Homework 11 and 12 Answers, 95.658 Spring 2011, Electromagnetic Theory II
Dr. Christopher S. Baird, UMass Lowell
Problem 1
Jackson 11.1
Two equivalent inertial frames
K
and
K
' are such that
K
' moves in the positive
x
direction with speed
v
as seen from
K
. The spatial coordinate axes in
K
' are parallel to those in
K
and the two origins are
coincident at times
t
=
t
' = 0.
(a) Show that the isotropy and homogeneity of space-time and equivalence of different inertial frames
(first postulate of relativity) require that the most general transformation between the space-time
coordinates (
x
,
y
,
z
,
t
) and (
x
',
y
',
z
',
t
') is the linear transformation,
x
'
=
f
v
2
x
−
v f
v
2
t ;
t
'
=
g
v
2
t
−
v h
v
2
x ;
y
'
=
y ;
z
'
=
z
and the inverse,
x
=
f
v
2
x
'
v f
v
2
t
'
;
t
=
g
v
2
t
'
v h
v
2
x
'
;
y
=
y
'
;
z
=
z
'
where
f
,
g
, and
h
are functions of
v
2
, the structures of the
x
' and
x
equations are determined by the
definition of the inertial frames in relative motion, and the signs of the inverse equation are a reflection
of a reversal of roles of the two frames.
SOLUTION:
First of all, there is no motion in the
y
or
z
directions, so that these coordinates must be identical:
y
'
=
y ;
z
'
=
z
Homogeneity of space-time means that points in space-time have the same density at all points as
observed in one particular frame. Mathematically, this requires linear relationships between all space
and time coordinates. (If, for instance,
x
' =
x
3
, than at locations marked by higher
x
, successive points in
x
' would be further and further spaced apart.) This gives:
x
'
=
f x
f
2
t ;
t
'
=
g t
h x ;
y
'
=
y ;
z
'
=
z
At this point,
f
,
f
2
,
g
, and
h
are arbitrary functions that cannot depend on
x
or
t
. The only thing left that
they could depend is the frame's velocity
v
and any universal constants that fall out.
We have inertial frames in relative motion, so that the shifting of the origins must be taken into
account. In Galilean relativity, this was taken into account by adding -
vt
to the spatial relation. We can
get this in the same general form by taking a -
v
out of the arbitrary
f
2
and
h
:
x
'
=
f x
−
v f
2
t ;
t
'
=
g t
−
v h x ;
y
'
=
y ;
z
'
=
z
Again, the homogenous nature of space-time means that as time marches on, the points cannot spread
out. This forces upon us
f
2
=
f
:
x
'
=
f x
−
v f t ;
t
'
=
g t
−
v h x ;
y
'
=
y ;
z
'
=
z