Re: Electromagnetic wave and photon spin
- From: "FrediFizzx" <fredifizzx@xxxxxxxxxxx>
- Date: Tue, 19 Sep 2006 00:21:51 -0700
"Sue..." <suzysewnshow@xxxxxxxxxxxx> wrote in message
news:1158619763.325412.221100@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
rds wrote:radiation?
First off I want to say that I am not physics major.
Can someone explain to me the details of how electromagnetic
a
Your spelling is better than mine so you should find this
a breeze:
http://farside.ph.utexas.edu/teaching.html
http://web.mit.edu/8.02t/www/802TEAL3D/teal_tour.htm
Actually occurs. I know the pat answers that the acceleration of
charged particles or change in direction, but that doesn't really
explain the process.
Sun electron shakes, eye electron shakes 8 minutes later.
~~ R.P Feynman
For that matter why does a magnetic field form when current flows at
aconstant rate (charge velocity)?
Magnetism is the geometric superposition of Coulomb
forces. The fundamentls are simple but the calculations
grow lots of hair because you can have to do them as
3 D volumes of space.
http://en.wikipedia.org/wiki/Multiple_integral
Where described as a moving charge
The electric field of the moving charges is interacting with other
matter which is ~stationary~ . At the minimum it is matter
in 'all of space' or 'free space'
permeability mu_0
permittivity eps_0
radiation resistance 377 ohms.
I notice in my car from 0 to 50 MPH , that when I accelerate I feel
thepull in the reverse direction, but when I reach the intended speed
backwards.pull subsides. In some way this is counter intuitive in that I would
assume that the 50 MPH pull would continually cause a drag
waves?In the same manner is this a parallel to the transmission of EM
orIn that I get a transmission (radiation) when charges accelerate
naturechange direction?
Not quite. The ~pull~ an electron feels is relatvive to its Coulomb
coupling to nearby matter. A force about 10^32 time larger than
those you feel in your automobile.
Lastly the description of EM wave as abstract photons (all energy no
mass) makes sense to me in that the EM wave propagates as an E field
(sin) + an M field (cos). Is there an actual difference in the
of these two fields? Or are they indistinguishable from one another
--more like two E-Fields with a 90 degree phase shift
In units where c = 1, the E field and B field of a monochromatic EM
radiation plane wave are identical except they are in phase and are
tranverse to the direction of motion. And 90 degrees rotated from each
other. However, that is an ideal case and nature is not usually so
perfect. ;-) But can be used as a good approximation in many cases.
Yes... You are spot on. The E component diminishes by 1/r^2
and is the farfield or radiative component. The magnetic
component diminishes by 1/r^3 and is effective in the nearfield
only.
You need to qualify your 1/r^3. That would only be true for certain
types of source configuration and/or boundary conditions.
Photons are a mathmatical abstraction not a propagation model.
Your question is to the heart of why Feynman puts wrist watches
on his photons to fix all the mirrors that QM broke. In QED,
the phase information and magnetic components are transported
on the tip of a mathematicians pencil. ;-)
Photons are definitely not a math abstraction. The are basically chunks
of propagating EM energy that don't disperse due to the nature of the
relativistic quantum "vacuum" medium. They have only helicity and some
value of momentum as intrinsic properties. All the EM associated with
them comes from the quantum "vacuum".
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com
.
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