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Unformatted text preview: alilean constraints
(space and time are the same for everyone in all inertial frames) lead directly
to the result that velocities transform by simple addition. Lorentz Transformation
To build a 2 × 2 transformation operator, we needed three constraints. The
three Galilean constraints led uniquely to the Galilean transformation operator
we played with in the last section.
Now, we’d like to put together a transformation operator built on Einstein’s
• The laws of physics are the same in all inertial frames of reference.
• The speed of light in vacuum, c, is the same in all inertial frames of
Since we only get three constraints, we’re going to have to replace two of the
Galilean constraints. Which ones? Well, I’ll give you a hint - the origin of the
rocket frame still moves through the lab frame with a velocity β cx. That’s right
- we can no longer guarantee that all observers will experience space and time
the same way - more on that later. The new constraints are as follows:
• i) Light (in vacuum) travels with the same speed, c, in all inertial frames
• ii) The laws of physics are the same for all observers.
• iii) The origin of the rocket frame (O ) was co-located with the origin of
the lab frame (O) at t = t = 0 and moves through the lab frame with a
velocity = β cx
Suppose a photon leaves the (mutual) origin at t = t = 0, traveling in the +ˆ
direction. According to the ﬁrst constraint, its position in the rocket at any
time t will be given by x = ct , and its position in the lab at any time t will be
given by x = ct. The spacetime coordinates that describe the photon in each
frame should look like:
ct r = ct
ct These two descriptions must be related by a Lorentz transformation:
L 0 ,0 ( β ) L 0 ,1 ( β )
L 1 ,0 ( β ) L 1 ,1 ( β )
so. . .
L 0 ,0 ( β ) + L 0 ,1 ( β ) = L 1 ,0 ( β ) + L 1 ,1 ( β )
If we send the photon in the −x direction instead, the spacetime coordinates
and they are related by:
L 0 ,0 ( β )
L 1 ,0 ( β ) L 0 ,1 (...
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