__Bohr's H Atom Revisited__

** Abstract.** A reason is suggested
why the electron in a Bohr Hydrogen atom doesn't radiate.

In1913 Niels Bohr introduced his model of the Hydrogen
atom. He suggested that the atom consists of a central, resting proton (nucleus)
with an electron orbiting in a circle. (In the lowest energy state the radius of
the orbit is relec = .529177E-10 meters.) In every orbit the electron's speed
is such that its angular momentum, m_{elec}v_{elec}r_{elec}, equals the constant h/2p.
This restriction dictates that only certain discrete angular momenta are
allowed. (For the lowest energy, v_{elec}=.2187542E7 meters/second.)

The orbiting electron in his model * does not radiate* ... a clear
contradiction to the time-honored proposition that a charge going in a circle
must radiate. Bohr's response to this objection was (reportedly) simply that, in the case of an atom, there is
no radiation in any stable
state (including the "ground" or lowest energy state).

In this article we take a more detailed look at Bohr's model.
The objective is to ascertain if there is a physical explanation for why his
model, suitably modified, does not radiate. Our strategy will be to demonstrate that the *total
tangential*
force on the electron ... its radiation reaction force and the
tangential component of the Lorentz force from the proton ... sum to zero. If
such is the case, then the total work done on the electron in any orbit would be
zero, and the atom should neither emit nor absorb radiant energy.

Now a legitimate objection is that in Bohr's
model (with its resting proton) * the Lorentz force has no tangential
component*, and thus the electron's
radiation reaction force would be *unopposed*. The atom should radiate (quite as
many complained when Bohr introduced his model). In meeting this objection we shall suppose
that, looking down from the positive z-axis, the electron and proton * both * circle
the Origin in a counterclockwise direction in the xy plane, and at a *common angular
speed,* w.
The radius of the more massive proton's circular orbit, r_{prot}, is of course smaller than that
of the electron (r_{elec}).

At any given
instant the two particles are diametrically opposed. At time t=0, the electron
is at (-r_{elec},0) and the proton is at (r_{prot},0). The
electron is moving at the speed v_{elec}, and the proton at the lesser speed
v_{prot}.

Bohr, as we know, suggested that the nucleus of the atom was at rest. But if the
*
linear* momentum of the entire atom is
zero at all times, then at any moment the proton must be moving in the opposite direction as the
electron. Owing to this motion there are delayed time and distance factors
that should be considered in computing the interactive force on the electron.
And when these factors are appropriately sized, the y-component of the Lorentz force on the electron
will not be zero! There is a small component that points in the same direction as the electron's velocity.
And that
direction * is * opposite to the radiation reaction force that the
electron experiences as a consequence of its circular motion.

The orbiting proton of course engenders a magnetic field at the electron. But the magnetic force on the electron is always perpendicular to its velocity, and plays no role in canceling the radiation reaction force.

A schematic illustration of the situation at time t=0 is provided in Appendix A. We know from Bohr's angular momentum rule what the electron's angular frequency is:

w=v_{elec}/r_{elec}=4.133856155E16/second.
(1)

And in order for the * total* momentum to be zero, we presumably have

m_{prot}wr_{prot}=m_{elec}wr_{elec}.
(2)

But we also have

r_{prot}=v_{prot}/w.
(3)

Dividing (2) by (3) and solving for vprot produces

v_{prot}=v_{elec}m_{elec}/m_{prot}
= 1193 meters/second. (4)

And

r_{prot}=v_{prot}/w=2.8820E-14
meters. (5)

At the delayed time, t_{d}, the proton is at an angle
q below the x-axis, where

q=w(-t_{d}).
(6)

We are now prepared to compute the Lorentz
force to find the y-component of the interactive force experienced by the electron.
In order to compute this, we shall invoke the formula for the proton's **E**
field. This formula is relativistically rigorous, and takes into account the
retarded effects. (Griffiths' text, **Introduction To Electrodynamics**,
provides an example of the formula.)

The y-component of this force is

F_{y}_{ }= -qE_{y}.
(7)

For the radiation reaction force we have

FRR=q^{2}w^{3}r_{elec}/(6pe_{0}c^{3}).
(8)

Using the computer we can "creep up" to just the delayed time needed to yield the result

**F**_{y}+**F**_{RR} __~__ 0.
(9)

Postscript: The modeling program was modified to compute Fy and
F_{RR} for the *electron's
force on the proton*. Again, the objective was to see if the y-component of
the interactive force on the proton is equal and opposite to the radiation
reaction force on that particle. It was found that F_{y}=1.16286192e-17, and
F_{RR}=-1.16286189e-17 ... again practically an equality. Thus together the electron
and proton, and hence the atom, do not radiate.

A
final note worth mentioning: The requirement that the atom's linear momentum
equal zero leads to a proton angular momentum * not* equal to h/2p.
Thus that restriction applies only to the electron in Bohr's atom.

__Appendix
A__

The Bohr Atom at a Typical Delayed Time

Appendix
B contains the Python code used to compute the values of FRR and F_{y}.

__Appendix
B__

#For the articls, Bohr Revisited.

import math

c=2.99792458e8 #Speed of light

h=6.626068e-34 #Planck constant

eps0=8.85418782e-12 #Permittivity constant

q=1.60217662e-19 #Elementary charge

ElectronMass=9.10938356e-31 #Electron rest mass

ProtonMass=1.6726219e-27 #Proton rest mass

N=0

while N<=10:

N=int(input('level=?(>10 to exit) '))

if(N<11):

print('Computing level ',N)

ElectronOrbitRadius1=.529177211067e-10

ElectronOrbitRadius=N**2*ElectronOrbitRadius1 #N'th level electron orbit radius

ElectronSpeed1=h/(2*math.pi*ElectronMass*ElectronOrbitRadius1)

ElectronSpeed=h/(2*math.pi*ElectronMass*ElectronOrbitRadius1*N) #Electron orbital speed

ElectronKineticEnergy1=ElectronMass*ElectronSpeed1**2/2

ElectronKineticEnergy=ElectronMass*ElectronSpeed**2/2 #Electron kinetic energy

ElectronOrbitalPeriod=2*math.pi*ElectronOrbitRadius/ElectronSpeed #Electron & proton orbital period

ElectronOrbitalFrequency=1/ElectronOrbitalPeriod #Orbital fequency

omega=2*math.pi*ElectronOrbitalFrequency #Angular frequency

ProtonOrbitalSpeed1=ElectronMass*ElectronSpeed1/ProtonMass

ProtonOrbitalSpeed=ElectronMass*ElectronSpeed/ProtonMass #Net momentum of atom is zero

ProtonOrbitalRadius1=ProtonOrbitalSpeed1/omega

ProtonOrbitalRadius=ProtonOrbitalSpeed/omega #Proton orbital radius

ProtonKineticEnergy=ProtonMass*ProtonOrbitalSpeed**2/2 #Proton kinetic energy

PE=-q**2/(4*math.pi*eps0*(ElectronOrbitRadius+ProtonOrbitalRadius)) #Potential energ of atom

TotalEnergy=ElectronKineticEnergy+ProtonKineticEnergy+PE #Total energy of atom

maxN=1e8 #Number of iterations in oebital computation

DelayedTime=(ElectronOrbitRadius1+ProtonOrbitalRadius1)/c #Time for signal to travel from prot to elec

dt=-DelayedTime/maxN #Time increment

loopN=-1 #Level no.

theta=0. #See diagram

RadiationReactionForce=q**2*omega**3*ElectronOrbitRadius/(6*math.pi*eps0*c**3) #Rad React force on elec

RetardedProtonSpeed=ProtonOrbitalSpeed #Proton travels at constant speed

ar=RetardedProtonSpeed**2/ProtonOrbitalRadius #Angular acceleration of proton

Fy=0. #Force of prot on elec

#Find parameters when forces on electron sum to zero

while RadiationReactionForce+Fy>=0:

loopN=loopN+1

#Print params and exit by clicking X in display

if loopN>maxN:

print('loopN,RadiationReactionForce,Fy',loopN,'; ',RadiationReactionForce,', ',Fy)

print('Infinite Loop; Aborting')

input('OK?')

#If necessary, keep 'inching up'

RetardedTime=-loopN*dt

theta=omega*RetardedTime

RetardedVelocityX=ProtonOrbitalSpeed*math.sin(theta) #See diagram

RetardedVelocityY=ProtonOrbitalSpeed*math.cos(theta)

RetardedAccelerationX=-(ProtonOrbitalSpeed**2)/ProtonOrbitalRadius*math.cos(theta)

RetardedAccelerationY=ProtonOrbitalSpeed**2/ProtonOrbitalRadius*math.sin(theta)

L=math.sqrt((ElectronOrbitRadius+ProtonOrbitalRadius)**2-(ProtonOrbitalRadius*math.sin(theta))**2) #See diagram

alpha=math.atan(ProtonOrbitalRadius*math.sin(theta)/L)

Lx=-L*math.cos(alpha)

Ly=L*math.sin(alpha)

#Compute electric field at electron (see Griffith's text)

ux=c*Lx/L-RetardedVelocityX

uy=(c*Ly/L-RetardedVelocityY)

u=math.sqrt(ux**2+uy**2)

ux=-u*math.cos(alpha)

uy=u*math.sin(alpha)

Ey=q/(4*math.pi*eps0)*L/(Lx*ux+Ly*uy)**3

Ey=Ey*(uy*(c**2-RetardedProtonSpeed**2)-(Lx*(ux*RetardedAccelerationY-uy*RetardedAccelerationX)))

Fy=-q*Ey #Interactive force on electron

#When Fy+RadiationReactionForce~0, display results.

print('RadiationReactionForce,Fy= ',RadiationReactionForce,', ',Fy)

#Do another level.