This preview shows page 1. Sign up to view the full content.
Unformatted text preview: What are
the constraints he would have imposed?
• i) Time is the same for all observers.
• ii) Space is 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.
Since the clocks were synchronized at t = t = 0, the ﬁrst constraint says t = t
(always), and. . .
L 0 ,0 ( β ) = 1 and L 0 ,1 ( β ) = 0 Since the origins were co-located at t = t = 0, the second constraint tells us
that x = x at t = t = 0. This forces
L 1 ,1 ( β ) = 1
Finally, we can describe O in the rocket frame. . .
The third constraint tells us that in the lab, the description of O looks like 5 r= ct
β ct From which we get. . .
L 1 ,0 ( β ) = β
Put it all together, and the Galilean transform looks like
L( β ) = 1
1 An Instructive Example
v =β c
u’ Lab Rocket Suppose a mouse leaves the origin of the rocket-ship at t = 0 and walks forward
with a velocity (in the rocket) u x . The position of the mouse (in the rocket)
would be given by x = u t and the spacetime coordinates of the mouse (in the
rocket) would look like:
What will the mouse’s motion look like in the lab? Since the origins coincide
at t = t = 0, the mouse will appear to leave the origin in the lab at t = 0 and
proceed forward with a velocity (in the lab) u x. The position of the mouse (in
the lab) at any instant would be given by x = u t and the spacetime coordinates
of the mouse (in the lab) would look like:
Let’s relate these two descriptions by the appropriate Galilean transform. . .
ct = ct 6 ut = (β c + u ) t
Divide the second equation by the ﬁrst. . .
u = u + β c
which is just another way of saying. . .
Vmouse,lab = Vmouse,rocket + Vrocket,lab
We recover the classical relative velocity equation! The G...
View Full Document