Derivation of the Lorentz transformation formulas:
The two reference frames are S and S’. S’ moves along the positive x-direction with a constant
speed v relative to S.
Let the origins O and O’ of the two frames coincide at t=t’=0. Hence the event (x,t)=(0,0)
should transform as (x’,t’)=(0,0). This means that the required linear transformation equations
don’t have a constant term:
x' = A ⋅ x + B ⋅ ct (1)
ct ' = D ⋅ x + E ⋅ ct (2)
Our task is to determine the constant factors A, B, D and E.
1. Consider a light pulse emitted from O at t=0 along the positive x-axis. The equation of the
trajectory of this light pulse is:
x=c·t. [light pulse to the right] (3)
The same light pulse – when observed from frame S’ – originates from O’ at t’=0. Since S’
measures the velocity of this light pulse to be c (just like S does) [this is a unique property of
light, completely unexplained by classical physics], the equation of the trajectory of the light
pulse in S’ must be:
x’=c·t’. [light pulse to the right] (4)
Combining (3) and (4) with (1) and (2) yields:
A + B = D + E. (5)
2. Now consider a similar light pulse – emitted from O at t=0 – but moving along the negative
x-axis.
The equation of the trajectory of this light pulse is:
x= −c·t. [light pulse to the left] (6)
Again, since S’ measures the velocity of this light pulse to be c (just like S does), the equation
of the trajectory of the same light pulse in S’ is:
x’= −c·t’. [light pulse to the left] (7)
Combining (6) and (7) with (1) and (2) yields:
A − B = −(D + E). (8)
By subtracting and adding eqs. (5) and (8) we get
B = D, (9)
and
A = E, (10)
respectively.
The Lorentz transformation equations (1) and (2) are thus simplified to the following
symmetrical form:
x' = A ⋅ x + B ⋅ ct (11)
ct ' = B ⋅ x + A ⋅ ct (12)
3. The inverse transformation equations are obtained from eqs. (11) and (12) by solving these
two equations for x and ct:
A ⋅ x'− B ⋅ ct '
x= (13)
A2 − B 2
− B ⋅ x'+ A ⋅ ct '
ct = (14)
A2 − B 2
We can reason in the following way: if S’ moves to the right with speed v relative to S, then S
moves to the left with speed (−v) relative to S’. Hence the inverse transformation equations
(13) and (14) and the transformation equations (11) and (12) should have a ’symmetrical
appearance’. Indeed, if the additional condition
A2 − B 2 ≡ 1 (15)
is satisfied, eqs. (13) and (14) take a form which is symmetrical to (11) and (12):
x = A ⋅ x'− B ⋅ ct ' (16)
ct = − B ⋅ x'+ A ⋅ ct ' . (17)
4. Next, let us describe the motion of O, the origin of frame S, as viewed from S’. On the one
hand, by substituting x = 0 into (11) and (12), we obtain
x' = B ⋅ ct , [motion of O] (18)
ct ' = A ⋅ ct , [motion of O] (19)
hence
B
x' = ⋅ ct ' . [motion of O] (20)
A
On the other hand, as viewed from S’, point O moves to the left with a speed v, and hence its
motion is described by
x’ = −v·t’. [motion of O] (21)
Comparing (20) with (21) yields the following relation between A and B:
v
B = −A⋅ (22)
c
5. Finally, solving eqs. (15) and (22) for A and B, we get
1
A= (23)
v2
1− 2
c
and
v
B=− c . (24)
v2
1− 2
c
Substituting these values into (11) and (12) we obtain the final form of the Lorentz
transformation equations:
x − v ⋅t
x' = , [Lorentz] (25)
v2
1− 2
c
v
t− ⋅x
t' = c2 . [Lorentz] (26)
v2
1− 2
c
