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In the earth frame, there is a star ten light years from earth, and at rest with respect to the earth. In the earth frame, a rocket is passing the earth at 0.999999 times the speed of light in a vacuum. In the earth frame the clocks on the earth, star, and rocket all display "zero" at this time.

In the rocket frame, the star is (due to Lorentz contraction), (because gamma is 707.1) 5.16 light days from the rocket and the earth, and the star clock displays a little under "ten years", due to time desynchronization. In the rocket frame, the star takes just over 5.16 days to reach the rocket (which regards its frame as being at rest), during which time the star clock changes by a negligible amount, due to time dilation being so great (gamma is 707) and the journey time in the rocket frame being so short (just over 5.16 days). So it should still read less than "ten years".

The trip in the earth-star frame has taken just over ten years, so the clock on the star must read just over ten years when the rocket arrives at the star.

In the rocket frame, when the rocket arrives at the star, the star clock displays just over ten years, because that's how long the trip took in the earth-star frame, and now that the rocket is at the star, the display of the star clock in the two frames are the same.

This is the paradox. In the rocket frame how does the star clock get from just under ten years to just over ten years during just 5.16 light days of highly time-dilated running?

It seems helpful to consider what happens if the rocket makes the trip at a slightly lower speed. The star clock, in the rocket frame, initially displays a quantity that is even further below ten years. The journey time in the earth-star frame is greater than ten years by an even greater amount. But in the rocket frame the star clock still doesn't have a much different chance to advance, due to the combination of Lorentz contraction and time dilation. I mean, the journey now takes maybe 10 days, in the rocket frame, so the star clock will only advance (because gamma is 707.1) about 240 hours divided by say 353 which is about 0.69 hours.

If there are calculations that can disprove my intuitions, I would be glad to see them.

Edit: I don't know how I got 0.69 hours. That might not be right. Maybe it's a back of an envelope calculation gone wrong.

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    $\begingroup$ Have you drawn a picture? $\endgroup$
    – WillO
    Jul 26 at 23:51
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    $\begingroup$ Pro tip: Choosing nice numbers (abs(v/c)=0, 3/5, 4/5, 5/13, … ) leading to Pythagorean triples will lead to easier calculations (simpler rational values, especially for the Doppler factor) so one can focus on the physics, rather than be obscured by the numerical arithmetic. $\endgroup$
    – robphy
    Jul 27 at 20:18
  • $\begingroup$ @robphy +1 That makes sense. I think I would want to continue using extreme cases though, and I don't know what sort Pythagorean triple (I like that phrase BTW) would be appropriate for 0.999999 of the speed of light, nor whether it would still simplify the arithmetic with such an extreme case. I like extremes because they seem easier to visualize and think about nonmathematically, as Lewis Epstein teaches, although that was always my inclination (no pun intended). $\endgroup$ Aug 2 at 17:54
  • $\begingroup$ Pythagorean triple examples: 3-4-5, 5-12-13… as sides of triangles…since $3^2+4^2=5^2$. In special relativity, we use them as $5^2-3^2=4^2$. A calculator may not be needed with these. In your case, you need a calculator (especially when you use different units (years and days)). It’s better to simplify the math and arithmetic so you can focus on the physics. Once you understand the physics, then use more “realistic” or “interesting” values. In my opinion, using your numbers and multiple units unnecessarily complicates matters… and probably reduces the audience for your question. $\endgroup$
    – robphy
    Aug 2 at 18:06
  • $\begingroup$ @robphy Sure, 25 + 144 = 169 is easily calculated. I can do that in my head. I like not only the phrase, but also the idea of Pythagorean triples. Does Pythagoras's theorem work the same way as usual in SR? Why did you write the exact same fact with the four squared alone on the far side of the equals sign? Your other points are being pondered. $\endgroup$ Aug 2 at 18:23

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This question is related to your view that clocks 'jump' in the twin paradox- which is a misunderstanding. Observers on the rocket and the Earth disagree about when the star begins its journey.

Consider this analogy. You and I are are timing how long it takes for a climber to scale a distant mountain, the base of which is ill defined. We each look at the climber with a telescope that has a spirit level attached, so that when we see the climber in the centre of our telescopes in the horizontal position (ie level with us), we can take that as the start of the climb. If your spirit level is not adjusted in the same way as mine, so that your telescope points slightly upwards while mine is truly level, you will decide that the climber started the ascent later than I did, and therefore that the climb to the top took less time. There is no paradox here, it is just that you took the start of the climb to be one point, while I took it to be another lower down.

Likewise with your example of the star. The Rocket takes the start of the stars movement to be a point in time 5 days ago, while the Earth observer takes it to be a point in time ten years ago. What the rocket takes to be the start of the trip and what the Earth observers takes to be the start of the trip are two separate events nearly ten years apart.

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  • $\begingroup$ "Observers on the rocket and the Earth disagree about when the star begins its journey." Are you sure about that? $\endgroup$ Aug 2 at 17:59
  • $\begingroup$ absolutely! It is the relativity of simultaneity. $\endgroup$ Aug 2 at 19:28
  • $\begingroup$ Observers on the earth don't see the star make a journey, surely? $\endgroup$ Aug 2 at 20:19
  • $\begingroup$ They see the rocket moving, not the star. But the point you are missing is that the date on the star on which the rocket's journey starts in the frame of the Earth is not the date on the star on which the star's journey begins in the frame of the rocket. They are two different dates. $\endgroup$ Aug 2 at 20:55
  • $\begingroup$ Imagine the experiment happened in reverse, with the rocket leaving the distant star to head to earth today in the Earth's frame. We would say that the rocket has 'now' left the distant star. However, in the Rocket's frame, the date on the Earth when the rocket blasts off is 2032. $\endgroup$ Aug 2 at 21:00
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There is a star ten light years from earth, and at rest with respect to the earth. A rocket is passing the earth at 0.999999c. The clocks on the earth, star, and rocket all display "zero" at this time.

This is where the problem is. Since the rocket is moving with respect to the Earth and star, what is meant by "this time" depends on which reference frame you're talking about, as seen in the spacetime diagram below.

spacetime diagram

The red dashed line shows what it considered simultaneous in the Earth-star frame, while the blue line shows what is seen as simultaneous in the rocket frame. The reason why you calculated the time measured on the star to be less, when looking at the rocket frame is because you're starting the clock later, in that case. In other words, you're not measuring the same thing, thus causing the alleged "paradox."


In your edit, you said

In the rocket frame, the star takes just over 5.16 light days to reach the rocket (which regards its frame as being at rest), during which time the star clock changes by a negligible amount, due to time dilation being so great (gamma is 707) and the journey time in the rocket frame being so short (just over 5.16 light days).

This is incorrect. The time measured on the star clock is the proper time (a Lorentz invariant quantity) of the star's trajectory, which does not depend on reference frame. The proper time is given by $\tau^2=\Delta t^2-\left(\frac{\Delta x}{c}\right)^2$. Assuming that the star clock starts where the red $t=0$ line intersects with the star's worldline, then it will read a little over 10 years when the rocket reaches the star, regardless of the frame in which you calculate this.

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  • $\begingroup$ I have edited the question to remove all ambiguity as to which frame of reference is being referred to. "The reason why you calculated the time measured on the star to be less, when looking at the rocket frame is because you're starting the clock later, in that case." Less than what? I didn't use word "less" in the question. It's not clear to me what you are saying. $\endgroup$ Jul 27 at 14:25
  • $\begingroup$ @MatthewChristopherBartsh When I said the time measured on the star would be less, I was referring to the part where you said "during which time the star clock changes by a negligible amount," in which you seemed to suggest that the star clock would show less time when calculated in the rocket frame than when calculated in the Earth-star frame. $\endgroup$
    – Sandejo
    Jul 27 at 15:44
  • $\begingroup$ I see what you mean now. Yes, my intuition is that the combined effects (product?) in the rocket frame of contraction of trip distance and slowing of the star clock by time dilation, with near light speed travel by the star as a given, should not allow the star clock to change by the amount needed so when the star arrives at the rocket, the star clock shows the ten years plus a bit (which I just realized is not as big as I thought, only a millionth part :) ) that it must show given that we know that in the earth-star frame a less than light speed journey crossing ten light years has happened. $\endgroup$ Jul 27 at 16:38
  • $\begingroup$ I think I see where I was going wrong. I imagined that in the rocket frame, the star clock was initially a lot less than ten years, and that finally, the star clock was definitely a lot more than ten years. But a millionth of ten years is only 5.259 minutes. I had thought that it would be at least about an hour. Five minutes change of display on the star clock during five days doesn't go against my intuition. $\endgroup$ Jul 27 at 16:55
  • $\begingroup$ I should really try doing the calculations or even some sort of proof that in the rocket frame the initial time desynchronization display value plus the amount the star clock runs during five days adds up to the same amount as the time of the trip in the earth-star frame. But math is not my strong point. $\endgroup$ Jul 27 at 16:55
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Here is the picture you should have drawn (with the earth frame in black, the rocket frame in blue, and the star in red). (This is Sandejo's picture with a bit more detail.)

enter image description here

In the rocket frame, the trip takes .014142146 years, during which time the star clock advances by .00002 years.

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  • $\begingroup$ "In the rocket frame, the trip takes .014142146 light years, during which time the star clock advances by .00002 light years." Light years or years? $\endgroup$ Jul 27 at 14:32
  • $\begingroup$ @MatthewChristopherBartsh : Sorry, yes, years. Will fix. $\endgroup$
    – WillO
    Jul 27 at 14:34
  • $\begingroup$ "the trip takes .014142146 light years". "takes" or "has a length of"? $\endgroup$ Jul 27 at 14:39
  • $\begingroup$ @MatthewChristopherBartsh: Takes. Again, I'll edit to "years". If you send me email (my address is in my profile) I can send you some notes I think you'll find helpful. $\endgroup$
    – WillO
    Jul 27 at 14:42
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In the earth frame the clocks on the earth, star, and rocket all display "zero" at this time.

Please note that in the rocket's frame the simultaneity is not the same as in the Earth frame. Hence, from rocket's perspective, at the time when it passes the Earth, the time on the star's clock is not "zero", but something about (10 years - 5.16 days).

In other words, the two events (a) clocks on Earth set to zero, and (b) clocks on star set to zero - are not simultaneous in the rocket's frame.

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  • $\begingroup$ What formula can I use to get ten years minus 5.16 days like you did? $\endgroup$ Aug 2 at 18:33
  • $\begingroup$ @MatthewChristopherBartsh see the graph by WillO: all events that are simultaneous to "rocket clock set to zero" in the rocket's frame, belong to the line $t^\prime = 0$. The line $t^\prime = 0$ crosses line $x=10$ (where star is located) at the point corresponding to $t=9.99999$, which is (10.00001-0.00002) years according to WillO. Hence, I maybe was wrong saying that it is (10 years - 5.16 days). But anyway it is not $0$ years, as you assume in your question. $\endgroup$ Aug 3 at 11:06
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Here's a spacetime diagram in the spirit of @WillO and @Sandejo for the case of $(v/c)=3/5$ and $L=3$, first using light-clock diamonds (so we can calculate by counting) and then using hyperbolic-trigonometry (which can be generalized to your case of $(v/c)=(1-10^{-6})$ and $L=10$). See the associated Desmos url.

You seem to be interested in segments $XS$ and $XY$ and $XN$

"Think trigonometrically"... but rather than circular-trig functions, you'll use hyperbolic-ones.


$$v_{rocket}/c=\tanh\phi=\frac{OPP}{ADJ}=\frac{TS}{OT}=\frac{3}{5}$$ $$L=OX=TS=3$$ So, $$\fbox{$XS=OT=\frac{1}{(v/c)}TS =\frac{1}{(v/c)} L$} \qquad=\frac{5}{3}L =\frac{5}{3}(3)=5$$

Furthermore, $$\mbox{[time dilation factor]}\quad \gamma=\cosh\phi=\frac{ADJ}{HYP}=\frac{OT}{OS}=\frac{5}{4}$$

An important relation (Minkowski's version of Pythagorean theorem for a timelike hypotenuse): \begin{align} (ADJ)^2-(OPP)^2&=(HYP)^2\qquad &(5)^2-(3)^2=(4)^2\\ \\ (ADJ)^2-(ADJ \tanh\phi)^2&=(HYP)^2\\ \left(\frac{ADJ}{HYP}\right)^2-\left(\left(\frac{ADJ}{HYP}\right) \tanh\phi\right)^2&=(1)^2\\ \left(\cosh\phi\right)^2-\left(\cosh\phi \tanh\phi\right)^2&=(1)^2 \quad\longrightarrow\quad & \cosh\phi=\frac{1}{\sqrt{1-\tanh^2\phi}} \\ && \gamma=\frac{1}{\sqrt{1-(v/c)^2}} \end{align}

robphy-RRGP-MCBartsh

Note also that since $OS$ is Minkowski-perpendicular to $OY$,
then Minkowski-right-triangle $OXY$ is similar to triangle $OTS$.
So, $$\frac{XY}{OX}=\frac{TS}{OT}=v/c$$ and $$\frac{OX}{OY}=\frac{OT}{OS}=\gamma$$

Thus, $$\fbox{$XY=(v/c) OX= (v/c) L$} \qquad=\frac{3}{5}(3)=\frac{9}{5}=1.8$$ and [length-contraction] $$OY=\frac{1}{\gamma}OX=\frac{1}{\gamma}L \qquad=\frac{1}{\left(\frac{5}{4}\right)}3=\frac{12}{5}=2.4$$ Check (Minkowski's version of Pythagorean theorem for a spacelike hypotenuse):: \begin{align} (XY)^2 -(OX)^2 &=(OY)^2\\ \left(\frac{9}{5}\right)^2 -\left(\frac{15}{5}\right)^2 &=-\left(\frac{12}{5}\right)^2\\ \end{align}

Since $ON$ is a light-signal, then $\frac{c}{c}=\frac{OPP}{ADJ}=\frac{XN}{OX}$. Thus, $$\fbox{$XN=(1)OX=L$}.$$


The calculations are summarized in https://www.desmos.com/calculator/srlbwlgt79
I now invite you to modify $v/c$ to $0.999999 = (1- 10^{-6})$ and $L=10$. (If you use $0.999999$, then the Desmos-$v$ is a slider.)

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