Kinetic and dynamic twin effect

In summary, the standard "twin paradox" in SR is actually a purely kinetic effect, solved by the obvious asymmetry of the twins and the time dilation. The acceleration is not necessary to solve the paradox, as the difference in time can be explained by the kinematic effect of time dilation. However, when considering a returning twin, the acceleration experienced will cause an additional time difference due to the equivalence principle. This time difference is not caused by pure time dilation, but rather by the effects of acceleration, similar to how a gravitational field can cause time dilation. The scenarios of slinging around a planet unaccelerated and bouncing off a spring have different effects on clocks, as the acceleration involved is different. Einstein's explanation of the one way
  • #1
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I just realized that the standard "twin paradox" in SR is actually a purely kinetic effect.
It is solve by the obvious assymetry of the twins and the time dilation.

However, the acceleration is not really needed in solving the paradox, since a third person could carry the clock backward after synchronization with the traveling twin. This shows it is only time dilation, a kinematic effect, that is at play there.

However, if one consider a returning twin, then this twin will also experience an acceleration. Because of the equivalence principle, I guess that this will cause an additional time difference which is not caused by the pure time dilation of SR. It would be the result of the acceleration instead, just like a gravitational field will do.

As I cannot develop that idea further, I would like your suggestions.
Maybe I am wrong and doing double accounting?
Also, I thought I could make a quick estimate using a equivalent potential for the acceleration. But on a second thought this looks strange to me, and I even question my understanding about why the potential appears in the gravity time dilation.

Thanks for you ideas.
 
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  • #2
lalbatros said:
I just realized that the standard "twin paradox" in SR is actually a purely kinetic effect. It is solve by the obvious assymetry of the twins and the time dilation.

However, the acceleration is not really needed in solving the paradox, since a third person could carry the clock backward after synchronization with the traveling twin. This shows it is only time dilation, a kinematic effect, that is at play there.
Right. Few seem to notice this.

However, if one consider a returning twin, then this twin will also experience an acceleration. Because of the equivalence principle, I guess that this will cause an additional time difference which is not caused by the pure time dilation of SR. It would be the result of the acceleration instead, just like a gravitational field will do.

There are two scenarios, slinging around a planet unaccelerated, and bouncing off a spring, say.
 
  • #3
lalbatros said:
a third person could carry the clock backward after synchronization with the traveling twin. This shows it is only time dilation, a kinematic effect, that is at play there.
This is right.

lalbatros said:
However, if one consider a returning twin, then this twin will also experience an acceleration. Because of the equivalence principle, I guess that this will cause an additional time difference which is not caused by the pure time dilation of SR. It would be the result of the acceleration instead, just like a gravitational field will do.
This is wrong. Acceleration will make two synchronized clocks attached to opposite ends of a solid object lose their synchronization because the clock at the rear has a higher velocity than the clock at the front during the acceleration phase. But in the twin paradox scenario, we're treating the astronaut twin as a point particle, i.e. as one clock. (Here I'm defining "front" and "rear" so that the acceleration is in the direction of an arrow pointing from the rear to the front).

The fact that the acceleration itself doesn't have any funny effects on an ideal clock (which can be described as a point particle) can't be derived from the mathematics of Minkowski space. In SR (and GR) it's a direct consequence of one of the postulates that tell us how to interpret the mathematics as predictions about the results of experiments: A clock measures the proper time of the curve in spacetime that represents its motion.
 
  • #4
Phrak said:
There are two scenarios, slinging around a planet unaccelerated, and bouncing off a spring, say.
There is acceleration even in the first of those scenarios, since (the direction of) the velocity is changing.
 
  • #5
Won't an (de)acceleration in free space slow down a clock like a gravity field would do it?
 
  • #6
What I described is how a gravitational field would do it. A clock attached to your floor is accelerating faster than a clock attached to your ceiling, and is therefore ticking at a slower rate. You can use the the equivalence principle to calculate how much slower, by pretending that you're in flat spacetime, in a room that's being accelerated by a rocket engine or something.
 
  • #7
Fredrik said:
There is acceleration even in the first of those scenarios, since (the direction of) the velocity is changing.

The two scenarios are different. In the comoving frame of the twin, the twin undergoes no acceleration.
 
  • #8
lalbatros,

It's worth pointing out that an inertial observer can track an accelerating object purely in terms of velocity-dependent time dilation. But an accelerating observer tracking any object (other than him/herself) needs to take account of (pseudo-)gravitational time dilation as well. As an observer (properly) accelerates, his/her definition of simultaneity is constantly changing and the rate of change is the "gravitational" component of dilation.

We usually consider the twin paradox in the absence of gravity, to avoid general-relativistic complications.
 
  • #9
Einstein confused many readers because of the way he described the one way time dilation problem between two spaced apart initially synchronized clocks in part IV of his 1905 paper. It seems to imply that to get a resulting time difference the two clocks "A" and "B" need to be first synchronized while at rest in the same frame and then one clock is put in motion to reach a steady speed - Einstein then asserts the clock which moved (The accelerated clock) will show less lapsed time when they meet - but if the initial conditions are such that two clocks are synched as they pass each in relative uniform motion -
and later they measure different times relative to a 3rd clock C which has remained in sync with either A or B (at rest in the fame of either A or B as it has not moved with respect to either A or B). everything is symmetrical - yet there is a difference.

Einstein added more confusion with his 1918 paper when he attempted to explain the time difference as due to a pseudo G field - this has led a number of authors to take the position "acceleration" is in someway at root in the twin scenerio... in someway there needs to be a physical difference between the two clocks e.g., which clock feels the turn around acceleration, etc.
 
  • #10
Continuation - there is no need to consider acceleration in these problems - the resolution is always in the way the experiment measures time and distance - when two clocks are used to measure the time lapse between a single clock that moves relative to the two clocks at rest wrt each other in the same frame - the single clock will always be measured to run slow by the two clocks
 
  • #11
yogi,

I agree that the twin paradox doesn't require taking acceleration into account to understand the main topic.
However, I was curious to know how an acceleration would contribute.
 
  • #12
yogi said:
Einstein confused many readers because of the way he described the one way time dilation problem between two spaced apart initially synchronized clocks in part IV of his 1905 paper. It seems to imply that to get a resulting time difference the two clocks "A" and "B" need to be first synchronized while at rest in the same frame and then one clock is put in motion to reach a steady speed - Einstein then asserts the clock which moved (The accelerated clock) will show less lapsed time when they meet - but if the initial conditions are such that two clocks are synched as they pass each in relative uniform motion -
and later they measure different times relative to a 3rd clock C which has remained in sync with either A or B (at rest in the fame of either A or B as it has not moved with respect to either A or B). everything is symmetrical - yet there is a difference.
It's not symmetrical at all. The distance between two clocks (A and B)at rest will always be greater in their rest frame than the distance between those clocks in any other inertial frame. And elapsed time equals distance/velocity.

This was the origin of the twins paradox. Except it's just a one way trip. Of course the result is the same for a one way trip, but that's rarely mentioned in resolutions.

Al
 
  • #13
ai68 -post 12. The situation is symmetrical in that it makes no difference which frame is put in motion - as per my post 10 -

I would agree that the origin of the twin paradox is the failure of most to popular writers to appreciate the account as two one-way trips - which immediately resolve when you add the third clock at the turn around point

Lalbatros - your post 11:
Yes - I see upon re-reading your initial post - and as a practical matter, the time difference during turn around would have to be taken into account by the location of the clock not undergoing velocity change - interestingly, you might see Born's treatment (Einstein's Special Relativity - paperback) of the issue where he arrives at the correct time difference upon the twins meeting only by considering the turn around acceleration and the distance between the two clocks. I personally think it is a good example of getting the right answer for the wrong reason, but who am I compared to Max Born
 
  • #14
yogi said:
ai68 -post 12. The situation is symmetrical in that it makes no difference which frame is put in motion - as per my post 10 -

I would agree that the origin of the twin paradox is the failure of most to popular writers to appreciate the account as two one-way trips - which immediately resolve when you add the third clock at the turn around point

Lalbatros - your post 11:
Yes - I see upon re-reading your initial post - and as a practical matter, the time difference during turn around would have to be taken into account by the location of the clock not undergoing velocity change - interestingly, you might see Born's treatment (Einstein's Special Relativity - paperback) of the issue where he arrives at the correct time difference upon the twins meeting only by considering the turn around acceleration and the distance between the two clocks. I personally think it is a good example of getting the right answer for the wrong reason, but who am I compared to Max Born

Hi yogi,

I'm not sure what you mean by this, but it's no coincidence that Born gets the same result that way.

Assuming that there is a clock at rest with Earth at the turnaround point synched with Earth's clock in Earth's frame, these clocks will be out of synch in the ship's frame by an amount proportional to the distance between them. As the ship's relative velocity decreases, the amount the clocks are out of synch decreases to zero (when the ship comes to rest with earth) then increases in the opposite direction as the ship accelerates. This results in Earth's clock "jumping ahead" or "running fast" in the ship's frame.

Einstein's GR resolution is similar, and has the same result, again not a coincidence. I think many view gravitational time dilation as fundamentally different from SR time dilation, but if gravity is viewed as a spacetime curvature instead of a force, gravitational time dilation is SR time dilation.

Al
 
  • #15
yogi said:
I would agree that the origin of the twin paradox is the failure of most to popular writers to appreciate the account as two one-way trips - which immediately resolve when you add the third clock at the turn around point
I agree, and would add that they also neglect to mention that a single one way trip with the ship just coming to rest with Earth and the ship's twin staying at the destination indefinitely would also have the same result.

Al
 
  • #16
yogi said:
ai68 -post 12. The situation is symmetrical in that it makes no difference which frame is put in motion - as per my post 10 -

I would agree that the origin of the twin paradox is the failure of most to popular writers to appreciate the account as two one-way trips - which immediately resolve when you add the third clock at the turn around point


The problem is that the layman is not asking about “two one-way trips”. The layman is asking about “one round-trip”. Its one thing to have the background to know that both problems give the same results. But to demonstrate that to a layman you have to solve both problems. Instead those purporting to explain the twins paradox are simply declaring the two problems equivalent and then solving the simpler one. You think the layman buys that? Let me assure you from personal experience that the answer is no. You want to know what I was thinking while one after another self-appointed expert confidently explained the “two one-way trips” to me? I was thinking this.

Suppose, instead of calculating the amount of time elapsed on the space trip, they are trying to help me calculate the amount of fuel consumed. Calculating the fuel consumed during the acceleration phases is a lot of work. So they are telling me to assume the spaceship changes velocity instantly, or nearly so. Then I could just skip the calculation for fuel consumption during acceleration. That gives me two one-way trips at constant velocity. No fuel consumption there either. So the spaceship makes the whole trip without using any fuel.

“There you go, you poor simpleton. Now do you understand how SR works?”
 
  • #17
AI68... I fully agree that its no coincidence as per your post 14 - a good example of the relationship revealed by the time differences measured during centrifuge experiments whether based on perhiperal velocity v = rw or acceleration v^2/r - they both give the same result

My criticism was based upon Born's assertion that twin time difference required the application of GR and it was not a problem to which the principles of SR were applicable. You find the same type of statements by such noteables as Dennis Sciama and others. Sciama asserted flat out that the difference in time was due to the fact that the traveling twin has accelerated with respect to all the other matter in the universe - whereas the stay at home twin has not.
 
  • #18
MikeLizza - it is usually easier to break a problem down into pieces - the many different rationale's for the difference in time between the home based twin and the returning twin get remarkable wrapped up in turn around acceleration, observation of distant clocks running at changed speed depending upon whether one is receding or approaching, the change in slope of the planes of simultaneity upon turn around etc - how confusing can it be to calculate the one way time difference and double the answer?
 
  • #19
MikeLizzi said:
The problem is that the layman is not asking about “two one-way trips”. The layman is asking about “one round-trip”. Its one thing to have the background to know that both problems give the same results. But to demonstrate that to a layman you have to solve both problems. Instead those purporting to explain the twins paradox are simply declaring the two problems equivalent and then solving the simpler one. You think the layman buys that? Let me assure you from personal experience that the answer is no. You want to know what I was thinking while one after another self-appointed expert confidently explained the “two one-way trips” to me?
Well, a "round trip" actually is comprised of two one way trips by definition. It's not just "declared equivalent". That's what "round trip" means.

Note that the issue of acceleration can't be ignored in a one way trip explanation like it commonly is in a two way trip explanation.

And if someone really understands a one way trip, they can't help but understand a two way trip.

Al

Edit: after re-reading your post, it seems like the explanations you got for a one way trip were woefully inadequate. Otherwise they would fully explain the result of a two way trip as well.
 
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  • #20
yogi said:
MikeLizza - it is usually easier to break a problem down into pieces - the many different rationale's for the difference in time between the home based twin and the returning twin get remarkable wrapped up in turn around acceleration, observation of distant clocks running at changed speed depending upon whether one is receding or approaching, the change in slope of the planes of simultaneity upon turn around etc - how confusing can it be to calculate the one way time difference and double the answer?

As Al68 is implying below, a “one-way trip” is correctly defined as;

a. starting in the Earth ref frame,
b. accelerating to ref frame moving with respect to earth,
c. staying in that ref frame for some period of time,
d. accelerating back to the Earth ref frame.

Given that definition, of course, one can calculate the value for one “one-way trip” and double it to get the answer for a round trip. However, what SR books (even textbooks) typically mean by a “one-way trip” is only part “c” in the above definition. And two of those “things” do not add up to a round trip.
 
  • #21
MikeLizzi said:
As Al68 is implying below, a “one-way trip” is correctly defined as;

a. starting in the Earth ref frame,
b. accelerating to ref frame moving with respect to earth,
c. staying in that ref frame for some period of time,
d. accelerating back to the Earth ref frame.

Given that definition, of course, one can calculate the value for one “one-way trip” and double it to get the answer for a round trip. However, what SR books (even textbooks) typically mean by a “one-way trip” is only part “c” in the above definition. And two of those “things” do not add up to a round trip.

That definition is exactly what I meant. And it's obviously not symmetrical. The way you spelled it out should make it obvious why the Earth ref frame ends up having the greatest elapsed proper time.

As far as textbook resolutions, while most are technically correct, the ones I've seen stress the unimportant aspects of the problem while ignoring what's important, IMHO.

Al
 
  • #22
lalbatros said:
I was curious to know how an acceleration would contribute.
I think what you are asking about is called the clock hypothesis. That acceleration and higher derivatives have no effect on time dilation.
 
  • #23
DaleSpam said:
I think what you are asking about is called the clock hypothesis. That acceleration and higher derivatives have no effect on time dilation.

It seems odd to say that acceleration (change in velocity) has no effect on time dilation.

We could even express time dilation as a function of an integral of acceleration for a given scenario. Not to mention gravitational time dilation is normally expressed as a function of a constant proper acceleration, even though it is derived from the time dilation due to a changing velocity relative to an inertial frame.

Or put another way, if time dilation is affected by relative velocity, and relative velocity is affected by acceleration, then time dilation must be affected by acceleration.

I would interpret the clock hypothesis to mean just that the equation for proper time is valid for any reference frame, regardless of its state of motion. Acceleration would affect the result, but not the equation that got the result.

Al
 
  • #24
MikeLizzi - your post Post 20 "As Al68 is implying below, a “one-way trip” is correctly defined as;

a. starting in the Earth ref frame,
b. accelerating to ref frame moving with respect to earth,
c. staying in that ref frame for some period of time,
d. accelerating back to the Earth ref frame."

That is not what I was referring to as a one-way trip unless what you mean by part "d" implies accelerating back to the Earth reference frame at some spatially distant point from where you started - in the clearly defined one way trip you would put a clock on Neptune that you sync using Einstein's method with an Earth clock and measure the START time as a spaceship flys by an Earth clock at constant speed and continues at the same speed to reach the Neptune clock where the spaceship clock and the Neptune clock are compared "on-the-fly" This is a one way trip w/o acceleration -

The spaceship clock can then be disconnected during turn around and started again as it passes the Neptune clock on the return flight to earth. This is the beginning of the second part of the round trip - but in actuality you do not have to take any further readings - to get the age difference between the twins when the meet (again on the fly)you simply double the answer between the spaceship clock and the Neptune clock made during the one way excursion - to me that is the least confusing way to resolve the twin fiasco
 
  • #25
yogi said:
MikeLizzi - your post Post 20 "As Al68 is implying below, a “one-way trip” is correctly defined as;

a. starting in the Earth ref frame,
b. accelerating to ref frame moving with respect to earth,
c. staying in that ref frame for some period of time,
d. accelerating back to the Earth ref frame."

That is not what I was referring to as a one-way trip unless what you mean by part "d" implies accelerating back to the Earth reference frame at some spatially distant point from where you started - in the clearly defined one way trip you would put a clock on Neptune that you sync using Einstein's method with an Earth clock and measure the START time as a spaceship flys by an Earth clock at constant speed and continues at the same speed to reach the Neptune clock where the spaceship clock and the Neptune clock are compared "on-the-fly" This is a one way trip w/o acceleration -

The spaceship clock can then be disconnected during turn around and started again as it passes the Neptune clock on the return flight to earth. This is the beginning of the second part of the round trip - but in actuality you do not have to take any further readings - to get the age difference between the twins when the meet (again on the fly)you simply double the answer between the spaceship clock and the Neptune clock made during the one way excursion - to me that is the least confusing way to resolve the twin fiasco

Sure that's less confusing, but it doesn't really explain anything. It doesn't explain why we compared the ship's clock to the Neptune clock instead of comparing Earth's clock to a clock(synched with the ship) at rest with (and following) the ship at a specified distance. (which would show ship time elapsed greater than Earth time elapsed).
 
  • #26
Al68 said:
It seems odd to say that acceleration (change in velocity) has no effect on time dilation.
...
I would interpret the clock hypothesis to mean just that the equation for proper time is valid for any reference frame, regardless of its state of motion. Acceleration would affect the result, but not the equation that got the result.
Here is everything I know about the clock hypothesis: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Clock_Hypothesis
 
  • #28
yogi said:
MikeLizzi - your post Post 20 "As Al68 is implying below, a “one-way trip” is correctly defined as;

a. starting in the Earth ref frame,
b. accelerating to ref frame moving with respect to earth,
c. staying in that ref frame for some period of time,
d. accelerating back to the Earth ref frame."

That is not what I was referring to as a one-way trip unless what you mean by part "d" implies accelerating back to the Earth reference frame at some spatially distant point from where you started - in the clearly defined one way trip you would put a clock on Neptune that you sync using Einstein's method with an Earth clock and measure the START time as a spaceship flys by an Earth clock at constant speed and continues at the same speed to reach the Neptune clock where the spaceship clock and the Neptune clock are compared "on-the-fly" This is a one way trip w/o acceleration -

The spaceship clock can then be disconnected during turn around and started again as it passes the Neptune clock on the return flight to earth. This is the beginning of the second part of the round trip - but in actuality you do not have to take any further readings - to get the age difference between the twins when the meet (again on the fly)you simply double the answer between the spaceship clock and the Neptune clock made during the one way excursion - to me that is the least confusing way to resolve the twin fiasco

Yes, I was implying that part "d" mean accelerating back to the Earth reference frame at some spatially distant point from where I started. Should have made that clear. With regard to your last paragraph, again, you are eliminating the acceleration portion of the round trip. So you are only solving part "c". As far as I am concerned, part "c" solved twice does not add up to a round trip solution. And I am not alone. Thousands of intelligent laymen all over the internet are trying to convey the same message.
 
  • #30
MikeLizzi said:
Yes, I was implying that part "d" mean accelerating back to the Earth reference frame at some spatially distant point from where I started. Should have made that clear. With regard to your last paragraph, again, you are eliminating the acceleration portion of the round trip. So you are only solving part "c". As far as I am concerned, part "c" solved twice does not add up to a round trip solution. And I am not alone. Thousands of intelligent laymen all over the internet are trying to convey the same message.

Hi Mike,

Are you looking for an explanation of differential aging for a one way trip, including d? I haven't bothered posting it because I thought we were just discussing how best to explain it to laymen.

The math is easy, for a 10 light year trip at 0.8c, the ship will reach the destination when the ship's clock reads 7.5 yrs, Earth's clock reads 12.5 yrs. Elapsed proper time equals distance/velocity, and the distance in the ship's frame is 6 light yrs, in Earth's frame it's 10 light years. It's as simple as that.

The objection is that in the outbound ship's frame, Earth's clock reads 4.5 yrs when the ship reaches the destination. But that isn't anyone's proper time for the trip. And after the ship comes to rest with earth, we have no clocks or twins in that frame. (But I'll mention a clock left in that frame later.)

Then we can double the results for a round trip. Ship's clock reads 15 yrs. Earth's clock reads 25 yrs.

Some point out that time dilation should be reciprocal. Well it is reciprocal between inertial frames. In either ship inertial frame (outbound or inbound) the elapsed proper time between the ship's departure and return to Earth is 41.66 yrs. So time dilation is reciprocal, we just didn't have a clock in either ship inertial frame for the entire trip. If we left a clock in the ship's outbound frame to keep going past the destination, it would read 41.66 yrs when the ship returned to earth.

I left out a lot due to laziness, and I don't have the best writing skills, but the bottom line is that the differential aging in the twins paradox doesn't actually depend on a turnaround or a reunion. The same effects occur for a one way trip, so we can just "resolve" each one way trip and add them together.

Al
 
  • #31
Al68 said:
Hi Mike,

Are you looking for an explanation of differential aging for a one way trip, including d? I haven't bothered posting it because I thought we were just discussing how best to explain it to laymen.

The math is easy, for a 10 light year trip at 0.8c, the ship will reach the destination when the ship's clock reads 7.5 yrs, Earth's clock reads 12.5 yrs. Elapsed proper time equals distance/velocity, and the distance in the ship's frame is 6 light yrs, in Earth's frame it's 10 light years. It's as simple as that.

The objection is that in the outbound ship's frame, Earth's clock reads 4.5 yrs when the ship reaches the destination. But that isn't anyone's proper time for the trip. And after the ship comes to rest with earth, we have no clocks or twins in that frame. (But I'll mention a clock left in that frame later.)

Then we can double the results for a round trip. Ship's clock reads 15 yrs. Earth's clock reads 25 yrs.

Some point out that time dilation should be reciprocal. Well it is reciprocal between inertial frames. In either ship inertial frame (outbound or inbound) the elapsed proper time between the ship's departure and return to Earth is 41.66 yrs. So time dilation is reciprocal, we just didn't have a clock in either ship inertial frame for the entire trip. If we left a clock in the ship's outbound frame to keep going past the destination, it would read 41.66 yrs when the ship returned to earth.

I left out a lot due to laziness, and I don't have the best writing skills, but the bottom line is that the differential aging in the twins paradox doesn't actually depend on a turnaround or a reunion. The same effects occur for a one way trip, so we can just "resolve" each one way trip and add them together.

Al

No, Al68, I am not looking for that calculation. I already know how to do it. I am trying to explain that your calculation, and that of others, is not the calculation for a one-way trip. But there is no point in diverting the subject matter of this thread any longer. My apologies to the original poster.
 
  • #33
MikeLizzi said:
No, Al68, I am not looking for that calculation. I already know how to do it. I am trying to explain that your calculation, and that of others, is not the calculation for a one-way trip. But there is no point in diverting the subject matter of this thread any longer. My apologies to the original poster.
Well, I doubled the one way trip to obtain results for a round trip. For a one way trip only:

For a 10 light year trip at 0.8c, the ship will reach the destination when the ship's clock reads 7.5 yrs, Earth's clock reads 12.5 yrs. Elapsed proper time equals distance/velocity, and the distance in the ship's frame is 6 light yrs, in Earth's frame it's 10 light years. It's as simple as that.

That's a one way trip. Assuming the ship's twin stays at 10 light yrs from earth, at rest with earth, he will be 5 yrs younger than the Earth twin forever.

The spacetime diagram would look like the first half of the twins paradox, then just parallel lines as long as they stay at rest.

Al
 

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