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Physics 325
Lecture 19
Let’s put the Lorentz transformations to work and see that they give us back the Lorentz
contraction and time dilation.
Suppose the primed frame moves to the right along the
x
axis at speed
v
with respect to
the unprimed frame.
v
G
S’
S
The Lorentz transformations that tell us how the primed axis is related to the unprimed
axis are (we’re just going to work in one spatial dimension for the moment)
( )
()
x
xv
t
tt
x
γ
γβ
′
=−
′
c
(19.1)
Now suppose an observer in S’ measures an object that is at rest in S’ and finds that it has
length
A
(this is its proper length since it is measured at rest).
Then the first equation
above tells us that if its length is measured in the unprimed frame at a fixed time (
)
the length will be
0
0
t
∆=
( )
( ) ( )
021
21
2
1
xx
xxv
γγ
′′
=
−
−−
≡
AA
because
t
2
t
1
=0
(fixed time in S).
Therefore
0
=
A
A
This is the Lorentz contraction.
Alternatively, suppose the object is at rest in S and measured by an observer in S’ at a
fixed time
in S’
.
We can still use Equations (19.1) to determine what S’ observes.
The
first equation gives us again
( )
( )
x
x
v
t
t
−=
−
(19.2)
but the times on the right hand side are no longer the same.
That’s because the
measurement is made at a fixed time in S’, not in S.
The second equation of (19.1) tells
us how to find the corresponding times in S.
We have that
0
t
′
∆
=
, therefore
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2 1
21
0
t t
tt
xx
cc
ββ
γ
⎛⎞
′′
−==
−−
−
⇒ −=
−
⎜⎟
⎝⎠
Inserting this into Equation (19.2) gives
2
0
1
1
v
xxv xx
β
γγ
γβ
−=
−
=
−
−
−
=−−=
=
A
⎛
⎞
⎜
⎟
⎝
⎠
So a Lorentz contracted length is measured from S’ this time because the proper length is
measured in S where the object is at rest.
We could have arrived at this result directly
using the inverse transformation
( )
x
xv
t
′
′
∆
=∆+
∆
where now we can insert
(the observation is made at a fixed time in S’).
It’s nice
to see it work both ways.
0
t
′
∆=
Now let’s put a clock at rest in S’.
A time difference
t
′
∆
measured in S’ is transformed to
in S using (19.1) as follows:
t
∆
(
c
⎛
−
⎜
)
⎞
⎟
1
(19.3)
The clock is at rest in S’, so
2
x
x
′
=
′
1
, but that does not imply that
2
x
x
=
(since S’ is
moving with respect to S, different times correspond to different locations).
Instead, we
calculate
x
x
−
from the first equation in (19.1)
2
1
0
x
x
v
t
tx
x
v
t
−− −
⇒ −= −
t
Inserting this result in Equation (19.3) gives:
()()
( )
2
1
v
c
−
−
=
−
−
=
This is time dilation, the moving clock in S’ is running slower than a clock in S.
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This note was uploaded on 10/06/2011 for the course PHYS 225 taught by Professor Makins during the Spring '10 term at University of Illinois, Urbana Champaign.
 Spring '10
 Makins
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