Re: Ages everyone?

David Klassen <klassen@xxxxxxxxx> wrote:
Keith Davies wrote:
David Klassen <klassen@xxxxxxxxx> wrote:
Flux is energy per area. If and distance X, a unit steradian
subtends area A=X*X, then at 2X the area is 2X*2X=4A. So
flux is 1/r^2. Assuming a point source---they radiate equally
in all directions.

A 2-d "flux" would, I presume, be defined as energy per
unit length crossed, which would then be 1/r. And if you
define a 4-d flux you'd get 1/r^3.

You can also get 1/r^3 for static dipoles or differential effects
of 1/r^2 systems (i.e. gravity is 1/r^2, but tides are 1/r^3).

I guess I'll have to take your word for it, I'm not a physicist.

(I am. Sort of... I prefer Planetary Astronomer since that's
what I "do", but the last degree was technically physics; and
that's what I teach.) But this isn't so much physics as geometery.

Just reading your description, it sounded like physics. I wasn't
suggesting you are a physicist, just admitting that I'm not.

If the signals went out 'linearly' (and ignoring outside effects) there

Exactly. Of course, that's "linearly" in a qualitative sense and
not a purly mathematical one. Since the energy goes as the square of
the amplitude of the wave. But it's really (almost) immaterial of
that. The idea is that the energy is evenly distributed in all
directions no matter the distance and then you don't have to think of
waves.

Or, you could think of it as photons shot out in all directions
equally. The further away you are, the fewer of them are going
to hit you, based solely on the geometry.

I was speaking loosely, because that's how it made sense. Shoot 'raser'
(like a laser, but radio), it goes straight out. Barring interference,
you should expect to see the same power at all points along the path.
'Linear', as it were.

Once you broadcast 'in a circle' or 'in a sphere', things change.

As you said. That's more or less the mental model I used.

should be 'no attenuation'. If they go out in two dimensions (surface
of a pond), the total wavefront would be proportional to the distance
from the signal source (circumference of a circle == $2r\pi$. To extend

So, what made you think I understood TeX/LaTeX notation?

Well, for one thing you appear to have a good idea of what you're
talking about, scientificly. How many trained scientists *don't*
recognize TeX?

Oh, wait, have I seen you posting in comp.text.tex?! I think
I remember seeing someone from here over there...

I do post there, very sporadically. Even more than here -- here I'll
miss a week or two, there I'll miss a few months at a time... I usually
just read there when I'm actually using (La)TeX.

that to three dimensions, the area of the wavefront is proportional to
the square of the distance traveled (${\frac{4}{3}}r\pi^2$).

Which, when you fixed it to $4\pi r^2$ was right.

.... yeah, I did fix that, in a followup to Eric. I don't know *where* I
got that one. Must've been the drugs.

I get it now.

Cool! Sometimes it just helps to "say it out loud".

Yeah.

I remember reading about (and could look up, I know I still have the
book it was mentioned it) a university where they had a grad student
debugger available to the students. However, before you were allowed to
see this person, you were required to describe your problem to the teddy
bear sitting in a chair outside his office.

Apparently this chopped the debugger's workload by more than 80%.


Keith
--
Keith Davies "Trying to sway him from his current kook-
keith.davies@xxxxxxxxxxxx rant with facts is like trying to create
keith.davies@xxxxxxxxx a vacuum in a room by pushing the air
http://www.kjdavies.org/ out with your hands." -- Matt Frisch
.

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