Monday, September 29, 2014

Not This Way...

The interaction between electron and photon is repulsive,


Consider a photon on approach to a electron, at a horizontal distance  dd  away,  what is the deflection?  We break the path of the photon into 2 parts, before it is level horizontally with the electron and after it is level with the electron.

Deflection before being level with the charge,

r=d2+(Doct)2

sin(θ)=dr=dd2+(Doct)2

Fh=Fsin(θ)=qpq4πεor2sin(θ)=qpq4πεo(d2+(Doct)2)dd2+(Doct)2

xd=t=Do/c0Fhmpdt=t=Doc0qpq4πmpεod(d2+(Doct)2)3/2dt=qpq4πmpεo{(ctDo)cdd2+(Doct)2}Do/c0

xd1=qpq4πmpεoc{Dodd2+D2o}

Deflection after being level with the electron,

r=d21+(ct)2

where  d1=xd1+d

sin(θ)=d1r=d1d21+(ct)2

Fh=Fsin(θ)=qpq4πεor2sin(θ)=qpq4πεo(d21+(ct)2)d1d21+(ct)2

xd=t=D/c0Fhmpdt=t=D/c0qpq4πmpεod1(d21+c2t2)3/2dt=qpq4πmpεo{td1d21+c2t2}D/c0

xd2=qpq4πmpεoc{Dd1d21+D2}

d1=xd1+d

xd2=qpq4πmpεoc{D(xd1+d)(xd1+d)2+D2}

If we estimate the optical path as the hypotenuse of a triangle  (It is usually wrong to estimated optical path to find path difference, because the actual path difference is already small.), the optical path difference between a straight photon and a deflected photon is,

OP=(Do+D)2+(xd1+xd2)2(Do+D)

For destructive interference,

(Do+D)2+(xd1+xd2)2(Do+D)=(2n+1)λ2

where  n=0,1,2,3...

(Do+D)2+(xd1+xd2)2Do+D1=(2n+1)λ2(Do+D)

(Do+D)2+(xd1+xd2)2Do+D=(2n+1)λ2(Do+D)+1

1cos(θ)=(2n+1)λ2(Do+D)+1

cos(θ)=2(Do+D)(2n+1)λ+2(Do+D)

Since,

2(Do+D)(2n+1)λ+2(Do+D)<1

we know that  θ  has solutions.

when  n is small,

2(Do+D)(2n+1)λ+2(Do+D)1,    θ=0


For constructive interference,

(Do+D)2+(xd1+xd2)2(Do+D)=nλ

n=0,1,2,3,...

1cos(θ)=nλDo+D+1

cos(θ)=Do+Dnλ+Do+D

when n  is small,

Do+Dnλ+Do+D1   θ=0

Both constructive and destructive interference mix and there is no pattern!  There is no observable interference pattern due to the interaction between the straight photons and deflected photons.