I was going over a question on my own, then I took a brief look at the solution...it's basically about

A rocket has a proper length of 250 m and travels at a speed v = 0.950c relative to the Earth. A missile is fired from the back of the rocket at a speed u′ = 0.900c relative to the rocket. Find the time it takes observers to calculate the time it takes for the missle to move to the front of the rocket.

When I was thinking about the question, it's pretty standard.

There are two ways about this question: Method 1:

-Work out the time required for the missle to reach the front relative to the rocket (so the rocket is the frame of reference) -Apply lorentz transform and find the time observed by people on Earth)

Method 2:

-Find gamma -Use addition laws to find rocket's speed relative to people on earth

Here's what I don't get:

I've worked round it and realized the correct way is to use Δx = γ(Δx′ + v Δt′). Then just use distance = speed * time to work out the time (which agreed with my original answer in method 1) What I don't understand is why do we not apply length contraction to the rocket when it clearly is travelling at relativistic speeds and hence appears shorter to observers on Earth? Technically the rocket should become shorter and hence time also shorterns. Is there something I'm clearly not understanding here?

  • $\begingroup$ Is it possible to think of it as both objects going through length contraction, ultimately, the measurement is the same in both cases? Since speed of both objects are still the same. $\endgroup$ – user51515 Apr 25 '16 at 19:20

There is no problem using length contraction, other than that it may take a couple of extra lines to arrive at the same result.

Just to have a reference, starting from $\Delta x = \gamma (\Delta x′ + v \Delta t′)$ and denoting ${\bar u}$ the velocity of the missile relative to the rocket, $u$ the velocity of the missile relative to Earth, and $L$ the proper length of the rocket, gives $$ t' = \frac{L}{\bar u}, \;\;\Delta x = \gamma (L + v \frac{L}{\bar u}) = ut \;\;\;\Rightarrow\;\;\; t = \gamma \frac{L}{u}\left( 1 + \frac{v}{\bar u}\right) $$ Now let's use the rocket's length contraction as seen from Earth. From the latter's frame, the distance traveled by the missile during time $t$, $\Delta x = ut$, is also the rocket's contracted length, $L/\gamma$, plus the distance traveled by the rocket's tip at velocity $v$ during the same time. This gives $$ \Delta x = \frac{L}{\gamma} + vt = ut \;\;\;\Rightarrow \;\;\; t = \frac{L}{\gamma (u - v)} $$ But from velocity addition $u = \frac{{\bar u} + v}{1 + \frac{{\bar u} v}{c^2}}$ and $$ u - v = \frac{{\bar u} + v}{1 + \frac{{\bar u} v}{c^2}} - v = \frac{{\bar u} + v - v - {\bar u} \frac{v^2}{c^2}}{1 + \frac{{\bar u} v}{c^2}} = \frac{\bar u}{\gamma^2} \frac{1}{1 + \frac{{\bar u} v}{c^2}} = \frac{\bar u}{\gamma^2} \frac{u}{{\bar u} + v} $$ Substitute the latter expression into $t$ and obtain $$ t = \frac{L}{\gamma (u - v)} = \frac{L}{\gamma}\frac{\gamma^2}{\bar u}\frac{{\bar u} + v}{u} = \gamma \frac{L}{u} \left( 1 + \frac{v}{\bar u}\right) $$ just as before.

  • $\begingroup$ thank you, udrv! I was going through a derivation similar to yours and I know where I went wrong now. $\endgroup$ – user51515 Apr 26 '16 at 7:39
  • $\begingroup$ Great! And welcome :) $\endgroup$ – udrv Apr 26 '16 at 13:55

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