe the words which are bold and underlined.

Once "B" reaches the turn around point, there will a number of signals from "A" still on the way. Since the speed was 0.8c, for a period of 12 years, then in _classical_ terms there would be about 9.6 years worth of signals inbound (and last signal from "A" should be "A has aged 876 days")

Of course 9.6 years and 876 days are wrong, classical physics doesn't apply.

Matheinste,

I think that fundamentally we agree, given your words:

[QUOTE+matheinste]A's world of course carries on as normal and even to B it is seen as a continuous forward time progression albeit very much speeded up at turnaround.[/QUOTE]

There is no real discontinuity as presented in Fredrik's diagram (only a mathematical discontinuity created under the conditions given above).

I do disagree that the progression is particularly sped up at the turnaround, since there is a limitation to the transmission of information from "A" to "B" (and indeed from "B" to "A"). "B" will receive more information from "A" on the way back home than on the outward leg. There will be no special bulk delivery of signals from "A" during the turnaround.

If you don't understand this, then I implore you to read the preceding posts to understand what is actually being discussed.

cheers,

neopolitan

 

[Jonathan Scott]

"B" may have to a new basis of determining, in his new frame, which distant events about which he receives information in the future are simultaneous with which local events (which may be in his past), but he won't reassess events which were simultaneous with events in his previous frame. All such events are in the absolute past and cannot be resurrected.

I do not think that a range of spatially colocated events at "A" (not colocated in time) can be simultaneous for "B", ever. That would violate causality. (For example, "B" cannot receive two events which one which is dependant on the other with a delay. Imagine an egg dropping to the floor under the influence of gravity, "B" cannot receive the event "egg whole" and "egg broken" together since there is a time component to the fall between.)

That's not the way relativity and Lorentz transformations work.

There's no problem with causality. Once B has changed velocity, this changes the relationship between space and time in such a way that light in the new frame is still seen to move at c. This affects calculated times and distances. Spacelike and timelike separations remain spacelike and timelike so there is no effect on causality. You can calculate this and see it on a space-time diagram. The change in velocity doesn't cause any sort of discontinuity of what B actually sees (although the rate at which information is received effectively becomes Doppler shifted). Anything that has been literally "seen" is in the timelike past. Changing direction affects the calculation of exactly how far it was in the timelike past and where it happened, relative to B's new frame of reference.

Your "turning the corner" simile only works if "B" changes speed, which he doesn't really ... except during the actual turnaround. "B" changes velocity, speed is the same, only the direction changes.

In the 2D analogy, turning the corner DOES have an effect on how far forward something is. If A and B are both moving at the same speed from the same point in different directions, but A then turns in the same direction as B, then before turning A will consider that it is "ahead" of B in whatever direction A is going, but after turning, it will definitely be "behind" B in the direction that A and B are going. (Because of the different signature, this is the reverse of the time/space case, where the one which turns has aged more).

 

[neopolitan]

I have no idea why you're talking about double delivery of signals and stuff like that.

Then you don't understand what has been written before. Think about it, consider hypersurfaces of simultaneity, which are equivalent to signals sent out at the speed of light.

A distant event which is simultaneous with a local event in my frame is one which arrives after a period of time defined by x/c where x is the separation between me and the event.

This is why, when "B" turns around and observes the last message from "A" implying an ageing of 4 years, "B" can calculate that that latest signal was sent 9.6 years ago in the "B-frame" and that that equates to 16 years ago in the "A-frame" and that that means that a signal sent simultaneously with "B" turning around will indicate that "A" has aged 20 years.

The double delivery comes in when you say that "B" slowing down will make "B" simultaneous with earlier events in the "A" frame. If one moment "B" is simultaneous with an event 25 years after departure (in the "A-frame") and the next is simultaneous with an event 22 years after departure, then "B" should rightly expect to receive two lots of signals between 22 years and 25 years. But I am saying that this does not happen.

Part of the problem, I guess, is trying to use information that you haven't got yet. Only once a signal has been received from "A" can "B" work out what event that signals transmission was simultaneous with.

What I can say, with great certainty is that there is no _real_ discontinuity.

cheers,

neopolitan

 

[matheinste]

If you don't understand this, then I implore you to read the preceding posts to understand what is actually being discussed.
olitan.

Quote:-

--If you don't understand this, then I implore you to read the preceding posts to understand what is actually being discussed.---

I have read them. all this has been discussed time and time again and the answer is well known and accepted. Your view is incoreect.

Matheinste.

 

[Fredrik]

Think about it, consider hypersurfaces of simultaneity, which are equivalent to signals sent out at the speed of light.

A distant event which is simultaneous with a local event in my frame is one which arrives after a period of time defined by x/c where x is the separation between me and the event.

What you're defining here is just the future light cone of the local event. The apex of the light cone is not simultaneous with any other event on the light cone.

This is why, when "B" turns around and observes the last message from "A" implying an ageing of 4 years, "B" can calculate that that latest signal was sent 9.6 years ago in the "B-frame" and that that equates to 16 years ago in the "A-frame" and that that means that a signal sent simultaneously with "B" turning around will indicate that "A" has aged 20 years.

Oddly enough, you're getting this part exactly right. It's as if you're using the correct definition of simultaneity for this part, even though you just defined it incorrectly.

The double delivery comes in when you say that "B" slowing down will make "B" simultaneous with earlier events in the "A" frame. If one moment "B" is simultaneous with an event 25 years after departure (in the "A-frame") and the next is simultaneous with an event 22 years after departure, then "B" should rightly expect to receive two lots of signals between 22 years and 25 years.

This is just wrong. The deliveries obviously only depend on B's location in spacetime, but simultaneity only depends on velocity. So a change of simultaneity isn't going to make anything too weird happen to the deliveries.

 

[neopolitan]

If you don't understand this, then I implore you to read the preceding posts to understand what is actually being discussed.
olitan.

Quote:-

--If you don't understand this, then I implore you to read the preceding posts to understand what is actually being discussed.---

I have read them. all this has been discussed time and time again and the answer is well known and accepted. Your view is incoreect.

Matheinste.

What, pray tell, is my view that is incorrect? I am just asking for a quick summary of what you think my view is, as expounded in this thread. Of course, it would help if you highlight areas which, when taken in context, indicate a huge misunderstanding.

I ask this because people often jump in and attack what they think is a misunderstanding, without trying to gain an understanding themselves.

Note also, that I have accepted the mathematical discontinuity which results from "B" changing frame. What I argue is that this is not any real discontinuity and is based on the rather unlikely scenario that "B" does not accept that there actually has been a change of frame.

This might derive from my being an engineer by training rather than a physicist. We use the calculations as tools, rather than laws. If I were designing a spaceship for "B", complete with a clock that shows "B" the time elapsed for "A", I would not plan on having that clock jump from 7.2 years to 32.8 years at turnaround. Nor would I have a clock which just quickly scrolled forwards to 32.8 years.

If you were designing such a clock, how would it operate?

cheers,

neopolitan

 

[granpa]

if you want to know the 'real' time on Earth then just put a string of synchronized clocks between Earth and the turn around point. all he has to do is look out the window and read the clock. each clock he observes will seem to him to tick at a rate of 1/gamma. but the total elapsed time as told by the nearest clock will be gamma times the clock on board his ship

when he stops at the turn oround point all the clocks will ogain be synchronized and ticking at the normal rate.

if all you want his clock to do is tell the 'real' time on Earth then by all means design it however you want. but if as an engineer you want his clock to be useful to him in timing events on board his ship then you should reconsider.

 

[Fredrik]

if you want to know the 'real' time on Earth then just put a string of synchronized clocks between Earth and the turn around point.

You are probably already aware of this, but the term "real" is very inappropriate. What these clocks would tell you is just what event on Earth your "now" is simultaneous with in A's frame. But A's frame is obviously not any more real than B's.

 

[granpa]

You are probably already aware of this, but the term "real" is very inappropriate. What these clocks would tell you is just what event on Earth your "now" is simultaneous with in A's frame. But A's frame is obviously not any more real than B's.

thats why it was in quotes