If K is a set of coordinates at rest, and K' is a set of coordinates
in motion relative to K in the +x direction with a speed of v, then
from K we see that x'=x-vt. Since we are also going to consider x and
x' from K', we will label values in the equation considered from K as
x1', x1, and t1, etc.
x(1)'=x(1) - vt(1)
Since there is relativity of time, from K' we have a different
perspective of distance between the origin of K and the origin of K'.
From K', K is in motion with speed v in the -x direction.
x(2) = x(2)' + vt(2)'
As seen from K' the distance between origins is vt(2)'.
So if we are talking about a distance common to both frames of
reference, ie., x(1)=x(2), then t(1) will not equal t(2)', meaning
vt(1)does not equal
vt(2)'. Consequently, to compare x and x', we have to use x as a
common distance in both frames of reference, and time as seen from K'.
The time in K' relates to time in K as it relates to x where x =ct(1)
= ct(2), the distance light travels in a time of t=t(1)=t(2).
So if a photon is emitted at x=0,x'=0 when the origins of K and
K' coincide, Then after a time of t, the photon will have gone a
distance of x=ct as seen from K. Suppose we place a mirror at a
distance of x= 1 light year from the origin of K, the we see that a
photon will reach the mirror at a time of 1 year. As seen from K', we
have
1 light year = ct(2)' + v t(2)'
1 = (1)t(2)' + vt(2)'
So suppose that v = .5c.
1= 1.5 t(2)'
t(2)' = 2/3 year.
So in K a photon emitted at x=0, t=0, reaches the mirror in 1 year, in
K', the same photon reaches the mirror in 2/3 year. The mirror is at
x(2)'= 2/3 light year when the photon reaches it.
So the photon is reflected and starts back toward the origins of
K and K'.
The photon has to travel a distance of 2/3 light year in K'. It will
take a time of t'= 2/3 year measured in K'.
Robert B. Winn
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