0
$\begingroup$

Suppose that i am stopped wrt earth, facing a straight line on the ground. A spaceship, whose proper length is $12m$, is approaching at a speed of $0,8c$, where $c$ is the speed of light. I want to know how long it takes to the spaceship cross the line, on both my frame and in the pilot's frame. For me, the spaceship is $7,2m$ long, so i would just measure: $\Delta t = \frac{7,2m}{0,8c} \approx 30 ns$. For the spaceship, the pilot would have measured $\Delta t' = \frac{12m}{0,8c} \approx 50ns$.

I'm having a problem determining the time intervals using a Lorentz transformation. I could just say that $30ns$ is the proper time, and so $ \Delta t' = \gamma \times 30 = 50ns$ where $\gamma$ is the Lorentz factor. But that leaves me with a question: Shouldn't the time be moving slower in the spaceship?

$\endgroup$
2
  • $\begingroup$ There is no problem : if time is moving slower in the pilot's frame, then the time interval measured in that frame will be smaller than the time interval measured in your frame. $\endgroup$
    – Toool
    Commented Nov 23, 2017 at 15:11
  • $\begingroup$ That is the problem. The time measured by him is $50ns$ which is greater than $30ns$. The time is moving faster for him. Why? $\endgroup$ Commented Nov 23, 2017 at 15:48

2 Answers 2

1
$\begingroup$

The two events that you are concerned with are

  1. The front of the ship passes the line, and
  2. The back of the ship passes the line

The proper time between two events is, by definition, the time between the events as measured by an observer for whom the events take place in the same location. The proper time is therefore measured by you, the stationary observer, rather than the observer on the ship.

"Time goes slower for people in motion" is a roughly correct basis for intuition, but you should always be able to frame a question like this in terms of spacetime events, which usually makes the solution fairly clear.

$\endgroup$
-1
$\begingroup$

Ok I just realized my previous answer was erroneous.

In the observer frame, the length of the spaceship is improper, and the time it takes to cross the line is proper.

So it's normal that the time interval in the observer frame is smaller than the time interval in the pilot's frame.

$\endgroup$
1
  • $\begingroup$ I would also like to know the reason to the downvote. Besides, can you give me a reference for this rule of yours? Thanks. $\endgroup$ Commented Nov 24, 2017 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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