Here is an example of a quantified energy momentum analysis.

Values for speed of light, Planck's constant, electron charge and mass:
c = 3.00e8 m/s
h = 6.63e-34 kg.m^2/s
e = 1.60e-19 C
m = 9.11e-31 kg

Initial photon, wavelength, energy, and momentum:
λ = 5.00e-7 m
Q = 3.97e-19 J (Q = hc/λ)
p = 1.325e-27 kg.m/s (p = h/λ)
This is the energy and momentum that must remain balanced. If numbers are not given adding up to these values, both for energy and momentum, then the balance has not been shown.

Recoil electron with kinetic energy K after absorbing momentum p
v = 1.46e3 m/s (v = p/m)
K = 9.64e-25 J (K = mv^2/2)

At this point in the analysis, we can balance the momentum, but (1-K/Q) = 99.99976% of the energy is still unaccounted for.

If this energy transfers to the rest of the plasma, then it will require over 400,000 more electrons with the same amount of energy to make up the balance (Q/K = 4.12e5). Energy transfers to the rest of the plasma occur as an electron moves through the electric fields; but the electron cannot transfer more than the energy K of its own motion. The photon is allegedly absorbed; it can't go on and interact with 400,000 more electrons itself.

Thus there is no possibility of energy balance at this point in the interactions.

In real physics, the energy of photo-absorption is actually taken up by excitation of an atom to a new energy level. This is how it occurs in French, and in every published source we have considered on photo-absorption. This is why photo-absorption in real science is for electrons bound to atoms; never for ionized electrons a meter or so away from any other particle.

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In his paper Lyndon then describes the recoiling electron as decelerating by interactions with the rest of the plasma and emitting a photon with energy K, by bremsstrahlung.

Emission of a CMB photon.
CMBλ = 2.06e-1 m (CMBλ = hc/K)

This is the only quantified interaction Lyndon gives with the rest of the plasma. It could occur, for example, in a close encounter with a heavy positive ion. Energy momentum can balance for the CMB emission, with energy K of the electron becoming energy of the CMB photon, and with momentum sqrt(2Km) of the electron absorbed into the heavy ion with a negligible change in energy.

This is no help for the 99.99975% of the initial energy Q still unaccounted.

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In his paper Lyndon then describes the electron as emitting a new redshifted photon, and recoiling again.

This energy is going to let the energy budget balance, but where did the energy come from? The electron only got energy K, and that's already gone into a CMB photon.

Basically, the original energy has to be stored somewhere, to power this emission. Lyndon has spoken of "storing" it in the rest of the plasma, and we've seen above that this is impossible. But even if it was stored in the rest of the plasma, that involves dissipation into enormous numbers of other particles. How is it then all brought back together into the original electron, so as to make physical sense of the electron emitting the photon? This is now also a violation of basic thermodynamics.

In real physics, the absorbing particle is an atom. The energy of photo-absorption gets stored in one of the well-defined fixed energy levels of the atom. This is why all the real absorption and emission reactions occur at very well defined energies; not spread out over the spectrum like the alleged Lyndon effect.

Another curious property of the emitted redshifted photon is that somehow the electron "remembers" the direction of the original photon, and emits it again with zero scatter angle!

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We can almost balance energy by combining absorption and emission into a single interaction. Lyndon hints at this on the top of page 4, when he speaks of the form factor f2 as follows: "One meaning that the photon has been absorbed and the electron remaining in an excited state and zero meaning that the photon was absorbed and an identical photon reemitted."

Actually, f2 being zero simply means no absorption at all; but let it pass. Also, Lyndon does not reemit an identical photon but one that is redshifted. Let that pass too. The question is if we can balance energy momentum if we consider the redshifted photon to be emitted again immediately.

The amount of redshift Lyndon invokes corresponds to loss of energy of 2K. Furthermore, the photon has no scatter angle. This is crucial to the impossibility of balancing energy momentum. If scatter was allowed, we could have simply used a Compton effect.

If we combined an incoming photon with energy Q, and an outgoing photon with energy Q-2K, then there is still 2K of energy to balance. Because Lyndon insists on no scatter angle, both momentum vectors are along the same line, and so there is still 2K/c momentum, in the direction of the original photon, still to be accounted for.

This is, in fact, exactly equivalent to absorbing a photon of energy 2K! The same analysis as above shows that an electron cannot absorb that amount of energy and momentum to balance the books. It can absorb momentum, but most of the energy 2K remains unaccounted.

How about if we include the CMB photons in the interaction? They have energy K, so two of them balances the energy budget, and also allows for spin-parity to be preserved.

This allows a balanced budget at last. The final energy budget is as follows. There is Q-2K for the redshifted photon, and 2K for the two CMB photons, and a negligible quantity of energy left over for the electron.

We still need to account for momentum, and the initial momentum is Q/c.

The red-shifted photon is (Q-2K)/c along the same line.
The CMB photons contribute K/c, twice, along any two lines we like

The total momentum of the three photons is a maximum when they are all along the same line, giving Q/c with nothing more required. It is a minumum when the CMB photons are backscattered, effectively subtracting from the momentum of the redshifted photon, giving a total of (Q-4K)/c. Other values for the photon momentum contribution can have magnitudes with anything between these two values, depending on the directions.

The momentum remaining to be balanced is thus anything from 0 to 4K/c, and this can be taken up by the electron, with a negligible difference to the energy.

That is, the books can balance if the interaction involves immediate re-emission of a redshifted photon with no scatter, and with the remaining energy taken up immediately by more photons, and with a small impulse to the electron of no more than 4K/c.

The velocity given to the electron with this impluse is 4K/mc, so the maximum allowed energy and velocity for the electron is
Electron maximum velocity: 4K/mc = 1.4e-2 m/s
Electron maximum energy: 8K^2/mc^2 = 9.1e-35 J

There is no possibility of a balanced energy momentum budget for Lyndon's reaction if the stationary electron ever exceeds a velocity of 14 millimeters per second. Any intermediate state with more velocity than this in the electron is bound to fail a simple energy momentum conservation test. All photons have to be emitted at once in the initial collision.

Lyndon won't accept this. He may accuse me of "forgetting" some contribution, but in reality I have considered and quantified the other sinks for energy or momentum and shown that they are not able to balance the books. He may accuse me of getting my "sums" wrong, but you can be sure he won't offer any quantified corrections, or balanced energy momentum analysis of his own.

Cheers -- Sylas

[[ Minor grammatical edits applied, and converted some formulae to simpler classical approximations; all numbers remain as original. ]]