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Lorentz Transformation

The document derives the Lorentz transformation from two assumptions: (1) the speed of light is the same in all inertial frames and (2) moving clocks run slower according to a factor γ. By considering light signals between two frames in motion and equating expressions at an event seen from both frames, the derivation obtains equations relating space and time coordinates that comprise the Lorentz transformation.

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115 views1 page

Lorentz Transformation

The document derives the Lorentz transformation from two assumptions: (1) the speed of light is the same in all inertial frames and (2) moving clocks run slower according to a factor γ. By considering light signals between two frames in motion and equating expressions at an event seen from both frames, the derivation obtains equations relating space and time coordinates that comprise the Lorentz transformation.

Uploaded by

Putu Eka
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Worlds Fastest Derivation of the

Lorentz Transformation
Alan Macdonald
Luther College, Decorah, IA 52101, U.S.A.
http://faculty.luther.edu/macdonal
macdonal@luther.edu
Assume:
(A) The speed of light is the same in all inertial frames. (Take c = 1.)
(B) A clock moving with constant velocity v in an inertial frame I runs at
a constant rate = (|v|) with respect to the synchronized clocks of I which it
passes.
Assumption (B) follows directly from the relativity principle. We do not
assume that the Lorentz transformation is linear.
In the figure, E is an arbitrary event
X
plotted in an inertial frame I, L+ and
E
L are the two light worldlines through
LL+
E, and O is the worldline of the spaF
tial origin of an inertial frame I 0 moving
O
T
with velocity v in I. On O,
X 0 = 0, X = vT, and T = T 0 .
Thus on O,
T + X = (1 + v)(T 0 + X 0 )
T X = (1 v)(T 0 X 0 ).

(1)
(2)

Since c = 1 in I, an increase in T along L is accompanied by an equal decrease


in X. Thus T + X is the same at E and F . Likewise, since c = 1 in I 0 , T 0 + X 0
is the same at E and F . Thus Eq. (1), which is true at F , is also true at E.
Similar reasoning using L+ proves Eq. (2) true at E. Add and subtract Eqs.
(1) and (2):
T = (T 0 + vX 0 )
X = (vT 0 + X 0 ).

(3)
(4)

For X = 0 in Eq. (4), X 0 = vT 0 ; the origin of I has velocity v in I 0 .


Thus, switching I and I 0 and using (B), the reasoning for Eq. (1) also gives
1
T 0 + X 0 = (1 v)(T + X). Substituting this in Eq. (1) gives = (1 v 2 ) 2 .

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