Jump to content

Special relativity: Can you explain the paradox?


Neil9327
 Share

Recommended Posts

Special relativity says that a clock on a spaceship moving relative to a stationary observer appears to be running more slowly than if it was stationary. So if this is true, what would you get in the following scenario?

 

Stationary observer Andy and moving observer Brian are both initially stationary and in the same location, and each has an atomic clocks measuring time in days, and displaying what they can see on a large monitor visible to both. In addition both clocks emit an electronic "pip" whenever one day passes to the next, and each records the number of pips made since reset. Day = 0, and the pip counter is set to 0.

 

Brian rapidly accelerates towards a distant star, and it takes him one second to reach around half the speed of light. After 100 days of travelling Brian stops in 1 second and accelerates to the same speed back towards Andy (in one second), and arrives back at Andy in a further 100 days.

Before they got out their clocks to compare what they showed, they each gave their opinion as to what they were expecting them to show.

 

 

Brian reports that Brian's clock shows that 200 days has elapsed, and he reported having received 200 pips from it.

Brian reports that Andy's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it.

 

Andy reports that Andy's clock shows that 200 days has elapsed, and he reported having received 200 pips from it.

Andy reports that Brian's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it.

 

When they both look at their clocks, and read their pip counters, what do they see?

 

Thanks

Link to comment
Share on other sites

I'll give this a try, but any good explanation of the twins' paradox will do.

 

Brian changed his frame of reference, Andy did not. Since Brian and Andy started in the same frame of reference and Brian accellerated (hence the new frame of reference) and returned to the previous frame of reference that Andy remained in the whole time, Brian will be the younger twin.

Edited by Paul Murphy
Link to comment
Share on other sites

Special relativity says that a clock on a spaceship moving relative to a stationary observer appears to be running more slowly than if it was stationary.

 

More precisely, the clock is running more slowly in a reference frame in which it is moving. Clock A and clock B have some relative velocity. In the reference frame in which A is at rest and B is moving, B is slower. In the reference frame in which A is moving and B is at rest, A is slower. Both are equally valid.

 

Anyway, as Paul Murphy said, the important thing is that the traveling twin has accelerated and changed rest frames 3 times, while the stay at home twin has not. Since simultaneity is also dependent on frame of reference, during those two seconds of turnaround, the traveling twin's present will change to one in which the stay at home twin is older.

Link to comment
Share on other sites

More precisely, the clock is running more slowly in a reference frame in which it is moving. Clock A and clock B have some relative velocity. In the reference frame in which A is at rest and B is moving, B is slower. In the reference frame in which A is moving and B is at rest, A is slower. Both are equally valid.

 

Anyway, as Paul Murphy said, the important thing is that the traveling twin has accelerated and changed rest frames 3 times, while the stay at home twin has not. Since simultaneity is also dependent on frame of reference, during those two seconds of turnaround, the traveling twin's present will change to one in which the stay at home twin is older.

 

So are you saying that during these one-second periods of acceleration of the travelling twin, that time speeds up? (the time of the stationary twin as observed by the travelling twin)

 

Below are some notes in italics I wrote, to follow up my question - am I on the right lines here?

 

OK so we are agreed that in my scenario that at the end of the journey, Brian and Andy (now standing next to each other) will look at their respective clocks and see that Brian's only shows 100 days (100 pips), and that Andy's shows 200 days (200 pips). And that this is because Brian's movement and heavy accelerations have caused Brian to age only half the time of Andy.

 

This is where it gets interesting. So does this mean that to make this happen, Brian must have had to take the decision to turn back after only 50 days (50 pips) (as he perceives it) such that his clock has only got to 100 by the time he returns? And when he returns, he sees that Andy's clock has got to 200 days (this being consistent with Andy's own experience of Brian taking 200 days for the whole trip). Am I right about this so far?

 

Assuming I am, then the following perhaps make sense: That for the portions of the journey when Brian is not accelerating, then for almost all of (from Brian's perspective) the 2 * 50 day journey, that Brian would look back at Andy, and see Andy's clock going at half speed, so resulting in only 25 days (25 pips) elapsing for the outbound journey, and 25 days elapsing for the return journey. 50 days total - as seen by Brian. Now since we know that on his return to Andy, Brian will see that Andy's clock has passed 200 days, there is a missing 150 days here. 150 days that needs to be added on to Brian's perception of Andy's clock at some point in the journey.

 

Would I be right in thinking that these 150 days are accounted for during Brian's three 1-second periods of heavy acceleration? i.e. during each of these short bursts of acceleration, Brian will see Andy's clock go forward by (say) 50 days? - that Brian will receive 50 pips from Andy's clock over the course of this one second (one every 20ms) - i.e. much speeded up?

 

If this is correct, then it is clear that while relative velocity of a target object (Andy in this case) will cause time at the target to slow down, acceleration of the observer can cause the observed time to speed up.

 

Following on from this, then perhaps one can see a scenario whereby you would see a clock running at normal speed where these two effects exactly cancel each other out; where you are (momentarily) travelling at a different velocity to the clock under observation, at some distance, and accelerating. Is there a formula that describes this?

 

If I am wrong, where above does my explanation depart from the reality?Thanks

Link to comment
Share on other sites

the twins paradox is non-intuitive enough without dragging doppler effect into it.

its all about relativity of simultaneity and how it changes whenever one accelerates. look it up

 

 

http://www.physicsforums.com/showthread.php?t=314080

 

Actually, Doppler effect can be quite useful in analyzing the twin paradox.

 

To wit: After Brian accelerates off, both He and Andy will receive pips from each other at the rate of:

 

[math] \sqrt {\frac{1-0.5}{1+0.5}} = 0.577[/math] per day. (assuming the pips were sent at the speed of light.)

 

When Brian reaches 100 days according to his clock, he has received 57.77 pips from Andy.

 

He heads back to Earth.

 

On the return, he receives pips at a rate of

 

 

[math] \sqrt {\frac{1+0.5}{1-0.5}} = 1.732[/math] per day

 

So after and additional 100 days (by his clock) he has received an additional 173.2 pips from Andy for a total of ~231 days

 

Thus he says the Andy's clock ticked off 31 more days than his did.

 

Andy also receives pips at a rate of 0.577 per day from Brian as Brian sped off.

 

After 100 days, he has received 57.7 pips from Brian. However, Brain kept going until his clock read 100 ticks, so Andy will not start receiving the return trip pips until he finishes receiving the away trip pips. At 0.577 pips per day, this takes an additional 73.31 days.

 

He then starts receiving pips at the rate of 1.732 pips per day. Again, he has to receive all the pips that Brian sent on the return trip, so this takes 57.7 days.

 

Thus he receives 200 total pips from Brian while counting 100 + 73.31 + 57.7 = ~231 days on his own clock. Exactly what Brian determined.

 

The difference here comes form the fact, that Brian, in turning around and heading back toward Andy started receiving pips at a fast rate immediately, while Andy had to wait until the information the Brian had turned around to reach him across the distance separating them.

 

I should also point out that the time difference between the two is equal to 200/231 = 0.866, the same result you get by using the time dilation formula.

Link to comment
Share on other sites

So are you saying that during these one-second periods of acceleration of the travelling twin, that time speeds up? (the time of the stationary twin as observed by the travelling twin)

 

Below are some notes in italics I wrote, to follow up my question - am I on the right lines here?

 

OK so we are agreed that in my scenario that at the end of the journey, Brian and Andy (now standing next to each other) will look at their respective clocks and see that Brian's only shows 100 days (100 pips), and that Andy's shows 200 days (200 pips). And that this is because Brian's movement and heavy accelerations have caused Brian to age only half the time of Andy.

 

This is where it gets interesting. So does this mean that to make this happen, Brian must have had to take the decision to turn back after only 50 days (50 pips) (as he perceives it) such that his clock has only got to 100 by the time he returns? And when he returns, he sees that Andy's clock has got to 200 days (this being consistent with Andy's own experience of Brian taking 200 days for the whole trip). Am I right about this so far?

 

Assuming I am, then the following perhaps make sense: That for the portions of the journey when Brian is not accelerating, then for almost all of (from Brian's perspective) the 2 * 50 day journey, that Brian would look back at Andy, and see Andy's clock going at half speed, so resulting in only 25 days (25 pips) elapsing for the outbound journey, and 25 days elapsing for the return journey. 50 days total - as seen by Brian. Now since we know that on his return to Andy, Brian will see that Andy's clock has passed 200 days, there is a missing 150 days here. 150 days that needs to be added on to Brian's perception of Andy's clock at some point in the journey.

 

Would I be right in thinking that these 150 days are accounted for during Brian's three 1-second periods of heavy acceleration? i.e. during each of these short bursts of acceleration, Brian will see Andy's clock go forward by (say) 50 days? - that Brian will receive 50 pips from Andy's clock over the course of this one second (one every 20ms) - i.e. much speeded up?

 

If this is correct, then it is clear that while relative velocity of a target object (Andy in this case) will cause time at the target to slow down, acceleration of the observer can cause the observed time to speed up.

 

Following on from this, then perhaps one can see a scenario whereby you would see a clock running at normal speed where these two effects exactly cancel each other out; where you are (momentarily) travelling at a different velocity to the clock under observation, at some distance, and accelerating. Is there a formula that describes this?

 

If I am wrong, where above does my explanation depart from the reality?Thanks

 

 

Here is how acceleration works under SR.

 

It is absolute motion.

 

From the accelerating frame, you will calculate Δt = c/a sinh( a T /c ) for the elapsed time in the stationary frame.

 

But, since it is absolute motion, it applies to both frames.

 

a - acceleration in the accelerating frame

T- accelerating time in the accelerating frame

 

according to

http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf

Edited by vuquta
Link to comment
Share on other sites

the rate and even the direction of ticking of the stationary clock from the accelerating twins point of view will depend on the distance between them.

 

Wrong, that is logically decidable. It was assumed they were co-located at the beginning.

The position at any time T in the accelerating frame is:

 

x(T) = c²/a [ cosh(a T/c) - 1 ]

 

Again, this is absolute moton and hence neither frame will disagree at the instant the acceleration stops.

 

In the paper we find the logic,

 

Remarkably the dependence on v(0) disappears. This follows from the fact that contrary to x0() and x1(), the quantity T () is a Lorentz invariant and as such should not depend on the choice of frame (i.e. the choice of v(0)).

http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf

 

As we can see, the choice of v is irrelevant meaning the numbers are absolute between the frames.

Edited by vuquta
Link to comment
Share on other sites

I am not a physicist, but I don't think anyone has answered clearly enough for you, so I will point out a couple of things i think you don't get yet:

 

OK so we are agreed that in my scenario that at the end of the journey, Brian and Andy (now standing next to each other) will look at their respective clocks and see that Brian's only shows 100 days (100 pips), and that Andy's shows 200 days (200 pips). And that this is because Brian's movement and heavy accelerations have caused Brian to age only half the time of Andy.

 

Actually no, your assuming that half the speed of light = time dilation of 2... this isn't how it works. At 1/2 c the time dilation factor is about 1.15

 

So if Brian's clock read 100 days at the end of his journey, Andy's clock would read about 115 days.

 

This is where it gets interesting. So does this mean that to make this happen, Brian must have had to take the decision to turn back after only 50 days (50 pips) (as he perceives it) such that his clock has only got to 100 by the time he returns? And when he returns, he sees that Andy's clock has got to 200 days (this being consistent with Andy's own experience of Brian taking 200 days for the whole trip). Am I right about this so far?

 

If you change Andy's clock to 115 days, then yes.

 

Assuming I am, then the following perhaps make sense: That for the portions of the journey when Brian is not accelerating, then for almost all of (from Brian's perspective) the 2 * 50 day journey, that Brian would look back at Andy, and see Andy's clock going at half speed, so resulting in only 25 days (25 pips) elapsing for the outbound journey, and 25 days elapsing for the return journey. 50 days total - as seen by Brian. Now since we know that on his return to Andy, Brian will see that Andy's clock has passed 200 days, there is a missing 150 days here. 150 days that needs to be added on to Brian's perception of Andy's clock at some point in the journey

 

You've got it completely wrong here. Lets assume that Brian does travel for 50 days out before turning around as you say and coming back. Brian would look back at Andy and see Andy's clock going slower on the outbound journey. On the inbound journey, Brian would see Andy's clock going faster. Brian would receive more pips from Andy on his journey back to Earth than he did on his journey away from Earth. But, in total, Brian would receive 115 pips from Andy, while only 100 pips on his own clock. Andy would receive 100 pips from Brian, while seeing 115 pips go by on his own clock. Brian and Andy would see the same thing, there is no paradox there. When they are back together, comparing clocks, Brians clock will say 100 days, and Andy's clock will say 115 days... which is exactly the number of pips they will both count during the journey.

Link to comment
Share on other sites

Special relativity says that a clock on a spaceship moving relative to a stationary observer appears to be running more slowly than if it was stationary. So if this is true, what would you get in the following scenario?

 

Stationary observer Andy and moving observer Brian are both initially stationary and in the same location, and each has an atomic clocks measuring time in days, and displaying what they can see on a large monitor visible to both. In addition both clocks emit an electronic "pip" whenever one day passes to the next, and each records the number of pips made since reset. Day = 0, and the pip counter is set to 0.

 

Brian rapidly accelerates towards a distant star, and it takes him one second to reach around half the speed of light. After 100 days of travelling Brian stops in 1 second and accelerates to the same speed back towards Andy (in one second), and arrives back at Andy in a further 100 days.

Before they got out their clocks to compare what they showed, they each gave their opinion as to what they were expecting them to show.

 

 

Brian reports that Brian's clock shows that 200 days has elapsed, and he reported having received 200 pips from it.

Brian reports that Andy's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it.

 

Andy reports that Andy's clock shows that 200 days has elapsed, and he reported having received 200 pips from it.

Andy reports that Brian's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it.

 

When they both look at their clocks, and read their pip counters, what do they see?

 

Thanks

 

I do not have the patience for numbers.

Here is your solution.

T is time of acceleration in the acceleration time

a is the acceleration in the accelerating frame.

 

Brian time

1) accel period

t1 = c/a sinh( aT/c)

2) You left off a decell period for the turn around. I will assume the same a and T

t2 = c/a sinh( aT/c)

3) Come back

t3 = c/a sinh( aT/c)

4) Decell back to the original frame

t4 = c/a sinh( aT/c)

 

The non inertial clocks beats slower and so Brian will be younger than Andy. You did not include a relative motion period as normal. But, you can plug the numbers and get the results.

Doesn't matter, accelerating clocks beat slower. However, if the relative motion period is long enough that you did not include, you can make this problem logically undecidable.

Edited by vuquta
Link to comment
Share on other sites

Wrong, that is logically decidable. It was assumed they were co-located at the beginning.

The position at any time T in the accelerating frame is:

 

x(T) = c²/a [ cosh(a T/c) - 1 ]

 

Again, this is absolute moton and hence neither frame will disagree at the instant the acceleration stops.

 

In the paper we find the logic,

 

Remarkably the dependence on v(0) disappears. This follows from the fact that contrary to x0() and x1(), the quantity T () is a Lorentz invariant and as such should not depend on the choice of frame (i.e. the choice of v(0)).

http://arxiv.org/PS_...1/0411233v1.pdf

 

As we can see, the choice of v is irrelevant meaning the numbers are absolute between the frames.

 

Actually vuquta, granpa has it right. The solution you set up in your last post is wrong.

 

The hyperbolic formula you posted assumes constant acceleration and the starting positions of the inertial and accelerating observer coincide. That works for the initial acceleration away from earth but not the turnaround because the acceleration is now in the opposite direction. So for the turnaround, we need to reset the parameters and now the starting positions don't coincide. For the turnaround we must use the more complete version of the formula you posted, the one that includes the difference in starting position.

 

Regardless, your conclusion is still correct. To accelerating observer, an inertial clock runs faster.

 

I'm not sure what you mean by the phrase "logically undecidable". We are always able to determine who will be younger.

Link to comment
Share on other sites

To understand what is going on in the Twins Paradox, I find it helpful to look at a picture. Also experiments in particlel accelerators verify that time indeed does slow for the "twin" who has undergone acceleration. See for example http://marksmodernphysics.com/Mark's%20Modern%20Physics/Musings/Musings%20-%20The%20Twins%20Paradox.doc

 

Sorry. Link has to be on separate line

 

http://marksmodernphysics.com/Mark's%20Modern%20Physics/Musings/Musings%20-%20The%20Twins%20Paradox.doc

Link to comment
Share on other sites

Actually vuquta, granpa has it right. The solution you set up in your last post is wrong.

 

The hyperbolic formula you posted assumes constant acceleration and the starting positions of the inertial and accelerating observer coincide. That works for the initial acceleration away from earth but not the turnaround because the acceleration is now in the opposite direction. So for the turnaround, we need to reset the parameters and now the starting positions don't coincide. For the turnaround we must use the more complete version of the formula you posted, the one that includes the difference in starting position.

 

Regardless, your conclusion is still correct. To accelerating observer, an inertial clock runs faster.

 

I'm not sure what you mean by the phrase "logically undecidable". We are always able to determine who will be younger.

 

Nope sorry, you both are wrong.

 

The sin hyperbolic is the same with the negative. Hence sinh(-a) = sinh(a), so the turn around does not matter since it is acceleration.

 

This is an error many make.

 

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

 

Check this link above and you will find it is consistent with my logic with one exception. It integrates from the inertial frame and I can prove that fails with a recent paper.

 

Therefore, it uses asinh, which is wrong.

 

Next, if you are not using a constant acceleration, then you are in GR strictly which was not part of the OP.

 

You cannot use SR to evaluate non constant acceleration.

 

Last, it does not matter.

 

Any accelerating frame regardless of it being constant or variable beats slower in time than the inertial clock, otherwise GPS fails since GPS uses this fact to pre-program the satellites.

 

As such, my conclusions are mathematically valid, variable or constant acceleration and backed up by the scientific evidence in GPS.

 

QED

 

Actually vuquta, granpa has it right. The solution you set up in your last post is wrong.

 

The hyperbolic formula you posted assumes constant acceleration and the starting positions of the inertial and accelerating observer coincide. That works for the initial acceleration away from earth but not the turnaround because the acceleration is now in the opposite direction. So for the turnaround, we need to reset the parameters and now the starting positions don't coincide. For the turnaround we must use the more complete version of the formula you posted, the one that includes the difference in starting position.

 

Regardless, your conclusion is still correct. To accelerating observer, an inertial clock runs faster.

 

I'm not sure what you mean by the phrase "logically undecidable". We are always able to determine who will be younger.

 

Oh, I forgot to address the logical decidability or absoluteness of the OP's twins proposal in terms of time dilation.

 

Since GPS pre-programs the satellites with GR gravity effects, this proves it is absolute or GPS would not work which is exactly what I said.

 

I think I covered erverything.

 

See Neil Ashby for GPS.

 

I'm not sure what you mean by the phrase "logically undecidable". We are always able to determine who will be younger.

 

 

OK, let's see.

 

I have two A and B in the same frame clocks synched.

 

 

A takes off and instantly acquires v.

 

After some time B takes of exactly the same way.

 

They will end up in the same frame.

 

Which is younger?

Edited by vuquta
Link to comment
Share on other sites

To understand what is going on in the Twins Paradox, I find it helpful to look at a picture. Also experiments in particlel accelerators verify that time indeed does slow for the "twin" who has undergone acceleration. See for example http://marksmodernphysics.com/Mark's%20Modern%20Physics/Musings/Musings%20-%20The%20Twins%20Paradox.doc

 

Sorry. Link has to be on separate line

 

http://marksmodernphysics.com/Mark's%20Modern%20Physics/Musings/Musings%20-%20The%20Twins%20Paradox.doc

 

I think I've got it now, after reading this .doc file. It is all to do with the fact that the accelerating twin starts to receive faster return time pips immediately after the change in velocity (the turnaround) whereas the stationary twin has to wait for <distance>/c until they see the turnaround and start getting the faster blips.

 

It is weird yet so simple to conceptualise once you've "got it".

 

I wrote my own version of how this happens using my Andy&Brian scenario without using worldlines, but I now see that Janus's post above explains it more succinctly than I have. Still I've posted it here (in italics) just so I can feel I haven't wasted 20 minutes of typing (OK I'll admit it - I wasted 20 minutes of typing)

 

Thanks very much for all your replies - much appreciated!

 

 

Andy and Brian start at time=0 at the start point.

Andy is stationary the whole time.

Each has a beacon that sends a signal (a pip) every hour.

At time=0, Brian moves out at half the speed of light towards a distant location.

While Brian is in steady state motion travelling out, due to dopplar shift alone, each receives one pip from the other every 1.5 hours.

 

Looking at the journey from Brian's perspective:

After 90 hours (according to Brian), Brian has received 60 pips from Andy. Brian has travelled a distance of 45 light-hours. Brian then stops and starts back towards Andy at the same speed.

Brian takes 90 hours (according to Brian) to get back to Andy.

While Brian is travelling back, he can see pips more frequently from Andy - every 40 minutes to be precise, due to dopplar shift alone.

So from Brian's perspective Brian receives a further 135 pips (90*1.5) from Andy.

So total number of pips received by Brian from Andy = 60 + 135 = 195 pips.

This is for Brian's own 180 pips.

 

Looking at the same journey from Andy's perspective:

On the outbound journey, Andy sees pips at the lower rate of one pip every 1.5 hours.

However since it takes 90hours * 0.5*speed of light for the light showing Brian's arrival at the turning point to reach Andy, then from Andy's perspective, it has taken 135 hours of his time (Andy's time) to reach that destination.

Over that 135 hours Andy receives 90 pips from Brian. And of course 135 pips from himself.

Once Andy sees Brian reach the destination and start back, then Andy starts receiving pips from Brian at the quicker interval (every 40 minutes).

We know that for the return journey that Andy must receive a further 90 pips from Brian because whatever relativity effects there are, the number of pips from Brian's machine in total for the trip must be the same whether measured by Andy or Brian.

The time taken (from Andy's perspective) to receive the 90 pips from Brian is 90/1.5 = 60 hours.

Add this to the 135 hours already spent watching the outbound journey gives a total number of hours for Andy from Andy's perspective of 195 hours (195 pips)

Total number of pips received by Andy from Brian = 90 + 90 = 180 pips.

This is for Andy's own 195 pips.

 

So in both of the above cases Andy has aged 15 days more than Brian. Cool!

 

And the reason why the situation is not symmetrical is, as indicated by the document you referenced, that the one that accelerates (Brian in this case) starts to see the change in frequency of receipt of pips as they complete the acceleration. But due to the distance between the two at that point, the one that has not accelerated (Andy in this case) has to wait for a few hours more perceiving the remote object to be still moving at its previous speed, before the light reaches him and he starts receiving pips at the faster rate. Since he thus receives them for a shorter period of time, he sees fewer pips, hence perceives the remote object to be ageing slower than him.

 

(Andy sees Brian travel a distance back of 45 light-hours in only 60 hours. This would make Andy appear to be travelling at 45/60*C = three quarters of the speed of light on return.)

Edited by Neil9327
Link to comment
Share on other sites

OK, let's see.

 

I have two A and B in the same frame clocks synched.

 

 

A takes off and instantly acquires v.

 

After some time B takes of exactly the same way.

 

They will end up in the same frame.

 

Which is younger?

 

In the new frame both will agree that B is younger.

Link to comment
Share on other sites

OK, let's see.

 

I have two A and B in the same frame clocks synched.

 

 

A takes off and instantly acquires v.

 

After some time B takes of exactly the same way.

 

They will end up in the same frame.

 

Which is younger?

 

In the new frame both will agree that B is younger.

 

Wrong answer, but what is your math and/or logic?

Link to comment
Share on other sites

if A and B are both at the origin at T=0 and A instantly accelerates to v at T=0 then you just switch frames immediately to the new frame.

In that frame it is B that is moving. Therefore it is B's clock that is ticking slower in that frame.

At T=1 B stops moving in that frame.

In that frame it is therefore B that ages less.

 

I suspect that you are just arguing with people to try to get them to do your homework for you.

I wouldnt have posted this answer but it was trivial and I couldnt resist the urge to point out how obviously wrong you are.

Edited by granpa
Link to comment
Share on other sites

if A and B are both at the origin at T=0 and A instantly accelerates to v at T=0 then you just switch frames immediately to the new frame.

In that frame it is B that is moving. Therefore it is B's clock that is ticking slower in that frame.

At T=1 B stops moving in that frame.

In that frame it is therefore B that ages less.

 

I suspect that you are just arguing with people to try to get them to do your homework for you.

I wouldnt have posted this answer but it was trivial and I couldnt resist the urge to point out how obviously wrong you are.

 

 

I think granpa has misunderstood the problem. Since "A" and "B" do not take off at the same time, the first traveler will always be younger. The problem can easily be solved from the point of view of a third observer who is always stays on earth.

 

The logic is as follows:

"A" and "B" start out at the same place on earth (an inertial reference frame).

 

Phase 1:

"A" accelerates away from earth until he reaches some another inertial reference frame.

"B" stays on earth.

Therefore "A" has lost some time on his clock relative to earth. "B" has not.

 

Phase 2:

"B" takes off with the same acceleration until he reaches the same inertial reference frame as "A".

"A" stays inertial.

Therefore "B" has lost the same amount of time on his clock (compared to earth) as "A" did in phase 1. "A" has lost some additional time. They are now in the same inertial reference frame and their clocks now run at the same rate.

 

Conclusion from the point of view of Earth: "A" is younger than "B". If you can't draw the same conclusion by analyzing the problem from the point of view "A" or "B" you are doing something wrong.

 

Link to comment
Share on other sites

from the point of view of the original frame yes thats obviously true.

vuquta asked:

They will end up in the same frame.

 

Which is younger?

He was asking who was younger from the new frame of A and B

In that frame it is as I said

Edited by granpa
Link to comment
Share on other sites

 

If you can't draw the same conclusion by analyzing the problem from the point of view "A" or "B" you are doing something wrong.

 

 

No, that's not true. Everyone has to agree if events occur, but on issues of simultaneity they most certainly do not have to agree, and I think this issue falls into the category of simultaneity. To analyze it properly, according to the question, one must do it from the perspective of an observer in the moving frame.

Link to comment
Share on other sites

from the point of view of the original frame yes thats obviously true.

vuquta asked:

 

He was asking who was younger from the new frame of A and B

In that frame it is as I said

 

Ouch.

Just like with the Twins Paradox, calculate who is youger from the point of view of one inertial observer and all other inertial observers must agree.

If that's not true there will paradoxes with causality.

Let's not get lost here folks.

 

No, that's not true. Everyone has to agree if events occur, but on issues of simultaneity they most certainly do not have to agree, and I think this issue falls into the category of simultaneity. To analyze it properly, according to the question, one must do it from the perspective of an observer in the moving frame.

 

 

Yes it is true. Think about it. Relativity of simultaneity means two observers in two different inertial reference frames will determine ONE event to occur at two different times. But you must never get the two observers thinking TWO events occured in a different sequence. Also see my last post.

Link to comment
Share on other sites

 

Yes it is true. Think about it. Relativity of simultaneity means two observers in two different inertial reference frames will determine ONE event to occur at two different times. But you must never get the two observers thinking TWO events occured in a different sequence. Also see my last post.

 

This contradicts the famous ladder (so-called) paradox.

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

Link to comment
Share on other sites

I think granpa has misunderstood the problem. Since "A" and "B" do not take off at the same time, the first traveler will always be younger. The problem can easily be solved from the point of view of a third observer who is always stays on earth.

 

The logic is as follows:

"A" and "B" start out at the same place on earth (an inertial reference frame).

 

Phase 1:

"A" accelerates away from earth until he reaches some another inertial reference frame.

"B" stays on earth.

Therefore "A" has lost some time on his clock relative to earth. "B" has not.

 

Phase 2:

"B" takes off with the same acceleration until he reaches the same inertial reference frame as "A".

"A" stays inertial.

Therefore "B" has lost the same amount of time on his clock (compared to earth) as "A" did in phase 1. "A" has lost some additional time. They are now in the same inertial reference frame and their clocks now run at the same rate.

 

Conclusion from the point of view of Earth: "A" is younger than "B". If you can't draw the same conclusion by analyzing the problem from the point of view "A" or "B" you are doing something wrong.

 

 

Actually, who is "younger" depends upon which frame you are considering it from.

 

Here's the space-time diagrams for the situation as seen from both the staring inertial rest frame of A and B and from their final inertial rest frame.

 

worldlines.gif

 

From the original frame, A (the blue world-line) ends up younger than B,(the red World-lines)

 

However, when you switch to the final rest frame, B is younger than A.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.