PHYSICS 603 - Winter/Spring Semester 20018 - ODU

Classical Mechanics - Problem Set 5 - Solution

Problem 1)

In the primed coordinate system,!∇( ) ' = ∂

∂x ', ∂∂ y ', ∂∂ z '

⎛

⎝⎜

⎞

⎠⎟ . According to the chain rule of

differentiation, each of these components (m = 1,2,3 for x’, y’, z’) can be expressed as

follows: ∂∂ rm '

=∂ rl∂ rm '

∂∂ rll=1

3

∑ . Now

!r( ) ' =R !r( )⇒ RT !r( ) ' =RTR !r( ) = !r( )⇒ rl = R lmT rm '

m=1

3

∑ (the transpose of the rotation

matrix is its inverse). It follows that ∂ rl∂ rm '

=R lmT =Rml and the “del operator” becomes

∂∂ rm '

=∂ rl∂ rm '

∂∂ rll=1

3

∑ = Rml∂∂ rll=1

3

∑ ⇒!∇( ) ' =R ⋅

!∇( ) , q.e.d.

Problem 2)

CD =1 0 00 cosθ sinθ0 −sinθ cosθ

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

cosφ sinφ 0−sinφ cosφ 00 0 1

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

cosφ sinφ 0−cosθ sinφ cosθ cosφ sinθsinθ sinφ −sinθ cosφ cosθ

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

BCD =cosψ sinψ 0−sinψ cosψ 00 0 1

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

cosφ sinφ 0−cosθ sinφ cosθ cosφ sinθsinθ sinφ −sinθ cosφ cosθ

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

cosψ cosφ − sinψ cosθ sinφ cosψ sinφ + sinψ cosθ cosφ sinψ sinθ−sinψ cosφ − cosψ cosθ sinφ −sinψ sinφ + cosψ cosθ cosφ cosψ sinθ

sinθ sinφ −sinθ cosφ cosθ

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

PHYSICS 603 - Winter/Spring Semester 20018 - ODU

Problem 3)

exp(M3ϕ ) = 1+M3ϕ +12M3M3ϕ

2 +16M3M3M3ϕ

3 + ...+ 1n!M3

nϕ n + ...

=1 0 00 1 00 0 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥+0 −1 01 0 00 0 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥ϕ +

12

−1 0 00 −1 00 0 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥ϕ 2 +

16

0 1 0−1 0 00 0 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥ϕ 3 +

124

1 0 00 1 00 0 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥ϕ 4 + ...

After this point, the same 4 matrices keep repeating themselves one after another. Gath-ering all terms yields

exp(M3ϕ ) =

1−ϕ2+ϕ 4

24− ... −ϕ +

ϕ 3

6− ... 0

ϕ −ϕ 3

6+ ... 1−ϕ

2+ϕ 4

24− ... 0

0 0 1

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

cosϕ −sinϕ 0sinϕ cosϕ 00 0 1

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

,

which is indeed the rotational matrix for an (active) rotation around the z-axis by an angle φ. This works because in some sense, the matrix M3 behaves like the imaginary number

i, which also gives -1 when squared. And in particular, the elements on the diagonal of M3

n are the real parts of in, and the elements on the off-diagonal are the (negative or posi-tive) imaginary parts. So in this sense , exp(M3ϕ ) is equivalent to

Re(eiϕ ) − Im(eiϕ ) 0

Im(eiϕ ) Re(eiϕ ) 00 0 1

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

cosϕ −sinϕ 0sinϕ cosϕ 00 0 1

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

.

In general, for any axis n̂ it is true that a rotation around that axis can be written as exp(n̂ ⋅

!Mϕ ) ; however, the calculation will be very difficult because the different M’s

don’t commute.