On 2 jun, 13:48, K1TTT wrote:
On Jun 2, 3:13*pm, Cecil Moore wrote:

On Jun 2, 9:31*am, K1TTT wrote:

easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation. *from Fields and Waves in
Communication Electronics, section 1.14. *a solution for the wave
equation in a transmission line (or other 1d form) must satisfy the
condition that the 2nd derivative wrt x and wrt t equal v-squared...
too hard to write in text. *but basically take the above and
differentiate twice wrt time and twice wrt distance and i don't think
you'll get anything that comes out to the velocity squared. *the
problem is that t and x must be part of the same sinusoid in the F(t-x/
v) so that the wave is truly traveling instead of stationary like your
equation provides.

in the basic maxwell equations in differential form you also run into
the same problem where the curl of the one field is proportional to
the time derivative of the other i think you'll end up with a
contradiction like k=w if you eliminate the dimensionless terms... i
may be wrong on that because its been so long since i've had to deal
with them on that level.

Oh!, I surrender :) The thread do not converge to a solution, it
seems run out toward a naked singularity! I stay tuned until this
good brainstorm calms down ..

Miguel