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Unformatted text preview: ) z
x = L1,0 (β ) ct + L1,1 (β ) x + L1,2 (β ) y + L1,3 (β ) z
y = L2,0 (β ) ct + L2,1 (β ) x + L2,2 (β ) y + L2,3 (β ) z
z = L3,0 (β ) ct + L3,1 (β ) x + L3,2 (β ) y + L3,3 (β ) z
So. . . we need to ﬁnd 16 matrix elements5 .
Simpliﬁcation
. . . or not. Consider the following experiment: orient the axes so that the rocket
travels in the +x direction in the lab. A sharp projection is extended from the
side of the rocket in such a way that it scrapes a very long wall that runs parallel
to the rocket’s motion through the lab. If that projection is mounted at y = a
in the rocket, where will the resulting mark appear in the lab? Remember  the
frames were coincident at t = t = 0, and the rocket frame is moving with a
velocity along the +x direction in the lab.
Relative motion does not aﬀect the transverse spacial coordinates of an event.
If the rocket is moving in the x direction, y = y and z = z . With the exception
of the identity submatrix in the transform, the rest of the elements relating to
transverse spacial components go to zero! The transform will look like:
ct
L 0 ,0 ( β ) L 0 ,1 ( β ) 0 0
ct x L1,0 (β ) L1,1 (β ) 0 0 x = y 0
0
1 0 y z
0
0
01
z
or. . .
ct = L0,0 (β ) ct + L0,1 (β ) x
x = L1,0 (β ) ct + L1,1 (β ) x
y = y
z = z
5 Note the conventional use of 0 to denote the temporal index. 4 Since there is not a lot happening in the transverse dimensions, they are frequently ignored. It is common practice to omit the transverse coordinates when
writing a transform. So  for relative motion along the x direction in the lab. . .
ct
L 0 ,0 ( β ) L 0 ,1 ( β )
ct
=
x
L 1 ,0 ( β ) L 1 ,1 ( β )
x
or. . .
ct = L0,0 (β ) ct + L0,1 (β ) x
x = L1,0 (β ) ct + L1,1 (β ) x
with the understanding that. . .
y = y
z = z Galilean Transformation
Let’s derive the (2 × 2) relativistic transformation for Galileo’s world....
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This note was uploaded on 12/03/2013 for the course PHYSICS 105A taught by Professor Corbin during the Winter '13 term at UCLA.
 Winter '13
 Corbin
 Inertia, Special Relativity

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