I thought of the above thought experiment and arrive on 2 conflicting conclusions. I can't seem to identify the flaw in my reasoning.

Suppose there is a star 4 light years from earth that has will explode and turn into a white dwarf in 3 years (as measured in the earth frame). A spaceship travels to the star at 86% the speed of light.

According to earth's frame of reference, the journey to the star will take 4.5 years so the star will have turned into a white dwarf.

according to the ship's frame of reference, however, the journey will only take 2.25 years. Moreover, since the star is travelling relative to the ship in its own frame, the event of the star exploding will actually take 6 years. So the spaceship will collide with a white dwarf instead of a star.

I thought this had something to do with simultaneity, but I know that the events must be same in all frames of reference. My collusions imply that collisions occur between different bodies depending on the frame of reference, which can't possibly be true, can it? Where am I going wrong?

  • $\begingroup$ "the event of the star exploding will actually take six years"...six years from when? $\endgroup$
    – WillO
    Dec 30, 2020 at 0:21
  • $\begingroup$ from the start of the journey. In the ship's frame of reference, ofcourse $\endgroup$ Dec 30, 2020 at 0:22
  • 2
    $\begingroup$ Then there is of course no reason for the time interval to be six years. You have been asking a lot of questions that would answer themselves if you took a moment to draw the appropriate spacetime diagram. It's a skill worth acquiring. $\endgroup$
    – WillO
    Dec 30, 2020 at 1:13
  • $\begingroup$ why is that interval not 6 years? are you referring to the round off from 5.87? I shouldn't really need a spacetime diagram for such a simple calculation. Maybe im confusing something, mind pointing that out? $\endgroup$ Dec 30, 2020 at 2:01
  • $\begingroup$ The interval is negative. In the traveler's frame, the star explodes before the ship starts its journey (approximately 1/3 of a year before). $\endgroup$
    – WillO
    Dec 30, 2020 at 2:06

3 Answers 3

  1. In the earth frame, there is a star 4 light years from earth, due to turn into a white dwarf in 3 years. A spaceship travels to the start at speed $v=\sqrt3/2\approx .866$. (All speeds are stated as multiples of the speed of light.)

  2. When the spaceship leaves earth, let's say the earth observer and the spaceship observer both set their clocks to zero. Here are three events:

A: The spaceship leaves earth.

B: The star becomes a white dwarf.

C: The spaceship reaches the star.

  1. In the earth frame, these have the following coordinates:

A: $t=0$, $x=0$

B: $t=3$, $x=4$

C: $t=4/v\approx 4.62$, $x=4$

  1. Lorentz transform to the ship frame:

A: $t'=0$, $x'=0$

B: $t'=6-4\sqrt3\approx -.928$, $x'=8-3\sqrt3\approx 2.804$

C: $t'=4/\sqrt3\approx 2.309$, $x'=0$

As you can see, the order of events in the earth frame is first $A$, then $B$, then $C$ (the ship leaves earth, then the star explodes, then the ship reaches the star). The order of events in the ship frame is first $B$ (occurring .928 years before $A$), then $A$, then $C$ (the star explodes, then the ship leaves earth, then the ship reaches the star).

So everyone agrees that the star explodes before the ship gets there. We should have known this even before we calculated because $B$ and $C$ are spacelike separated.

This is essentially an elementary homework problem and I'm mildly ashamed of myself for answering it, but I want you to see that if you took half the time you've been spending on these questions and spent it learning relativity, you'd easily be able to do this for yourself.

  • $\begingroup$ I am in high school and the special relativity taught here is only uptil the 2 equations for length contraction and time dilation(along with relativistic dynamics). Can this question not be answered using them only? Is learning lorentz transformation necessary for this question? $\endgroup$ Jan 1, 2021 at 6:58
  • $\begingroup$ Moreover, the spaceship started out in a frame of reference where the star hadn’t exploded and ended up in a frame where it already had exploded. Since the transition can’t be instantaneous, does thus mean that an observer in the spacehip saw the lifetime of the star in “fast-forward” during acceleration? $\endgroup$ Jan 1, 2021 at 7:00
  • $\begingroup$ 1) Yes, you really need the Lorentz transformations. Fortunately, they are easy: $x'=(x-vt)/\sqrt{1-v^2}$ and $t'=(t-vx)/\sqrt{1-v^2}$. You can check that these are the formulas I used to get from the earth coordinates to the ship coordinates. $\endgroup$
    – WillO
    Jan 1, 2021 at 7:21
  • $\begingroup$ 2) Suppose you are facing north, and the north pole is 3000 miles straight ahead. Then you turn ninety degrees, and suddenly the north pole is 3000 miles to the right. You started off using a frame where the pole was in front of you and ended up with a frame where the pole was to your right. Does that mean that you saw the pole travel thousands of miles during your turn-around? I guess in some sense the answer is yes, but that's generally NOT a useful way to think of it. It's simpler to say that you didn't see the pole move; you just changed the frame you use to describe it. $\endgroup$
    – WillO
    Jan 1, 2021 at 7:23
  • $\begingroup$ 2) continued. What's happening with the spaceship and the star is exactly like what's happening with you and the pole. The star explodes. You describe that as happening in the future, just as you describe the pole as being in front of you. You change frames. (Changing your velocity in spacetime is exactly analogous to facing a new direction in space.) Now you have a new language for describing the times of events just as when you turn around on earth, you have a new language for describing where the pole is. If you want to call that the pole moving, that's up to you. $\endgroup$
    – WillO
    Jan 1, 2021 at 7:25

It is a synchronization problem. When the two observers synchronize their clocks at earth (Let us assume the traveler accelerates almost instantaneously so it reaches full speed still at earth), the star that is current for the traveler is in the future for the person on earth. So for the traveler, the time left to explode is smaller than otherwise. Imagine the star becomes a red giant before exploding. This means that at synchronization time, it is possible that for the earth observer the star is still in normal stage, but for the traveler it will be already in red giant stage. When they synchronize their clock at earth, far away objects will still be out of synchrony. So the star will explode before the spaceship will collide, for both observers, of course. Both observers will agree on a single event in spacetime, such a the collision between two objects.

  • $\begingroup$ So in about 2.25 years, the ship observed an event that took 3 years in the earth's frame. This implies that time flowed faster for the ship's frame s compared to the proper frame and goes against my intuition of time dilation slowing down time. Moreover, does this mean that at ultra-relativistiic velocities, I could essentially see the entire universe unfold in a matter of years, days or even minutes? $\endgroup$ Dec 30, 2020 at 0:47
  • $\begingroup$ time dilation does slow time, but there is also an initial lack of synchrony that overcomes it. It is easy to see in a minkowski diagram, do you understand them? you have a dilated time, but that time is already shorter, when the traveller starts the trip, to him the star is already a red giant. His present is not the same than for people at earth $\endgroup$
    – user65081
    Dec 30, 2020 at 1:02
  • $\begingroup$ unfortunately, I do not. I have only an introductory knowledge of SR $\endgroup$ Dec 30, 2020 at 1:14
  • $\begingroup$ Actually---the numbers in the OP don't quite make sense (the assumption about velocity is not exactly consistent with the assumption about length of journey, etc) but if they are even approximately correct, then in the traveler's frame, the star has already exploded even before the traveler leaves earth, not just before he gets to where the star used to be. $\endgroup$
    – WillO
    Dec 30, 2020 at 1:40
  • $\begingroup$ @WillO thanks, I hadn't check the numbers $\endgroup$
    – user65081
    Dec 30, 2020 at 1:52

Rule #1 when you can't figure out what's going on in a special relativity problem: draw a spacetime diagram. This is, by far, the best way to understand what's going on in most special relativity paradoxes. If you want to actually understand what's going on with special relativity, rather than simply learning a few counterintuitive formulas, I highly recommend learning about spacetime diagrams. Here's a good set of lecture notes from Stanford about how to construct them and how to interpret them. They're not hard (a high-school student such as yourself should be able to understand them), and they make everything so much clearer.

For your situation, we denote the events as follows:

  • Event $A$: spaceship passes Earth. (We'll assume, for simplicity, that it's zipping past Earth rather than accelerating from rest.)
  • Event $B$: the star collapses into a white dwarf.
  • Event $C$: the spaceship passes the location of the star/white dwarf.
  • Event $D$: the "countdown" until the star's collapse begins. In the Earth's frame, this countdown takes three years.

enter image description here

What we see, when we construct this diagram, is that the events you're concerned about happen in a different order in the spaceship frame. In the Earth's frame, they happen as follows:

  1. The spaceship passes the Earth (Event $A$). Simultaneously, the countdown to the star's collapse begins (Event $D$).

  2. The star collapses (Event $B$.)

  3. The spaceship passes the location of the star, now a white dwarf (Event $C$.)

But in the spaceship's frame, the events happen in the following order:

  1. The countdown to the collapse begins (Event $D$.)

  2. The star collapses into a white dwarf (Event $B$.)

  3. The spaceship passes the Earth (Event $A$.)

  4. The spaceship passes the location of the star, now a white dwarf (Event $C$.)

In the ship's reference frame, the "countdown" to the collapse does in fact take much longer than the journey from the Earth to the star's location. We can see this from the diagram — there's much more distance along the $t'$ axis between $D$ and $B$ than there is between $A$ and $C$. But in the ship's reference frame, the countdown also starts (and ends!) well before the ship passes the Earth. So there's no paradox with events happening "out of order".

  • $\begingroup$ When you refer to the "ship's frame" at the end, do you mean its final intertial frame or referring to the total time felt by the spaceship(including acceleration). What bugs me the most is that the ship started out in a frame where the star hadn't exploded and ends up in one where it already has. Since this process is gradual, not instantaneous, does that mean the an observer on the ship essentially saw the evolution of the start in a smaller time interval than an observer on earth? $\endgroup$ Jan 7, 2021 at 0:40
  • $\begingroup$ @OVERWOOTCH: my answer assumes that the spaceship never accelerated — it was always moving at a fixed velocity, and passed the Earth at $x=t=0$. Talking about accelerated observers, and what they actually observe, is much trickier. In particular, defining which events are simultaneous according to an accelerating observer is much harder; see the “accelerated observers” section in this Wikipedia article. ... $\endgroup$ Jan 7, 2021 at 1:41
  • $\begingroup$ ... But in that regime, neither the formulas you’ve learned nor the Lorentz transformations directly apply. $\endgroup$ Jan 7, 2021 at 1:42
  • $\begingroup$ I see. All of these questions arise from a hard sci-fi novel (tau zero) where a ship accelerates closer and closer to lightspeed and reaches the end of the unverse in a couple of decades. It essentially sees the universe grow old in fast forward, which is completely goes against my intuition of time dilation slowing down time for the observer. $\endgroup$ Jan 7, 2021 at 10:10
  • $\begingroup$ Overwootch: Regarding what your accelerated observer "sees", you have to be careful about what "see" means. Do you really mean what the observer is actually seeing, or do you mean what the observer is inferring from what he sees? In relativity, "seeing" is, perhaps confusingly, generally mean to mean the latter, but you seem (I think) to be using it to mean the former. In that case, multiple observers at the same event --- some moving quickly relative to others, some accelerated, others not, whatever you want to assume --- all see exactly the same thing (CONTINUED) $\endgroup$
    – WillO
    Jan 8, 2021 at 0:02

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