Will this spaceship collide with the star? (time dilation)

Question

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?

"the event of the star exploding will actually take six years"...six years from when? — WillO, Dec 30 '20 at 00:21
from the start of the journey. In the ship's frame of reference, ofcourse — Vulgar Mechanick, Dec 30 '20 at 00:22
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. — WillO, Dec 30 '20 at 01:13
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? — Vulgar Mechanick, Dec 30 '20 at 02:01
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). — WillO, Dec 30 '20 at 02:06
You need to put more effort into this question to make it answerable, otherwise, it puts too much burden on the answerer. I suggest defining an Earth coordinate $(t, x)$, and say: the star explodes at (2023, 4) (while it's still (2020,0) here). Then we have something to Lorentz transform into the ship's frame. — JEB, Dec 30 '20 at 02:42
i am unfamiliar with coordinates like that. Could you please edit? — Vulgar Mechanick, Dec 30 '20 at 07:23
Where does 5.87 come from? I drew a Spacetime diagram (which really does clarify what is happening, even in the apparently simple cases) and pretty much agree with the conclusions of @WillO . — robphy, Jan 01 '21 at 05:22
Helpful hints. As @WillO strongly suggests, draw a position-vs-time graph (a spacetime diagram). It is an underused way to organize what is going on, as well as give proper meanings to formulas [which are often misused]. Additionally, when possible, avoid velocities like (1/2)c or (sqrt(3)/2)c and instead choose velocities like (3/5)c and (4/5)c , which lead to triangles with pythagorean triples and thus easier arithmetic. Otherwise, there is an unnecessary numerical layer (like 5.87) that obscures the physics. — robphy, Jan 03 '21 at 00:59

score 4 · Answer 1 · answered Jan 01 '21 at 06:04

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.)
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.

In the earth frame, these have the following coordinates:

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

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

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

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.

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? — Vulgar Mechanick, Jan 01 '21 at 06:58
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? — Vulgar Mechanick, Jan 01 '21 at 07:00

score 1 · Answer 2 · 2020-12-30T00:33:21.673

1

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.

edited Dec 30 '20 at 00:33

answered Dec 30 '20 at 00:27

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? – Vulgar Mechanick Dec 30 '20 at 00:47
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 – Dec 30 '20 at 01:02
unfortunately, I do not. I have only an introductory knowledge of SR – Vulgar Mechanick Dec 30 '20 at 01:14
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. – WillO Dec 30 '20 at 01:40
@WillO thanks, I hadn't check the numbers – Dec 30 '20 at 01:52

Michael Seifert · Answer 3 · 2021-01-06T16:49:13.670

1

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.

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:

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

The countdown to the collapse begins (Event $D$.)
The star collapses into a white dwarf (Event $B$.)
The spaceship passes the Earth (Event $A$.)
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".

edited Jan 06 '21 at 16:49

answered Jan 06 '21 at 16:43

Michael Seifert

48,772

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? – Vulgar Mechanick Jan 07 '21 at 00:40
@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. ... – Michael Seifert Jan 07 '21 at 01:41
... But in that regime, neither the formulas you’ve learned nor the Lorentz transformations directly apply. – Michael Seifert Jan 07 '21 at 01:42
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. – Vulgar Mechanick Jan 07 '21 at 10:10
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) – WillO Jan 08 '21 at 00:02
(CONTINUED) --- they see the photons that are arriving at that event. All you can ever see is a photon that's hitting your eye, and if you and I are at the same place and time, then the photons that hit our eyes are all going to come from the same place and time. So we will see exactly the same thing, and again it doesn't matter who says who is moving, who is accelerated, etc. Your frame has to do not with what you actually see but with how you interpret what you see. – WillO Jan 08 '21 at 00:02

Will this spaceship collide with the star? (time dilation)

3 Answers3