Re: Question about light clock and derivation of time dilation
- From: "Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 03 Jul 2005 15:20:40 GMT
<john_doe_ph_d@xxxxxxxxx> wrote in message news:1120401164.004302.67760@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Time dilation is typically the first SR effect derived in text books,
> and it is usually done with the example of a light clock as follows,
> where S is the source and M is a mirror. Since it appears to be so
> important, I'm trying to understand every aspect of this example.
>
> --- M
>
> ^
> |
> |
> |
> |
> |
> |
>
> | | S
>
> The well-known derivation is based on the fact that, to an observer B
> moving horizontally w.r.t. the clock, the light is taking a longer path
> from S to M as shown below:
>
> --- --- ---
> ^
> |
> ^
> |
>
> ^
> |
> | | | | | |
>
> But the speed of light is constant and dimensions perpendicular to the
> direction of motion don't change. Therefore, B concludes that the time
> for the light to go from S to M is longer than the time determined by
> an observer stationary w.r.t. the clock.
>
> Fine, but I am trying to reconcile B's observation of the light path
> with the fact that the light source is pointed in the vertical
> direction. Doesn't that tell B that the light is "really" just moving
> up and down? Wouldn't B have to see the light source tilted to make
> sense of his observation of the light path?
>
> To make this more concrete, let's suppose that the source S and mirror
> M are connected with a narrow fiber optic (with some scattering
> material included so that B can still observe the light). How can B
> reconcile the observation that the light path is tilted with his
> knowledge that the light is confined to move in the fiber? Doesn't he
> "really" know that the light is just moving straight up and down?
According to B, each light signal follows a titlted path,
but the fiber and the bean remain vertical.
With the standard setup the Lorentz transformation is given by
Transformation equations:
t' = g ( t - v x )
x' = g ( x - v t )
y' = y
or the inverse:
t = g ( t' + v x' )
x = g ( x + v t' )
y = y'
where g = 1/sqrt( 1 - v^2 )
(1) World line of upward light signal in rest frame of clock:
x = 0
y = c t
This is a vertiacal line in the (x,y)-plane
Transformed world line of light signal in restframe of B:
g ( x' + v t' ) = 0
y' = c g ( t' + v x' )
or, cleaned up and cast into standard formulation:
x' = -v t'
y' = c / g t'
or, with the parameter t' eliminated:
y' = -v c/g x'
This is a tilted line with slope -v c / g in the (x',y') plane.
(2) Equation of the beam (or the fiber) in the restframe of the clock:
x = 0
This is a vertical line in the (x,y)-plane.
Transformed equation of beam in restframe of B
g ( x' + v t' ) = 0
or, cleaned up and cast into standard formulation:
x' = -v t'
This is the equation of a vertical line in the (x',y') plane.
Dirk Vdm
.
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