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?

Answer 1

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.

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

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

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

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

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.

Answer 2

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.

Answer 3

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:

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".