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Take an earth-star and a spaceship frame. The spaceship is heading at near the speed of light towards the star from the earth. Now let's say that in the earth-star frame, there are three events

  1. the event that the spaceship passes the earth
  2. the event that the spaceship is destroyed by the exploding star
  3. the event that a clock on the star hits 10 years

We assume that in the earth-star frame, the spaceship is destroyed as soon as it reaches the star, and that exactly 10 years have passed by in this frame since event 1. That is, the spacetime interval is $(10,10v)$ between events 1) and 2). Moreover, the spacetime interval is $(0,0)$ between events 2) and 3).

Now here's the paradox. Let's switch to the spaceship frame. The spacetime interval is now $(\frac{10}{\gamma},0)$ between 1) and 2) and again $(0,0)$ between two and three. The spaceship is blown up exactly when the star clock reads 10 years. We have a problem now.

For 10 years passing by on the star clock in the earth-star frame, the spacetime interval is $(10,0)$, so in the spaceship frame, the interval is $(10 \gamma,10\gamma v)$. The clock on the star and earth moves by one factor of gamma slower. So at time of passing by, the star hasn't aged enough to explode, in fact it has only aged by $\frac{10}{\gamma^2}$ years.

What has gone wrong?

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2 Answers 2

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I'm not sure how you get from $10\gamma$ to $10/\gamma^2$. From the rocket's frame, the star's clock goes slower, which means that it registers less time than the rocket, which means that you divide by $\gamma$, which gets you back to $10$ years. You're just applying the inverse Lorentz transformation.

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  • $\begingroup$ it takes $10 \gamma$ years in the spaceship frame for the star clock to read $10$ years, yes? But the travel time according to the spaceship is $\frac{10}{\gamma}$. The total time elapsed on the star clock would then be $\frac{10}{\gamma^2}$ $\endgroup$
    – user277610
    Commented Oct 30, 2020 at 18:40
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First of all, from the ship's point of view, everything about the star is proceeding at a reduced rate of speed, relative to what the ship would expect for a star at rest in its frame. Not only is the clock running slow, but all physical processes, including the decay of the star, are running slow as well. This effect has been observed for muons. Cosmic ray muons traveling at relativistic speeds have a longer lifetime (in our reference frame) than muons produced at rest in a lab, due to time dilation.

Second, let's think about the state of the star when the ship passes the earth. Due to relativity of simultaneity, the state of the star "at the same time" is different in these two frames. In the earth frame, the star is "10 years before exploding" just as the ship passes. In the ship frame, the "star at the time the earth passes the ship" is actually at an earlier) time (ie, the clock on the star will be earlier than 10 years before exploding).

Finally, from the ships point of view, more than 10 years pass between passing earth and reaching the star (as you showed).

If you carefully combine the above three effects, you will see there is no contradiction and you can give a consistent account in any frame.

Edit Before I had that the star would be at a later time, but it should be earlier.

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  • $\begingroup$ how do you find this later time a priori? $\endgroup$
    – user277610
    Commented Oct 30, 2020 at 18:42
  • $\begingroup$ In the earth-star frame the clocks are synchronised so the interval is $(0, 10v)$ in the spaceship frame it would be $(0,-\frac{\gamma10v^2}{c^2})$. Does this make sense? $\endgroup$
    – user277610
    Commented Oct 30, 2020 at 18:48
  • $\begingroup$ Start from the ship frame. There is a spacelike interval between (a) the ship passing the earth, and (b) the star at the same instant in time. We can say this interval is $(0,L/\gamma)$, where $L$ is the distance from earth-to-star in the earth frame. Apply a Lorentz transformation to boost by $-v$ to go from the ship to the earth frame (where $v$ is the velocity of the ship). This leads to an interval $(-\gamma v L, L)$. In other words, the star is a distance $L$ away (of course), but its clock is at $-\gamma v L$. As you can see I got the timing wrong, and the clock is at an earlier time. $\endgroup$
    – Andrew
    Commented Oct 30, 2020 at 18:50
  • $\begingroup$ this just makes it worse. it only takes $\frac{10}{\gamma}$ years for the journey to occur in the spaceship frame. As I said, the star ages slower so the time elapsed is no way near 10 years when the journey end $\endgroup$
    – user277610
    Commented Oct 30, 2020 at 18:53
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    $\begingroup$ Ah yes, the clock will take $10 \gamma$ years to hit $10$ years. But it will have started at $t = -\frac{10 \gamma v^2}{c^2}$ so the sum of those times will be $\frac{10}{\gamma}$ as expected $\endgroup$
    – user277610
    Commented Oct 30, 2020 at 19:30

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