6 Classical Mechanics
© 2010 Taylor & Francis Group, LLC
Solution:
We have
′
=
=
∑
j
λ
1
3
with, for such a rotation,
λφλφλ
11 11 22 22 33 3
=⋅
′
==⋅
′
==⋅
′
ˆˆ
cos
ˆˆ
cos
ˆˆ
ee ee ee
33
12 21 21 12
1=
=⋅
′
==⋅
′
=−
,
ˆˆ
sin
ˆˆ
sin.λφλφee ee,
All other components of λ
ij
are zero. The transformation equations take the form
′
=+
′
=− +
′
=xx xxxx
11 2212
cossin sincosφφ φφ
and the transformation matrix is
λ
φφ
φφ=−
cossin
sincos
0
0
001
.
The orthogonal conditions are obviously satisfied:
λλ λλ φφ
λλ λλ φ
11 11 21 21
22
22 22 12 12
2
1+=+=
+=+
cossin
cosssin
cossin sincos .
2
1
0
φ
λλ λλ φφ φφ
=
+
It was mentioned earlier that one advantage of using the matrix notation is that successive
transformations can be easily handled by matrix multiplication. We now demonstrate this b ...