Scenery:
an object in reference frame S' has a length dx' = L0.
This reference frame S' move with velocity +V with respect to reference frame S.
S and S' have their x,y,z axes aligned and furthermore x = x' = 0 at time t = t' =0.
define:
G = gamma(V) = 1 / sqrt( 1 - b^2)
b = V/c
From the Lorentz transformation LT, we know:
(A) c dt = G ( c dt' + b dx')
(B) dx = G ( b c dt' + dx')
For length measurement in S, the two positions x1 and x2 (which are in S!) need
to be measured at the same time (in S!), so dt must be 0: c dt = 0
which in (A) means:
c dt' = -b dx'
then, in (B) you fill this in:
dx = G ( b (-b dx') + dx')
dx = G ( 1 - b^2) dx'
dx = G ( 1 / G^2) dx'
dx = dx' / G
since G >= 1, dx will be smaller than dx'
The length dx of a moving object, is always smaller than its length in the
reference frame of the object itself. The length in the reference frame of the
object itself, is called the "proper length" of the object.
TN2612_2015 lecture 2
Two inertial frames:
Frame S’ moves with velocity V in the positive x-direction
Galilei transformation:
inverse GT:
x′= x − Vt
y′ = y
z′ = z
t′ = t
x= x′ + Vt ′
y = y′
z = z′
t = t′
Grafical representation
for the frame moving in the positive direction, you need to subtract that
speed.
viceversa for the negative direction
Lorentz Transformation:
x ′ γ ( x − β ct )
=
y′ = y
z′ = z
ct ′ γ ( ct − β x )
=
Grafical representation
inverse LT:
x γ ( x ′ + β ct ′)
=
y = y′
z = z′
ct γ ( ct ′ + β x ′)
=
β=
=
γ
V
<1
c
1
V2
1− 2
c
>1
Consequences of the Lorentz transformation
1. length contraction (moving objects shrink)
Essential for measuring a length: determine the begin and end point of the object at the
same time.
Measuring the length of the train:
S’ in the frame of the moving train. Frame S is the platform. The length of the train in its
own rest frame S’ is l ′ = l0 , l0 being the ‘proper length’.
Find two observers in S: one sees the front of train passing by at time t, the second sees the
end passing by at the same time t. → distance between observers is the length L of the
train.
Define two events:
ev1 (front of the train at time t): ( ct , x f )
ev2 (end of the train at same time): ( ct , xe )
=
x ′f γ ( x f − β ct )
=
xe′ γ ( xe − β ct )

 → l ′ = x ′f − xe′ = γ ( x f − xe ) = γ l

In S the train is ‘seen’ shorter: l =
Famous barn and pole paradox:
l0
γ
2. time dilation (moving clocks run slow)
Flash of light in S’ from z1′ to z2′ and back.
2L
with L= L=′ z2′ − z1′ and t0 the proper time.
The time interval needed ∆t ′ = t0 =
c
2
1

In S: the total path length is 2  V ∆t  + L2 =∆
c t
2


4 L2
→ ∆t =
V 2 ∆t 2 + 4 L2 = c 2 ∆t 2 → ∆t 2 = 2
c −V 2
leading to the time dilation ∆t =γ t0
alternative derivation: Sec 11.5
2L c
1−
V2
c2
3. simultaneity is relative.
Two events in S:
ev1: ( ct1 , x1 ) , ev2: ( ct2 , x2 ) . In S simultaneous: → t1 =
t2
in S’:
=
ct1′ γ ( ct1 − β x1 )
=
ct2′ γ ( ct2 − β x2 )
Thus, ∆t ′ = t1′ − t2′ = 0 only if x1 = x2
because it's in a higher
position of ct, meaning at a
later time
t1-t2=0 cause theyre simultaneous

 → c ( t1′ − t2′ )= γ c ( t1 − t2 ) + γβ ( x1 − x2 )= γβ ( x1 − x2 )

so two events are only simultaneous for people who are standing in the same exact spot
USS Enterprise under attack
GT:
x ′ = x − ( −V ) t = x + β ct
LT:
=
x ′ γ ( x + β ct )
=
ct ′ γ ( ct + β x )
ct ′ = ct
GT:
ev1:
LT:
ev1:
ct ′ =−2 ∆s
x ′ =0 + β ( −2 ∆s ) =−2 β∆s
ct ′ =γ ( −2 ∆s + β 0 ) =−2γ∆s
x ′ =γ ( 0 + β ( −2 ∆s ) ) =−2 βγ∆s
ev2:
ev2:
S
( ct
ev1
ev2
ev3
ev4
ev5
ev6
∆s , x ∆s )
( −2,0 )
( −1,1)
( 0,0 )
( 0,1)
(1,0 )
( 2,1)
=
β V=
c 0.6
γ = 1 1 − 0.62 = 1 0.8 = 1.25
ct ′ = −∆s
x ′ = ∆s + β ( −∆s ) = (1 − β ) ∆s
etc.
ct=′ γ ( −∆s + β∆s=
) γ ( −1 + β ) ∆s
x=′ γ ( ∆s + β ( −∆s )=
) γ (1 − β ) ∆s
S’
GT
( ct ′ ∆s , x′ ∆s )
LT
( ct ′ ∆s , x′ ∆s )
( −2, −1.2 )
( −1,0.4 )
( 0,0 )
( 0,1)
(1,0.6 )
( 2,2.2 )
( −2.5, −1.5)
( −0.4,0.5)
( 0,0 )
( 0.75,1.25)
(1.25,0.75)
( 3.25,2.75)
etc.
as seen by Enterprise (frame S)
as seen by Mission Control (frame S’) using Galilei Transformation
as seen by Mission Control (frame S’) using Lorentz Transformation