Erasmus said:
. . . you have equations governing an electric field, but it gives nothing with which to simulate the relevant magnetic effects.
Ah, but given the electric force from a stationary charge Q acting on a test charge q (which could be moving) is enough. Once that is established, along with the assumption that this force is independent of q's velocity, all electric and magnetic fields are defined. Following Jackson, if in the inertial frame K' the equation of motion is

, then the equation of motion in the inertial frame K, moving at velocity

is :
and
And all electric and magnetic forces follow. A computer would have no trouble with these.
Erasmus said:
. . . non-conservation of momentum seems to keep cropping up.
We have tunneling right? This allows a non-conservation of energy at the strictly microscopic level, right? Well, why then is it so surprising that there is no strict conservation of momentum during microscopic tunneling?
Quote
. . . without the exclusion principle I don't think we can explain the specific heats of metals.
In a system at temperature T, the average thermal energy will be around kT. Hence, as you cool a system you expect all the electrons to spiral into the lowest possible state.
Also, we expect any quadratic degree of freedom to participate in the specific heat of an object. For metals, this isn't true. Most of the conduction electrons don't actually participate. The answer to both questions is the pauli exclusion principle. I fail to see how your theory can answer this.
You are correct. There is no Pauli exclusion principle in this New Theory. However, Boltzmann's constant
k comes from ideal gases:
So
k is defined as the average kinetic energy of an ideal gas molecule per degree Kelvin per degree of freedom. Multiplying by a mole N
o gives
So the specific heat of an ideal gas at constant volume becomes
OK, fine. But for solids, my statistical mechanics text gives that the "quantum" specific heat for solids is:
where θ
E is an arbitrary constant chosen to match experimental data.
So I have admitted that QM does a good job at experiment matching. I am not too concerned about QM's ability to experiment match. Again, this does not satisfy the curiosity of the curious.
The text goes on to say that "Experimentally, the specific heat approaches zero more slowly than this, indeed

as

. The reason for this descrepancy is . . . " (some hand waving follows).
I would expect that any application of Boltzmann's constant to electron energy levels in metals would also include some "quantum fudge factor" needed to match experimental data. This is QM's
modus operandi.
For metals, my Statistical Mechanics text gives that
where

and

are arbitrary constants chosen to match experimental data.
Se la vi.
Qfwfq said:
. . . and the curl of B is zero [for radial fields]. I was talking about what you said about a radially symmetric non-zero curl, which you seemingly said wouldn't be a problem. The only confusion here appears to be between the two things, charge moving in and out, or just magically appearing and disappearing, so we should try to be clear which we're talking about.
Ok, now that we have that radially pulsating charge does not radiate at the macroscopic level, it is easier to understand the microscopic level of this New Theory.
The two appropriate Maxwell equations are:
If we heuristically apply these equations to the New Theory's blinking charge, then the first equation yields that since the
E field is radial, then the curl of any radial function is 0, therefore there are no changing magnetic fields (actually no magnetic fields at all).
Now the second equation. For our blinking charge, there are no magnetic fields. Thus, the first term in the equation is 0. So we are left with:
And we see that there are displacement currents. Charge, as defined by Maxwell, is where electric field lines begin (positive) and end (negative). Thus, when this New blinking charge blinks, there are areas of space that have short
E field segments. For a positron, the head of this blinking
E segment has negative "displacement charge" and the tail has positive "displacement charge". This yields a net "displacement charge" of zero moving outward, and we have no net charge escaping to infinity.
Qfwfq, does this answer the question?
Andrew A. Gray