Is the Twin Paradox Truly Unresolvable from Speedo's Perspective?

[Michio Cuckoo]

According to Lorentz, time is always dilated.So think of two twins, Speedo and Goslo.

- Goslo stays on Earth and drinks tea.
- Speedo gets into a rocket, zooms off into outer space and zooms back.Imagine YOU are Speedo. You zoom off, and when you return to Earth, you find that Goslo is 20 years older than you.

Both of you were the same age when you left, but now he's way older. This means he must have aged faster. But how could Goslo appear to age faster, when time is always dilated?!.
Discuss this but only consider Speedo's point of view. Because if you are Speedo, what would you see?

 

[Jonathan Scott]

Let's look at an equivalent scenario in ordinary space, involving two ants. Both ants start at A in this diagram:

 A
 |\
 | \
 |  *1
 *2  \
 |    B
 |   /
 |  /
 | /
 |/
 C

Initially ant *1 is crawling towards B and ant *2 at the same speed towards C. If either ant looks at the other, it sees the other ant progressing more slowly in its own forward direction than itself. Then ant *1 reaches B and turns towards C. After the turn, both ants STILL see the other ant moving more slowly than themselves in their own direction. Ant *2 reaches C first, but there is no contradiction involved. Note that as ant *1 turned the corner at B, it saw ant *2 going from being behind it to being ahead of it.

Apart from a minus sign in the relationship between time and space, the maths of the twin paradox is very similar to this.

 

[Michio Cuckoo]

Let's look at an equivalent scenario in ordinary space, involving two ants. Both ants start at A in this diagram:

 A
 |\
 | \
 |  *1
 *2  \
 |    B
 |   /
 |  /
 | /
 |/
 C

Initially ant *1 is crawling towards B and ant *2 at the same speed towards C. If either ant looks at the other, it sees the other ant progressing more slowly in its own forward direction than itself. Then ant *1 reaches B and turns towards C. After the turn, both ants STILL see the other ant moving more slowly than themselves in their own direction. Ant *2 reaches C first, but there is no contradiction involved. Note that as ant *1 turned the corner at B, it saw ant *2 going from being behind it to being ahead of it.

Apart from a minus sign in the relationship between time and space, the maths of the twin paradox is very similar to this.

So if I'm Speedo, as I turn my rocket back to Earth, Goslo will "suddenly" become older? Is that what you're saying?

I want people to think about this from Speedo's point of view.

 

[ghwellsjr]

Speedo will see Goslo's clock ticking slower than his own during his outbound portion of the trip. At the same time, Goslo will see Speedo's clock ticking slower than his own. They each see the other one's clock ticking slower than their own by exactly the same amount.

Then when Speedo turns around, he immediately sees Goslo's clock ticking faster than his own. So exactlly half the time Speedo sees Goslo's clock ticking slower than his own and exactly half the time he sees Goslo's clock ticking faster than his own.

However, Goslo does not see Speedo's clock go from ticking slow to ticking fast at the half-way point of the trip, because he has to wait for the image of Speedo turning around to propagate from that distant location to himself, so he continues to see Speedo's clock ticking slower than his own for much more than half the trip. Eventually, near the end of the trip, he sees Speedo turn around and now he sees Speedo's clock ticking faster than his own until Speedo gets back to him, In fact, they both see each other's clock ticking faster than their own by exactly the same amount during this last portion of the trip.

Now when Speedo gets back, since he watched Goslo's clock tick slow and fast for exactly half the trip, and since Goslo watched Speedo's clock tick slow for most of the trip and fast for a very short time, Goslo will see that Speedo is much younger than himself when they reunite.

It's really very simple, OK?

 

[Jonathan Scott]

Speedo will see Goslo's clock ticking slower than his own during his outbound portion of the trip. At the same time, Goslo will see Speedo's clock ticking slower than his own. They each see the other one's clock ticking slower than their own by exactly the same amount.

Then when Speedo turns around, he immediately sees Goslo's clock ticking faster than his own. So exactlly half the time Speedo sees Goslo's clock ticking slower than his own and exactly half the time he sees Goslo's clock ticking faster than his own.

However, Goslo does not see Speedo's clock go from ticking slow to ticking fast at the half-way point of the trip, because he has to wait for the image of Speedo turning around to propagate from that distant location to himself, so he continues to see Speedo's clock ticking slower than his own for much more than half the trip. Eventually, near the end of the trip, he sees Speedo turn around and now he sees Speedo's clock ticking faster than his own until Speedo gets back to him, In fact, they both see each other's clock ticking faster than their own by exactly the same amount during this last portion of the trip.

Now when Speedo gets back, since he watched Goslo's clock tick slow and fast for exactly half the trip, and since Goslo watched Speedo's clock tick slow for most of the trip and fast for a very short time, Goslo will see that Speedo is much younger than himself when they reunite.

It's really very simple, OK?

Although this description appears to be correct, there is a big difference between what we actually see (by means of light or similar) and how we measure time and space, and that could cause confusion.

What we see varies continuously, as described above. However, we plot the location of events in our own time and space by assuming that what we see was delayed by the time that light took to reach us, or by doing equivalent calculations based on earlier events.

When the moving twin turns around, the main effect this has is that "now" at a distant location as calculated by light travel time in his own new frame of reference jumps suddenly, as described by the Lorentz transformation. This means that the point in the other twin's life (or "world line") that corresponds to "now" for the turning twin is suddenly shifted, in the same way that the point on ant *2's path that is level with ant *1's progress suddenly switches so that ant *2 is now ahead.

Yes, it works as if Speedo effectively sees Goslo as suddenly being older after he turns, because the point in Goslo's life which corresponds to "now" from Speedo's point of view has suddenly shifted. (Be careful of which way round "older" and "younger" go for the space-time version, as the switched signs mean that the straight line in the space-time case corresponds to the most proper time, not the least).

 

[Michio Cuckoo]

Then when Speedo turns around, he immediately sees Goslo's clock ticking faster than his own.

But how is that possible? Isn't time always dilated?

 

[Michio Cuckoo]

Although this description appears to be correct, there is a big difference between what we actually see (by means of light or similar) and how we measure time and space, and that could cause confusion.

Let's assume Speedo and Goslo can perceive each other using insta-beams, which travel at infinite speed.

So Speedo and Goslo can perceive instant time-dilation of the other twin.

If Speedo is using insta-beams, he will always witness time dilation, NEVER time compression, so he will always see Goslo as younger.

 

[Jonathan Scott]

But how is that possible? Isn't time always dilated?

As I just said, what you see isn't the same as what you calculate to be happening. When you are moving towards a signal or away from it, you see the signal sped up (known as "blue shift" ) or slowed down (known as "red shift").

 

[Jonathan Scott]

Let's assume Speedo and Goslo can perceive each other using insta-beams, which travel at infinite speed.

So Speedo and Goslo can perceive instant time-dilation of the other twin.

If Speedo is using insta-beams, he will always witness time dilation, NEVER time compression, so he will always see Goslo as younger.

By "infinite speed" you presumably mean in such a way that the beam takes exactly zero time to pass between two points in space. However, this property depends on the velocity of the frame of reference, in the same way that being level with the other ant depends on the direction of travel. A beam which is takes exactly zero time to travel between two points in one frame of reference (which means their separation is "purely spacelike") can take a positive time to traverse the same distance in another frame of reference and a negative time in another (so it is received before it is sent).

 

[Dale]

According to Lorentz, time is always dilated.

This is not true. According to Lorentz, moving clocks are always time dilated in an inertial frame.

Do you understand the importance of that distinction?

 

[Dale]

I want people to think about this from Speedo's point of view.

Speedo is non-inertial, so there is no standard definition of what his point of view is. You have to define it explicitly. In particular, you have to define what convention speedo adopts for determining simultaneity in his frame and for determining distance in his frame.

 

[ghwellsjr]

Speedo will see Goslo's clock ticking slower than his own during his outbound portion of the trip. At the same time, Goslo will see Speedo's clock ticking slower than his own. They each see the other one's clock ticking slower than their own by exactly the same amount.

Then when Speedo turns around, he immediately sees Goslo's clock ticking faster than his own. So exactlly half the time Speedo sees Goslo's clock ticking slower than his own and exactly half the time he sees Goslo's clock ticking faster than his own.

However, Goslo does not see Speedo's clock go from ticking slow to ticking fast at the half-way point of the trip, because he has to wait for the image of Speedo turning around to propagate from that distant location to himself, so he continues to see Speedo's clock ticking slower than his own for much more than half the trip. Eventually, near the end of the trip, he sees Speedo turn around and now he sees Speedo's clock ticking faster than his own until Speedo gets back to him, In fact, they both see each other's clock ticking faster than their own by exactly the same amount during this last portion of the trip.

Now when Speedo gets back, since he watched Goslo's clock tick slow and fast for exactly half the trip, and since Goslo watched Speedo's clock tick slow for most of the trip and fast for a very short time, Goslo will see that Speedo is much younger than himself when they reunite.

It's really very simple, OK?

Although this description appears to be correct, there is a big difference between what we actually see (by means of light or similar) and how we measure time and space, and that could cause confusion.

Wow, this is the first time I ever heard anyone apologize for the Relativistic Doppler explanation of the Twin Paradox. It exactly and precisely describes the Point of View of each twin and is fundamentally the raw measurement that each one makes of the other ones clock. Any other explanation requires arbitrary assumptions about how we measure time and space and different assumptions lead to different "measurements". So please don't denigrate my perfectly valid explanation of what each Twin sees. Isn't that fundamentally what "Point of View" means?

What we see varies continuously, as described above. However, we plot the location of events in our own time and space by assuming that what we see was delayed by the time that light took to reach us, or by doing equivalent calculations based on earlier events.

Yes, all based on assumptions, we get out what we add into the situation.

When the moving twin turns around, the main effect this has is that "now" at a distant location as calculated by light travel time in his own new frame of reference jumps suddenly, as described by the Lorentz transformation. This means that the point in the other twin's life (or "world line") that corresponds to "now" for the turning twin is suddenly shifted, in the same way that the point on ant *2's path that is level with ant *1's progress suddenly switches so that ant *2 is now ahead.

Your ant explanation claims that just because I turn my head, I can cause another person's age to change abruptly, even to the point of him getting younger.

Yes, it works as if Speedo effectively sees Goslo as suddenly being older after he turns, because the point in Goslo's life which corresponds to "now" from Speedo's point of view has suddenly shifted. (Be careful of which way round "older" and "younger" go for the space-time version, as the switched signs mean that the straight line in the space-time case corresponds to the most proper time, not the least).

And what if Speedo turns around again, does Goslo suddenly become younger?

Now I'm not saying that your explanation is invalid because it's not, but it is based on assumptions that are totally arbitrary and totally unnecessary in an explanation based on Einstein's assumptions (postulates) and his definition of a Frame of Reference. There is no need to have Speedo change frames just because he changes his speed or direction of travel.

So what causes confusion is claiming that one explanation is not as valid as another. They can all work as long as we understand what the assumptions and definitions are that we use in each of them. But every explanation will agree with the Relativistic Doppler explanation.

 

[ghwellsjr]

Then when Speedo turns around, he immediately sees Goslo's clock ticking faster than his own.

But how is that possible? Isn't time always dilated?

Haven't you ever experienced Doppler shift when, for example, an emergency vehicle passes you and you suddenly hear the pitch of the siren drop? A similar thing would happen if you could take a high speed video of a fast moving object passing close by you. You would see the colors change and if there was a visible clock, you would see it change its tick rate. You would never see it jump in time though.

Time dilation is an explanation that is provided in Special Relativity based on a clock's motion in an inertial (non-accelerating and non-changing Frame of Reference). During the first half of the trip, if you analyze the situation from a frame in which Goslo is at rest, his clock ticks normally while Speedo's is time dilated and ticks slower. If you analyze the same situation from a frame in which Speedo is at rest, then his clock ticks normally while Goslo's time is dilated and his clock ticks slower. If you continue analyzing in the same frame, you will get a consistent result.

 

[jartsa]

Speedo will see Goslo's clock ticking slower than his own during his outbound portion of the trip. At the same time, Goslo will see Speedo's clock ticking slower than his own. They each see the other one's clock ticking slower than their own by exactly the same amount.

Then when Speedo turns around, he immediately sees Goslo's clock ticking faster than his own. So exactlly half the time Speedo sees Goslo's clock ticking slower than his own and exactly half the time he sees Goslo's clock ticking faster than his own.

However, Goslo does not see Speedo's clock go from ticking slow to ticking fast at the half-way point of the trip, because he has to wait for the image of Speedo turning around to propagate from that distant location to himself, so he continues to see Speedo's clock ticking slower than his own for much more than half the trip. Eventually, near the end of the trip, he sees Speedo turn around and now he sees Speedo's clock ticking faster than his own until Speedo gets back to him, In fact, they both see each other's clock ticking faster than their own by exactly the same amount during this last portion of the trip.

Now when Speedo gets back, since he watched Goslo's clock tick slow and fast for exactly half the trip, and since Goslo watched Speedo's clock tick slow for most of the trip and fast for a very short time, Goslo will see that Speedo is much younger than himself when they reunite.

It's really very simple, OK?

If we use non-relativistic Doppler shift of light here, then the result is incorrect: no time dilation.

If we use relativistic Doppler shift here, then the result is correct: time dilation.

The difference between non-relativistic Doppler shift and relativistic Doppler shift is that in relativistic Doppler shift time dilation is taken account.

EDIT: Or should say: non-relativistic frequency shift of ticking, and relativistic frequency shift of ticking.

EDIT2: I mean: If we calculate clock ticks assuming the clock is time dilated, then we get less ticks, if calculate clock ticks assuming the clock is not time dilated, then we get more ticks, and therefore a Doppler shift based explanation is not really an explanation.

 

[Jonathan Scott]

Your ant explanation claims that just because I turn my head, I can cause another person's age to change abruptly, even to the point of him getting younger.

Note that "turning" in the ant case is equivalent to a velocity change in the space-time case. This has the effect of changing the point in the other person's time line which is considered equal to "now" locally. You can't see someone "now" anyway, so the concept of what their age is "now" is a projection anyway.

And what if Speedo turns around again, does Goslo suddenly become younger?

Yes, if by "turns around" you mean changes velocity to move in the original direction. Nothing happens to Goslo of course, but Speedo's "now" mapped on to Goslo's time line has changed.

 

[Nugatory]

So if I'm Speedo, as I turn my rocket back to Earth, Goslo will "suddenly" become older? Is that what you're saying?

I want people to think about this from Speedo's point of view.

You will find a really excellent explanation at http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/TwinParadox/twin_paradox.html.

As others have already pointed out, you have to be careful with that phrase "Speedo's point of view". When Speedo says that he "saw" Goslo and Goslo's clock, he is actually saying that light rays from Goslo and Goslo's clock hit his retina; these light rays tell Speedo about Goslo's state of affairs when the rays left Goslo, and by the time they get to Speedo, something completely different could be going on with Goslo.

Ghwellsjr has given you a spot-on accurate explanation of what Speedo "sees"; this is basically the "Doppler Shift Analysis" of the FAQ I linked to above.

We can also analyze the situation by having both twins try to figure out what was going on for each twin at various points during the experiment. The best way to do this is for each twin to keep precise records ("When my clock read T, I saw my twin moving at speed S towards/away from me, and the distance between us was D"), and when they get back together they can compare notes and come up with a consistent view of events in which Speedo ended up experiencing less time than Goslo. This leads to the "Spacetime Diagram Analysis" in the FAQ I linked to above, and is pretty much what Jonathan Scott was explaining in his first post. (I think it's also what you were trying to do with your "insta-beams", but that's a topic for a different post).

[And, just to stave off any possible misunderstanding: The two analyses are not conflicting viewpoints, competing theories, or a suggestion that there is any disagreement about what's going on in the twin paradox. They are both using the exact same physics and understanding of how the universe works - one just spends more time analyzing the behavior of light rays passing between the two twins than does the other]

 

[Michio Cuckoo]

This is not true. According to Lorentz, moving clocks are always time dilated in an inertial frame.

Do you understand the importance of that distinction?

So if I undergo acceleration, I can actually witness Time Compression?!

Enlighten me, master.

 

[Q-reeus]

So if I undergo acceleration, I can actually witness Time Compression?!

Enlighten me, master.

If by time compression you mean 'everyone else's clocks go faster' then sure - hop onto the periphery of a rotating carousel. Your clock unambiguously ticks slower than for 'stationary' observers.

 

[Janus]

So if I undergo acceleration, I can actually witness Time Compression?!

Enlighten me, master.

Yes. For example, If you were in the tail of an accelerating rocket, you would see a clock in the nose run fast compared to your own. The longer your rocket, the greater the difference in time rate. Conversely, if you were in the nose, you would see the clock in the tail as running slower.
This also applies to clocks outside of the rocket and not accelerating with you. Clocks in the direction you are accelerating will run faster, and ones in the opposite direction will run slower. This is in addition to any time dilation caused by relative velocities. The difference here is that these "outside" clocks will not see any additional effect on you due to your acceleration and will only see the time dilation due to your relative velocity.

 

[SinghRP]

I am late. Looks like the discussion has gone well beoynd the scope.
Please see the Pseudogravity thread and my statements.
George Gamow in his popularization series "Gravity" (Dover, 2002; p. 126) addresses this twin paradox. (Here he missed a point though. He ignored the effect of the Earth's own gravity.) Good luck.

 

[Nugatory]

Let's assume Speedo and Goslo can perceive each other using insta-beams, which travel at infinite speed.

So Speedo and Goslo can perceive instant time-dilation of the other twin.

If Speedo is using insta-beams, he will always witness time dilation, NEVER time compression, so he will always see Goslo as younger.

Although we use this "Let's assume..." style of though experiment all the time, there is a pitfall: Whatever we assume has to be at least theoretically possible, or the results of the argument will be very misleading. For example, if I start with "Let's assume that 2+2=5, ..." I can come up with all sorts of strange weird paradoxes that prove nothing except that my starting assumption is inconsistent with reality.

The problem with your hypothetical insta-beam is that if it's going to do what (I think) you want it do, all observers have to see the transmission and the reception happen at the same time. That's no more possible than 2+2=5, so logical reasoning about what people using instabeams would observe is bound to lead to impossible paradoxes.

I suspect that what you're trying to do is get rid of the distracting arguments about how hard it is for the two twins to compare the passage of time when they're in different places and forced to communicate with light traveling at a finite speed. If that's what you want, you'll be better served by the space-time diagram approach I mention in my previous post.

 

[Michio Cuckoo]

I suspect that what you're trying to do is get rid of the distracting arguments about how hard it is for the two twins to compare the passage of time when they're in different places and forced to communicate with light traveling at a finite speed. If that's what you want, you'll be better served by the space-time diagram approach I mention in my previous post.

Thanks. That's what I was trying to do.

A lot of people have given me all sorts of crazy explanations, the Doppler Shift, and that the Doppler Shift is equivalent to Time Compression; that Time Compression is actually possible in an accelerating frame, despite what Lorentz said; and that Goslo suddenly becomes older when you turn around.

I need help.

 

[Nugatory]

Thanks. That's what I was trying to do.

A lot of people have given me all sorts of crazy explanations, the Doppler Shift, and that the Doppler Shift is equivalent to Time Compression; that Time Compression is actually possible in an accelerating frame, despite what Lorentz said; and that Goslo suddenly becomes older when you turn around.

I need help.

Read that FAQ that I pointed you at... That will get you through this...

The Doppler Shift is a spot-on accurate explanation of what the twins "see", as it describes how the light rays passing between the two twins behave.

Goslo doesn't "suddenly become older", he sees the days and weeks and months and years of his life ticking steadily by at a constant rate just as we'd expect. Speedo has the same experience with his life. What's happening at the turnaround is that Speedo is changing reference frames, hence changing his notion of how readings on Goslo's clock relate to readings of his own clock.

Don't sweat the gravitation and acceleration stuff until you've got the basic space-time diagram and Doppler effect pictures of the paradox down cold.

 

[Dale]

So if I undergo acceleration, I can actually witness Time Compression?!

If you use a non inertial coordinate system you can indeed "witness time compression".

Good examples have already been mentioned, rotating reference frames and Rindler coordinates.

 

[Dale]

A lot of people have given me all sorts of crazy explanations, ... that Time Compression is actually possible in an accelerating frame, despite what Lorentz said;

Lorentz never said what you claim he said.

 

[ghwellsjr]

Speedo will see Goslo's clock ticking slower than his own during his outbound portion of the trip. At the same time, Goslo will see Speedo's clock ticking slower than his own. They each see the other one's clock ticking slower than their own by exactly the same amount.

Then when Speedo turns around, he immediately sees Goslo's clock ticking faster than his own. So exactlly half the time Speedo sees Goslo's clock ticking slower than his own and exactly half the time he sees Goslo's clock ticking faster than his own.

However, Goslo does not see Speedo's clock go from ticking slow to ticking fast at the half-way point of the trip, because he has to wait for the image of Speedo turning around to propagate from that distant location to himself, so he continues to see Speedo's clock ticking slower than his own for much more than half the trip. Eventually, near the end of the trip, he sees Speedo turn around and now he sees Speedo's clock ticking faster than his own until Speedo gets back to him, In fact, they both see each other's clock ticking faster than their own by exactly the same amount during this last portion of the trip.

Now when Speedo gets back, since he watched Goslo's clock tick slow and fast for exactly half the trip, and since Goslo watched Speedo's clock tick slow for most of the trip and fast for a very short time, Goslo will see that Speedo is much younger than himself when they reunite.

It's really very simple, OK?

If we use non-relativistic Doppler shift of light here, then the result is incorrect: no time dilation.

If we use relativistic Doppler shift here, then the result is correct: time dilation.

The difference between non-relativistic Doppler shift and relativistic Doppler shift is that in relativistic Doppler shift time dilation is taken account.

EDIT: Or should say: non-relativistic frequency shift of ticking, and relativistic frequency shift of ticking.

EDIT2: I mean: If we calculate clock ticks assuming the clock is time dilated, then we get less ticks, if calculate clock ticks assuming the clock is not time dilated, then we get more ticks, and therefore a Doppler shift based explanation is not really an explanation.

You have to read carefully what I said. I didn't mention time dilation at all. I didn't even mention relativity. I didn't say it was Doppler, relativistic or normal. I was explaining what each twin sees or views. Isn't that the fundamental meaning of "Point of View"?

Now if it was normal Doppler then what I said would be incorrect, wouldn't it? Because I said that at the end of the trip, they each see the other ones clock having accumulated a different amount of time and that wouldn't be true for normal Doppler.

I'm trying to get Michio Cuckoo to understand what the scenario is before he tries to create non-existent problems for it. I'm presenting the raw data measurements that any viable theory has to explain.

 

[jartsa]

You have to read carefully what I said. I didn't mention time dilation at all. I didn't even mention relativity. I didn't say it was Doppler, relativistic or normal. I was explaining what each twin sees or views. Isn't that the fundamental meaning of "Point of View"?

Now if it was normal Doppler then what I said would be incorrect, wouldn't it? Because I said that at the end of the trip, they each see the other ones clock having accumulated a different amount of time and that wouldn't be true for normal Doppler.

I'm trying to get Michio Cuckoo to understand what the scenario is before he tries to create non-existent problems for it. I'm presenting the raw data measurements that any viable theory has to explain.

I see. We are talking about what Speedo sees. OK I trust that simply putting Speedo's speed into relativistic doppler shift formula, tells us what frequencies Speedo sees.

 

[Michio Cuckoo]

You have to read carefully what I said. I didn't mention time dilation at all. I didn't even mention relativity. I didn't say it was Doppler, relativistic or normal. I was explaining what each twin sees or views. Isn't that the fundamental meaning of "Point of View"?

Now if it was normal Doppler then what I said would be incorrect, wouldn't it? Because I said that at the end of the trip, they each see the other ones clock having accumulated a different amount of time and that wouldn't be true for normal Doppler.

I'm trying to get Michio Cuckoo to understand what the scenario is before he tries to create non-existent problems for it. I'm presenting the raw data measurements that any viable theory has to explain.

So as Speedo undergoes acceleration, he sees the rest of the stationary universe undergo Time Compression; and that is why Goslo ends up older?

 

[ghwellsjr]

So as Speedo undergoes acceleration, he sees the rest of the stationary universe undergo Time Compression; and that is why Goslo ends up older?

I don't say that. I say that as Speedo undergoes acceleration at the half-way point of his trip, he immediately sees an increase in the rate at which Goslo's clock ticks but this is only because he is moving toward the images of Goslo's clock that were already in transit toward him. It's just a Doppler shift. But Goslo doesn't see any such change until way later when the images of Speedo undergoing acceleration finally reach him. It's this imbalance in the time that each one sees the other ones clock ticking slower then faster that allows them to see a difference in the accumulated times on each others clock.

 

[Michio Cuckoo]

I don't say that. I say that as Speedo undergoes acceleration at the half-way point of his trip, he immediately sees an increase in the rate at which Goslo's clock ticks but this is only because he is moving toward the images of Goslo's clock that were already in transit toward him. It's just a Doppler shift. But Goslo doesn't see any such change until way later when the images of Speedo undergoing acceleration finally reach him. It's this imbalance in the time that each one sees the other ones clock ticking slower then faster that allows them to see a difference in the accumulated times on each others clock.

So Goslo's clock ticking faster as seen by Speedo is a result of the Doppler effect.

This means that from Speedo's reference frame, he can still apply Time Dilation to Goslo.

Therefore, the only explanation as to how Goslo got older is that as Speedo changed his direction, he also switched to another world line, and he observes a time gap, i.e. Goslo suddenly becomes older.

Amirite?

 

[Dale]

This means that from Speedo's reference frame, he can still apply Time Dilation to Goslo.

First, you have to define Speedo's reference frame. As I mentioned in post 11 there is no standard way of doing so, which means that you have to be explicit.

What exactly do you mean when you say "Speedo's reference frame".

 

[A.T.]

Therefore, the only explanation as to how Goslo got older is that as Speedo changed his direction, he also switched to another world line, and he observes a time gap, i.e. Goslo suddenly becomes older.

Comparing distant clocks depends on certain conventions. According to the usual one, time flows at different rates at different positions in an accelerating frame (just like in a gravitational field, see Equivalence principle). So yes, in the accelerating frame of Speedo time is running very fast at the very distant position of Goslo, allowing Goslo to age quickly.

This observed non-uniform aging of Goslo is just a consequence of the non-inertial frame of Speedo. Describing the world from non-inertial frames is always "strange", even in classical mechanics where suddenly inertial forces appear. In Relativity non-inertial become a bit more "strange".

 

[ghwellsjr]

I don't say that. I say that as Speedo undergoes acceleration at the half-way point of his trip, he immediately sees an increase in the rate at which Goslo's clock ticks but this is only because he is moving toward the images of Goslo's clock that were already in transit toward him. It's just a Doppler shift. But Goslo doesn't see any such change until way later when the images of Speedo undergoing acceleration finally reach him. It's this imbalance in the time that each one sees the other ones clock ticking slower then faster that allows them to see a difference in the accumulated times on each others clock.

So Goslo's clock ticking faster as seen by Speedo is a result of the Doppler effect.

This means that from Speedo's reference frame, he can still apply Time Dilation to Goslo.

Therefore, the only explanation as to how Goslo got older is that as Speedo changed his direction, he also switched to another world line, and he observes a time gap, i.e. Goslo suddenly becomes older.

Amirite?

Again, please note that I have been talking about what each twin actually sees, not how you can resolve what happens according to an arbitrary non-inertial reference which to me is a pointless exercise and shouldn't be pursued until you thoroughly understand how you can resolve what happens from an arbitrary inertial reference frame. Do you understand how to analyze the situation from an inertial reference frame, for example, one in which Goslo remains at rest throughout the scenario?

 

[AntonL]

And which twin aged?

Suppose the two twins are each in their respective spaceships traveling at some or other high speed as measured in reference system C, so A and B stay young while C ages.

A & B do not know about C so let's forget about C and just look at A & B in their respective reference frames

A & B sync their clocks:
1) A sees B depart at some high speed and return, A aged and B remained young
but also at the same time
2) B sees A depart at some high speed and return, B aged and A remained young

So whoever ages depends into which spaceship you place the reference frame.

Now if above is all gibberish then SR is to blame. Sorry I rephrase: The science must be correct it is just that I do not understand SR. Maybe someone can give me a nice logical explanation and proof to explain to me which clock actually slowed down when A and B meet again.

PS:Now let's go back to reference system C and relativity says A & B clocks tick slower than C's clock.

3) Now A sees B depart but actually B slows to C so B clock is faster than A, but in reference system A, B clock is slower than A

As I said, I am totally but I am sure their must be an explanation forthcomming

 

[Dale]

A & B sync their clocks:
1) A sees B depart at some high speed and return, A aged and B remained young
but also at the same time
2) B sees A depart at some high speed and return, B aged and A remained young

So whoever ages depends into which spaceship you place the reference frame.

This is not true. All inertial frames will agree which is younger when they return. Once you have completely specified the problem in anyone inertial frame you can determine what happens in any other.

The science must be correct it is just that I do not understand SR. Maybe someone can give me a nice logical explanation and proof to explain to me which clock actually slowed down when A and B meet again.

Just use the following formula to calculate the time accumulated for A and for B. The formula is valid in any inertial frame:
http://en.wikipedia.org/wiki/Proper_time#Mathematical_formalism

PS:Now let's go back to reference system C and relativity says A & B clocks tick slower than C's clock.

3) Now A sees B depart but actually B slows to C so B clock is faster than A, but in reference system A, B clock is slower than A

As I said, I am totally but I am sure their must be an explanation forthcomming

Just pick any inertial frame, use the formula above, and look at the result. Then, Lorentz transform to any other inertial frame, and do the same. Compare the results.

 

[ghwellsjr]

A & B sync their clocks:
1) A sees B depart at some high speed and return, A aged and B remained young
but also at the same time
2) B sees A depart at some high speed and return, B aged and A remained young

So whoever ages depends into which spaceship you place the reference frame.

Now if above is all gibberish then SR is to blame. Sorry I rephrase: The science must be correct it is just that I do not understand SR. Maybe someone can give me a nice logical explanation and proof to explain to me which clock actually slowed down when A and B meet again.

I already did that in post #4. Did you read it? Do you understand it?

Please note that my explanation is not based on SR, it is based only on the Principle of Relativity, Einstein's first postulate, but not on his second postulate. His second postulate is also needed in order to construct a Frame of Reference. So even without resorting to a reference frame, you can still prove unambiguously which twin ages less and which twin ages more. You can't blame SR or your lack of knowledge of SR for the fact that we can determine the age difference between the two twins in either of your two scenarios.

You do realize that you have described two different scenarios even though you said "but at the same time", don't you? What matters is which twin is the one that "returns", that is, which one fires his rockets and turns around. If B is the one that does the returning as you said in scenario 1, then B ages less. If A is the one that does the returning as you said in scenario 2, then A is the one that ages less. As I pointed out in post #4, whichever twin is the one that returns is the one that immediately sees the image of the other ones clock go from ticking slow to ticking fast while the other twin does not see that until much later.

Like I said in post #4, it's really very simple.

 

[stevendaryl]

According to Lorentz, time is always dilated.


So think of two twins, Speedo and Goslo.

- Goslo stays on Earth and drinks tea.
- Speedo gets into a rocket, zooms off into outer space and zooms back.


Imagine YOU are Speedo. You zoom off, and when you return to Earth, you find that Goslo is 20 years older than you.

Here's a related "paradox" in good old planar Euclidean geometry.

Suppose you are walking along a road and you come to a second road that intersects the first at an angle theta (for definiteness, assume that it is 8.05°). Suppose that each road has distance markers, one every meter.

While you are traveling on your road, you see that when you pass marker number 100 on your road, you look straight to your left and see marker number 101 on the other road. When you pass marker number 200, you see marker number 202 on the other. Etc. When you get to marker number 500, the other road makes a turn toward your road. You continue to compare markers: when you pass marker number 600, the corresponding marker on the other road is 606, etc. Finally, when you reach marker number 1000, the two roads come together again, and the second road shows distance marker 1010.

Letting N be the marker number on your road, and N' be the marker number on the other road, we can show geometrically that the formula relating the two distances is:

ΔN' = ΔN √(1+m²)
where m is the slope (tangent of 8.05°), which is .1414. For our choice of m, this turns out to be:
ΔN' = 1.01 ΔN

This is not too surprising: a straight line is the shortest distance between two points, so it makes sense that the "bent" road is slightly longer than the "straight" road. But now let's look at things from the point of view of a traveler on the "bent" road. When he comes to marker number 101, he looks straight to his right and sees that the corresponding marker on your road is not 100, but 102. When he comes to maker number 202, he looks straight to his right and he sees, not marker number 200, but marker number 204. The relationship that he notices is just the opposite of the pattern you noticed. Instead of

ΔN' = 1.01 ΔN

he finds

ΔN = 1.01 ΔN'

When the traveler on the "bent" road comes to the bend, which is at marker number 505, he looks straight to his right, and sees marker number 510 on your road. So far, from the point of view of the other traveler, it looks like your road is going to be the longer one, not his. But then he turns the corner...

As he is turning, he is looking to his right at your road, and he sees distance markers on your road moving past. Before the turn, he sees marker number 510, but then as he turns, he sees to his right distance markers number 509, 508, ..., 500, 499, ... 490. Immediately after making the turn, when he looks to his right, he sees marker number 490 on your road. From then on, he sees your numbers increase by the same formula as before:

ΔN = 1.01 ΔN'

So when he travels another 505 meters (for a total of 1010 meters counting both legs of the trip), he sees your road's markers advance by 510 meters, for a total of 490 + 510 = 1000.

So, the moral of the story is that only the "straight-line" traveler can use the formula
ΔN' = ∫dN √(1+m²)
to compute the length of another road. A traveler along a "bent" road will get the wrong answer using that formula, unless he takes into account the effects of turning a corner.

The twin paradox is very similar. Any inertialtraveler can use the formula
tau = ∫√(1-(v/c)²) dt
to compute the amount of aging for another traveler. But a noninertial traveler cannot use that formula, unless he takes into account the effects of turning around.

Proper time in SR is analogous to distance along a path in Euclidean geometry.
Relative speed in SR is analogous to relative slope in Euclidean geometry.
Changing velocity in SR is analogous to turning a corner in Euclidean geometry.

 

[Dale]

Here's a related "paradox" in good old planar Euclidean geometry...

Excellent analogy!

 

[ghwellsjr]

Stevendaryl, isn't your explanation identical to the one Jonathan Scott presented in post #2? And I have the same complaint about yours as I did about his: it doesn't address the question that Michio Cuckoo asked:

Discuss this but only consider Speedo's point of view. Because if you are Speedo, what would you see?

You talk about what your two travelers would see in your analogy but it isn't at all like anything that Speedo (and Goslo) can see, is it? What they see is what I described in post #4 and is all you need to show that Speedo ages less than Goslo, don't you agree?

 

[stevendaryl]

Stevendaryl, isn't your explanation identical to the one Jonathan Scott presented in post #2? And I have the same complaint about yours as I did about his: it doesn't address the question that Michio Cuckoo asked:

You talk about what your two travelers would see in your analogy but it isn't at all like anything that Speedo (and Goslo) can see, is it? What they see is what I described in post #4 and is all you need to show that Speedo ages less than Goslo, don't you agree?

I agree that the Doppler shift explanation explains the difference in the two ages, but I think it leaves a bit of a puzzle as to why can the inertial twin use the formula:

τ= ∫√(1-(v/c)² dt

to compute the age τ of the noninertial twin, but the noninertial twin cannot use that formula to compute the age of the inertial twin. Why not? After all, the noninertial traveler is only noninertial for a tiny moment; how can that tiny moment make such a big difference?

That formula for proper time is exactly analogous to the formula for path length in planar Euclidean geometry:

L = ∫√(1+m² dx

Trying to apply the time dilation formula to compare the ages of the twins is very much like trying to apply the length formula to compare the lengths of two roads.

 

[Nugatory]

Stevendaryl, isn't your explanation identical to the one Jonathan Scott presented in post #2? And I have the same complaint about yours as I did about his: it doesn't address the question that Michio Cuckoo asked:

You talk about what your two travelers would see in your analogy but it isn't at all like anything that Speedo (and Goslo) can see, is it? What they see is what I described in post #4 and is all you need to show that Speedo ages less than Goslo, don't you agree?

IMHO: That's a fair criticism, but there's something else going on that's also worth noting:

Online discussions of the twin paradox tend to veer back and forth between "What do we see?" and "How can that be?". The former is Michio Cuckoo's original question, best answered by your post #4 and Dopplerish explanations; the latter is often better answered with a spacetime diagram analysis, and both Jonathan Scott's and Stevendaryl's answers are of that flavor.

 

[Mentz114]

I agree that the Doppler shift explanation explains the difference in the two ages, but I think it leaves a bit of a puzzle as to why can the inertial twin use the formula:

τ= ∫√(1-(v/c)² dt

to compute the age τ of the noninertial twin, but the noninertial twin cannot use that formula to compute the age of the inertial twin. Why not?

Anyone can calculate the proper length of any worldline if they know it.

After all, the noninertial traveler is only noninertial for a tiny moment; how can that tiny moment make such a big difference?

Yes, it kinks the worldline and reduces its proper length.

That formula for proper time is exactly analogous to the formula for path length in planar Euclidean geometry:

...

Trying to apply the time dilation formula to compare the ages of the twins is very much like trying to apply the length formula to compare the lengths of two roads.

The proper length uses Lorentzian distance, but otherwise the two calculations are the same.

See https://www.physicsforums.com/showthread.php?t=587949&page=8&post=126

(the link isn't right so you need to scroll down the page)

 

[ghwellsjr]

I agree that the Doppler shift explanation explains the difference in the two ages, but I think it leaves a bit of a puzzle as to why can the inertial twin use the formula:

τ= ∫√(1-(v/c)² dt

to compute the age τ of the noninertial twin, but the noninertial twin cannot use that formula to compute the age of the inertial twin. Why not? After all, the noninertial traveler is only noninertial for a tiny moment; how can that tiny moment make such a big difference?

Of course the noninertial twin can use that formula, just like you can use the formula or anybody else can, we just have to use it from any arbitrary inertial Frame of Reference. The only reason why you say that the inertial twin can use the formula is because you have bought into the false notion that the twins can only use a frame in which they are at rest which leads to another false notion that the noninertial twin must use a noninertial reference frame.

As I pointed out in post #13, we can use a frame in which Goslo remains at rest throughout the scenario and all the time dilation falls on Speedo making the explanation trivially simple. But we could also use a frame in which Speedo is at rest during the first half of the trip in which all the time dilation falls on Goslo. But during the second half of the trip, Speedo will have even more time dilation while Goslo's continues the same. However, this inertial frame is more complicated to analyze because we have to calculate both twins' time dilation to see which one ends up with more.

Please don't get me wrong, I'm not saying that noninertial frames are invalid, I'm just saying that they are very complicated to analyze and don't offer anything more than any inertial frame or no frame at all (Doppler analysis) and most significantly, they don't provide the twins with any more insight into what is happening to the other twin--none whatsoever--there is no advantage to using a noninertial frame except to demonstrate the mathematical prowess of the presenter.

 

[Samshorn]

My explanation is not based on SR, it is based only on the Principle of Relativity, Einstein's first postulate, but not on his second postulate. His second postulate is also needed in order to construct a Frame of Reference. So even without resorting to a reference frame, you can still prove unambiguously which twin ages less and which twin ages more.

Your explanation tacitly invokes the light-speed postulate, and also a frame of reference. Remember, Galilean relativity also satisfies the principle of relativity, but it doesn't imply unequal aging of the twins. So the unequal aging can't be explainable in terms of ONLY the principle of relativity. When you talk about Doppler effects that each twin will see, you are tacitly assuming things about the propagation of light being unaffected by the motion of the source, and so on, things that are tantamount to the light-speed postulate, in addition to the principle of inertia (i.e., the principle of relativity), and these things collectively are sufficient to construct inertial coordinate systems. You can't get any of the unique effects of special relativity (as distinct from Galilean relativity) purely from the relativity postulate alone. The light-speed postulate (or something equivalent to it) is necessary.

 

[stevendaryl]

Of course the noninertial twin can use that formula, just like you can use the formula or anybody else can, we just have to use it from any arbitrary inertial Frame of Reference.

Well, that's the issue: why must it be an inertial frame of reference? Why can't it be a "piecewise inertial" frame of reference? That is, why can't you use

τ= ∫√(1-(v/c)2 dt

where t is proper time for the "traveling" twin, and v is the speed of the "stay-at-home" twin relative to the instantaneous rest frame of the traveling twin? You can't because there are pieces of the worldline of the stay-at-home twin that are left unaccounted for. (In the Euclidean analog, the problem is that there are pieces of the path of the "straight" road that are counted twice by the corresponding formula for length).

 

[AntonL]

This is not true. All inertial frames will agree which is younger when they return. Once you have completely specified the problem in anyone inertial frame you can determine what happens in any other ... Compare the results.

really?


and

I already did that in post #4. Did you read it? Do you understand it?
.

No I don't understand post 4

Nor have you seemed to understood the problem I posed. Let's reformulate the "Which twin aged?" problem.

Let each twin be in a separate spaceship A and B orbiting a planet C in circular orbits. We assume the planet to be non rotating. A orbits clockwise and B orbits anticlockwise. Both A and B clocks as seen from C are ticking at the same rate but at a reduced rate to that of C due to the orbit velocity and some increased rate due to gravitational effects as compared to if they where stationary on planet C. each time A and B pass (twice per orbit) the clocks have the same time. The GPS satellites pride themselves as proving SR as the clocks are corrected for both gravitational and velocity aspects as described.

Lets go into spaceship A and set our reference frame in this spaceship A. That is A is now experienced as stationary an B is moving relative to A and planet C is orbiting around A. Please describe what we expect from clock B now relative to A.

As I said I am totally confused as I cannot find a logical pleasing answer.

 

[Mentz114]

Nor have you seemed to understood the problem I posed. Let's reformulate the "Which twin aged?" problem.

Let each twin be in a separate spaceship A and B orbiting a planet C in circular orbits. We assume the planet to be non rotating. A orbits clockwise and B orbits anticlockwise. Both A and B clocks as seen from C are ticking at the same rate but at a reduced rate to that of C due to the orbit velocity and some increased rate due to gravitational effects as compared to if they where stationary on planet C. each time A and B pass (twice per orbit) the clocks have the same time. The GPS satellites pride themselves as proving SR as the clocks are corrected for both gravitational and velocity aspects as described.

Lets go into spaceship A and set our reference frame in this spaceship A. That is A is now experienced as stationary an B is moving relative to A and planet C is orbiting around A. Please describe what we expect from clock B now relative to A.

As I said I am totally confused as I cannot find a logical pleasing answer.

You are making it difficult for yourself. Firstly, the time on any clock, irrespective of what is going on elsewhere is given by a simple formula which relies *only* on information about the worldline of the clock in question. So if you want to compare the elapsed time for different clocks, just calculate the proper time of the worldline between the events for each worldline, then compare them. As easy as measuring shoes.

 

[stevendaryl]

Let each twin be in a separate spaceship A and B orbiting a planet C in circular orbits. We assume the planet to be non rotating. A orbits clockwise and B orbits anticlockwise.

If your problem involves orbits, then you can't solve it using Special Relativity.

 

[Dale]

really?

Really, really. It is a pretty typical homework problem. I recommend working through it for your own instruction.

Suppose in some inertial frame that A leaves Earth at v=.6c and reaches a planet 3 ly away in a time of 5 y and then immediately turns around at v=.6c returning 10 y later, while B stayed at Earth the whole time. Calculate the age of the twins at the reunion:
1) in the frame where the Earth is at rest the whole time
2) in a frame where the Earth is moving at .333 c the whole time

 

[Nugatory]

No I don't understand post 4

Nor have you seemed to understood the problem I posed. Let's reformulate the "Which twin aged?" problem.

Let each twin be in a separate spaceship A and B orbiting a planet C in circular orbits...

Unfortunately, that reformulated problem is waaaay harder than the flat spacetime problem in the original formulation of the twin paradox. Your best bet is to work through post 4 and the FAQ at http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/TwinParadox/twin_paradox.html until you have nailed the flat spacetime problem, then think about the problem of two traveling twins and one stay-at-home twin, still in flat spacetime with no gravity, no circular orbits, travel in a straight line, ...