Twin paradox: who decided who is the younger one

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SUMMARY

The twin paradox illustrates the effects of time dilation as described by Einstein's Special Relativity. In this scenario, one twin remains stationary while the other travels at 0.99c, experiencing acceleration and deceleration. The traveling twin ages less due to the asymmetrical nature of their journeys, as they change inertial frames, while the stationary twin remains in a single inertial frame. The discussion emphasizes that time dilation occurs due to high speeds in an inertial frame, not merely acceleration.

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  • #61
harrylin said:
Easy: if one accelerates relative to nothing, there would also be nothing to cause an effect from it.
In that sense an empty universe has still the geometrical property to define geodesics. So you feel acceleration w.r.t. these geodesics (I guess this is not what Mach had in mind).
 
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  • #62
bobc2 said:
Hi, ghwellsjr. Here is the short story. Example a) is a spacetime diagram with black rest frame and blue frame moving relative to rest frame. But you cannot compare times between the black frame and the blue frame. Example b) uses the hyperbolic calibration curves which allow you to compare times between t and t'. And you can see how much time dilation there is for the blue guy looking at a clock along black's world line (t axis). When blue's calendar says 30 years, he "sees" (correcting for light travel time, etc.) black's calendar showing about 26 years. You can measure the slope of blue's time axis to see how fast he is moving with respect to the black rest system.

By the way, notice that the X1 axis of blue is tangent to the hyperbolic curve at the time point of interest.
Hyperbolic_Calib3.jpg
Thanks again, Bob, for putting your time into making these graphics. I now understand what the calibration curve is for and how it is used.

I get the impression that back in the "old" days, before computers or even calculators, there must have been preprinted Minkowski diagrams available with the calibration curves already in place so that the user could label the black axes, draw in his sloping blue axis for whatever β he was interested in, and then he could easily use the calibration curves to label his blue axis--all without doing any calculation except determining the slope of the blue axis, which he could get from a lookup table (along with γ and its reciprocal).

But we have computers now which I'm sure you used to calculate and plot the calibration curves which is more work than simply plotting the blue axis with its appropriate labels.

You point out that the tangent of the calibration curve at the blue axis allows you to easily see the time dilation on the black axis but it is even easier to see if you start at the 30-year point on the black axis and just look at the horizontal (tangent) line going over to the blue axis and see the time dilation there 26 years. And once you know that, you also know that at 30 years for blue, he will "see" 26 years for black.

If the whole purpose of this is to graphically show on a Minkowski the reciprocal nature of time dilation, then why didn't you point this out?

However, all the same things can be shown using just the Lorentz Transform (which is the source of the information that gets drawn on a Minkowski diagram) so why not just stick with the exact numbers that you get from the Lorentz Transform, now that we all have computers and calculators? They work even when the values of β are close to zero or close to one where the Minkowski diagram becomes very difficult to evaluate.
 
  • #63
ghwellsjr said:
Thanks again, Bob, for putting your time into making these graphics. I now understand what the calibration curve is for and how it is used.

And thanks for your ideas on this subject.

ghwellsjr said:
I get the impression that back in the "old" days, before computers or even calculators, there must have been preprinted Minkowski diagrams available with the calibration curves already in place so that the user could label the black axes, draw in his sloping blue axis for whatever β he was interested in, and then he could easily use the calibration curves to label his blue axis--all without doing any calculation except determining the slope of the blue axis, which he could get from a lookup table (along with γ and its reciprocal).

But we have computers now which I'm sure you used to calculate and plot the calibration curves which is more work than simply plotting the blue axis with its appropriate labels.

I'll bet you are right about that. And yes, I used MatLab to do the math, then copied and pasted into Microsoft Paint to add a couple of things.

ghwellsjr said:
You point out that the tangent of the calibration curve at the blue axis allows you to easily see the time dilation on the black axis but it is even easier to see if you start at the 30-year point on the black axis and just look at the horizontal (tangent) line going over to the blue axis and see the time dilation there 26 years. And once you know that, you also know that at 30 years for blue, he will "see" 26 years for black.

A really good point. Thanks for pointing that out.

ghwellsjr said:
If the whole purpose of this is to graphically show on a Minkowski the reciprocal nature of time dilation, then why didn't you point this out?

I don't know. I guess I was originally more focused on using the diagram to emphasize it's representation of the 4-dimensional continuum and possibility of viewing objects as 4-dimensional as well.

ghwellsjr said:
However, all the same things can be shown using just the Lorentz Transform (which is the source of the information that gets drawn on a Minkowski diagram) so why not just stick with the exact numbers that you get from the Lorentz Transform, now that we all have computers and calculators? They work even when the values of β are close to zero or close to one where the Minkowski diagram becomes very difficult to evaluate.

You have a good point there. If space-time diagrams don't do anything for folks, then just stick to the calculations as you say.
 

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