How does reflection work?
T Murray
Can anybody explain in fairly simply terms how photons are reflected at a
plane metal surface. I can imagine photons interacting with free
electrons, but why are they ejected at a specific angle.
I hope this isn't too obvious, I haven't studied physics for ~15 years.
Tim Murray
Patrick van Esch
T Murray (tdmu...@dra.hmg.gb) wrote:
: Can anybody explain in fairly simply terms how photons are reflected at a
: plane metal surface. I can imagine photons interacting with free
: electrons, but why are they ejected at a specific angle.
: I hope this isn't too obvious, I haven't studied physics for ~15 years.
You're saying "photons" so I guess you want a quantummechanical
picture. If you would have said "electromagnetic wave" I would
have presumed you want to see things classically.
Quantum mechanically, photons simultaneously take many paths,
and according to certain rules, phases are calculated for every
path. (this is the path integral formalism as proposed by Feynman
and shown to be equivalent to standard quantum mechanics).
So a photon can simultaneously scatter off electron number 1,
electron number 2 etc...
these are to be considered as different alternatives that "happen"
simultaneously.
Now when a photon is supposed to be detected, it gets out its
pocket calculator and calculates the probability to be at a certain
point. This probability is given by the sum of contributions along
every path. It turns out that for endpoints that do not lie on
the "reflected track" the phases coming from the scattering from
electron 1 and electron 2 etc... are all different and hence tend
on the average to cancel one another.
OTOH, right ON the track of the reflected beam, all these phases,
are, well, eh, in phase and add together. So the probability of doing
that is very much higher than going somewhere else, and the photon
- after all these calculations - tosses a few coins and decides
to appear right there where the probability is high, eg. on the
reflected track.
hope this is what you were looking for !
cheers,
Patrick.
--
Patrick Van Esch
mail: van...@dice2.desy.de
for PGP public key: finger van...@dice2.desy.de
Doug Craigen
T Murray wrote:
>
> Can anybody explain in fairly simply terms how photons are reflected at a
> plane metal surface. I can imagine photons interacting with free
> electrons, but why are they ejected at a specific angle.
>
> I hope this isn't too obvious, I haven't studied physics for ~15 years.
Without knowing much about your physics background, its hard to know where
to start.
Did you ever study reflections of waves on a string? This is a fairly
common reference point for anybody who took intro physics.
Imagine you send a pulse down a string which is fixed at the other end.
_
_____/ \___________|
|
When the pulse reaches the end, it doesn't move. This can be described by
considering the string to be very long, with the opposite pulse coming the
other way:
_ ->
______/ \____________________ ________
\_/
<-
As the one pulse disappears into the fixed point, the other emerges from it,
and at all times they cancel at it. Hence, the fixed point generates a
reflection.
*****
Photon reflection can be understood in much the same terms. A metal is a
"fixed point" with regard to electric fields - E in a metal is zero. Hence
as a photon disappears into the metal surface, the surface behaviour is to
generate an opposite photon.
I'm not sure how to explain beyond this without some math and vector
diagrams...
|++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++|
| Doug Craigen |
| |
| If you think Physics is no laughing matter, think again .... |
| http://cyberspc.mb.ca/~dcc/phys/humor.html |
|++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++|
me...@cars3.uchicago.edu
In article <01bbd793$b09c0580$23155092@unknown>, "T Murray" <tdmu...@dra.hmg.gb> writes:
>Can anybody explain in fairly simply terms how photons are reflected at a
>plane metal surface. I can imagine photons interacting with free
>electrons, but why are they ejected at a specific angle.
>
>I hope this isn't too obvious, I haven't studied physics for ~15 years.
>
people. To see where it is coming from, you should think in terms of
waves. Imagine a plane wave impinging on surface and each surface
element, in turn, radiating in all directions. Now, the amplitude in
any direction is obtained by adding all the rereadiated wavelets which
ar going in this direction and, when you go through the details (not
very complicated but I'm not about to put it in ASCII) you find out
that in any direction you get waves of all phases, which cancel one
another. The sole exception is the specular reflection angle (i.e. an
angle equal to the incoming angle)> Here everything adds in phase and
you get your reflection.
The above is for a flat, featureless, reflecting surface. Now, what
happens if the surface has a periodic structure in it, i.e. the
reflectivity varies in a periodic way. Then you find that there are
additional directions in which you can get reflections (the directions
depend on the wavelength and the period of the structure). You just
got a diffraction grid.
In general the reflected amplitude, as a function of angle, is related
to the Fourier transform of the surface. with a flat mirror you've a
Fourrier transform of a constant, i.e. a delta function (single
reflection angle). Once you get features in the surface, it becomes
more interesting.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Brice Wightman
Patrick van Esch (van...@jamaica.desy.de) writes:
> T Murray (tdmu...@dra.hmg.gb) wrote:
> : Can anybody explain in fairly simply terms how photons are reflected at a
> : plane metal surface. I can imagine photons interacting with free
> : electrons, but why are they ejected at a specific angle.
>
> : I hope this isn't too obvious, I haven't studied physics for ~15 years.
>
>
> You're saying "photons" so I guess you want a quantummechanical
> picture. If you would have said "electromagnetic wave" I would
> have presumed you want to see things classically.
>
> Quantum mechanically, photons simultaneously take many paths,
> and according to certain rules, phases are calculated for every
> path. (this is the path integral formalism as proposed by Feynman
> and shown to be equivalent to standard quantum mechanics).
>
> So a photon can simultaneously scatter off electron number 1,
> electron number 2 etc...
> these are to be considered as different alternatives that "happen"
> simultaneously.
> Now when a photon is supposed to be detected, it gets out its
> pocket calculator and calculates the probability to be at a certain
> point. This probability is given by the sum of contributions along
> every path. It turns out that for endpoints that do not lie on
> the "reflected track" the phases coming from the scattering from
> electron 1 and electron 2 etc... are all different and hence tend
> on the average to cancel one another.
> OTOH, right ON the track of the reflected beam, all these phases,
> are, well, eh, in phase and add together. So the probability of doing
> that is very much higher than going somewhere else, and the photon
> - after all these calculations - tosses a few coins and decides
> to appear right there where the probability is high, eg. on the
> reflected track.
For an extended version of this explanation with figures, read "QED" by
Richard Feynman.
--
===========================================================================
Brice Wightman am...@freenet.carleton.ca
Ottawa, Canada VE3EDR
===========================================================================
Bill Oertell
It might be worth mentioning that objects don't "reflect" radar
energy; they absorb and reradiate it. This is one of the reasons for
the Stealth fighter's unique wings. They're designed to focus most of
the radar energy aft.
--
Bill
------------------------------------
| If everything is possible, |
| nothing is knowable. Be skeptical.|
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