1
$\begingroup$

In a constant acceleration travel like depicted here, it is said that after some months of acceleration at 1g, in the ship reference frame, we would be going faster than C relative to Earth, potentially much faster:

a rocket could travel the diameter of our galaxy in about 12 years ship time, and about 113,000 years planetary time

If in the ship reference frame we are going at a far greater speed than C relative to Earth, does this mean that we actually see objects like stars and planets passing by the ship at a far greater speed than C ?

If yes, how can the rule "we always see the light going at C in our own reference frame" still be true ? Because that would mean we would see photons crossing us slower than the other objects we are passing by.

What am I missing ?

$\endgroup$
  • $\begingroup$ What am I missing ?" - length contraction in the spaceship's frame (the diameter of the galaxy is much smaller in direction of relative motion). $\endgroup$ – Alfred Centauri Jun 28 '18 at 12:28
  • $\begingroup$ @AlfredCentauri If you're traveling around the galaxy then the galaxy isn't in the direction of motion, it's normal to the direction of notion by the definition of circular motion. Images off to the side like this would be warped around the field of view and be subject to motion blur, but the trajectory at any given moment would still be subject to the effects that would prevent one from accelerating past C. $\endgroup$ – user158288 Jun 28 '18 at 12:39
  • $\begingroup$ @RayOfHope, I honestly don't see the relevance of your comment to my comment, or to this question, or to the linked article. $\endgroup$ – Alfred Centauri Jun 28 '18 at 12:51
0
$\begingroup$

Imagine that in the initial frame of reference, before the rocket begins its constant (linear) acceleration, there is a long line of regularly spaced 'mile posts' spaced 1 light-year apart in the direction of the spaceship's acceleration.

As the spaceship accelerates, the mile posts are observed from the spaceship to have ever increasing speed and ever decreasing spacing (due to length contraction).

Note: while the speed of the mile posts relative to the spaceship is ever increasing, the rate of increase of speed decreases such that relative speed approaches but never reaches $c$.

So, after having accelerated for some time, it can certainly be that 100,000 mile posts have passed by the spaceship after just 12 years have elapsed on the spaceship's clock.

Now, you seem to be concluding from this that, according to those in the spaceship, they have traveled 100,000 light-years in 12 years which implies a speed far greater than $c$.

But remember, due to length contraction, the mile posts are separated by far less than 1 light-year according to the rulers at rest in the spaceship's frame of reference. In other words, according to those in the spaceship, they've covered a distance less than 12 light-years in that 12 years of elapsed time.

From the linked article:

From the ship's frame, the acceleration would continue at the same rate. However, due to Lorentz contraction, the galaxy around the ship would appear to become squashed in the direction of travel, and a destination many light years away would appear to become much closer. Traveling to this destination at subluminal speeds would become practical for the onboard travellers. Ultimately, from the ship's frame, it would be possible to reach anywhere in the observable universe, without the ship ever accelerating to light speed.

$\endgroup$
  • $\begingroup$ Thank you, very clear answer ! I was supecting lenght contraction to be the answer but I wasn't sure... So in reality in a constant acceleration space travel, we never have the impression to travel faster than C relative to objects, we have the impression to travel at 0.99... C but we know that we are travelling faster (in ship time reference) because we are aware the distances are squished. $\endgroup$ – Kupay Jun 28 '18 at 13:36

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.