As in S we have
eiω(t+(x/c) cos α+(y/c) sin α) ,
ω
where ~k = − (cos α, sin α, 0) ,
c
and so
tan α − tan α′
1 + tan α tan α′
γ(cos α + v/c) − cos α
= sin α
γ cos α(cos α + v/c) + sin2 α
1 + v/2c cos α + O(v 2 /c2 )
v
sin α
=
c
1 + v/c cos α + O(v 2/c2 )
tan(α − α′ ) =
or
2 v
v
v
cos α + O 2
tan(α − α ) = sin α 1 −
.
c
2c
c
′
Non-relativistic (Bradley)
α − α′
α′
α
As tan(α − α′ ) ≈ α − α′ then
α − α′ =
7.4
v
sin α .
c
Minkowski (Space–Time) diagrams
Portray x, y, z, t as a point in a four-dimensional space-time: (ct, ~r) ≡ (ct, x, y, z).
• Every point P represents an event in space-time
• Under a Lorentz transformation for S → S ′
s2 = c2 t2 − ~r2 ,
~r2 = x2 + y 2 + z 2 ,
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is a Lorentz invariant, ie s2 = s′2 (after a few lines of algebra). So

 > 0 time-like
2
= 0 light-like
s =

< 0 space-like
This decomposition remains the same in all inertial frames (ie after any Lorentz
transformation)
Minkowski or Space–time picture (wlog two dimensional)
ct
P
Time like
FUTURE
Space like
PRESENT
Space like
PRESENT
O
x
Time like
PAST
light cone
Worldline
• Light signal define a cone (at 45◦ as choose ct for y-axis)
• All time-like points lie within light-cone
• All space-like points lie outside light-cone
• World-lines are motion of a particle (starting at t = 0, x = 0). World-lines of
a photon lie on the light cone
Consider two space-time points P1 (ct1 , ~r1 ) and P2 (ct2 , ~r2 ). Then
s212 = c2 (t1 − t2 ) − |~r1 − ~r2 |2 ,
also a Lorentz invariant.
90
Space-like separation
From s212 < 0, then x1 − x2 > c(t1 − t2 ), ie events not connected by a light signal
no causal connection
Can find a Lorentz transformation to S ′ where both events P1 and P2 are at the
same time, as
v
c(t′1 − t′2 ) = γ c(t1 − t2 ) − (x1 − x2 ) .
c
Choose
v=c
c(t1 − t2 )
< c,
x1 − x2
so S ′ possible
Hence can always find Lorentz transformations so that the order of space-like events
is interchanged.
Time-like separation
From s212 > 0, then c(t1 − t2 ) > x1 − x2 , ie events connected by a light signal
causal connection possible
Cannot find a Lorentz transformation to S ′ where both events P1 and P2 are at the
same time, as would need
v=c
c(t1 − t2 )
> c,
x1 − x2
so S ′ not possible
So cannot interchange cause and effect. However as
x′1 − x′2 = γ ((x1 − x2 ) − v(t1 − t2 )) .
can find a frame S ′ where x′1 = x′2 , events happen at the same place
Define proper time by
c2 τ 2 = s2 ,
or τ =
r
t2 −
~r2
.
c2
So here we have
τP1 − τP2 =
r
(t1 − t2 )2 −
|~r1 − ~r2 |2
.
c2
ie time measured on a clock moving with uniform speed |vecr1 −~r2 |/(t1 −t2 ) between
the two events, P1 and P2 .
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ct
ct′
lightcone
hyperbola
ctP
P
ct′P
x′
x′P
1′
xP
x
1
7.4.1
Diagramatic relation between S and S ′
We now briefly consider the relationship between S and S ′ on a Minkowski diagram
Co-ordinate systems same at t = 0 = t′
• S ′ time-axis defined by x′ = 0 = γ(x − vt) or
ct =
1
x,
v/c
ie straight line, gradient > 1, ie between S time-axis and light signal
• S ′ space-axis defined by t′ = 0 = γ(t − v/c2 x) or
ct = v/c x ,
ie straight line, gradient < 1, ie between S space-axis and light signal
Scaling of axes
The Lorentz transformation S → S ′ changes the scale of the axes.
Choose a length unit by s̄2 = −1. Then
x2 = (ct)2 + 1 ,
– ie hyperbola, all points on it have length s̄2 = −1. In particular t = 0 cuts x-axis
at x = 1.
But as s2 is a Lorentz invariant, then this is equivalent to
x′2 = (ct′ )2 + 1 ,
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