1. Lorentz transformation
Transformation of space and time
t = g (t' + x' v/c²) t' = g (t – x v/c²)
x = g (x' + vt') x' = g (x – vt)
y = y' y' = y
z = z' z' = z
Four-vector ct x y z
g g v/c 0 0
gv/c g 0 0
ct'
x '
y '
z '
X = = = X' Lv
X = Lv X'
Spacetime interval:
For two events
e0 = (t0, x0, y0, z0) = (t0', x0', y0', z0') and
e1 = (t1, x1, y1, z1) = (t1', x1', y1', z1')
we find that the spacetime interval
s2 = (cDt)2 – (Dx2 + Dy2 + Dz2)
= (cDt')2 – (Dx'2 + Dy'2 + Dz'2)
is Lorentz-invariant
(Dt = t1 – t0 ; Dx = x1 – x0 ; …)
Relativistic spacetime diagram x ct
ct' x'
Light 1
Unit hyperbola:

2. Velocity transformation
ux = dx/dt
dx = g (dx' + vdt')
dt = g (dt' + dx' v/c²)
 dx/dt = (dx'+ vdt')/ (dt'+ dx' v/c²) = (dx'/dt'+ v)/ (1 + dx'/dt' v/c²)
dy/dt = dy' / g (dt'+ dx' v/c²) = dy'/dt' / ( 1 + dx'/dt' v/c²)
 ux = (ux'+ v)/(1 + ux'v/c²) reverse transformation: ux' = (ux – v)/(1 – uxv/c²)
uy = uy'/ g (1 + ux'v/c²) replace v by –v: uy' = uy/ g (1 – uxv/c²)
uz = uz'/ g (1 + ux'v/c²) uz' = uz/ g (1 – uxv/c²)
special case u = ux: (that is: u is parallel to the relative velocity v of the reference frames)
u = (u'+ v)/(1 + uv/c²) u' = (u – v)/(1 – uv/c²)
 dt'
t = g (t' + x' v/c²) t' = g (t – x v/c²)
x = g (x' + vt') x' = g (x – vt)
y = y' y' = y
z = z' z' = z ct x y z
g g v/c 0 0
gv/c g 0 0
ct' x' y' z'
X = = = X' Lv

3. Four-Velocity Velocity transformation:
ux = (ux'+ v)/(1 + ux'v/c²) reverse transformation: ux' = (ux – v)/(1 – uxv/c²)
uy = uy'/ g (1 + ux'v/c²) replace v by –v: uy' = uy/ g (1 – uxv/c²)
uz = uz'/ g (1 + ux'v/c²) uz' = uz/ g (1 – uxv/c²)
U (τ is proper time of the object)
X X X
U U'
Coordinate transformation:
X = Lv X'  d X = Lv d X'
τ is Lorentz-invariant
 U = Lv U'

4. Structure of spacetimex
Past
Future ct y
Elsewhere x t y
s² = 0
Light (c = Δx/ Δt) x ct
ct'
s² > 0
Future x'
t > 0
1 s
t' < 0 x'
Elsewhere
3e8 m
1 m
s² < 0
Past

5. The Twin Paradox
A and B are twins of age 20. B decides to join an expedition to a nearby planet. The distance between Earth and the planet is d = 20 light years and the space ship used for the expedition travels at a speed of v = 4/5 c.
- What time is it on Earth and on the spaceship
respectively when B arrives at the planet?
After a short stop at the planet the spaceship
returns with the same speed.
- How old are A and B at B’s return?
t = 0
t = t1
t = 2t1
t" = t2'
t" = 2t2'
Time passing on Earth while spaceship turns around
t'= t1'=t2'
t1 : x1 = xplanet ;x1' = 0  t1 = 25 y
t1' :x1 = xplanet ;x1' = 0  t1' = 15 y
Earth
planet
spaceship
t' = 0
t2' : x2 = xplanet ;x2' = 0  t2' = 15 y
t2 : x2 = 0 ; t2' = 15 y  t2 = 9 y
t = t2
t = g (t' + x' v/c²) t' = g (t – x v/c²)
x = g (x' + vt' ) x' = g (x – vt)

6. The Twin Paradox and Doppler Shift
Earth
planet

7. Relativistic Doppler Effect
fR = fE [ (1 – v/c) / (1 + v/c) ]½ for source moving away from receiver (or vice versa)
fR = fE [ (1 + v/c) / (1 – v/c) ]½ for source moving towards receiver (or vice versa)
(fR : frequency of receiver in receiver’s frame; fE frequency of emitter in emitter’s frame)
“longitudinal Doppler shift”
Transversal effect due to time dilation:
fR = fE g x 1 ct 1
ct' x'
Plain wave emitted source at rest in S'
Plain wave emitted source at rest in S

8. Cosmic Microwave Background
Planck (2009)
WMAP (2001)
COBE (1989)
Cosmic Microwave Background
(Nobel prize in physics 2006: George Smoot, John Mather)
COBE (COsmic Background Explorer) satellite
2.721 K
2.729 K
0 K
4 K
T = 0.02 mK
(Dipole subtracted)