Re: Weirdo's square of negative mass.
- From: spintronic <spintronic@xxxxxxxxxxx>
- Date: Thu, 12 Nov 2009 20:05:08 -0800 (PST)
On 13 Nov, 00:21, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:
On Nov 9, 2:03 pm, spintronic <spintro...@xxxxxxxxxxx> wrote:..
..
The laws of conservation of energy and momentum leave no choice : W' -
W = h (1 - cos A)/M(t) c
That's not right.
That's Compton's formula which you accept when it suits your needs.
The W' was on a different line, I didn't see it.
And h/mC is actually the atomic wavelength of the subatomic particle.
Obviously the wavelength is shorter for relativistic momentum = h/(m_0/
gamma)C
Actually, m_0 is greater for relativistic momentum.
I said the "Wavelength was shorter"
Because m' is increased.
m_0 (In this case) is constant. m' is greater, but m' = m_0/sqrt(1-
v^2)
The scattering from the electrons was easier to detect, so that's what
he looked for.
If the photons bounce off of a free ELECTRON, conservation of energy
and momentum requires using the mass of the ELECTRON.
If the photons bounced off of a free PROTON, conservation of energy
and momentum requires using the mass of a PROTON in the equations.
(one of the requirements for Compton scattering is that the particle
is free to move)
Which Iv'e been saying all along.
Until the data shows your 'model' is full of sh*t, in which case you
start screaming badly spelled insults at people ...
Eh????
1) My model is sound
2) What part of "Which Iv'e been saying all along" confused you?
So if the proton struck a carbon NUCLEUS, conservation of energy and
momentum requires using the mass of the object hit in the equations
(in this case, the energy would rattle around amongst the neutrons and
protons until another photon is ejected, with energy in accordance to
Compton's equations).
I'm glad you brought this up. I argued in 2001, that a hydrogen atom
can only emmit in the direction of the photon that excited it
I can prove it, if you wan't another thread.
Nope - it can eject a photon IN ANY DIRECTION.
No. It can't.
Photon hits atom
() <~~~
v=0
()
--|--
|
/ \
ObseverA: Moving left
v>0
()
--|--
|
/ \
ObseverB: Stationary
v=0
******************************
Atom gets excited
( )
Relative obsever A atom v = 0.
()
--|--
|
/ \
()
--|--
|
/ \
ObseverB: moving (relatively) right
v>0
**********************************
Observer B saw the atom in it's ground state.
With energy E.
The photon hit it now has energy E'= E+E_y
It now has energy & momentum moving left.
If the atom emmits a photon to the right, it now has
Energy E (AGAIN), BUT HAS MOMENTUM 2P.
(which is impossible)
.
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