Solutions of Problems 8
Solution (8.1)
Consider the rotation
1 10
x
x
0
→
x2
x2
Then we have that
0
x1
0
2x2
=
− sin θ
cos θ
cos θ
sin θ
x1
x2
.
x1 cos θ − x2 sinθ
=
2(x1 sin θ + x2 cos θ)
1 cos θ − sin θ
x
6=
.
sin θ cos θ
2x2
So (x1 , 2x2 ) is not a vector. Next we observe that
[(x1 )2 + (x2 )2 ]x1
[(x1 )2 + (x2 )2 ]x2
1 2
2 2
= [(x ) + (x ) ]
x1
x2
Then note that
(x1 )2 + (x2 )2
is a scalar, i.e.
0
0
(x1 )2 + (x2 )2 = (x1 )2 + (x2 )2 .
This is true because (x1 )2 + (x2 )2 is the length square of the vector (x1 , x2 ) and as we have
shown this in invariant under O(2). Thus we have
0
0
0
[(x1 )2 + (x2 )2 ]x1
0
0
0
[(x1 )2 + (x2 )2 ]x2
10 2
20 2
0
x1
0
x2
= [(x ) + (x ) ]
10 1
x
x
1 2
2 2
1
2
2
2
0
= [(x ) + (x ) ]
= [(x ) + (x ) ]A
x2
x2
1 2
[(x ) + (x2 )2 ]x1
=A
,
[(x1 )2 + (x2 )2 ]x2
for every A ∈ O(2). So
[(x1 )2 + (x2 )2 ]x1
[(x1 )2 + (x2 )2 ]x2
is a vector. This can be written as
|x|2 x .
Next consider
0
0
(x1 + x2 )2 = (x1 cos θ − x2 sin θ + x1 sin θ + x2 cos θ)2 6= (x1 + x2 )2 .
1
So this is not a scalar. Similarly
0
0
(x1 + 2x2 )2 = (x1 cos θ − x2 sin θ + 2x1 sin θ + 2x2 cos θ)2 6= (x1 + 2x2 )2
and so it is not a scalar either. Finally,
0
0
(x1 )2 + (x2 )2
2
= (x1 cos θ − x2 sin θ)2 + (x1 sin θ + x2 cos θ)2
2
= (x1 )2 + (x2 )2
2
and so it is a scalar.
Solution (8.2)
First consider the rotation
0
xi =
X
Rij xj .
j
Then
and
Thus
d
d i0 X
x =
Rij xj
dt
dt
j
d2 i0 X
d2 j
x
=
R
x .
ij
dt2
dt2
j
X
d2 i0
d2 j
i0
x
x − xj
−
x
=
R
ij
2
2
dt
dt
j
and so it is invariant under rotations. However under translations
0
xi = xi + ai
we have
and
d i0
d
x = xi
dt
dt
(1)
d2 i0
d2 i
x = 2x .
dt2
dt
(2)
Thus
d2 i0
d2 i
d2 i
i0
i
i
i
x
−
x
=
x
−
(x
+
a
)
=
x
−
x
− ai
2
2
2
dt
dt
dt
and therefore it is not invariant under translations and therefore invariant under the Euclidean group.
To continue, first observe that
X dxj dxj
dt dt
j
2
is invariant under rotations. Indeed
X dxj 0 dxj 0
X
dxk dx`
=
Rjk Rj`
dt dt
dt dt
j
j,k,`
=
X
R̃kj Rj`
j,k,`
X
dxk dx`
dxk dx`
=
δk`
dt dt
dt dt
k,`
=
X dxj dxj
.
dt dt
j
Using this we have
0
X dxj 0 dxj 0 dxi0
d2 xi
+
dt
dt
dt
dt2
j
=
X dxj dxj X
dxk X
d2 xk
Rik
+
Rik 2
dt dt
dt
dt
j
=
X
k
k
k
X dxj dxj dxk
d2 xk Rik
+
.
dt dt dt
dt2
j
Thus this equation is invariant under rotations. It is also invariant under translations
because of equations (1) and (2). So this equation is invariant under the full Euclidean
group.
Solution (8.3)
Consider the rotation
0
xi =
X
Rij xj .
j
Then
X
0
Rik xi =
X
R̃ki Rij xj =
X
i
=
X
Rik Rij xj
i,j
i,j
δkj xj = xk
j
Thus we have using the chain rule that
∂xj ∂
∂
=
0
∂xi
∂xi0 ∂xj
∂
∂
.
0 = Rij
i
∂x
∂xj
Thus
∂
∂xi
3
transforms as a vector. Thus if P is scalar, we have
0
X
∂
∂
d2 xi
d2 xj X
0
−
P
−
R
P
=
R
ij
ij
0
2
j
dt2
∂xi
dt
∂x
j
j
=
X
Rij
j
d2 xj
∂ −
P .
2
dt
∂xj
So the equation is invariant under rotations.
0
Under translations xi = xi + ai , we have found in the previous problem that
0
d2 xi
d2 xi
=
.
dt2
dt2
Using again the chain rule, we also find that
∂
∂
.
0 =
i
∂x
∂xi
Thus if P is invariant under translations
0
d2 xi
d2 xi
∂
∂
0
−
−
P
P
=
0
2
i
2
dt
∂x
dt
∂xi
and so the equation is invariant as well.
A law which is invariant under the Euclidean group is that of Newton’s gravity.
Solution (8.4)
To show that T × V transforms as a vector, we have
X
X
(T 0 × V 0 )i =
ijk (T 0 )j (V 0 )k =
ijk Rj` Rkm T` Vm ,
jk
j,k,`,m
where Rij is a 3 × 3 real matrix. Next we use the equation
X
ijk Ri` Rjm Rkn = detR`mn
ijk
(this follows from the definition of the determinant and that of ) which in turn implies
that
X
X
−1
−1
ijk R`t
Ri` Rjm Rkn = detR
R`t
`mn .
ijk`
`
Applying this in the transformation of V × T , we find that
X
X
X
−1
−1
(T 0 × V 0 )i = detR
R`i
`jk T j V k = detR
R`i
(T × V )` .
`
jk
`
4
Demanding that (T × V ) transforms as a vector, we have that
−1
detRR`i
= Ri`
which implies that
R̃R = detR ,
where R̃ is the transposed of R. Taking the determinant of the above relation we have
(detR)2 = (detR)3
which implies that detR = 1 and so
R̃R = 1 .
Thus T × V transforms as a vector only when R is an orthogonal rotation and R ∈ SO(3),
i.e. detR = 1, and not otherwise.
This can be generalized as follows:
X
jn−1
(Y 0 )i =
ij1 j2 ...jn−1 (X 0 )j11 (X 0 )j22 . . . (X 0 )n−1
j1 ,j2 ,...,jn−1
=
X
X
j1 ,j2 ,...,jn−1
k1 ,k2 ,...,kn−1
Next we use the identity
X
k
n−1
ij1 j2 ...jn−1 Rj1 k1 Rj2 k2 . . . Rjn−1 kn−1 X1k1 X2k2 . . . Xn−1
.
j1 ,...jn Rj1 k1 . . . Rjn kn = detR k1 ,...,kn
j1 ,...,jn
which follows from the definition of the determinant of a matrix as above. Then
X
X
Rk−1
R
.
.
.
R
=
detR
Rk−1
k1 ,...,kn .
j1 k1
jn kn j1 ,...jn
1i
1i
k1 ,j1 ,...,jn
k1
It follows that
X
Rj2 k2 . . . Rjn kn i,j2 ,...jn = detR
j2 ,...,jn
X
Rk−1
k1 ,...,kn .
1i
k1
Substituting this back in the transformation of Y , we find that
X
(Y 0 )i = detR
Rik Y k .
k
provided that
R̃R = detR .
This implies that
(detR)n−2 = 1 .
So if n is odd, detR = 1 and R̃R = 1. Thus
R ∈ SO(n) .
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