Undergrad Does the "space twin" benefit from length contraction?

[Herman Trivilino]

All the time the rover twin is traveling uniformly, he will see the clocks on the Earth ticking more slowly than his (and vice versa).

There is no such notion as "all the time" as you are using it here. If one twin says that during the time I was getting married you were reading page 12, the other twin will say that during the time you were getting married I was reading a different page.

That extra time on Earth must have transpired during his deceleration! When he gets home he will be younger yet except for the noninertial parts of the trip each twin will agree (correctly!) the other's biological clocks are running slower.

To your use of "when" the same comment as above applies. The notion of "when" upon which the twins will agree occurs at the departure and reunion. Therefore it's possible to say that they agree when they share the same location.

 

[hutchphd]

Just looked at Parok's post...sorry but I don' t understand what he is doing. The description from every reference frame will be consistent within. the special theory.

This is not true if you mean "see" literally. Due to the Doppler effect, the traveling twin will see the stay-at-home's clock ticking slowly on the outbound leg and quickly on the inbound leg. The stay-at-home will see the traveller's clock ticking slowly for the 80% of the trip time the traveller can be seen to be heading outwards, and quickly for the 20% of the time the traveller is seen to be inbound. (Edit: the 80/20 figure assumes a speed of 0.6c and that my mental arithmetic is reliable.)

I don't understand the argument. If we use this definition of "see" then each twin will agree instead that an equal time has passed for the other (is that what you are saying?) and yet Rover is still younger than Homer when they meet ! Do they agree that they were not actually born twins? What are you saying here?

 

[PeroK]

Just looked at Parok's post...sorry but I don' t understand what he is doing. The description from every reference frame will be consistent within. the special theory.

I don't understand the argument. If we use this definition of "see" then each twin will agree instead that an equal time has passed for the other (is that what you are saying?) and yet Rover is still younger than Homer when they meet ! Do they agree that they were not actually born twins? What are you saying here?

There is the "Doppler" explanation of the paradox, which analyzes what each twin observes directly. Rather than what they measure in their respective reference frames. The Doppler explanation, in particular, does not need all the "weirdness" to happen during one or more acceleration phases.

 

[hutchphd]

I thought you were giving the Doppler explanation. Where is the asymmetry in this explanation?t

 

[PeroK]

I thought you were giving the Doppler explanation. Where is the asymmetry in this explanation?t

It's the change of direction.

 

[hutchphd]

So what does Rover observe about Homer's clock during the "change in direction" ?

 

[PeroK]

So what does Rover observe about Homer's clock during the "change in direction" ?

Nothing. The change of direction can be instantaneous - or near instantaneous.

More precisely it's the timing of the change of direction. For example, if the relative velocity is $\frac35 c$ and the distance is $2.4$ light years (Earth frame), then at the turn around point, Rover's clock reads $4$ years and the light from Homer's second birthday has just reached him.

Whereas at the turn around point Homer's clock reads $5$ years etc.

 

[PeroK]

In fact, to put it more simply. In Homer's frame, Rover travels for 5 years and a greater distance before he turns round.

Whereas, in Rover's frame Homer will travel for only 4 years and a shorter distance before turning round.

So, the asymmetry is inherent in the problem from the outset - owing to length contraction, at least.

 

[hutchphd]

For clarity let us assume that each of the Travelers is emitting a 1 Hz signal and counting this signal is how they characterize each others time (correctly or not...that is how they do it) . The only time where they will disagree about the rate of tick tock of each others clock is during the deceleration where they must measure very different things. What does each measure from that interval ??.

 

[PeroK]

For clarity let us assume that each of the Travelers is emitting a 1 Hz signal and counting this signal is how they characterize each others time (correctly or not...that is how they do it) . The only time where they will disagree about the rate of tick tock of each others clock is during the deceleration where they must measure very different things. What does each measure from that interval ??.

There is no deceleration. We are hypothesising an instantaneous change of direction.

 

[hutchphd]

In fact, to put it more simply. In Homer's frame, Rover travels for 5 years and a greater distance before he turns round.

Whereas, in Rover's frame Homer will travel for only 4 years and a shorter distance before turning round.

So, the asymmetry is inherent in the problem from the outset - owing to length contraction, at least.

Does this not imply that Rover will expect Homer to be younger?

 

[PeroK]

Does this not imply that Rover will expect Homer to be younger?

No.

 

[hutchphd]

There is no deceleration. We are hypothesising an instantaneous change of direction.

You mean you are hypothesizing an infinite deceleration...not a zero. Not a good idea to introduce infinities. I would prefer to call it large but finite and re-ask the same question.

 

[jbriggs444]

[

For clarity let us assume that each of the Travelers is emitting a 1 Hz signal and counting this signal is how they characterize each others time (correctly or not...that is how they do it) . The only time where they will disagree about the rate of tick tock of each others clock is during the deceleration where they must measure very different things. What does each measure from that interval ??.

The word "during" for two separated observers in relative motion is a non-starter.

Edit: further, there is no time at which either observer will either "see" or "observe" the other's tick rate to be 1 Hz -- with the possible exception of an instant at turnaround where the traveller might be momentarily at rest in the stay-at-home twin's frame and "see" 1 Hz. Or a pair of instants at turnaround where the traveler "observes" a tick rate momentarily at 1 Hz. The stay at home twin would "see" a 1 Hz tick rate momentarily significantly after the halfway time and would "observe" a 1 Hz tick rate momentarily exactly at the halfway time.

 

[PeroK]

You mean you are hypothesizing an infinite deceleration...not a zero. Not a good idea to introduce infinities. I would prefer to call it large but finite and re-ask the same question.

Nothing special happens to what they see at the turn around. See the link in post #46 for a Doppler analysis.

In any case, the underlying implication of your questions that only acceleration can cause the differential ageing is flawed.

Many analyses can happily assume an instantaneous ( or nearly if you prefer) turn around.

 

[hutchphd]

OK I will rephrase. Each twin carries a machine that plots the received signal on a graph. When they get back together they can look at these graphs and the number of ticks each sees for the other must correspond to their relative ages (events do not disappear). We have already agreed about symmetry during the inertial parts of the journey. So there will be a section of the graphs that do not agree. What does that look like??

 

[jbriggs444]

OK I will rephrase. Each twin carries a machine that plots the received signal on a graph. When they get back together they can look at these graphs and the number of ticks each sees for the other must correspond to their relative ages (events do not disappear). We have already agreed about symmetry during the inertial parts of the journey. So there will be a section of the graphs that do not agree. What does that look like??

Let's get more specific. This machine has a pen that moves up and down to reflect the signal state and a strip of paper moving right to left under the pen. I.e. a chart recorder.

We agree that the two strips of chart paper will have different numbers of peaks. But until we specify how the paper moves under the two pens, we cannot know where the two strips will differ. [Yes, there is a natural assumption that might be made -- but it's still an assumption]

 

[PeroK]

We have already agreed about symmetry during the inertial parts of the journey.

No, we haven't. See post #68, for example. The experiment is fundamentally asymmetric.

Rover, as we have already explained, will record a shorter outbound journey than Homer.

 

[hutchphd]

It is a normal strip recorder bought by for each twin at birth from acme strip recorder company and each works perfectly.

 

[Nugatory]

OK I will rephrase. Each twin carries a machine that plots the received signal on a graph. When they get back together they can look at these graphs and the number of ticks each sees for the other must correspond to their relative ages (events do not disappear). We have already agreed about symmetry during the inertial parts of the journey. So there will be a section of the graphs that do not agree. What does that look like??

This is covered in some detail in the "Doppler Shift Analysis" section of the FAQ that I linked to above.

The earthbound twin receives ticks from the traveller at a rate less than his own tick rate for more than half the time between separation and reunion. Shortly before the reunion, the rate at which the earthbound twin receives ticks increases to be greater than his own. This transition from receives-slow to receives-fast does not happen at the same time as the deceleration for turn around, it happens much later.

The space twin receives ticks at the slow rate until the turnaround, and after the turnaround receives ticks at the faster rate. So the two graphs disagree only in where the transition happens: after the midpoint for earthbound, and at the midpoint for the traveller.

 

[hutchphd]

Got it. That's a very satisfying explanation. I greatly appreciate your patience and hope it has been edifying for others as well!

 

[Herman Trivilino]

Got it. That's a very satisfying explanation. I greatly appreciate your patience and hope it has been edifying for others as well!

If you haven't done so yet, you might consider going to YouTube and doing a search for "Hewitt Twin Trip". He does the analysis in a concrete way that makes it clear that no matter which twin is the receiver of signals sent from the other twin, they will both agree based on their own counts how much more proper time passed for the staying twin.

By the way, the relative speed in that video is $\frac{3}{5} c$ making $\gamma$ equal to $\frac{5}{4}$.

Edit: Oh, and that makes the Doppler shift factor $\sqrt{\frac{1+\beta}{1-\beta}}$ equal to $2$.

 

[hutchphd]

Ah yes the old 3,4,5 triangle. I fear I am about to be addicted to this site. Great stuff thanks

 

[hutchphd]

While the "who is older" analysis does not strictly require it, I am still interested in what Romer will experience during a real life (finite non instantaneous) turn around. Let me imagine his space filled with a field of clocks which he synchronizes at birth (while sitting next to Homer). On his travels the clocks will accompany him.
If I understand the Rindler coordinate transform, as soon as Romer feels any acceleration his field of clocks will:
1) no longer be synchronized...they will run at different rates during the acceleration depending upon their position along the acceleration axis.
2) In fact I guess each clock will experience a slightly different acceleration?
v3) When returned to Homers inertial frame the deviations from acceleration will sum to zero and the Romer clocks will again be synchronized (but still disagree with Homer's clocks!)
Again I appreciate your indulgence...Is this correct?

 

[PeroK]

While the "who is older" analysis does not strictly require it, I am still interested in what Romer will experience during a real life (finite non instantaneous) turn around. Let me imagine his space filled with a field of clocks which he synchronizes at birth (while sitting next to Homer). On his travels the clocks will accompany him.
If I understand the Rindler coordinate transform, as soon as Romer feels any acceleration his field of clocks will:
1) no longer be synchronized...they will run at different rates during the acceleration depending upon their position along the acceleration axis.
2) In fact I guess each clock will experience a slightly different acceleration?
v3) When returned to Homers inertial frame the deviations from acceleration will sum to zero and the Romer clocks will again be synchronized (but still disagree with Homer's clocks!)
Again I appreciate your indulgence...Is this correct?

If you consider an acceleration phase at the beginning, then a clock at the front of Rover's ship will run faster than one at the rear. During the journey, therefore, the ship clocks may be out of sync. Then, if there is a deceleration phase at the turnaround, the clocks will end up back in sync again.

The clocks don't experience a different acceleration, but their different positions in the accelerating frame leads to time dilation between them. There is a heuristic argument for gravitational time dilation based on this and the equivalence principle.

 

[hutchphd]

The clocks don't experience a different acceleration, but their different positions in the accelerating frame leads to time dilation between them. There is a heuristic argument for gravitational time dilation based on this and the equivalence principle.

If I am riding with Romer some distance down the axis and feel (measure) the same acceleration but my clock reads a different time than must I not be traveling at different speed ??

 

[PeroK]

If I am riding with Romer some distance down the axis and feel (measure) the same acceleration but my clock reads a different time than must I not be traveling at different speed ??

Not necessarily. You get internal time dilation in an accelerating reference frame.

The factor is $1 + \frac{ah}{c^2}$ where $h$ is the distance in the direction of acceleration $a$.

The clock to the rear runs slow by this amount, in the accelerating frame.

Note that if the ship is accelerating homogeneously in the initial IRF. I.e the front and rear always have the same velocity in that frame, then the clocks will stay synchronised in that frame. And, if at the end of the acceleration phase, they are in sync in the initial IRF, then they must be out of sync - by $hv/c^2$ in the ship frame.

This ties in with the "leading clocks lag" rule, if you think about it.

 

[Nugatory]

If I am riding with Romer some distance down the axis and feel (measure) the same acceleration but my clock reads a different time than must I not be traveling at different speed ??

It takes some care to properly describe an extended object (such as your collection of clocks moving along with Romer) undergoing acceleration. The basic problem is that when we say that all parts of the object are experiencing the same acceleration at the same time... we have to assign some meaning to "at the same time" and "the distance between the clocks at any given moment", and of course both of these notions are frame-dependent.

This has become a good time to mention two other google searches: "Born rigid motion" and "Bell spaceship paradox".

 

[PeroK]

It takes some care to properly describe an extended object (such as your collection of clocks moving along with Romer) undergoing acceleration. The basic problem is that when we say that all parts of the object are experiencing the same acceleration at the same time... we have to assign some meaning to "at the same time" and "the distance between the clocks at any given moment", and of course both of these notions are frame-dependent.

This has become a good time to mention two other google searches: "Born rigid motion" and "Bell spaceship paradox".

Yes. Note that the heuristic argument in post #87 ignores the length contraction of the ship. But, if acceleration is modest, then the effect of a few metres contraction over, say, one light year of travel results in a tiny difference in velocity across the ship in the IRF.

 

[hutchphd]

I don't disagree with any of these details, but my simple question should have a simple answer I believe. Me and Romer synchronize our clocks to zero before the acceleration and agree after the acceleration that we are still attached to the same spacecraft but disagree as to the time for acceleration. Our delta V is manisfestly the same yet our times for it differ...does that not require the acceleration we measure (eg with our buttocks) to be different?