Notes on Relativity and Cosmology - Physics Department, UCSB

88 CHAPTER 4. MINKOWSKIAN GEOMETRY
and if I compare coordinate systems (with one rotated relative to the other) I
find
∆x 2 1 + ∆y 2 1 = ∆x 2 2 + ∆y 2 2.
Let’s think about an analogous issue involving changing inertial frames. Consider,
for example, two inertial observers. Suppose that our friend flies by at
speed v. For simplicity, let us both choose the event where our worldlines intersect
to be t = 0. Let us now consider the event (on his worldline) where his
clock ‘ticks’ t f = T. Note that our friend assigns this event the position x f = 0
since she passes through it.
What coordinates do we assign? Our knowledge of time dilation tells us that
we assign a longer time: t us = T/ √ 1 − v 2 /c 2 . For position, recall that at
t us = 0 our friend was at the same place that we are (x us = 0). Therefore,
after moving at a speed v for a time t us = T/ √ 1 − v 2 /c 2 , our friend is at
x us = vt us = Tv/ √ 1 − v 2 /c 2 .
x = 0
s
t =
us
1
1 - v 2/c2
x = v t
us us
t = 1 sec
us
t = 1 sec
s
x = 0 us
t =0
us
Now, we’d like to examine a Pythagorean-like relation. Of course we can’t just
mix x and t in an algebraic expression since they have different units. But,
we have seen that x and ct do mix well! Thinking of the marked event where
our friend’s clock ticks, is it true that x 2 + (ct) 2 is the same in both reference
frames? Clearly no, since both of these terms are larger in our reference frame
than in our friend’s (x us > 0 and ct us > ct f )!
Just for fun, let’s calculate something similar, but slightly different. Let’s compare
x 2 f − (ct f) 2 and x 2 us − (ct us ) 2 . I know you don’t actually want to read
through lines of algebra here, so please stop reading and do this calculation
yourself.
⋆⋆ You did do the calculation, didn’t you?? If so, you found −c 2 T 2 in both
cases!!!!