Understanding twin paradox without math

In summary: The legend of the diagram says that the red line is the clock, not the blue lines. The blue lines says "Earth"
  • #1
obtronix
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TL;DR Summary
Understanding twin paradox without math
Using a simple time clock in the horizontal position you can see the solution to the twin paradox. In this graph the Time clock tics once every half year on earth. 5 years pass on Earth 4 years pass on the spaceship. You don't have to worry about mysterious simultaneity or time jumps, acceleration vs inertial frames.

Note: I received a warning about the source of this graph, I created it with Excel.

 
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  • #2
@robphy will probably be interested in this as it is a variation of his approach.
 
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  • #3
Nice. But one could argue that you're still using math: geometry.
 
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  • #4
Is the Earth located at 0.,25 ly and -0.25 ly?
I do not understand the vertical blue lines.
 
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  • #5
malawi_glenn said:
Is the Earth located at 0.,25 ly and -0.25 ly?
I do not understand the vertical blue lines.
Think of the blue lines as the extremes of position of a yellow swinging pendulum. The moving person also has a yellow pendulum, but with white lines as the extremes of motion.
 
  • #6
anorlunda said:
Think of the blue lines as the extremes of position of a yellow swinging pendulum.
I think the intent of the OP is that they are the opposing mirrors of a light clock.
 
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  • #7
malawi_glenn said:
Is the Earth located at 0.,25 ly and -0.25 ly?
I do not understand the vertical blue lines.

Earth is located at zero.
Vertical blue lines is the horizontal clock.
 
  • #8
obtronix said:
Earth is located at zero.
Vertical blue lines is the horizontal clock.
I presume both clocks are half a light second long on their rest frames, so tick once per year. How did you set the width of the white light clock in the frame shown?
 
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  • #9
Ibix said:
I presume both clocks are half a light second long on their rest frames, so tick once per year. How did you set the width of the white light clock in the frame shown?
For this diagram I just used the exact lorentz factor, but my point was any length contraction will show what's going on.

I suppose a non-math length contraction explanation is a topic for another post😃
 
  • #10
Mister T said:
Nice. But one could argue that you're still using math: geometry.

The better description is that it doesn’t require the usual algebraic calculations with gamma factors.

If you click on my avatar enough times, you can see the animated version of it.
 
  • #12
It shows what the Earthbound twin thinks happens, but I don't believe there was any confusion about that. The confusion is why the same sort of diagram, from the traveling-twin-centered view does not work. The fact that the traveling twin does not remain in a single inertial frame is given as a reason that his diagram does not apply. But IMO, that does not really explain why it is guaranteed to be wrong. The real proof that it must be wrong is that the Earthbound solution does apply and gives a different answer. (There are other, more complicated, methods of calculation from the traveling-twin-centered view that agree with the Earthbound answer.) EDIT: Still, in this simple analysis to prefer the Earthbound version over the traveling-twin-centered version requires a reference to "mysterious simultaneity or time jumps, acceleration vs inertial frames."

UPDATE: @obtronix shows what my mistake is in the next post. Thanks!
 
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  • #13
FactChecker,

Here's the spacecraft version, since the spacecraft is not in a single inertial frame you need two graphs, one from the outgoing spacecraft 's point of view and one from the incoming spacecraft 's point of view, you get the same answer in both.


 
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  • #14
obtronix said:
FactChecker,

Here's the spacecraft version, since the spacecraft is not in a single inertial frame you need two graphs, one from the outgoing spacecraft 's point of view and one from the incoming spacecraft 's point of view, you get the same answer in both.



I see. If I understand what you are saying, if you pick either the outgoing or incoming frames and stay with that, you get an answer that agrees with the Earthbound answer. Thanks!
 
  • #15
obtronix said:
Vertical blue lines is the horizontal clock

How can one object have two worldlines?
 
  • #16
malawi_glenn said:
How can one object have two worldlines?
Sure, if it is not a point object. Here the object has a worldline for the front and a worldline for the back.

Of course, one problem with this drawing is the discontinuity in those worldlines.
 
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  • #17
Dale said:
Here the object has a worldline for the front and a worldline for the back.
The legend of the diagram says that the red line is the clock, not the blue lines. The blue lines says "Earth"
Perhaps its just the labelings that confuses me.
 
  • #18
malawi_glenn said:
The legend of the diagram says that the red line is the clock, not the blue lines. The blue lines says "Earth"
The labeing is not very precise. The red line is the light ray that is bouncing back and forth between two mirrors in the clock. The blue lines are the mirrors of the Earth clock.
 
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  • #19
PeterDonis said:
The labeing is not very precise. The red line is the light ray that is bouncing back and forth between two mirrors in the clock. The blue lines are the mirrors of the Earth clock.
Yeah that makes more sense
 
  • #20
FactChecker said:
I see. If I understand what you are saying, if you pick either the outgoing or incoming frames and stay with that, you get an answer that agrees with the Earthbound answer. Thanks!
Yes as long as you take the measurements from an inertial frame anywhere you'll get the same result. Measurements from non-inertial frames are invalid.
 
  • #21
obtronix said:
Measurements from non-inertial frames are invalid.
This is not correct. Non-inertial frames work fine as long as you use the correct metric.
 
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  • #22
FactChecker said:
I see. If I understand what you are saying, if you pick either the outgoing or incoming frames and stay with that, you get an answer that agrees with the Earthbound answer. Thanks!
You can use any IRF and get the same answer. You can simply appeal to the invariance of the spacetime interval.
 
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  • #23
obtronix said:
Measurements from non-inertial frames are invalid.
Not invalid, simply messier to interpret and easy to get wrong through double counting some parts of spacetime and/or missing out others.
 
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  • #24
Measurements are made using measurement devices (such as clocks). They are not based on particular frames, inertial or not.
 
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  • #25
My takeaway is that there are good ways to use IRFs from either twin's perspective to get the correct answer. The confusion occurs when a purely mathematical use of higher-order derivatives are confused with the physical acceleration. Mathematically, there is complete symmetry between the Earth-centered and the traveling-twin-centered second derivative of position. The higher-order derivatives are just as "relative" as the velocity is. The only thing that really distinguishes the two perspectives is the acceleration of the traveling twin.
So there are good ways to get the correct answer that does not reference "mysterious simultaneity or time jumps, acceleration vs inertial frames". But that does not address the confusion of the paradox, which is caused by mistaking the second derivative of position (which is completely relative) with physical acceleration. The answer to the paradox does involve "mysterious simultaneity or time jumps, acceleration vs inertial frames".
 
  • #26
obtronix said:
Here's the spacecraft version, since the spacecraft is not in a single inertial frame you need two graphs, one from the outgoing spacecraft 's point of view and one from the incoming spacecraft 's point of view, you get the same answer in both.
Very clear, thank you!
For your diagrams v/c=0.6?
How did you set the horizontal spacing between the mirror lines ?
 
  • #27

Kairos said:
Very clear, thank you!
For your diagrams v/c=0.6?
How did you set the horizontal spacing between the mirror lines ?
yes, .6c, spacing between mirrors was what the inertial Observer observed using length contraction, then I manually drew the Light Beam at a 45° angle until it hit one of the mirrors
 
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  • #28
Kairos said:
Very clear, thank you!
For your diagrams v/c=0.6?
How did you set the horizontal spacing between the mirror lines ?
You can figure it out by having the Earth twin bounce two pulses off the traveling twin. If they are emitted time ##\delta t## apart, are reflected ##\delta t'## apart and return ##\delta T## apart (primed quantity measured by the traveller, others by the stay-at-home) then the principle of relativity says that $$\frac{\delta t'}{\delta t}=\frac{\delta T}{\delta t'}$$Then you declare that ##\delta t'=\sqrt{\delta t\delta T}## is your unit of time. Then draw mirrors for the traveller so that the bounce time is the same as the spacing between the reflection events, and draw mirrors for the stay at home so that the bounce time is the square root of the product of the spacings between emission and reflection.

Not quite no maths, but apart from the product and the square root I think everything else can be done with a ruler.
 
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  • #29
obtronix said:
FactChecker,

Here's the spacecraft version, since the spacecraft is not in a single inertial frame you need two graphs, one from the outgoing spacecraft 's point of view and one from the incoming spacecraft 's point of view, you get the same answer in both.



Hi @obtronix
The outgoing frame shows both departure and reunion of the twins and the kink in the worldline would be the traveling twin jumping to a passing incoming ship is that correct?
 
  • #30
DAH said:
Hi @obtronix
The outgoing frame shows both departure and reunion of the twins and the kink in the worldline would be the traveling twin jumping to a passing incoming ship is that correct?
No, the perspective is from a third party who's always in the outgoing inertial frame observing both the Earth twin and the spaceship twin.

He is observing the twin in the spaceship make a turn, the kink is instantaneous acceleration of .6c + .6c relative to his reference frame.
 
  • #31
obtronix said:
the kink is instantaneous acceleration
Which is not physically possible; that might be why it's confusing people. The twin's proper acceleration at the turnaround can't be infinite, but that's what a kink in the worldline describes. A finite proper acceleration requires a smooth, rounded-off "corner" in the worldline. So strictly speaking, if you're going to draw a kink in your diagram, it should be because you are assuming that the proper acceleration at turnaround is so large that, at the scale of the diagram, the smooth rounded-off corner looks like a kink.
 
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  • #32
PeterDonis said:
Which is not physically possible; that might be why it's confusing people. The twin's proper acceleration at the turnaround can't be infinite, but that's what a kink in the worldline describes. A finite proper acceleration requires a smooth, rounded-off "corner" in the worldline. So strictly speaking, if you're going to draw a kink in your diagram, it should be because you are assuming that the proper acceleration at turnaround is so large that, at the scale of the diagram, the smooth rounded-off corner looks like a kink.
Literally every twin paradox example assumes infinite acceleration that should not confuse anybody, realistic acceleration just complicates the whole issue.
 
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  • #33
I think that is OK. We often assume instantaneous acceleration. I think that the discontinuity is a bigger problem.
 
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  • #34
Dale said:
I think that is OK. We often assume instantaneous acceleration. I think that the discontinuity is a bigger problem.
What discontinuity?
 
  • #35
obtronix said:
realistic acceleration just complicates the whole issue
I'm not saying you need to actually do the math or draw the diagram using a realistic acceleration. I'm just saying that, since the kink in the traveling twin worldline seems to be confusing some people, it might help to make clear that it's not meant as a claim that the twin's acceleration is literally infinite, only that it is large enough that, for purposes of this discussion, the kink in the worldline is an acceptable approximation to the actual rounded corner.
 
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