Phy 515 class notes for Spring, 1999
Phy 515, Week 14
26 Apr
Work out some detailed examples of the adjoint of an operator. In
particular, let L = d2/ dx2. The domain of L is
the set of functions on 0<x<L with the boundary conditions
u1( 0 ) = 0 = u1'( 0 ) =
u1( d ) = u1'( d ).
I had found the adjoint of this Friday; it was the same differential
operator, but on functions u2 that are bounded, with bounded
derivatives at the endpoints.
What is the eigenvalue problem for this adjoint operator? The Sturm-Liouville
equation (with w = 1) is
d2 u2 / dx2 =
l u2.
The solutions are just exponentials,
exp( i l1/ 2 x ). If you look at the
boundary conditions, you see that they are satisfied no matter what the
values of l are. The set of such
l is the entire complex plane.
For comparison, look at the boundary conditions on the original
operator, where the function and the derivative vanish at both
endpoints. The only solution that does this is the constant, zero, and
that's not very helpful in doing Fourier expansions.
For comparison, look at a finite-dimensional version of this problem.
If a rigid body is rotating, its angular momentum is a function of its
angular velocity. The angular momentum of a point mass is just
r ¥ p. The linear velocity
of a point within the body obeys
v = w ¥ r. Put these
together, and you have the total angular momentum of a rotating object:
This defines L in terms of w.
That is,
L = I( w ). This function is
linear, meaning that it satisfies the equation
I( aw1 +
bw2 ) =
a I( w1 ) +
b I( w2 ).
The scalar product
that appeared in the context of Sturm-Liouville theory has its analog in
the ordinary three-dimensional dot product, and you can manipulate
everything much the same way. It's much simpler in the the latter case
though, because you don't have to worry about boundary conditions and
about convergence.
The first thing is to establish that this operator is self-adjoint. In
finite dimensions, this is easy.
< w2 ,
I( w ) >
=
< I†( w2 ) ,
w1 >
This is
|
< w2, I( w1 ) > = w2* · |
ó
õ
|
dm r ¥( w1 ¥r ) |
|
|
= w2* · |
ó
õ
|
dm [ w1 r2 -
r ( w1 ·r ) ] |
|
|
= |
ó
õ
|
dm [ w2*·w1 r2 -
( w2* ·r )(
w1 ·r ) ] = < I( w2 ), w1 > |
|
In finite dimensions, this is all that is needed to show that I is
self-adjoint.
The derivation that I went through
for the differential equations,
leading to the orthogonality relationship, has a parallel construction
here. In the differential equation case, I wrote the equation for two
values of l, complex conjugated one, multiplied
one equation by the second solution, and vice versa, subtracted and
integrated. After that came a lot of partial integration.
The manipulations for the inertia operator precisely parallel these.
I( w1 ) =
l1
w1,
and
I( w2 ) =
l2
w2
After the parallel manipulations, you get
( l2*
-
l1 )
w2*
·w1 =0.
From this equation, you deduce that the l's are
real, and that for different l's, the
corresponding w's are orthogonal.
28 Apr
Do the problem from Schaum: summing the series
S ( z -
np )-2.
Look further at the tensor of inertia, and tensors in general. Show how
to find the components of a tensor with respect to some basis. For this
example, use the common orthonormal basis. The key observation is that
for a vector
w, the value of the function,
I( w ), is a
vector. This is why you have to distinguish between the function, I,
and the value of the function evaluated for some specific argument. The
former is a tensor, the latter is a vector. The key property you need
to pay attention to is the linearity of this function as stated above:
I( aw1 +
bw2 ) =
a I( w1 ) +
b I( w2 )
When you write the equation
L = I( w ), you can write all the
vectors in terms of basis vectors. For now, just take them to be the
common unit vector x, y, and z, though this
restriction isn't really needed.
Lxx + Lyy + Lzz =
I( wxx +
wyy +
wzz )
= wx I( x ) +
wy I( y ) +
wz I( z )
The three expressions, I( x ), I( y ), I( z ), are
all vectors. This means that they in turn can be expressed in terms of
this basis.
I( x ) = Ixx x + Iyx y +
Izx z
The three numbers Ixx, Iyx, Izx are the
components of this vector. The notation for the index denotes which
component you are dealing with as well as which argument you are using
in the function I. Of course there are two other similar equations for the
components of I( y ) and I( z ).
Put these together with the equation above for L to get
Lxx + Lyy + Lzz =
wx
( Ixx x + Iyx y +
Izx z )
wy
( Ixy x + Iyy y +
Izy z )
wz
( Ixz x + Iyz y +
Izz z ).
When two vectors are equal, you can equate like components. This gives
three equations for the different components.
Lx =
Ixx wx +
Ixy wy +
Ixz wz
Ly =
Iyx wx +
Iyy wy +
Iyz wz
Lz =
Izx wx +
Izy wy +
Izz wz
This is more commonly written as a matrix equation, and that is the
reason for the odd-looking way that matrix multiplication is defined.
It's just representing the components of this function.
Note: Nothing in the above derivation used the fact that the
basis vectors are orthogonal or normalized. No scalar product appeared
in the construction of the components of the tensor. In order to
understand these basic relationships between tensors (linear operators)
and their components, you should not use scalar products.
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