Re: Tracking

: Erik Max Francis <max@xxxxxxxxxxx>
: I still don't follow where this imaginary photon wavelength is coming from.

That would be from the factor of gamma in relativistic doppler.
For classical doppler, the frequency becomes negative.
Of course, we are ignoring the fact that those expressions
are supposing velocity of emitter is less than velocity of signal.
But, oh well.

: When the photon is emitted it travels at c. It never travels at any
: other speed.

Right. Which has nothing to do with this issue.

: If it has finite energy when it is emitted, it has finite wavelength
: (or frequency).

Shrug. And if doppler says it'd have imaginary energy when emitted[1],
then it'd have imaginary wavelength. Or vice versa. I don't know why
this should be puzzling. It may be unphysical, but not puzzling.

[1] Yes, I realize that this is cheating a bit, given what doppler
is supposed to be calculating. But only a bit. The cheating is,
however, related to the fact that photons do not have, intrinsically,
wavelength of energy. Those things depend on how fast the object
that absorbs them is moving. Similar for kinetic energy, etc, etc.
Conserved but not invariant, and all that.
signal propogation, so


Wayne Throop throopw@xxxxxxxxx http://sheol.org/throopw
.

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