The Red Shift Mechanism

[Lazarus]

Red shift of light can be calculated from the changing energy required to go from one atomic shell to another.

 

Consider the electron in the elliptical Bohr hydrogen atom going from shell 2 to shell 3. With the atom at rest a photon released will have a wave length of 656 nm and energy of 30e-20 joules.

 

The when the atom is moving the latent energy doesn't change but the kinetic energy of the electron does. That means the energy of a photon released at the apogee of the ellipse will diminish as the velocity of the atom increases. The energy of the photon is equal to the change in the latent energy minus the change in kinetic energy.

 

The similarity of the computed energy and wave length to the currently used values is remarkable.

 

The condition of the calculated table following is:

 

The elliptical Bohr atom, shells 2 and 3

 

Atom velocity

* * * * * * * >* * * * * * * down

* *p * * >* e * e *

* * * * <* * Electron clockwise * Photon velocity

* * up

* * * * * * * <* * * * * *

 

 

The initial conditions are:

The semi-major axis of the ellipse is the same as the circular Bohr radius

The energy of the photon is the accepted one for change of level.

The velocity of the electron shell 1 electron at the apogee was chosen to match the wave length of a photon from an atom at rest.

 

The table below was generated by:

1 Calculating the energy required for an electron to go from the apogee of shell 2 to the apogee of shell 3, using various atom velocities. The light generated is red .

2 Calculating Z from the calculated wave lengths.

3 Calculating the atom velocity from the value of Z.

 

The calculated velocities are quite similar to the accepted velocities for matching wave lengths.

 

 

Redshift Calculation from Elliptical Bohr Hydrogen atom

Red light Wave length = 656 nm Wave Photon

Atom Shell 2 Shell 3 Calculated Length Energy

Velocity Velocity Velocity Vel from Z Z in nm *e-20

0 3,325 2,217 0 0.000 656.112 30.2966

100,000 103,325 102,217 100,000 0.000 656.331 30.2865

400,000 403,325 402,217 400,201 0.001 656.988 30.2563

1,000,000 1,003,325 1,002,217 1,001,503 0.003 658.306 30.1957

2,000,000 2,003,325 2,002,217 2,006,338 0.007 660.515 30.0947

3,500,000 3,503,325 3,502,217 3,519,837 0.012 663.856 29.9432

5,600,000 5,603,325 5,602,217 5,651,326 0.019 668.591 29.7312

8,400,000 8,403,325 8,402,217 8,516,142 0.029 675.009 29.4485

12,000,000 12,003,325 12,002,217 12,237,762 0.042 683.446 29.0850

16,500,000 16,503,325 16,502,217 16,950,182 0.058 694.292 28.6306

22,000,000 22,003,325 22,002,217 22,800,477 0.079 708.026 28.0753

28,600,000 28,603,325 28,602,217 29,951,341 0.105 725.241 27.4088

36,400,000 36,403,325 36,402,217 38,583,267 0.138 746.697 26.6213

45,500,000 45,503,325 45,502,217 48,895,735 0.179 773.391 25.7024

56,000,000 56,003,325 56,002,217 61,106,276 0.229 806.665 24.6422

68,000,000 68,003,325 68,002,217 75,445,441 0.293 848.380 23.4305

81,600,000 81,603,325 81,602,217 92,144,384 0.374 901.198 22.0573

96,900,000 96,903,325 96,902,217 111,409,641 0.477 969.071 20.5124

114,000,000 114,003,325 114,002,217 133,376,657 0.613 1058.140 18.7858

133,000,000 133,003,325 133,002,217 158,029,811 0.796 1178.492 16.8673

154,000,000 154,003,325 154,002,217 185,073,456 1.054 1347.944 14.7469

177,100,000 177,103,325 177,102,217 213,740,068 1.440 1601.199 12.4144

202,400,000 202,403,325 202,402,217 242,537,915 2.073 2016.056 9.8598

230,000,000 230,003,325 230,002,217 268,988,479 3.283 2810.402 7.0730

260,000,000 260,003,325 260,002,217 289,497,775 6.492 4915.623 4.0438

292,500,000 292,503,325 292,502,217 299,620,447 38.747 26078.370 0.7622

 

 

So why do the electrons stay in their orbit?

 

All that is needed for the distinct fixed orbits is for the nucleus to have a periodically changing or rotating magnetic field. I thought that would be highly unlikely until I found that the Handbook of Nuclear Properties of 1997 mentions nuclei with magnetic forces having a rotational character.

 

[swansont]

The Bohr model is wrong. How does this affect your results?

[Lazarus]

Hi Tom,

 

The Bohr atom is really irrevelant.

 

The electron orbiting a proton is all that counts.

 

I am fully aware that it goes against lots of things

but I would appreciate any suggestions of the

logical fallacy involved.

 

Bob

[swansont]

The fallacy is that the electron has an orbit.

[Lazarus]

The fallacy is that the electron has an orbit.

 

Everything else can orbit so why can't an electron?

[Greg H.]

 

Everything else can orbit so why can't an electron?

 

It's not that they can't, it's that they don't. Do some reading on the electron cloud model.

[swansont]

 

Everything else can orbit so why can't an electron?

 

For the reasons that the Bohr model is wrong. Having a classical trajectory means it should continually radiate, and any trajectory will have angular momentum, but we know the S states (like the ground state of hydrogen) have zero orbital angular momentum.

[Lazarus]

It would seem that an electron should only radiate on a change of kinetic

or latent energy., not a change of momentum. Otherwise, there is

creation of energy.

Edited July 25, 2013 by Lazarus

[Greg H.]

Changes in momentum cause changes in kinetic energy by:

[math]E_k = \frac{p^2}{2m}[/math]

 

where

 

p is momentum

m is mass of the body.

Edited July 25, 2013 by Greg H.

[Lazarus]

A ball going east at 1 mile per hour and rebounded

at 1 mile per hour going west would have the same

kinetic energy but different momentum.

[Greg H.]

What you just stated is a physical impossibility unless the ball somehow gained mass by bouncing off a wall it passed none of its kinetic energy to.

[Lazarus]

Try this.

 

A comet goes by the earth. At a distance of on milliion miles

coming and going it's kinetic and latent energy are the same.

However, it's momentum is different.

[swansont]

It would seem that an electron should only radiate on a change of kinetic

or latent energy., not a change of momentum. Otherwise, there is

creation of energy.

 

And you are wrong. Accelerated charges, including those moving in a circle, radiate. It's seen in cyclotrons. You can see this if you look at what their E & B fields do during an acceleration.

[Greg H.]

Try this.

 

A comet goes by the earth. At a distance of on milliion miles

coming and going it's kinetic and latent energy are the same.

However, it's momentum is different.

 

If its kinetic energy stayed the same, I would expect its momentum to change, since a comet will lose mass as it passes the sun due to out-gassing. The change in momentum would be required to maintain the same [math]E_k[/math].

 

That doesn't make your statement about the ball any less ludicrous as the math simply doesn't support your statement.

 

You cannot change the momentum of the ball without the mass also changing, and have [math]E_{ki} = E_{kf}[/math]. They're called equations for a reason - they have to be equal.

Edited July 26, 2013 by Greg H.

[Lazarus]

 

And you are wrong. Accelerated charges, including those moving in a circle, radiate. It's seen in cyclotrons. You can see this if you look at what their E & B fields do during an acceleration.

 

You are right. I jumped the gun on that. If curature of the path of the eledtron

causes a photon to be fired off in the forward direction the electron would loose

kinetic energy.

 

But doesn't that imply that a steady current in a closed loop would contantly

be radiating?

 

If its kinetic energy stayed the same, I would expect its momentum to change, since a comet will lose mass as it passes the sun due to out-gassing. The change in momentum would be required to maintain the same [math]E_k[/math].

 

That doesn't make your statement about the ball any less ludicrous as the math simply doesn't support your statement.

 

You cannot change the momentum of the ball without the mass also changing, and have [math]E_{ki} = E_{kf}[/math]. They're called equations for a reason - they have to be equal.

 

Greg,

 

Let's move earth and the comet out to empty space and try again.

[swansont]

You are right. I jumped the gun on that. If curature of the path of the eledtron

causes a photon to be fired off in the forward direction the electron would loose

kinetic energy.

 

But doesn't that imply that a steady current in a closed loop would contantly

be radiating?

Yes.

[Lazarus]

I see that the justification for being unable to detect

radiation from a steady current loop is a mathematical

cancellation of radiation.

 

The physical loop has electrons that emit photons in

the direction that the electrons are traveling. That

should mean that the photons spread out like a

sprinkler so the photons would not interfere with

each other.

 

Is the radiation really detectable or does the math

match what actually happens with the photons?

[swansont]

http://en.wikipedia.org/wiki/Synchrotron_radiation

[Lazarus]

Thank you very much for the reference.

 

 

Electrons radiate when passing through an magnetic

field. That happens in a synchrotron and in space.

 

The electron in a current loop must be in a zero

magnetic field because just outside of the loop

with the lines of force going north the inside of

the loop will have the lines of force going south.

 

If there is no detectable radiation that should

mean that the curvature of the electron path is

not a cause of radiation.

[swansont]

For synchrotron radiation magnetic fields are used to curve the path because it's most convenient. The power radiated depends on the square of the acceleration, so for a circle it depends on v⁴. Particles in an accelerator are moving at ~10⁸ m/s, while particles in a wire are moving at ~10^-2 m/s. So, all else being equal, the amount of radiation is down by 10^-40 . That's why it would be hard to detect the radiation from electrons in a normal current loop.

[Lazarus]

Don't you at least consider it a bit strange that the

velocity of galaxies and stars can be computed by

just assuming an electron orbiting a proton?

[John Cuthber]

As far as I am aware, they can't.

Please show us the nature of the calculation.

Edited July 28, 2013 by John Cuthber

[Lazarus]

As far as I am aware, they can't.

Please show us the nature of the calculation.

I am so glad you asked.

 

The discription of the calculations is at the head

of this thread.

 

The code to do the calculations and comparisons

follows:

 

' shellsx.bas

defdbl a-z

 

' Relate Bohr atom speed to energy of red or blue shifted photon

 

' Calculate Bohr electron energy change from one shell to another

 

redblue=0' 0 = redshift, 1 = blueshift

littlen=2' Lower number Shell

bign=3' Higher number Shell

velfact=.0030394' To adjust electron velocity at apogee of elipse

incremt=0' Initial atom velocity increment

delta=0' To change velocity

 

print "Enter for red, 1 for blue"

input redblue

 

print "Little shell, Enter for 2 or 1 - 9"

input a

if a<>0 then littlen=a

 

print "Big shell, Enter for 3 or 1 - 9"

input a

if a<>0 then bign=a

 

print "Velocity factor, Enter for .0030394 or .001 to 1"

input a

if a<>0 then velfact=a

' Formulas

 

' c=f*l Speed of light = frequency times wave length

' e=h*f Energy of photon = frequency times Plank's

' 1/l=Rh*(1/n1*n1 - 1/n2*n2) l=wave length, Rh=Rydberg, n=shell #

' F=m*a Force = mass times acceleration

' Fc=v*v/r Centripital force = vel squared over radius

' ee=k/r Escape energy is proportional to inverse of r

' ke=m*v*v*.5 Kinetic energy=mass times vel squared times .5

' z=wobserve/wemit-1' z = observed wave length over emitted minus 1

' v=(((z+1)**2-1)/((z+1)**2+1)) * c' Yeah

' d=v/h' Distance = velocity divided by Hubble

 

open "shells.txt" for output as #8

 

hubble=4.675e-17' Hubble constant

rh=1.0973731e7' Rydberg constant

plank=6.626e-34' Plank's constant

c=3e8' Velocity of light in meters

shell1v=2.188e6' Velocity of electron in shell 1

v1=shell1v*velfact' Modified velocity of electron in shell 1

m=.911e-30' Mass of an electron

facte7=1e7' 10 to the 7th power

facte9=1e9' 10 to the 9th power

facte20=1e20' 10 to the 20th power

z=0' Red shift factor

 

vlittle=v1/littlen' Velocity of electron in smaller shell #

vbig=v1/bign' Velocity of electron in bigger shell #

 

r1sqr=1/(littlen*littlen)' Reciprical of little shell # squared

r2sqr=1/(bign*bign)' Reciprical of big shell # squared

rlamda=rh*(r1sqr-r2sqr)' Reciprical of wave length

lamda=1/rlamda' Wave length of photon in meters, stopped

freq=c/lamda' Frequency in Hz

ienergy=freq*plank' Initial energy of photon in Joules

 

vatom=0' Initial velocity of the Bohr atom

kelittle=m*vlittle*vlittle*.5'Initial kinitic energy of little shell

kebig=m*vbig*vbig*.5' Initial kinitic energy of big shell

ideltake=kelittle-kebig' Initial kinetic energy change

wave7=lamda*facte7' Adjust wave length for printing

 

if redblue<>0 then print#8," Blueshift vs velocity"

if redblue=0 then print#8," Redshift vs velocity"

print#8," Eliptical Bohr Hydrogen atom

print#8," Wave length =";using"###.##";wave7;

print#8,"e-7"

print#8," Atom Shell";

print#8,littlen;

print#8," Shell";

print#8,bign;

print#8," length energy"

'''''print#8," velocity velocity velocity *e20 in nm %";

'''''print#8," Z Zv/c";

print#8," velocity velocity velocity Zv/c Z in nm";

print#8," *e20 %";

print#8,

 

 

''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''

 

for i=1 to 60

 

if redblue<>0 then' Doing blue shift

av1=vlittle-vatom'Absolute velocity of little shell electron

av2=vbig-vatom' Absolute velocity of big shell electron

end if

 

if redblue=0 then' Doing red shift

av1=vlittle+vatom'Absolute velocity of little shell electron

av2=vbig+vatom' Absolute velocity of big shell electron

end if

 

ake1=m*av1*av1*.5' Current kinetic energy of little shell electron

ake2=m*av2*av2*.5' Current kinetic energy of big shell electron

 

deltake=ake1-ake2' Current change of kinetic energy

 

energy=ienergy-(deltake-ideltake)' Adjust energy for change in KE

 

freq=energy/plank' Frequency

 

wave=lamda*(ienergy/energy)' New wave length

 

energy20=energy*facte20' Mult by 10 to the 20th

wave9=wave*facte9' Mult by 10 to the 9th

 

percent=(energy/ienergy)*100' Calc percent wave change

if redblue<>0 then percent=(ienergy/energy)*100' Reverse for blue

 

distance=vatom/hubble' distance

 

z=wave/lamda - 1' calc z

vz=(((z+1)*(z+1)-1)/((z+1)*(z+1)+1)) * c' Velocity from z in m/s

 

print#8,using"####,###,###";vatom;av1;av2;

print#8,using"####,###,###";vz;

print#8,using"###.###";z;

print#8,using"#######.###";wave9;

print#8,using"#####.####";energy20

''''' print#8,using"#####";percent

''''' print#8,distance

 

delta=delta+100000

incremt=incremt+delta

vatom=vatom+incremt' Speed up the atom

''''' vatom=vatom+100000' Speed up the atom

if vatom>c goto c9999

next i

c9999:

[Greg H.]

 

Greg,

 

Let's move earth and the comet out to empty space and try again.

 

You can move them to Nirvana with a side trip to Purgatory for all I care. Math doesn't change, and it doesn't lie (unless you use it wrong). If the Kinetic Energy remains the same, then either BOTH the mass AND the momentum change, or neither does. You can't have it any other way if you expect the equations to balance.

[Lazarus]

Gref, my friend,

 

I was wrong about the electron. If the current idea is

correct a curving path for and electron would release

a photon so the electron would lose some of it's

kinetic energy.

 

I don't think a larger body would lose kinetic energy.

 

Your said:

 

Posted 25 July 2013 - 12:52 PM

Changes in momentum cause changes in kinetic energy by:

 

where

 

p is momentum

m is mass of the body.

 

Since momentum is mv so your equation becomes

 

E=mmvv/2m or mvv/2

 

which is kinetic energy only. That doesn't prove

much about momentum.

 

In fact p=square root of 2mE. Since 2mE is 2mmvv/2

you have to know v and m to solve the equation for

momentum.

 

Also, to determine the results of a collision of

2 objects the total kinetic energy AND the

total momentum must remain the same.

 

Bob