Re: Antonym of 'evanescent'?

In article <joegwinn-A99B23.10413022122007@xxxxxxxxxxxxxxxxxxxxxxxx>,
Joseph Gwinn <joegwinn@xxxxxxxxxxx> wrote:

In article <siegman-DA76C3.17434221122007@xxxxxxxxxxxxxxxxx>,
AES <siegman@xxxxxxxxxxxx> wrote:

In article <joegwinn-F339D3.18334521122007@xxxxxxxxxxxxxxxxxxxxxxxx>,
Joseph Gwinn <joegwinn@xxxxxxxxxxx> wrote:


Not to be a cold blanket, but how about "real"? If I recall, the
evanescent fields are (or can be) modeled mathematically as imaginary
(as in square root of minus one).

Joe Gwinn

Nope, sorry. The fields in different regions all can have in the most
general description the general form

e[x,y,z] = Exp[ - j kVec . rVec + gVec . rVec ]

where rVec is the vector rVec = {x,y,z] ; kVec (sometimes called the
"propagation vector" or "the k vector") is a vector with in general kx,
ky, kz components; and ditto for the "gain" vector gVec = {gx,gy,gz}
(which can also be a loss vector if you change its sign).

In general kVec and gVec need not be parallel. In the usual, elementary
lossless evanescent wave case kVec and gVec are at 90 degrees.

Hmm. So kVec and gVec are in quadrature in (lossless) evanescent waves.
That's probably the source of my distant memory, which regarded the
specific case of frustrated total internal reflection, at optical and
microwave frequencies. For instance, the gVec is imaginary as compared
to the kVec.

Joe Gwinn

As written above, kVec and gVec are both ***purely real*** vectors.
.

Relevant Pages

  • Re: Antonym of evanescent?
    ... AES wrote: ... Joseph Gwinn wrote: ... In general kVec and gVec need not be parallel. ...
    (sci.optics)
  • Re: Antonym of evanescent?
    ... Joseph Gwinn wrote: ... evanescent fields are modeled mathematically as imaginary ... In general kVec and gVec need not be parallel. ...
    (sci.optics)