3. Diagonalization of the off-diagonal matrix
In the last section, matrix was simplified as much as possible. All the way to the final form we have seen how to incorporate some atmospheric gradients into the exponential solution, and at the end we have required the commutation condition to still hold with minimum freedom restrictions. We can then calculate the matrix in the exponent of the solution; but we must compute the exponential of this matrix too. The last can be done by using the matrix series (4), but we prefer the comfort of calculating the exponential of scalars. We can achieve this purpose by diagonalizing . In fact, we only need to diagonalize . We solve the eigenvalue equations for this matrix:
The four eigenvalues are given by:
which is a biquadratic equation, so first we solve
and the four eigenvalues are therefore:
We introduce the notations:
Using this notation, one finds that the corresponding eigenvectors are given by
We may now calculate matrices and which diagonalize :
These transformation matrices and just derived can be applied in general, save for those exceptions where they become singular. These particular cases are:
1. The case when .
To solve this singularity we suggest the following transformation matrix and its inverse as a substitution of the previous ones:
with the resultant diagonalization of :
2. The case when .
With no loss of generality one can take and then (and subsequently ) becomes the identity matrix. For the diagonalization transformation, the eigenvalues are the following:
Here is reduced to
and the diagonalizing matrix is:
3. The case when , and .
In this case, ; is imaginary if . The eigenvalues are given as follows:
and the transformation matrix
4. The case when , and .
This is the only case where matrix cannot be diagonalized. It will be reduced by using
and its inverse, so that we get
Note that the resulting matrix, although not diagonal is simple enough as to have its exponential calculated as
where we have used the notation . Back to the general case, we diagonalize (or apply the equivalent transformation in case 4) to the RTE. It is a new transformation of the equation, as were , and , and as in these 3 rotations, a term of the form
appears in the transformed equation. However, as all the entries of , and subsequently of , are constant, this term is immediately zero, in perfect agreement with the statement that expression (27) is a necessary and sufficient condition to write a scalar-like exponential solution. 1
For the general case, we define the new generalized Stokes and emission vectors after the diagonalizing transformation for the RTE as
is the diagonalized matrix. For the last singular case, the equivalent reduced matrix should be used.
© European Southern Observatory (ESO) 1999
Online publication: December 22, 1998