3
$\begingroup$

I am currently studying length contraction in special relativity (and having the hardest time getting the harder exercises in my book right) and there is the following exercise in my book is driving me crazy (I apologize if the translation I am making isn’t the best, the book is in French):

A hollow spaceship A whose proper length is 300m travels at a speed of 0.8c towards the right as seen from planet B. A photon (v = c) is emitted from the front of the spaceship to the back (so towards the left) and a sensor detects it when it “hits” the other side of the spaceship.
Determine the time duration of the photon’s trajectory as seen by the planet, as well as the the length of this trajectory.

According to the answer key, because the photon and the spaceship are travelling in opposite directions, the speeds of each would add up, meaning we would do $c + 0.8 c = 1.8c$ and then simply divide the spaceship’s contracted length by this speed to get the time duration.

The fact that we are dealing with speeds greater than light greatly bothers me. Could someone please explain how this is an acceptable method?

$\endgroup$
  • 1
    $\begingroup$ If I see something traveling to the left at speed $c$ and something else traveling to the right at speed $c$, then I see the distance between them decreasing at a rate of $2c$, right? No issues here $\endgroup$ – Aaron Stevens May 27 at 5:06
4
$\begingroup$

Essentially, special relativity states that any body's speed with respect to you cannot exceed $c$. There are no restrictions on the relative velocities of 2 bodies from your frame of reference.

If we observed the photon from the frame of the sensor, it would still be $c$ - then, their relative velocities would not be added. Instead, length contraction occurs to give us the same result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.