giovedì 19 giugno 2014

Space-time diagram: The "explosion problem": relativistic Doppler effect

Suppose a situation as the one depicted here. Red dashed line is the path of light, the blue axes is the high speed frame, and the black axes is an Inertial frame of reference; green dots represent the signal:

Assumptions:

☛ Two reference systems: S (β=cost; inertial) , and S' (β=cost; β≅c)

 ☛ S' emits a light pulse every τ meters of time

☛ S' is moving wrt S with speed β≅c

 ☛ S listens to the signal emitted by S' (which, as said, has a fixed period τ), and measures different periods, depending on β (the speed of S').

☛ We need only τ, whose value as a coordinate refers to the S' system of reference, and β :

First, we set S time and space coordinates to {0,0} and consider that a first flash of light is sent by S' at that space-time position; so this "first flash" will happen at coordinates {0,0} of both S and S'. Now, let :


 be the 4-position vector, in the S' system, of the event "Light pulse sent out", which we will call "A";

 and also let :


be the 4-position of event A in the S system.

To get the coordinates of A {τ,0} in the S system, we just need to apply a "reverse" Lorentz transformation from S' to S:




BUT the first component of X is still NOT the time that A measures!

To get the total time, one has to add the time the light takes to cover the (increasing due to
β) distance from S.

So one can write:



This is because, as light speed is always =1, it takes exactly x time to get to position x (refer to the picture!); so light traces a 45° inclined worldline.

Now, rearranging the previous equation, we may write


 

that just resembles the acoustic Doppler effect:

this is the relativistic Doppler effect

We can get two very important relations from this picture:


-the shift of the period, from
τ to t, seen by S; it is also a shift in frequency. It is not fixed!
-the velocity β of S' as a function only of t and τ:
 

 Final, synthetic way of writing the whole thing:




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