Acceleration-free twin paradox

[jethomas3182]

I find the discussion about acceleration confusing. I want to avoid that.

So try this scenario: Lucy is traveling at .6c relative to Bob. Just when they reach their point of closest approach, one kilometer, they synchronize clocks.

As their distance increases, each of them sees a red shift. Each estimates the other's time is dilated by a factor of gamma, that is 1/sqrt(1-.6^2)=1/sqrt(.64)=1/.8

After some distance, Lucy meets Betsy who is traveling the opposite direction at .6c relative to Bob. When they reach their point of closest approach, one kilometer, Lucy gives Betsy her clock's current time and Betsy sets her clock by it. Betsy and Bob will both see each other's light is blue-shifted. They both calculate the other's time is dilated by a factor of gamma = 1/.8.

When Betsy and Bob reach their point of closest approach, one kilometer, they exchange times. Which clock will be ahead, and by how much?

 

[jerromyjon]

I believe that scenario is pretty simple..

After some distance,

I also believe that is crucial, so we will say 5 minutes has passed on Bob's clock since Lucy passed and synced her clock, we can call this sync 00:00. At this time 4 minutes has passed on Lucy's time dilated clock and she syncs Betsy's clock to 00:04. Since she would be the same distance from Bob it will take another 5 minutes on Bob's clock until she passes him. At that point her clock reads 00:08 and Bob's clock would read 00:10.

 

[jethomas3182]

I believe that scenario is pretty simple..

we will say 5 minutes has passed on Bob's clock since Lucy passed and synced her clock, we can call this sync 00:00. At this time 4 minutes has passed on Lucy's time dilated clock and she syncs Betsy's clock to 00:04. Since she would be the same distance from Bob it will take another 5 minutes on Bob's clock until she passes him. At that point her clock reads 00:08 and Bob's clock would read 00:10.

Let's say that 4 minutes has passed on Lucy's clock when she sync's Betsy's clock.

Lucy's situation is exactly equivalent to Bob's, and she has exactly as much reason to think that Bob's time is dilated compared to hers, as Bob has to think her time is dilated compared to his. Her clock reads 4 minutes so she can calculate Bob's proper time, it is 3.2 minutes.

Independent of Lucy, Betsy has every reason to think that Bob's travel at .6 c will cause his time to dilate. If her clock reads 4 minutes when she reaches Bob, it is predictable that Bob's clock will have advanced 3.2 minutes.

Since both of these independent inertial frames are as valid as Bob's, and they are clearly sequential, one starting at precisely the time and place that the other ended, Betsy is justified to conclude that Bob's proper time is 6.4 minutes.

But of course Bob's actual proper time is not 6.4 minutes or 8 minutes but 10 minutes.

Why is it that Lucy's view and Betsy's view are both wrong?

 

[jerromyjon]

Because their velocity distorts their perception of spacetime.

 

[jethomas3182]

Ah! I have an alternative view. Betsy need not take Lucy's view of things. From Betsy's point of view, Bob is traveling toward her at .6 c. Lucy is traveling toward her at .88 c.

So from her point of view, Lucy's time is dilated even more than Bob's. If I read it right, not just .8, but .47.

So Betsy will figure that Bob's clock goes 3.2 seconds when Betsy's clock goes 4 seconds, but she will also figure that while Lucy's clock went 4 seconds Bob's clock went .8/.47 seconds, about 6.8 seconds! So she should calculate his proper time at approximately 10 seconds! My answer was a bit off from that, but I rounded and maybe the right answer will be exactly 10 seconds.

 

[jerromyjon]

This is what I imagined by your description. Lucy and Betsy would pass at 1.2c relative to each other...

 

[jethomas3182]

View attachment 75333
This is what I imagined by your description. Lucy and Betsy would pass at 1.2c relative to each other...

Yes, but you can't just add velocities. When they're in the same or opposite directions, you add them like this:

V1 + V2 = (V1 + V2)/(1+V1V2) Where V1 and V2 are fractions of lightspeed. Like .6 c.

 

[jerromyjon]

My calculator says that 299,792,458 m/s (c) times 0.6 = 179,875,474.8 m/s. Since Lucy and Betsy are presumably headed in almost the opposite direction their velocity relative to each other has to be nearly double so 179,875,474.8 m/s times 2 = 359,750,949.6 m/s which is undoubtedly > c.

 

[PeterDonis]

When Betsy and Bob reach their point of closest approach, one kilometer, they exchange times. Which clock will be ahead, and by how much?

Bob's clock will be ahead; as you say in a subsequent post, his clock will read 10 minutes, and Betsy's will read 8 minutes. Of course this result is simple to derive in Bob's frame.

If you try to analyze the scenario in Lucy's or Betsy's frame, the key thing you have to include, that it doesn't seem like you're including, is relativity of simultaneity. Bob, Lucy, and Betsy all have different simultaneity conventions, and that fact must be included when you are comparing readings on their clocks.

Since Lucy and Betsy are presumably headed in almost the opposite direction their velocity relative to each other has to be nearly double

You also need to add velocities correctly using the relativistic formula, which jethomas3182 gave you. If V1 = V2 = 0.6, then their relativistic "sum" (i.e., the relative velocity of Lucy and Betsy) is (0.6 + 0.6) / (1 + 0.6 * 0.6) = 1.2 / 1.36, or about 0.88.

 

[jerromyjon]

(0.6 + 0.6) / (1 + 0.6 * 0.6) = 1.2 / 1.36, or about 0.88

That is the first I've ever heard of it and I am quite curious where it was derived from...

 

[jethomas3182]

Bob's clock will be ahead; as you say in a subsequent post, his clock will read 10 minutes, and Betsy's will read 8 minutes. Of course this result is simple to derive in Bob's frame.

The way I read it, Betsy's clock will have advanced 4 minutes since she left Lucy. She might not feel it appropriate to count Lucy's 4 minutes, but might instead translate them to the time that seems right for her own frame. But it's a picky point.

If you try to analyze the scenario in Lucy's or Betsy's frame, the key thing you have to include, that it doesn't seem like you're including, is relativity of simultaneity. Bob, Lucy, and Betsy all have different simultaneity conventions, and that fact must be included when you are comparing readings on their clocks.

I don't see what to do about that. The events I look at are Bob and Lucy together at the same time and place, Lucy and Betsy together at the same time and place, and Betsy and Bob together at the same time and place. We never have to think about any other simultaneity.

 

[Nugatory]

That is the first I've ever heard of it and I am quite curious where it was derived from...

Google for "relativistic velocity addition".

If A's speed relative to C is $u$ and B's speed relative to C is $v$, then A's speed relative to B will be $(u+v)/(1+uv)$ (measuring time in seconds and distance in light-seconds so that $c=1$). If it weren't for this, the speed of light would not come out to be $c$ for all observers - try setting either $u$ or $v$ to one to see what I mean.

 

[PeterDonis]

The way I read it, Betsy's clock will have advanced 4 minutes since she left Lucy. She might not feel it appropriate to count Lucy's 4 minutes, but might instead translate them to the time that seems right for her own frame.

According to the OP, Betsy sets her clock to read the same as Lucy's when she passes Lucy. That's why I said her clock would read 8 minutes when she passes Bob.

 

[PeterDonis]

That is the first I've ever heard of it and I am quite curious where it was derived from...

It's a basic fact about relativity; you can Google for it, as Nugatory suggested, or check any relativity textbook.

 

[Filip Larsen]

I find the discussion about acceleration confusing. I want to avoid that.

It is possible to reduce or abstract the situation in the twin paradox to 3 space-time events (2 clock synchronization events and 1 clock comparison event) giving rise to 2 space-time intervals [1] and still resolve the paradox without any reference to or use of acceleration. By resolve I mean both in a qualitative manner in that the apparent symmetry between the twins, which gives rise to the paradox in the first place, can be seen to be an asymmetric situation, and in a quantitative way giving the difference in clock reading at the last event.

Of course, without acceleration the traveling twin cannot turn back home, so one has to settle for three non-accelerated clocks instead. Alternatively the 3-event model can be said to give a lower limit for the situation where the traveling twin uses arbitrarily high acceleration to turn around.

[1] http://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals

 

[A.T.]

Of course, without acceleration the traveling twin cannot turn back home,

Which removes the core of the "paradox". If both twins stay inertial, and just exchange information via light signals or other travelers, the situation between them is symmetrical, and there is no "paradox".

 

[harrylin]

I find the discussion about acceleration confusing. I want to avoid that.
[..] Lucy gives Betsy her clock's current time and Betsy sets her clock by it. [..] Which clock will be ahead, and by how much?

If you understood the correct answers in the "acceleration" thread (but I'm afraid that not all answers there were correct), you would know that the calculation is just the same as for the extreme case of nearly instantaneous turn-around.
In fact, you replaced one break of symmetry but another but equivalent break of symmetry. For the calculations it is irrelevant how you realize the change in velocity. But, there is of course no harm in elaborating that simple fact. :)

 

[Filip Larsen]

Which removes the core of the "paradox". If both twins stay inertial, and just exchange information via light signals or other travelers, the situation between them is symmetrical, and there is no "paradox".

On the contrary. The paradox stems directly from the "symmetry" of time dilation where two identical and non-accelerating clocks moving relative to each other will measure the rate of the other clock as slower than their own. The twin paradox situation is a "naive" attempt to construct a thought experiment using that symmetry of time dilation so that it leads to (apparent) inconsistency. A more careful analysis of course shows that there are no inconsistency and that the situation is not symmetrical since the traveling twin is at rest in different reference frames at different times. You can either use acceleration to make that shift or, since the original poster expressed interest in an acceleration-free model, just use a single event.

By the way, note that the 3-event model, by definition, uses space-time events; there is no need for light signals. The traveling twin is still represented by a world line, although its not a smooth one as if finite acceleration is used. By replacing the traveling twin with two clocks which are at least as capable of measuring time as the twin is (drama aside its not really the actual biological age of the twins that are interesting, its how we measure and compare time in a relativistic framework) I think the resolution to the paradox is more clear. Since the symmetry breaking in the twin paradox comes from the shift of reference frame as the traveling twin undergo acceleration, and in the 3-event model that shift of reference frame is replaced with a single space-time event synchronizing a clock in one reference frame with a clock in a another reference frame (clock synchronization is the basic building block of any thought experiment in SR) you end up with a much simpler model. To me this one event resolves the paradox just as fine as acceleration would, but I understand if others perhaps are uncomfortable leaving out acceleration.

 

[Dale]

The paradox stems directly from the "symmetry" of time dilation

I don't think that one person keeping track of time on a single clock is symmetrical to two people exchanging timekeeping information between two clocks.

I understand your motivation, but I don't see how it is pedagogically supposed to resolve the student's issue.

 

[Filip Larsen]

I don't think that one person keeping track of time on a single clock is symmetrical to two people exchanging timekeeping information between two clocks.

To the extend that it is the time interval measured by the clocks that are interesting it should not matter whether we use one or two properly synchronized clocks if we want to know the total time interval. Disregarding acceleration also mean that you do not have to care if the rate of an accelerated clock is independent of acceleration or not, which is nice if you are new to special relativity (and people pondering the twin paradox usually are).

I understand your motivation, but I don't see how it is pedagogically supposed to resolve the student's issue.

It was my intention to make him aware of the use of space-time intervals as way to very easily come up with the equations needed to answer his question. In fact, they are a very useful tool for any problem in special relativity that can be modeled with space-events, a tool that I think is often overlooked.

 

[jerromyjon]

Still don't get how either of these two travelers can be believed to observe Bob's clock tick slower as Bob would realistically see their clocks as both slow. If they could see "Bob's reality" moment by moment his clock is ticking faster.

 

[PeterDonis]

The events I look at are Bob and Lucy together at the same time and place, Lucy and Betsy together at the same time and place, and Betsy and Bob together at the same time and place.

But there is no frame (and can't be) where all three of these events happen at the same place. So in order to analyze all three events in any frame, simultaneity will come into play. The only way to avoid it is to just look at the lengths of each of the sides of the triangle formed by these three events; and those side lengths are 10 (Bob/Lucy to Bob/Betsy), 4 (Bob/Lucy to Lucy/Betsy), and 4 (Lucy/Betsy to Bob/Betsy). This leads to a "traveling" result of 8 (4 + 4) and a "stay-at-home" result of 10, as has already been said.

 

[PeterDonis]

If they could see "Bob's reality" moment by moment his clock is ticking faster.

For anyone who is not co-located with Bob, doing this requires a simultaneity convention. And Lucy and Betsy have two different simultaneity conventions. That's where the "gap" is. The Usenet Physics FAQ article on the twin paradox (which is well worth reading in its entirety) calls your objection the Time Gap Objection, and explains how it is resolved.

 

[jethomas3182]

Still don't get how either of these two travelers can be believed to observe Bob's clock tick slower as Bob would realistically see their clocks as both slow. If they could see "Bob's reality" moment by moment his clock is ticking faster.

No, I don't think that's right. Both of them see Bob's clock ticking slower. Each of them sees Bob's clock ticking 3.2 seconds while their own ticks 4 seconds.

This is required by SR.

But we can't just add 4+4 to get 8, and we can't just add 3.2 + 3.2 to get 6.4.

Arithmetic doesn't work that way in SR, just as we can't add velocity .6 c to velocity .6 c to get velocity 1.2 c.

 

[jethomas3182]

But there is no frame (and can't be) where all three of these events happen at the same place. So in order to analyze all three events in any frame, simultaneity will come into play. The only way to avoid it is to just look at the lengths of each of the sides of the triangle formed by these three events; and those side lengths are 10 (Bob/Lucy to Bob/Betsy), 4 (Bob/Lucy to Lucy/Betsy), and 4 (Lucy/Betsy to Bob/Betsy). This leads to a "traveling" result of 8 (4 + 4) and a "stay-at-home" result of 10, as has already been said.

I don't think that works.

Try this -- somebody is traveling at a speed so large that one of his seconds is equal to 1000 of yours. He arrives at your location and you are interested in something that happened 4 second ago by his clock. Then you wait another 4 seconds by your clock.

What kind of calculation would you be doing where it's appropriate to add his four second to your four seconds?

4+4=8

There might be a situation where that's appropriate, but I don't understand it yet.

 

[PeterDonis]

Each of them sees Bob's clock ticking 3.2 seconds while their own ticks 4 seconds.

But those two intervals of 3.2 seconds do not cover the entire length of Bob's worldline between the two meeting events; i.e., they do not cover the the entire length of the third side of the triangle in spacetime (see below). There is a gap in between them, 3.6 seconds long, corresponding to the change in simultaneity conventions between the outgoing and ingoing legs of the traveling twin's trip (or between Lucy and Betsy, if you look at it that way).

But we can't just add 4+4 to get 8,

Sure we can. Each 4 is the length of one side of a triangle in spacetime. Adding the two together gives the combined length of the two sides. This can then be compared with the length of the third side (10). It's just geometry.

What kind of calculation would you be doing where it's appropriate to add his four second to your four seconds?

A calculation comparing the sum of the lengths of two sides of a triangle in spacetime with the length of the third side, as above. Again, it's just geometry.

Physically, the sum of the lengths of the two sides gives a very good approximation to the proper time experienced by a traveling twin who turns around very sharply at the vertex of the triangle between the two sides of length 4. Understanding that this calculation is just geometry, and doesn't depend on your choice of coordinates, can help to drive home the point that the traveling twin will be younger than the stay-at-home twin when they meet up again, and that that fact is independent of your choice of coordinates; it's an invariant fact that arises from the geometry of spacetime.

 

[jethomas3182]

But those two intervals of 3.2 seconds do not cover the entire length of Bob's worldline between the two meeting events; i.e., they do not cover the the entire length of the third side of the triangle in spacetime (see below). There is a gap in between them, 3.6 seconds long, corresponding to the change in simultaneity conventions between the outgoing and ingoing legs of the traveling twin's trip (or between Lucy and Betsy, if you look at it that way).

I'm no saying you're wrong, but I don't understand it.

What I see is that when you convert from Lucy's time to Betsy's time it converts to 6.8 seconds. Betsy thinks Lucy's clock is slow relative to hers, even slower than Bob's clock is relative to Lucy's clock. There's no gap, just a conversion.

Sure we can. Each 4 is the length of one side of a triangle in spacetime. Adding the two together gives the combined length of the two sides. This can then be compared with the length of the third side (10). It's just geometry.

Usually when I add two vectors I don't just add their lengths. There are times when that's appropriate, like when you want to measure a perimeter. But I don't yet see that it's appropriate now.

 

[PeterDonis]

I'm no saying you're wrong, but I don't understand it.

Look at figure 4 of this page from the Usenet Physics FAQ article. The left (vertical) side of the triangle corresponds to Bob's worldline. The lower right side corresponds to Lucy's worldline. The upper right side corresponds to Betsy's worldline. The two sets of blue lines are the lines of simultaneity corresponding to Lucy and Betsy. Notice the gap in the middle of Bob's worldline, where no blue lines intersect it.

when I add two vectors I don't just add their lengths.

The numbers in question (the two 4's and the 10) are not vectors. They are lengths of sides of a triangle in spacetime.

There are times when that's appropriate, like when you want to measure a perimeter.

Which is effectively what is being done here.

 

[A.T.]

I don't think that one person keeping track of time on a single clock is symmetrical to two people exchanging timekeeping information between two clocks.

It's just replacing light signals with a slower communication method, a third traveler. But as long as you don't bring the twins together, there is no frame invariant age difference between them, so no paradox.

 

[GTOM]

You also need to add velocities correctly using the relativistic formula, which jethomas3182 gave you. If V1 = V2 = 0.6, then their relativistic "sum" (i.e., the relative velocity of Lucy and Betsy) is (0.6 + 0.6) / (1 + 0.6 * 0.6) = 1.2 / 1.36, or about 0.88.

I don't really understand this. Let's suppose i send a probe from Earth to a light second away in one direction.
Another probe in the opposite direction a light second away.
Now i send two relativistic protons in opposite direction. Will they hit the two probes (two light seconds away from each other) in one second according to my clock or not?
If the first (two light seconds in one second), than what is their speed compared to each other from my viewpoint?