Special relativity two moving frames

[whatisreality]

Homework Statement

I'm finding this very hard to get my head round! There's earth, and a star which is 6ly away, in the same reference frame. A starship sets out from the star, and another ship leaves Earth at the same time. Each one has $v=0.6c$. What is the relative speed of the starship as seen by the Earth ship?

The starship explodes after it's a quarter of the way there. When do passengers on the Earth ship see the explosion? Their clock starts counting at departure.

Homework Equations

The Attempt at a Solution

For the first bit, from the velocity addition formula $v'=0.88c$.
Second bit: Really struggling. According to a hint we were given , to solve this, I need to work out when in the Earth and star's frame the Earth ship it sees the explosion, and where it is at that moment.

If $v'=0.88c$, then $\gamma = \frac{1}{1−0.88^2}$, and it sees 6ly as $\frac{6ly}{\gamma}$. So in the Earth ship's frame the starship traveled 0.3384ly of 1.3536ly when it explodes. The speed of light is invariant, so the light from the explosion reaches the Earth ship 0.3384 years = $1.07\times 10^7$ seconds later.

So this was what I had done before I was given the hint, and the hint says to do something completely different, because my way is wrong. Why is it wrong? And how am I supposed to work out when the Earth ship sees the explosion in the Earth's frame? Thanks for any help!

 

[PeroK]

Your initial distances and velocities are given in the Earth frame. It's easier, therefore, to do any calculation in this frame, then transform it to a different frame.

If you use the Earth ship frame, you'll first have to set out the problem in terms of distances and velocities in that frame. That's why the hint is a good one.

 

[phinds]

Just to add to what PeroK said, I have found it to not be helpful to think in terms of "moving frames". Basically, frames DON'T move. Things IN a frame move. So if your frame is a starship, then the Earth is moving and the other planet is moving and the other ship is moving and it all gets more complicated than if you just do things in the Earth's frame. I'm basically just restating what PeroK said, but slightly differently to emphasize that frames don't move.

 

[Doc Al]

If $v'=0.88c$, then $\gamma = \frac{1}{1−0.88^2}$, and it sees 6ly as $\frac{6ly}{\gamma}$.

That's true.

So in the Earth ship's frame the starship traveled 0.3384ly of 1.3536ly when it explodes.

That's the distance the starship traveled from its starting point in the Earth ship frame. OK.

The speed of light is invariant, so the light from the explosion reaches the Earth ship 0.3384 years = $1.07\times 10^7$ seconds later.

Rethink that calculation. You need to know how far the light traveled to reach the Earth ship, so you need to know how far from the Earth ship the explosion took place (not how far the ship was from its starting point). Then you could calculate the light travel time. You'd have to add in the time the explosion took place, according to Earth ship clocks. (Lots to consider, doing it that way. Don't forget the relativity of simultaneity.)

(I agree with PeroK that solving from another frame would be easier.)