Does the "space twin" benefit from length contraction?

[laymanB]

There's a difference between a rest frame and an IRF. In a previous thread I analyzed this problem from the traveller's outbound IRF. In this frame, the Earth moves with constant velocity. On the outward leg, the traveller is at rest in this frame. But, on the homeward leg the traveller is also moving in this frame.

The thing the traveller cannot do is associate their rest frame with a single IRF throughout.

Thanks, I think you are hitting on where my problem is at. Could you provide a link to that thread?

 

[PeroK]

Thanks, I think you are hitting on where my problem is at. Could you provide a link to that thread?

Search for "Einstein or Resnick" in the title. It's post #75 from Nov-15.

It's a useful exercise in SR if nothing else.

But, looking forward to your study of GR, you need to think more and more about the geometry of spacetime and follow the posts of @PeterDonis.

 

[laymanB]

Search for "Einstein or Resnick" in the title. It's post #75 from Nov-15.

It's a useful exercise in SR if nothing else.

But, looking forward to your study of GR, you need to think more and more about the geometry of spacetime and follow the posts of @PeterDonis.

Thanks. I very much enjoy both of your posts. Thanks to you and @PeterDonis for your patience.

 

[PeterDonis]

This would be the spacetime diagram drawn in the Earth twin reference frame, correct?

Yes.

I was envisioning that if the spacetime diagram drawn in the reference frame of the traveling twin who considers himself at rest the entire time, that is image would be symmetric along the ct axis.

I think you mean, that it would look like a reflection of the Earth twin diagram about the ct axis. The answer to that is probably not, but that answer by itself does not address the key problem.

The key problem is that, as has already been said, any "frame" in which the traveling twin is at rest the whole time is not an inertial frame. That means that, no matter what the spacetime diagram in that frame turns out to look like, you can't draw any conclusions based on it the way you can draw conclusions based on a spacetime diagram drawn in an inertial frame. For example, in a spacetime diagram drawn in a frame in which the traveling twin is at rest the whole time, the traveling twin's worldline will look like a straight line. But you cannot conclude from that that the traveling twin does not feel acceleration, or that his proper time between the start and end events is maximal; whereas you can conclude those things about the Earth twin from the diagram you showed, drawn in the Earth twin's rest frame, because that frame is an inertial frame and you can draw conclusions like that from a spacetime diagram drawn in that frame.

I find that most places that teach about the twin paradox make no mention of how to solve it without resorting to acceleration. Including the link you provided in this thread.

Read that article again, carefully. In particular, read the spacetime diagram analysis, carefully. Nothing in that analysis "resorts to" acceleration; the only time acceleration is mentioned at all is in postulating that it doesn't affect the mechanism of clocks.

What that analysis does require is that the spacetime diagram is drawn in an inertial frame, because, as I said above, that is what justifies drawing the sorts of conclusions that are drawn. The analysis in the article draws them algebraically, by manipulating formulas that are valid for inertial frames. But you can also draw them, at least qualitatively, by looking at the diagram and seeing that the Earth twin's worldline is a single straight line, while the traveling twin's is not. But that requires that the diagram is drawn in an inertial frame, just as the formulas used in the article require using coordinates assigned in an inertial frame.

In post #27, you did say that the acceleration is not ignored, just idealized.

Yes, but only in the sense that the "corner" in the traveling twin's worldline is there no matter what. You don't need to know anything about the magnitude of the acceleration, or use it in any formulas, or assume that it takes any finite time; the basic conclusion, that the traveling twin is younger, does not require any of that. It just requires recognizing that the traveling twin's worldline is not a single straight line from start to finish, while the Earth twin's worldline is.

 

[laymanB]

The key problem is that, as has already been said, any "frame" in which the traveling twin is at rest the whole time is not an inertial frame. That means that, no matter what the spacetime diagram in that frame turns out to look like, you can't draw any conclusions based on it the way you can draw conclusions based on a spacetime diagram drawn in an inertial frame. For example, in a spacetime diagram drawn in a frame in which the traveling twin is at rest the whole time, the traveling twin's worldline will look like a straight line. But you cannot conclude from that that the traveling twin does not feel acceleration, or that his proper time between the start and end events is maximal; whereas you can conclude those things about the Earth twin from the diagram you showed, drawn in the Earth twin's rest frame, because that frame is an inertial frame and you can draw conclusions like that from a spacetime diagram drawn in that frame.

I think you and @PeroK have found where my confusion lies. It seems to be this misconception that I can equally analyze the problem from both the perspective of the Earth twin and traveling twin as if they are both in inertial frames the entire time and then compare the results. I have a sneaking suspicion that many others like myself are tripping over this same point. Time to get a better handle on frames. Thanks.

 

[GrayGhost]

My post was redundant to LaymanB's post #7 ...

https://www.physicsforums.com/threa...-from-length-contraction.933918/#post-5898964

... so I deleted my post here. Sorry 'bout that.

Best Regards,
GrayGhost

 

[pervect]

Non-inertial reference frames in special relativity have somewhat different properties than they do in Newtonian mechanics. The general advice I'd give (and the approach most textbooks take) is to analyze the problem in an inertial reference frame.

Rindler coordinates for a uniformly accelerating observer illustrate some of the unexpected and non-intuitive features of accelerated frames in special relativity. Rindler coordinates are useful in problems such as a uniformly accelerating space-ship, or a uniformly accelerating elevator (usually called Einstein's elevator).

Some key features: Clocks don't run at the same rate depending on their position. A clock higher in the elevator runs faster. This is not the fault or any feature of the clocks themselves, it's really a feature of the use of the RIndler coordinates. Furthermore, the frame cannot be consistently defined to fill all of space-time, there is an apparent "horizion" in the Rindler coordinates. It can be described as the point where clocks get so low in the elevator that they stop, but this is somewhat misleading depiciton, as this is not any fault of the clocks. Rather, the coordinates themselves are not defined - there's nothing wrong with the clocks, it's a limitation on the coordinates themselves.

The mathematical methods needed to rigorously analyze non-inertial reference frames properly are rather advanced, and usually taught in GR courses. If one does not have these advanced methods available, the best solution is not to use non-inertial coordinates at all. Without the use of such mathematics, the above informal descritption is the best I can do in describing how non-inertial frames work in special relativity, but it's really no substitute for a full formal treatment, and there's a high possibility of confusion in trying to actually apply this informal summary to work real problems.

[add]
Gravitation, by Misner, Thorne, Wheeler (MTW) has a good treatment of the accelerated observer using advanced (tensor) methods.

 

[PeterDonis]

Time to get a better handle on frames.

You might also consider becoming more familiar with the geometric viewpoint. For example, earlier I said it should be obvious that the traveling twin's frame is non-inertial. The geometric viewpoint makes it easy to see why.

Consider that in the worldlines of the Earth twin and the traveling twin we have two different curves between the same two points in spacetime (A and B). There can only be one curve between any two points in spacetime that is a straight line (for the same reason there can only be one curve between any two points in ordinary Euclidean space that is a straight line). We are told in the problem specification that the Earth twin's worldline is a straight line; therefore the traveling twin's worldline cannot be one. And if the traveling twin's worldline is not a straight line, the frame in which he is at rest--which means his worldline has constant spatial coordinates--cannot be an inertial frame, since only a straight line can have constant spatial coordinates in an inertial frame.

 

[laymanB]

@PeterDonis that is very helpful. Thanks

 

[PeroK]

Thanks, I think you are hitting on where my problem is at. Could you provide a link to that thread?

In fact, here is the scenario analyzed from the traveller's inbound frame.

Two common misconceptions are the role of acceleration and that it's the traveller's time, and not the Earth's, that is really dilated. This analysis should dispel those ideas.

We have a ship traveling towards Earth at $\frac35 c$. We will analyze the problem from this frame. So:

The Earth moves towards the ship at $\frac35 c$ throughout. The Earths gamma factor is $\frac54$.

The traveller leaves the Earth at $\frac35c$ in the Earths frame and moves towards the ship at $\frac{15}{17}c$ in the ship frame(*). It's gamma factor is $\frac{17}{8}$ in the ship frame.

(* You use the velocity addition formula here to get the traveller's velocity in the ship frame.)

Let's assume that the Earth is initially a distance of $7.5$ light years from the ship.

According to the ship:

The traveller takes $8.5$ years to reach the ship. At which point the travellers clock reads $4$ years.

The Earth takes $12.5$ years - a further $4$ years - to reach the ship, at which point the Earth clock reads $10$ years.

Now, we adopt the scenario where the traveller, instead of turning round, asks the ship to record the time of the return leg. After all, the "incoming" ship is traveling at the correct speed relative to Earth and represents exactly the idealised change of motion required for the experiment.

When they pass the traveller's clock read $4$ years. This reading was transferred to a spare clock on the ship, which recorded a further $4$ years before the Earth arrived.

This confirms that if the traveller could instantaneously change direction, the results would be the same as obtained from the Earths frame.

In one sense, these calculations are pointless. If you have confidence in the physics, then you analyse the problem from the Earth frame and that's that.

But, perhaps it's good to see the same result from a frame where the Earth is moving and time dilated throughout. And, where there is no need for acceleration.

 

[laymanB]

Read that article again, carefully. In particular, read the spacetime diagram analysis, carefully. Nothing in that analysis "resorts to" acceleration; the only time acceleration is mentioned at all is in postulating that it doesn't affect the mechanism of clocks.

You are right. This website is quite good if it is read carefully and you understand the implications of the changing IRF. If you don't, like I didn't initially, it seems like mathematical slight of hand.

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[PeroK]

If you don't, like I didn't initially, it seems like mathematical slight of hand.

It's actually "sleight of hand", but that's another matter.

 

[laymanB]

It's actually "sleight of hand", but that's another matter.

See, I learned something else new today. I really never did realize that sleight in that phrase was spelled like that.

I'm working through your example right now.

 

[laymanB]

This confirms that if the traveller could instantaneously change direction, the results would be the same as obtained from the Earths frame.

Nicely done. Thanks for taking the effort to write this up.

 

[hutchphd]

I used to teach relativity to freshman engineering students for fun, so here is just one more thought to slightly mess with your head. The traveling twin could start his "life" already at speed 0.6c passing Earth (hence no acceleration) and the final comparison upon his return could by made by taking a snapshot as he zoomed past Earth on the return (again no acceleration). The crux is that the traveling twin has to experience acceleration in order to make a return. So if he makes a linear trip it is during the interval when he decelerates from +0.6c to -0.6c where all of the asymmetry in aging must take place! So while he is turning around some weird stuff happens...

 

[Ibix]

The crux is that the traveling twin has to experience acceleration in order to make a return. So if he makes a linear trip it is during the interval when he decelerates from +0.6c to -0.6c where all of the asymmetry in aging must take place! So while he is turning around some weird stuff happens...

Add a third party, one who travels at 0.6c in the opposite direction. So we have a stay-at-home, an outbound traveller, and an inbound traveller. The outgoing traveller never stops, simply reports their clock reading to the inbound traveller when they pass each other. When the inbound traveller passes the stay-at-home they report the outbound's outbound time plus their own time since they passed the outbound. This must give the same answer as the twin paradox, but there is no acceleration whatsoever

The acceleration is necessary for two twins to meet up again. But it can't be the cause of the age difference. (I believe PeroK has already pointed this out).

 

[hutchphd]

The acceleration is both necessary and sufficient for again meeting. It is the only such asymmetry in the problem. It therefore must be the "cause" of the age difference. QED

 

[Ibix]

The acceleration is both necessary and sufficient for again meeting. It is the only such asymmetry in the problem. It therefore must be the "cause" of the age difference. QED

I just described a variant of the experiment with no acceleration and the same result. So "necessary" is false.

It's also possible to construct experiments where twins undergo the same acceleration phases in the same order, just with different duration inertial segments. They end up having different ages. So it is very difficult to conclude that the acceleration is causing the age difference.

 

[hutchphd]

What number comment describes that variant?...I missed it...sorry

 

[Ibix]

What number comment describes that variant?...I missed it...sorry

#51. See also PeroK's #45 for a longer discussion.

 

[hutchphd]

This must give the same answer as the twin paradox

Oh I found it. In fact this is exactly my point...the asymmetry comes from the change in direction. 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). Imagine his surprise when his Earth twin is older! 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. There are pieces of general relativity that deal with acceleration but I am not able to reproduce them out of memory and too lazy to work it out right now...

 

[Nugatory]

There are pieces of general relativity that deal with acceleration...

This is a very common misconception. Special relativity handles problems involving acceleration just fine as long as the spacetime is flat, as it is in this discussion. General relativity is needed only if the spacetime is not flat because there are gravitating bodies involved, and then it's needed even if no acceleration is involved.

For an example of special relativity properly handling acceleration, google for "Rindler coordinates".

More generally, acceleration just means that the worldline of the accelerated object is curved instead of straight, so you may have to do a line integral along the curve instead of just calculating $-\Delta{t}^2+\Delta{x}^2+\Delta{y}^2+\Delta{z}^2$ from the endpoints of a line segment. This additional mathematical complexity is why many introductory SR texts don't discuss the acceleration cases, inadvertently feeding the misconception that SR cannot handle them.

 

[Nugatory]

That extra time on Earth must have transpired during his deceleration!

If you are not already familiar with the Twin Paradox FAQ at http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html, give it a try, and pay particular attention to the part about the "Time gap objection".

 

[Ibix]

In fact this is exactly my point...the asymmetry comes from the change in direction

You said it came from acceleration in #50 & 52, which is a different point.

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).

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.)

It's only once the twins subtract out the effects of the lightspeed delay that they calculate that the other's clock ticks slowly. Time dilation is not directly observable.

Imagine his surprise when his Earth twin is older!

He won't be surprised at all if he's actually been watching the other's clock.

That extra time on Earth must have transpired during his deceleration!

This is true in a limited sense. As I noted above, the slow running of the other's clock is not directly observable. It's something that each twin calculates based on their direct observations plus some assumptions about how one should synchronise separated clocks. If the traveller does that process naively before and after the turn around then they find that what they were calling "now, on Earth" just before the turn around is not the same as what they now call "now, on Earth" just after the turn around. The difference in "now" will account for the extra time.

But this is not an effect of acceleration. It's not even an effect of changing direction. It's an effect of changing which clocks the traveling twin chose to regard as synchronised with theirs. In other words, it boils down to using one definition of "during" for the outbound trip, a different definition for the inbound trip, and ascribing the resulting inconsistency to something that happened "during" the turnaround.

For one possible non-naive resolution to this, see https://arxiv.org/abs/gr-qc/0104077

There are pieces of general relativity that deal with acceleration

As Nugatory noted, general relativity is not needed to analyse this. You can do an analysis (for the finite acceleration case, not the instantaneous turn around) using gravitational time dilation, but that has always seemed over-complicated to me.

 

[PeroK]

I used to teach relativity to freshman engineering students for fun, so here is just one more thought to slightly mess with your head. The traveling twin could start his "life" already at speed 0.6c passing Earth (hence no acceleration) and the final comparison upon his return could by made by taking a snapshot as he zoomed past Earth on the return (again no acceleration). The crux is that the traveling twin has to experience acceleration in order to make a return. So if he makes a linear trip it is during the interval when he decelerates from +0.6c to -0.6c where all of the asymmetry in aging must take place! So while he is turning around some weird stuff happens...

The fundamental flaw in your logic is this.

Imagine a) the scenario where the traveller has only one acceleration phase, at the turn around.

b) The scenario where the traveller has two acceleration phases: one at the start and one at turn around.

c) The scenario where the traveller has a third acceleration phase at the end.

All of these scenarios result in approx the same differential ageing.

If "weird stuff" happens when you accelerate that caused differential ageing, then more "weird stuff" would happen in case c) than in b) etc.

Moreover, if the acceleration caused the differential ageing, then it wouldn't matter how far the traveller cruised though space. The ageing would depend only on the acceleration profile, not on the overall journey.

In fact, the opposite is the case: the differential ageing depends simply on the velocity profile, but not on the acceleration profile.

 

[Mister T]

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.