Paradox within the twin paradox

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  • #126
"Simultaneously" in whose frame? Of course you could design things so that B will reset itself to read some time T "simultaneously" with A reading T as simultaneity is defined in B's rest frame, but in that case A and B will not read T simultaneously in A's rest frame. Likewise you could design things so they both read T simultaneously in A's rest frame, but then they wouldn't read T simultaneously in B's rest frame. So simultaneity hasn't been "ensured" in any absolute sense, only relative to some particular choice of reference frame. And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.

This is confusing, does "in its own reference frame" refer to A's frame or B's? Certainly when B receives the signal, it could figure out what count A has reached simultaneously with B receiving the signal using the definition of simultaneity in B's frame, or it could figure out what count A has reached simultaneously with B receiving the signal using the definition of simultaneity in A's frame. Which one are you talking about here, if either?

I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading T in A's frame, then they do not read T simultaneously in B's frame. Likewise if B sets itself so that they both read T simultaneously in B's frame, they do not read T simultaneously in A's frame. You seem to be suggesting it could be a "fact" in both frames that they are both at the same count T the moment after B receives the first signal and resets itself, but that's impossible.
There is a problem with what you have said here.

Firstly, the answer to your question about what 'count' B is setting itself to was already explained here:

When counter B receives its signal, it will be able to do a relativistic calculation to find out the precise count that counter A has reached at the instant in its own reference frame that is simultaneous to the instant in B's reference frame for which it received the signal.
In other words, B is using the definition of simultaneity in A's reference frame to set itself.

Now, here is the problem with what you have said:

And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.
By making this statement, you are only repeating what I have already said. The fact that they get "progressively more out-of-sync" is the paradox!

Are you following me?

1. B calculates what the count is for A in A's reference frame.
2. B perceives A as going slower than itself.
3. When A reaches B, B sees that A is the same as itself.

Wait! no. What if A is not the same as B when they meet?

If A is not the same as B when they meet, then how could they be symmetrically relative?

That is the paradox.

I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading T in A's frame, then they do not read T simultaneously in B's frame.
Nothing is being "read". B is doing a calculation, which is not the same as "reading". "Reading" implies relativistic transformations brought about by spacial differences. Calculations are not dependent on location. Imagine that A and B are touching in space and that B is reading from A instantly. This would be the equivalent. Remember what you told me?

Yes, that's exactly right. Disagreements between frames about simultaneity only occur for events that happen at different spatial locations, if two events happen at the same time and the same position in one frame, then all other frames agree those events happened at the same time and position.
To understand why doing a relativistic calculation from B's frame is the same as if they were touching and exchanging the same information derived from the calculation, all you have to do is realize that the calculation is being done in B's frame, so the information that B derives about A's frame is not dependent on space or time differences (assuming the calculation is done instantly for B in B's frame. If the calculation is not done instantly, the same argument applies, but just with a slight deviation).

You seem to be suggesting it could be a "fact" in both frames that they are both at the same count T the moment after B receives the first signal and resets itself
You are making an absolute statement. What "moment" are you referring to? The moment in A's frame or the moment in B's frame? I thought you didn't believe in absolutes? ;-) Joking.
 
  • #127
Teachmemore, when describing two light pulses, you finally settled on:

Are you thinking that the faster something travels, the slower time goes for that thing and does this have anything to do with your statement that these two light pulses traveling in opposite directions from a common starting point eventually 'become "trapped" in the same frame of reference'?
I'm not sure what you mean here.

In what you refer to the "what I finally settled on", all I was doing was defining "fixed distance".

The answer to your second question is no.
 
  • #128
Teachmemore, I have been trying to get you to use just one frame of reference and now you want to use no frame of reference. When describing these thought experiments, you get to play "god". In fact you must play "god". You must define counter A's position and velocity starting from time zero and you must define counter B's position and velocity starting from time zero in a single frame of reference. You must define when the counters get reset as defined by the one frame of reference. You cannot talk about the reference frame for A and a different reference frame for B as if each reference frame applies only to one counter. All reference frames apply to all objects all the time.

I explained all this and gave you an example back in post #31. Please study and understand this, and follow its advice before you try to explain another thought experiment. If you don't understand it, please ask for clarification:
You completely mis-interpreted the statement for which you just quoted.

I understand your post #31 quite clearly. If I had any issue with it, I would not hide that from you ;-)

If you want to understand what I mean by "independent of any godly knowledge", please refer to posts #74 or read through the entire discussion on page 5 of this thread.
 
  • #129
JesseM
Science Advisor
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There is a problem with what you have said here.

Firstly, the answer to your question about what 'count' B is setting itself to was already explained here:
When counter B receives its signal, it will be able to do a relativistic calculation to find out the precise count that counter A has reached at the instant in its own reference frame that is simultaneous to the instant in B's reference frame for which it received the signal.
In other words, B is using the definition of simultaneity in A's reference frame to set itself.
As I said, "in its own reference frame" wasn't clear about whether it was referring to A or B which was why the quoted explanation was unclear. But OK, B is using A's definition of simultaneity to set itself. Let me make sure I understand this correctly with a numerical example. Suppose in B's rest frame, B is at rest at position x=0, while A is moving at 0.6c in the +x direction and crosses A's path at t=0. Also suppose that at t=-100 in this frame, A's own clock reads tA=-80, and it's running slow by a factor of [tex]\sqrt{1 - 0.6c^2/c^2}[/tex] = 0.8 so 100 seconds later at t=0 A's clock will have elapsed a time of 80 seconds, so it'll read tA=0 when A meets B (though because of the way B sets itself, B will not read tB=0 at t=0 in its own frame, instead it'll read t=-10 as we'll see below).

Now suppose at t=-100 in this frame when A reads tA=-80, it sends its first signal towards B, and since it's 60 light-seconds away from B at this moment, the signal will reach B 60 seconds later at t=-40. If the event of the signal reaching B happens at coordinates x=0, t=-40 in B's frame, then we can use the Lorentz transformation to figure out the position and time of this event in A's frame:

x' = gamma*(x - vt)
t' = gamma*(t - vx/c2)

With gamma = [tex]\frac{1}{\sqrt{1 - v^2/c^2}}[/tex], in this case 1/0.8 = 1.25.

So, for x=0, t=-40 in B's frame, the corresponding x' and t' coordinates in A's frame would be:

x' = 1.25*(-0.8*-40) = 40
t' = 1.25*(-40) = -50

So, the event of A's first signal reaching B happens at a time coordinate of t'=-50 in A's frame, and A's clock is running in sync with the time coordinate in his frame, so if B wants to set itself to read the same time as A in A's frame, then B would calculate that it must set itself to a time of tB=-50 seconds at that moment. Is this what you had in mind?

Note that if this is right, then in B's own frame it does not read the same time as A when it resets itself--if B sets itself to read tB=-50 seconds at t=-40 in its own frame, at this moment in its own frame A actually reads tA=-32 (A's clock is running slow by a factor of 0.8 in this frame, so if A reads tA=-32 at t=-40 in this frame, that ensures that 40 seconds later at t=0 in this frame, A will read tA=-32 + 0.8*40 = 0 as originally assumed). It is only in A's frame that the event of B setting itself to read tB=-50 is simultaneous with the event of A reading tA=-50.
teachmemore said:
Now, here is the problem with what you have said:
JesseM said:
And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.
By making this statement, you are only repeating what I have already said. The fact that they get "progressively more out-of-sync" is the paradox!
When I said "even if they both read T simultaneously in one of these frames", the subsequent comment about them getting progressively more out-of-sync was only meant to apply to the one frame where they were both initially reading the same time. In my example above, they both read a time of -50 seconds simultaneously in A's frame, so they get progressively more out-of-sync in that frame. But in B's frame, when B sets itself to read tB=-50 this is simultaneous with the event of A reading tA=-32, so in this frame A starts out ahead of B, which means that if A is running slow in this frame by a factor of 0.8 that actually means they get closer to being synchronized as time goes on! Although B's time will not have quite caught up with A's time when they meet, the gap will have narrowed, with A reading tA=0 and B reading tB=-10. And 50 seconds later in B's frame, B will read tB=-10 + 50 = 40, while A is running at 0.8 the normal rate in this frame so it reads tB= 0.8*50 = 40, so this is the time when they have momentarily become perfectly synchronized in this frame.
teachmemore said:
Are you following me?

1. B calculates what the count is for A in A's reference frame.
2. B perceives A as going slower than itself.
3. When A reaches B, B sees that A is the same as itself.

Wait! no. What if A is not the same as B when they meet?
They won't be the same, not if B set itself so that its time was the same as A's time in A's rest frame, that must mean that B will be behind when they meet since in A's frame B was initially set to the same time as A but was running slow thereafter. You can see this is true in my example above, where at t=-40 in B's frame it must set itself to tB=-50 in order for it to be synchronized with A in A's frame at that moment, then when A and B meet, A will read tA=0 while B will read tB=-10.
teachmemore said:
If A is not the same as B when they meet, then how could they be symmetrically relative?
What do you mean by "symmetrically relative"? Their rates are symmetrically relative in the sense that A is running slow in B's frame while B is running slow in A's frame. In A's frame, at t'=-50 seconds A reads tA=-50 seconds and at this moment B resets itself to read tB=-50 seconds, then they meet 50 seconds later, and A has elapsed a full 50 seconds in this time but B is running slow by a factor of 0.8 so it has only elapsed a time of 0.8*50 = 40 seconds, meaning A will read tA = -50 + 50 = 0 while B will read tB = -50 + 40 = -10. Meanwhile, in B's frame, at t=-40 seconds A reads tA=-32 and at this moment B sets itself to read tB=-50, then 40 seconds later when they meet, B has elapsed a full 40 seconds so it reads tB=-50+40=-10, while A is running slow by a factor of 0.8 so it's only elapsed a time of 40*0.8=32 seconds, so it reads a time of tA=-32+32=0. You can see that both frames agree that A reads tA=0 when they meet and B reads tB=-10 when they meet, but the two frames disagree on whether A or B has elapsed more time since the moment B received A's first signal and reset itself to read tB=-50 seconds.
JesseM said:
I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading T in A's frame, then they do not read T simultaneously in B's frame.
teachmemore said:
Nothing is being "read". B is doing a calculation, which is not the same as "reading". "Reading" implies relativistic transformations brought about by spacial differences.
Huh? How does it "imply" that? By "read" I just meant the normal idea of asking what a clock "reads" at a given moment, i.e. what time the clock is displaying on its clock face (or digital readout or whatever) at any given coordinate time. In my example above A "reads" a time of tA=0 at t=0 in B's frame, and it is running slow by a factor of 0.8, so at t=-10 A "reads" tA=-8, at t=-20 A "reads" tA=-16, etc. etc. Unless English is not your native language, I assume you have heard people talk about "reading time" on a clock.
teachmemore said:
To understand why doing a relativistic calculation from B's frame
What "relativistic calculation", the one to reset the clock? You said before this calculation was done so their times would match in A's frame, not B's frame: "B is using the definition of simultaneity in A's reference frame to set itself." That was the assumption I was using in my calculation above, hopefully you're not changing the assumption here.
teachmemore said:
is the same as if they were touching and exchanging the same information derived from the calculation,
"The same" in what sense? It obviously is not the same the important sense that if B resets itself so that its reading at that moment matches A's reading in A's frame, then if B does this when they are far apart its reading will not match A's from the perspective of B's own frame, whereas if B resets itself this whey when they are touching then all frames will agree they show the same time at that moment.
JesseM said:
You seem to be suggesting it could be a "fact" in both frames that they are both at the same count T the moment after B receives the first signal and resets itself
teachmemore said:
You are making an absolute statement. What "moment" are you referring to?
Sorry if it was confusing, but I was saying that the corresponding version of that statement couldn't be true in "both frames". In other words, it's impossible that it could be true in frame A that both clocks are "at the same count T the moment after B receives the first signal and resets itself" (with 'moment' defined here in terms of A's definition of simultaneity) and be true in frame B that both clocks are "at the same count T the moment after B receives the first signal and resets itself" (with 'moment' defined here in terms of B's definition of simultaneity). One or the other of those statements could be true, but they can't both be true in the same physical scenario.
 

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