B Twin paradox explained for laymen

[PeterDonis]

The point that I have been trying to make this entire time is that there is a logical symmetry between the two twins unless some other fact (acceleration, light color shift, external objects, some other physical fact) is brought into the problem.

Some other fact besides what? Basically you're saying that there is a logical symmetry between the twins if we don't know any physical facts about them. Which I suppose is true, but seems rather pointless. Obviously we need to have some physical fact that differentiates the twins. I don't think anyone disputes that. I certainly don't.

Essential parts (like inertial reference frame) of the Twins Paradox can not even be defined without talking about acceleration or some representation of it.

This, however, is wrong. I've given you an explicit counterexample.

very good scientists (including Einstein) used acceleration to explain the twin paradox

Arguments from authority are not valid and carry no weight here.

it is exactly at the point of the turn-around that one twin departs from an inertial reference frame due to acceleration that distinguishes him from the other twin

This is one possible physical fact you could have to distinguish the twins. But it's not the only one. The traveling twin can tell he is turning around even if he has no way to measure or feel his proper acceleration, by looking at the change from redshift to blueshift of the light signals that are coming from the stay-at-home twin.

In other words, it would be fine to say that acceleration is one possible way to distinguish the twins. But you are taking the position that it's the only way, or that it's a necessary way, that we can't distinguish the twins without it. That position is wrong.

 

[PAllen]

This is one possible physical fact you could have to distinguish the twins. But it's not the only one. The traveling twin can tell he is turning around even if he has no way to measure or feel his proper acceleration, by looking at the change from redshift to blueshift of the light signals that are coming from the stay-at-home twin.

Well, then why wouldn’t the stay at home twin think they turned around when same change occurs for them, at a later proper time?

 

[PeterDonis]

why wouldn’t the stay at home twin think they turned around when same change occurs for them, at a later proper time?

Because the stay-at-home twin isn't firing his rockets. I should have clarified that the traveling twin knows he is turning around because he sees the frequency shift change at the same time he is firing his rockets, and he can tell this even if he has no way of measuring or sensing his proper acceleration.

Of course in this particular case there's an even simpler physical fact about the traveling twin that distinguishes him from the stay-at-home twin--that he fires his rockets. But the frequency shift method generalizes to, for example, the curved spacetime scenario where the traveling twin slingshots around a distant planet or star--he sees the frequency shift change during the slingshot maneuver--even though there is no firing of rockets and no proper acceleration. (Whereas the stay-at-home twin is just sitting there in deep space, doing nothing, and not close to any planet or star, when he sees the frequency shift change.)

Even in these other cases, it's true that one can always point to some other physical fact that has to be correlated with the frequency shift in order to spot a turnaround. But the same is true of proper acceleration: for it to signal a turnaround, it has to be accompanied by some other change of state, such as firing the rocket engines. A proper acceleration that is not accompanied by some other change of state--such as the stay-at-home twin being actually on the actual Earth, instead of floating in free space--does not signal a turnaround.

 

[FactChecker]

This is one possible physical fact you could have to distinguish the twins. But it's not the only one. The traveling twin can tell he is turning around even if he has no way to measure or feel his proper acceleration, by looking at the change from redshift to blueshift of the light signals that are coming from the stay-at-home twin.

In other words, it would be fine to say that acceleration is one possible way to distinguish the twins. But you are taking the position that it's the only way, or that it's a necessary way, that we can't distinguish the twins without it. That position is wrong.

You are defining an IRF by how clocks (in the form of light color shifts) behave. I am not sure that I find that very satisfying. It seems like circular logic to say that time is different in other reference frames because observed clocks (in the form of light color shifts) behave differently. In any case, given that definition of IRF, I think that it would be interesting to say how the color phase shift makes one twin different at the moment when he is turning around. At other times, the twins are both following inertial paths and their situation is symmetric. Einstein proposed a difference in potential pseudo-gravity associated with acceleration. That explanation has the benefit of making the effect greater when observers are farther from each other. Is there an equivalent theory using light color shift?

 

[PeterDonis]

You are defining an IRF by how clocks (in the form of light color shifts) behave.

No, I'm not. In my redshift/blueshift example, I'm not defining an IRF at all. There is no need to. You can of course do physics using an IRF, but you do not need to. That was my point.

I am not sure that I find that very satisfying.

The fact that you don't find it very satisfying doesn't mean it's false. You like to analyze problems using an IRF. That's fine. But it doesn't mean it's the only possible way of doing it.

It seems like circular logic to say that time is different in other reference frames because observed clocks (in the form of light color shifts) behave differently.

I have said no such thing. Again, my frequency shift example does not use an IRF at all.

Is there an equivalent theory using light color shift?

Yes. The time elapsed on the traveling twin's clock before the turnaround is longer, so the traveling twin observes redshift for a longer time by his clock, and then blueshift for a longer time by his clock. This will lead him to predict a larger difference in aging.

 

[PAllen]

Because the stay-at-home twin isn't firing his rockets. I should have clarified that the traveling twin knows he is turning around because he sees the frequency shift change at the same time he is firing his rockets, and he can tell this even if he has no way of measuring or sensing his proper acceleration.

Of course in this particular case there's an even simpler physical fact about the traveling twin that distinguishes him from the stay-at-home twin--that he fires his rockets. But the frequency shift method generalizes to, for example, the curved spacetime scenario where the traveling twin slingshots around a distant planet or star--he sees the frequency shift change during the slingshot maneuver--even though there is no firing of rockets and no proper acceleration. (Whereas the stay-at-home twin is just sitting there in deep space, doing nothing, and not close to any planet or star, when he sees the frequency shift change.)

Even in these other cases, it's true that one can always point to some other physical fact that has to be correlated with the frequency shift in order to spot a turnaround. But the same is true of proper acceleration: for it to signal a turnaround, it has to be accompanied by some other change of state, such as firing the rocket engines. A proper acceleration that is not accompanied by some other change of state--such as the stay-at-home twin being actually on the actual Earth, instead of floating in free space--does not signal a turnaround.

In SR, if a path has a change of tangent vector = proper acceleration, then there must exist a path with longer proper time. Compared the path which maximizes proper time, the existence proper acceleration distinguishes any other path. Whether you choose to measure it, and how (e.g. using auxiliary observers) is irrelevant and cannot remove the existence of proper acceleration. It is still a necessary and sufficient condition in SR to know longer proper time paths exist.

In GR, things are more complex, but not that much. The existence of proper acceleration on a path is still sufficient to know there exist longer proper time paths. However, its absence no longer guarantees anything. All you can say is that global proper time maximizing path must have no proper acceleration. Or that absence of proper acceleration is a necessary but not sufficient condition for maximizing proper time.

 

[FactChecker]

I have said no such thing. Again, my frequency shift example does not use an IRF at all.

You are saying that the traveling twin is younger because his clock ticked less.

Yes. The time elapsed on the traveling twin's clock before the turnaround is longer, so the traveling twin observes redshift for a longer time by his clock, and then blueshift for a longer time by his clock. This will lead him to predict a larger difference in aging.

One could say that is "begging the question".
I guess this might be a legitimate approach if the problem is framed in a way that "inertial paths" are maximal (minimal?) elapsed time paths. I would like to think about that for a while. I might just be going "down the rabbit hole" with that.

 

[PeroK]

@FactChecker here is an interesting asymmetry, which is related to the Doppler/light signals explanation.

First, note there is an asymmetry in that we have two objects in the Earth's rest frame: the Earth and the distant planet, but only one object in the traveller's rest frame. Let's remove this asymmetry by assuming:

1) The traveller is moving at speed $0.8c$ when he/she passes the Earth.

2) There are two travellers. separated by 4 light years in their rest frame.

Let's call the travellers $B_1$ and $B_2$ and the Earth $A_1$ and the distant planet $A_2$.

The first event is when $B_1$ passes $A_1$ (the Earth). There is complete symmetry. In the A-frame we have $B_1$ and $B_2$, separated by a length contracted distance of 2.4 light years. And, in the B-frame we have $A_1$ and $A_2$ the same.

Now, when $B_1$ and $A_2$ pass: $B_1's$ clock reads $3$ years and $A_2's$ clock reads $5$ years. Likewise, we must have symmetry when $A_1$ passes $B_2$, $A_1's$ clock reads 3 years and $B_2's$ clock reads $5$ years.

If there is no turnaround, we have complete symmetry and travellers all continue on their way. But, if the $B$ travellers turn round when $B_1$ reaches $A_2$, then they do so at $t_B = 3$ years and $B_2$ never reaches $A_1$. Because that event was due to take place at $t_B = 5$ years. Or, if the $A$ travellers turn round when $A_1$ reaches $B_2$, they do so at $t_A = 3$ years and $B_1$ never reaches $A_2$. Because that event was due to happen at $t_a = 5$ years.

Therefore, if there is a turnaround an asymmetry appears in which events took place! If $B_1$ reached $A_2$, then it must have been the $B's$ who turned round. And, if $A_1$ reached $B_2$, then it must have been the $A's$ who turned round.

Isn't that interesting? Where did that asymmetry come from?

PS note that whatever pair turned round would have agreed in advance to turn round after 3 years. There was no signalling.

PPS If both pairs decided in advance to turn round after 3 years, then only the first meeting of $A_1$ and $B_1$ would take place. That's why the traveller knows he/she must have turned round.

PPPS In your scenario IF the Earth and distant planet had turned round, then the traveller would never have reached the distant planet.

 

[PeterDonis]

You are saying that the traveling twin is younger because his clock ticked less.

Of course; those two statements are just different ways of saying the same thing. But see below.

One could say that is "begging the question".

It is no such thing. See below.

I guess this might be a legitimate approach if the problem is framed in a way that "inertial paths" are maximum elapsed time paths.

You are missing the point. In the version of the scenario I am giving, the only information anyone has is clock readings and frequency shifts. There is nothing else. Nobody needs to know what is "inertial" and what is not. Nobody needs to know anything about acceleration.

Of course if you want to correlate the frequency shifts with all that other stuff, you need to have all that other information available. But you don't need to do that in order to know which twin will have aged more in the version of the scenario I am giving.

Also, if we compare scenarios in which the traveling twin takes different times on the outbound leg before turning around (firing his rocket and observing the frequency shift in light signals from the stay-at-home twin), that difference in elapsed times on the outbound leg is part of the specification of the scenario. Obviously you have to specify some difference between two scenarios in order to make them different. But note that that specification only talks about the traveling twin's elapsed time, and only on the outbound leg of the trip. It says nothing about the stay-at-home twin's elapsed time, or the traveling twin's elapsed time on the inbound leg. Those things are derived in the solution of the scenario, and they can be derived purely from the information I specified. So it is not a case of specifying the answer as part of the scenario.

 

[PeterDonis]

The traveling twin is younger because less time (as measured by frequency) has passed. I accept that as true. But it is a platitude.

No, the platitude is that the traveling twin is younger because less time by his clock has passed.

There is no such thing as time (as measured by frequency). You can use the frequency shifts to calculate which twin will have aged less, but that doesn't mean the frequency shifts are measuring who ages less.

(If you were to use that logic, I could equally well say that it's a platitude that the traveling twin is younger because less time, as measured by proper acceleration, has passed.)

 

[FactChecker]

I feel that these concepts have been discussed in the past by far greater minds than mine, including Einstein's. So although I feel that I am in good company, I can not add anything to this discussion.

 

[FactChecker]

Twin B sees light signals coming from twin A to be redshifted for almost all his (Twin B's) trip, then blueshifted for a very short time at the end of his trip.

This surprises me. I thought there would be a blue shift during B's entire return trip, when the relative distance is decreasing.

 

[Ibix]

This surprises me. I thought there would be a blue shift during B's entire return trip, when the relative distance is decreasing.

The traveller sees blue shift for all the return leg. The stay-at-home doesn't see blue shift until light from the turnaround reaches him, which doesn't happen until long after the turnaround. Sketch a Minkowski diagram and you'll see it.

 

[PeterDonis]

I thought there would be a blue shift during B's entire return trip, when the relative distance is decreasing.

See the spacetime diagram corresponding to the "Doppler Shift Explanation" in the Usenet Physics FAQ entry on the twin paradox:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler

The whole FAQ is worth reading as well.

 

[FactChecker]

The traveller sees blue shift for all the return leg. The stay-at-home doesn't see blue shift until light from the turnaround reaches him, which doesn't happen until long after the turnaround. Sketch a Minkowski diagram and you'll see it.

EDIT: I may have switched A and B here. I have rewritten it without As & Bs to be clear.
I see. But other stationary observers in the stationary IRF who are close to the turn-around point would see the blue shift much earlier. Using their stationary observations, there is no need to complicate the issue with the travel time of light all the way back to the starting point. I think that stationary observers would agree that the traveler's light was blue-shifted as soon as he turned around and that it just took a long time to get back to the starting point.

 

[Ibix]

I see. But other stationary observers in A's IRF who are close to the turn-around point would see the blue shift much earlier. Using their stationary observations, there is no need to complicate it with the travel time of light all the way to A. I think that stationary observers would agree that B's light was blue-shifted for a fairly long time and that it just took a long time to get to A.

Sure. But so what? It's not these other observers' clock readings we're trying to comprehend. It's A's clock readings, and it's the differences between the pulse frequencies received by A and B that is interesting.

 

[PeterDonis]

other stationary observers in A's IRF who are close to the turn-around point would see the blue shift much earlier.

Much earlier, but not at the same time the traveling twin sees them. And those stationary observers will also see much less differential aging between themselves and the traveling twin, from the point where that twin passes them on the way out, and the point where the twin passes them on the way back.

The stationary observer who happens to be exactly at the turnaround point will see no change from redshift to blueshift at all in the light signals from the traveling twin. He will see a change from blueshift to redshift. And of course he will see it right at the instant of the turnaround, which is the only time he is co-located with the traveling twin; so his "differential aging" with that twin is zero.

The other stationary observers will also see a change from blueshift to redshift of the traveling twin's light signals--they will see it twice, once when the twin passes them on the outbound leg, then again when the twin passes them on the inbound leg. In other words, a pair of "blueshift to redshift" changes marks the starting and ending points of the "differential aging" period--the period between two successive meetings where elapsed times can be directly compared. And there must be a "redshift to blueshift" change in between, whose timing, as seen by the stationary observer, is related to the amount of differential aging. The stationary observer right at the turnaround point is simply the degenerate case of this where the differential aging period is reduced to zero.

 

[FactChecker]

Sure. But so what? It's not these other observers' clock readings we're trying to comprehend. It's A's clock readings, and it's the differences between the pulse frequencies received by A and B that is interesting.

It makes all the difference in the world. If observers in the same stationary reference frame can not agree, then something is wrong. They can observe and report back. Their observations should agree after taking distance and the travel time of light into account. IMHO, it is a mistake to unnecessarily complicate these things with the travel time of light to different observers.

EDIT: I will have to think about this in the context of @PeterDonis 's response above. It may require some real re-thinking on my part.

 

[FactChecker]

Much earlier, but not at the same time the traveling twin sees them. And those stationary observers will also see much less differential aging between themselves and the traveling twin, from the point where that twin passes them on the way out, and the point where the twin passes them on the way back.

The stationary observer who happens to be exactly at the turnaround point will see no change from redshift to blueshift at all in the light signals from the traveling twin. He will see a change from blueshift to redshift. And of course he will see it right at the instant of the turnaround, which is the only time he is co-located with the traveling twin; so his "differential aging" with that twin is zero.

The other stationary observers will also see a change from blueshift to redshift of the traveling twin's light signals--they will see it twice, once when the twin passes them on the outbound leg, then again when the twin passes them on the inbound leg. In other words, a pair of "blueshift to redshift" changes marks the starting and ending points of the "differential aging" period--the period between two successive meetings where elapsed times can be directly compared. And there must be a "redshift to blueshift" change in between, whose timing, as seen by the stationary observer, is related to the amount of differential aging. The stationary observer right at the turnaround point is simply the degenerate case of this where the differential aging period is reduced to zero.

This may make me rethink the whole thing. Thanks. I will now go away and ponder this.

 

[Ibix]

It makes all the difference in the world. If observers in the same stationary reference frame can not agree, then something is wrong.

Of course they can correct for the travel time of the light. But the point is that the raw observations of the twins are different, independent of whether they noticed the acceleration or not and, importantly, without constructing any reference frame.

 

[FactChecker]

Of course they can correct for the travel time of the light. But the point is that the raw observations of the twins are different, independent of whether they noticed the acceleration or not and, importantly, without constructing any reference frame.

Aha. I see. Thanks.

 

[pervect]

The approach I recommend is to take the twin paradox as a given feature of special relativity. If you have two clocks, and they take different paths through space-time, the one that reads the longest time will be the clock that undergoes inertial motion.

This can, in fact, be used to define inertial motion in special relativity. This is similar to the way that Newtonian physics can be regarded as equivalent to the principle of "least action". However, explaining this relationship in more detail unfortunately starts to go beyond the knowledge that a layman can be expected to have, so I will not say more about that statement unless asked.

It is worthwhile to point out that it is just as non-paradoxical for the twins clocks to read differently when they re-unite as it is non-paradoxical for two cars, traveling different routes between two cities, to have different odometer readings when they reunite. In the case of the cars, it can be regarded as a definition of straight line motion to say that the care that travels in a straight line has the shortest distance on their odomoeter.

The analogy is not perfect, because the cars odometer reading is the shortest for the linear motion, while the proper time of the two twins (the clock being analogous in space-time to the odometer in space) is the longest rather than the shortest.

This simple explanation gets more complicated if one has to deal with gravity. Rather than get into these complexities. It's simplest to avoid them. If one insists on dealing with them, the minimal level of complexity is to argue that the amount of gravitational time dilation for an object on the Earth's surface compared to an object "at infinity" is known to be very small, less than 1 part in a billion, rather than getting into the details of how general relativity works.

 

[FactChecker]

The approach I recommend is to take the twin paradox as a given feature of special relativity. If you have two clocks, and they take different paths through space-time, the one that reads the longest time will be the clock that undergoes inertial motion.

I see. I think you mean the greatest time among all paths between the two points, not just two paths. Comparing only two paths, it may be that neither is inertial.

This can, in fact, be used to define inertial motion in special relativity. This is similar to the way that Newtonian physics can be regarded as equivalent to the principle of "least action". However, explaining this relationship in more detail unfortunately starts to go beyond the knowledge that a layman can be expected to have, so I will not say more about that statement unless asked.

It is worthwhile to point out that it is just as non-paradoxical for the twins clocks to read differently when they re-unite as it is non-paradoxical for two cars, traveling different routes between two cities, to have different odometer readings when they reunite. In the case of the cars, it can be regarded as a definition of straight line motion to say that the care that travels in a straight line has the shortest distance on their odomoeter.

Suppose the traveling twin flies inertially to and from the far point and makes a fast, non-inertial turnaround. Also, suppose he is mistaken and attempts to use the SR equations to calculate the age of the stationary twin. He knows that the stationary twin aged slower during the entire time that the traveling twin was following an inertial flight path to and from the turn-around point. Given that, is there a good, intuitive explanation (other than pseudo-gravitational potential) of why that short turn-around either caused the stationary twin to age a lot or invalidated the calculations that he did during the inertial parts of his flight?

 

[PeroK]

I see. I think you mean the greatest time among all paths between the two points, not just two paths. Comparing only two paths, it may be that neither is inertial.Suppose the traveling twin flies inertially to and from the far point and makes a fast, non-inertial turnaround. Also, suppose he is mistaken and attempts to use the SR equations to calculate the age of the stationary twin. He knows that the stationary twin aged slower during the entire time that the traveling twin was following an inertial flight path to and from the turn-around point. Given that, is there a good, intuitive explanation (other than pseudo-gravitational potential) of why that short turn-around either caused the stationary twin to age a lot or invalidated the calculations that he did during the inertial parts of his flight?

This all assumes that symmetric time dilation represents a real physical process that must have a cause. Nothing anyone else does can affect how the stationary twin ages. The stationary twin takes an inertial path through spacetime. Full stop. Any other (non-inertial) path through spacetime is shorter. But, someone taking a shorter path does not in any way "cause" the non-inertial traveller to age more. There are no physical causes and effects here. It's geometry.

 

[Dale]

If you're not going to do that, you might as well drop the "rest frame" idea altogether and just do the calculation in the most convenient inertial frame for the problem

That actually is the whole point of relativity. Ideally you should not be doing much in the way transformations. You should simply pick the most convenient coordinate system and use it. That seems to get lost in these discussions.

 

[PeroK]

@FactChecker another way to analyse this that like is as follows:

1) Pick any IRF in which the Earth is moving at some arbitrary speed $v$ and analyse the problem in that frame. Or:

2) A special case is to analyse the entire scenario in the traveller's oubound IRF. In this frame:

The Earth moves with constant speed $v = 0.8c$, say.

The traveller is stationary for the outbound leg and then moves at speed $V$, which can be calculated using velocity addition, during the second leg, and eventually catches up with the Earth. If you calculate things in this frame (purely for the inertial legs) then (as you must) you get the same answer. The Earth twin has aged 10 years and the traveller 6 years when they reunite.

In this frame, there is only inertial motion for both twins; the acceleration phase(s) can be neglected (as usual, assuming they are short and sharp), yet the twins age differently between their meetings.

Now, if you try to add a physical cause and effect for the acceleration phase, that must throw the calculations. If acceleration does anything it must do it in this IRF as well? Or, do something at least. But, it can't do anything because you've already got the right answer from the inertial phases alone.

 

[PAllen]

For those who insist on a pseudogravity approach, it is only applicable in coordinates for which the noninertial path has fixed position the whole time. It doesn’t apply in any inertial frame.

 

[pervect]

I see. I think you mean the greatest time among all paths between the two points, not just two paths. Comparing only two paths, it may be that neither is inertial.Suppose the traveling twin flies inertially to and from the far point and makes a fast, non-inertial turnaround. Also, suppose he is mistaken and attempts to use the SR equations to calculate the age of the stationary twin.

I'm not following. If he's mistaken, he gets the wrong answer. I'm not too inclined to go into this further, unless I can understand your motivation for sayin he's making mistakes.

Note that a person can tell if they are accelerating or not by physical experiments. It's not a matter of opinion, or convention. There are physical effects when one accelerates. To make a mistake about accelerating or not accelerating is to get the physics wrong. Perhaps that's your motivation for assuming he's making a "mistake"? If so, it is IMO misguided.

The idea that the acceleration itself causes the differential aging may be what's confusing you. Let's go back to the road trip analogy.

One driver takes a straight route, all of the time, and gets the shortest distance on his odometer.

The other driver travels in a straight line, "most of the time", but his course is off. He realizes this, so he makes a turn, correcting his course. The turn is sharp, it doesn't take him much time to make the turn. Then he travels in a straight line for the rest of the trip.

Travelling in a straight line "most of the time" simply is NOT good enough to have the shortest distance on the odometer. One has to travel in a straight line ALL the time, not just most of it.

One can say that if the angle of the course correction is small, while he'll drive further, it won't be too much further. The analogous space-time analogy to the angle of the turn is the total velocity change. This can be seen by drawing a space-time diagram. If the total velocity change is small, then there's not much differential aging. If there is a large change in the velocity, the space-time equivalent of a large angle, then there will be a large impact on the differential aging. The amount of time it takes to change the velocity is not critical to the aging calculation.

 

[PeterDonis]

I think you mean the greatest time among all paths between the two points, not just two paths. Comparing only two paths, it may be that neither is inertial.

This is a valid point. The correct statement is that, in flat spacetime, of all possible timelike paths between two fixed events that are timelike separated, the one with the longest elapsed time will be the inertial path.

 

[PeterDonis]

I'm not following.

See my post #179 just now. The correct general statement of the principle you are using is given there.

If you want to specialize that statement to just two paths, you have to also specify that one of the paths is inertial. If you have two paths and neither one is inertial, there is no general rule for which path will have the longer elapsed proper time; you have to look at the specific paths. I don't think you intended to discuss that case, but the way you worded your statement didn't rule it out.

If he's mistaken, he gets the wrong answer.

He means, if neither twin is inertial the whole time, then one twin will make a wrong prediction if he assumes that the other twin is inertial the whole time. Which, again, I don't think is what you intended, but your wording didn't rule it out.