3

Say I am in a spaceship. Another ship moves relative to me along a direction directly away from me, with a relative velocity(comparable to c) which I determine by measuring the doppler effect. My spaceship comes with a very large coordinate system attached to it, with the coordinates written at some of the points, and synchronized clocks kept at these points. The other ship moves between two such points.

The passing of the ship by these points constitute two events, and I directly obtain the time interval and space interval between these points.

So now If I use the velocity = displacement/time formula to indirectly obtain the time, do I get the proper time(which I think a clock on the other ship could measure) or will I get the improper time(which I think my system of clocks will have given me)? I am yet to start university physics so I may be wrong about some things. Feel free to ask for any clarification.

Raghav Gupta
  • 123
  • 5

1 Answers1

6

Let's call the two points $A$ and $B$, and we'll take them to be a distance $dx$ apart as measured in your frame.

In your frame you measure the other spaceship to take a time $dt$ to get from $A$ to $B$, and since the other spaceship is travelling at a velocity $v$ relative to you that time is given by:

$$ dx = v~dt \tag{1} $$

The proper time, $d\tau$, for the trajectory between the two points is given by the Minkowski metric:

$$ c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 \tag{2} $$

and we will assume all motion is along the $x$ axis so $dy = dz = 0$. Then substituting $dx = vdt$ from (1) we find the proper time is simply:

$$ d\tau = dt\sqrt{1 - v^2/c^2} = \frac{dt}{\gamma} \tag{3} $$

So as I'm sure you suspected just dividing the distance by the velocity does not give you the proper time, it gives you the coordinate time.

It is worth considering what happens in the rest frame of the moving spaceship. In this frame the spaceship is stationary at the origin so it doesn't move in space and $dx'=0$. The point $A$ passes the spaceship at some point $(t' 0)$ and the point $B$ passes at the same point in space but a later time $(t' + dt', 0)$. Then using the metric (2) to calculate the proper time simply gives:

$$ d\tau = dt' \tag{4} $$

So the proper time between the two spacetime points is equal to the time recorded by the spaceship. This is a general result in relativity: the proper time is the elapsed time for the inertial observer travelling between the two points.

And one last point while we're here. If we equate the expressions (3) and (4) for the proper time we get:

$$ dt' = \frac{dt}{\gamma} \tag{5} $$

which is the well known expression for the time dilation of a moving observer.

John Rennie
  • 367,598