Acceleration-free twin paradox

  • #26
PeterDonis
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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.
 
  • #27
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.
 
  • #28
PeterDonis
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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.
 
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  • #29
A.T.
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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.
 
  • #30
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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 dont really understand this. Lets 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?
 
  • #31
PeterDonis
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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?
No, because the protons can't move at the speed of light. It will take them longer than one second to travel one light second. If we suppose that the protons are each moving at, say, 0.9c relative to you, then it will take them each 1 / 0.9 = 1.11 seconds to reach their respective probes, according to your clock.

what is their speed compared to each other from my viewpoint?
(0.9 + 0.9) / (1 + 0.9 * 0.9) = 1.8 / 1.81, or about 0.994c.
 
  • #32
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No, because the protons can't move at the speed of light. It will take them longer than one second to travel one light second. If we suppose that the protons are each moving at, say, 0.9c relative to you, then it will take them each 1 / 0.9 = 1.11 seconds to reach their respective probes, according to your clock.



(0.9 + 0.9) / (1 + 0.9 * 0.9) = 1.8 / 1.81, or about 0.994c.
Sorry, i knew not exactly one second, but something close to it. Of course almost light speed or light speed isnt the same.
However if their speed compared to each other is 0.994c, then how can they cover two light seconds distance in 1.11 second? Ok with a speed of sqrt(3)/2 you can get a length contradiction of 2, however, theoretically they could reach a speed closer to c, and get bigger length contradiction, but the time to reach still will be more than a second.
 
  • #33
PeterDonis
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However if their speed compared to each other is 0.994c, then how can they cover two light seconds distance in 1.11 second?
In a reference frame in which either one is moving at 0.994c, they don't. You are comparing speeds in one frame with distances in a different frame, and that's going to give nonsensical results. Also, you're ignoring relativity of simultaneity, which is also going to lead to nonsensical results. You need to transform the coordinates of all the events of interest from one frame to the other, using the Lorentz transformation, and then analyze the scenario in the new frame.

In the frame in which the Earth is at rest, each proton covers 1 light second in 1.11 seconds, for a speed of 0.9c. The distance between the two protons, in this frame, does increase by 2 light seconds in 1.11 second, but that distance divided by that time does not correspond to the speed of anything, because speeds don't add linearly in relativity (they do in Newtonian physics, but experiments have shown that Newtonian physics does not correctly describe reality in this regime, relativity does), so it's not a problem that this calculation gives a result greater than c.

with a speed of sqrt(3)/2 you can get a length contradiction of 2, however, theoretically they could reach a speed closer to c, and get bigger length contradiction, but the time to reach still will be more than a second.
First, it's length "contraction", not length "contradiction".

You are still ignoring relativity of simultaneity. You can't correctly analyze this kind of scenario just by applying length contraction and time dilation factors. As I said above, you need to transform all the event coordinates into the new frame (a frame in which one of the protons is at rest), and then analyze the scenario in the new frame.
 
  • #34
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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.
There is a "convention" that defines simultaneity within any specific inertial frame (e.g. the Einstein convention). A convention is necessary to establish a time coordinate throughout the spatial dimensions of an inertial frame. The same convention is used in each inertial frame to establish simultaneity specifically for that frame. Thus simultaneity is not global, but relative to the frame of motion. Without this convention we cannot measure the time that elapses between two events in an inertial frame that occur at different positions in that frame. Without it we cannot even assign time coordinates as is routinely done in Minkowski diagrams.

In the paradox, two things happen absolutely simultaneously (because they happen at the same place) when the traveler instantaneously reverses direction. The traveler moves from an outgoing inertial frame, where his clock defines current time throughout the spatial extent of that frame, to an ingoing inertial frame. The traveler's clock does not change at the instant of turnaround. Thus the traveler must assign the value on his clock as the current time throughout the ingoing inertial frame. By definition of simultaneity in the two different inertial frames, the simultaneous time on the home clock abruptly changes in the traveler's frame of motion.

The article on the "Time Gap Objection" says "the apparent "gap" [on the home clock] is just an accounting error, caused by switching from one frame to another.

This is more or less true, but slightly misleading. To be clear it is not really an "error". It is quite true given the definition of simultaneity. Furthermore it can be shown to be a gap by physical experiment (however difficult that experiment may be).

An important thing about this "gap" is to realize that the home clock does not "jump" in reality. That is, an observer at rest near the home clock will not witness any such a jump. The jump only occurs from the perspective of the traveler's changing frame of motion. That is it happens in this changing frame. (This is somewhat analogous to the change in position of every object in your field of vision when you turn you head.)

Also, there is nothing mathematically inconsistent about analyzing the twin paradox in this way in order to determine what the value on the two clocks will be when the traveler returns. If there were, there would be an inconsistency in special relativity as described by Einstein.

This analysis has been attacked as incorrect with misunderstandings and tangential or discursive references in the previous thread. If someone feels that this analysis is invalid, that it contradicts SR, then it would be enlightening for everyone interested for that someone to demonstrate precisely why it is not a valid way to determine the time difference upon return of the traveler (in addition to application of the Lorentz transformation of time on the outgoing and incoming legs).
 
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  • #35
PeterDonis
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two things happen absolutely simultaneously (because they happen at the same place) when the traveler instantaneously reverses direction. The traveler moves from an outgoing inertial frame, where his clock defines current time throughout the spatial extent of that frame, to an ingoing inertial frame. The traveler's clock does not change at the instant of turnaround.
The first "happening" is not really a happening, as you describe it; changing your state of motion does not "move" you from one inertial frame to another. It just changes how your worldline is described in all inertial frames. What physically happens at the turnaround is that the traveling twin experiences proper acceleration: he fires rockets or otherwise exerts a force on his ship to change its motion. If the traveling twin chooses to switch his description of physics from one inertial frame to another when he turns around, that's a choice he makes which is perfectly valid but not required by any physics.

The second "happening" is fine, but it only applies to the traveling twin's clock. It does not apply to other clocks that are spatially separated from him (such as the stay-at-home twin's clock). If the traveling twin chooses to change which inertial frame he uses to describe his motion when he turns around, then that change automatically changes the reading on the stay-at-home twin's clock that is simultaneous with a given reading on the traveling twin's clock. I don't think you disagree with this because you say the sme thing yourself, just in slightly different words.

it can be shown to be a gap by physical experiment (however difficult that experiment may be).
Can you describe such an experiment?

the home clock does not "jump" in reality.
"Reality" is a loaded word. A better way of stating this is: events which are close together on the home clock's worldline will have the home clock displaying time readings that are close together. But different inertial frames may pick different events on the home clock's worldline as happening "now" (i.e, simultaneously with some faraway event), and those different events don't have to be close together.

The jump only occurs from the perspective of the traveler's changing frame of motion.
This is equivalent to saying that it happens because the traveler changes his simultaneity convention, which is what I was saying. Changing frames, in the sense you are using the term (at least, as you appear to be using it most of the time--see my next comment below), necessarily includes changing simultaneity conventions.

That is it happens in this changing frame.
If "in this changing frame" refers to an inertial frame, then this is not quite right, because there are two different inertial frames, involved, not one. it happens because the traveler switches from one frame to another for purposes of describing what's going on. As I said above, there is nothing in the physics that requires him to do this; it's just a choice he is free to make (or not to make).

If "in this changing frame" refers to a non-inertial frame in which the traveler is always at rest, then we are back to all the issues with non-inertial frames that have already been discussed in this thread, and I don't see the point of rehashing them. All of this is a question of words anyway, not physics; we all appear to agree on the physics, as in the actual observations that the different observers make. The only discussion we are having is about how to describe those observations in words.

there is nothing mathematically inconsistent about analyzing the twin paradox in this way
Of course not. As long as you correctly calculate the gap, and include it in your prediction of what the home clock will read when the traveler returns, you will get the right answer this way, yes.
 
  • #36
PeterDonis
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The OP's question has been answered, so this thread is closed.
 

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