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PHYS3550 General Relativity
Einstein’s 1920 derivation of the Lorentz Transformations
John K. Webb
School of Physics, UNSW, Sydney, NSW 2052, Australia
(Dated: March 16, 2015)
Einstein’s 1920 derivation of the Lorentz Transformations [1] is fine up to a certain point
(indicated using a red line below), but then seems to lose clarity and even perhaps permit
inconsistencies (unless of course I have made a mistake. Let me know if so please!).
In my lecture slides for PHYS3550 I give a different derivation. It’s the same as the first
part of Einstein’s derivation but uses interval invariance to make the final part simpler.
I.
EINSTEIN’S DERIVATION
Here I follow Einstein’s derivation, matching his equation numbering, using the convention c = 1 and using a different notation. We take the usual situation of an intertial reference
frame O moving with velocity v along the positive x-axis relative to another intertial reference frame O.
The goal is to find general expressions that express coordinates measured by O in terms
of those measured by O (or vice versa), ensuring consistency with the Relativity Principle.
A light ray traveling in the +x direction is described by O as x = t i.e.
x − t = 0 . . . . . . . (1)
and is described by O as
x¯ − t¯ = 0 . . . . . . . (2)
Equations (1) and (2) satisfy the general condition
(¯
x − t¯) = λ(x − t) . . . . . . . (3)
where λ is an unknown constant. Repeating the argument above for a light ray traveling in
the −x direction gives
(¯
x + t¯) = µ(x + t) . . . . . . . (4)
where µ is another unknown constant.
Now add (or subtract) equations (3) and (4) and for simplicity use
a=
λ+µ
. . . . . . . (4a)
2
b=
λ−µ
. . . . . . . (4b)
2
This gives
x¯ = ax − bt . . . . . . . (5a)
t¯ = at − bx . . . . . . . (5b)
All we have done so far is to prove that the coordinates measured by O are a linear
transformation of those measured by O, which, arguably, we could have assumed as our
starting point (what else could it be?). We need to find the constants a and b.
Equations (5) must be consistent with all events in O. We can make use of the fact that
O’s origin is permanently at x¯ = 0. Equation (5a) then gives
x=
bt
. . . . . . . (5c)
a
But the velocity of O as measured by O is x/t, so
v=
b
. . . . . . . (6)
a
We now invoke the Relativity Principle: The length of a rod (unit length, lying along the
x-direction) at rest in O as viewed by O must be the same as the length of the same rod at
rest-in O as viewed by O. In other words, each observer must measure the rod to have unit
length when the rod is stationary in their own rest-frame.
We can imagine verifying this in practice by taking a snapshot from O of the moving bar.
Taking a snapshot “freezes” the measurements at a specific time, e.g. for convenience, at
t = 0. Therefore we can put t = 0 in (5a), which gives
x¯ = ax . . . . . . . (6a)
2
Since we have specified the proper length of the rod be 1, the two events corresponding
to the rod-ends are thus unit distance apart when measured by O, but when measured by
O, they are separated by
∆x =
1
. . . . . . . (7)
a
Now apply the same argument as above to the reverse situation, i.e. where the same
rod is stationary in O and is viewed from O. The snapshot is then taken from O at t¯ = 0.
Equation (5b) then gives at = bx, or
bx
. . . . . . . (7c)
a
t=
Next, Einstein substitutes (7c) into (5a) and uses (6) to derive an expression for x¯ at
t¯ = 0:
x¯ = ax − bt = ax −
b2 x
= ax(1 − v 2 ) . . . . . . . (7d)
a
but since we have a unit-length bar at rest in O, the length of the bar as measured by O is
∆¯
x = a(1 − v 2 ) . . . . . . . (7a)
But to be consistent with the Relativity Principle, it must be that equation (7) = equation
(7a) (that is, that the proper lengths measured by both observers are the same - it’s the
same rod). Therefore, we have
a= √
1
. . . . . . . (7b)
1 − v2
Note we have discarded the negative solution in (7b) as being unphysical. See the explanation given in my PHYS3550 lecture slides.
We have thus determined the constant a, from which b follows, and the Lorentz Transformations follow trivially by substitution back into (5a) and (5b):
x − vt
. . . . . . . (8a)
x¯ = √
1 − v2
t − vx
t¯ = √
. . . . . . . (8b)
1 − v2
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II.
A POSSIBLE PITFALL
However, here’s a potential pitfall: Let’s restart the derivation from immediately before
equation (5c):
Equation (5b) then gives at = bx, or
b=
at
. . . . . . . (5c0 )
x
Next, substitute (5c0 ) into (5a) and use v = x/t (as was done in going from (5c) to (6)). We
then get
1
at2
= ax 1 − 2
. . . . . . . (7d0 )
x¯ = ax −
x
v
and recalling we have a unit-length bar in O, as before, we have
∆¯
x = a(1 −
1
) . . . . . . . (7a0 )
v2
This is horrible! (7a0 ) can only be consistent with (7a) if v = 1. Einstein’s derivation after
equation (6) seems convoluted and the method appears to permit inconsistencies. However,
deriving the Lorentz Transformations is easy using interval invariance - see my PHYS3550
lecture slides.
[1] Einstein, Albert. Relativity: The Special and General Theory. New York: Henry Holt, 1920,
Appendix I.
http://www.bartleby.com/173/a1.html
or
http://newt.phys.unsw.edu.au/~jkw/phys3550/Papers/AE_LT_derivations_1920.pdf
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