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In natural units, the Dirac Equation is $$i \frac{\text{d}}{\text{d}t} \psi = \left[\vec \alpha \cdot \vec P +\beta m + e \Phi\right]\psi.$$ I use Pauli-Dirac basis for matrices, \begin{align*} \vec \alpha = \begin{pmatrix} 0 & \vec \sigma\\ \vec \sigma & 0 \end{pmatrix} && \text{and} && \vec \beta = \begin{pmatrix} I & 0\\ 0 & -I \end{pmatrix}. \end{align*}

When the fields and momentum are $0$, the states and energies are

(1) $\tilde \phi=a (e^{-imt},0,0,0)$ and $b (0,e^{-imt},0,0)$ corresponding to a positive energy $m$

(2) $\tilde\chi=c (0,0,e^{imt},0)$ and $d (0,0,0,e^{imt})$ corresponding to a negative energy $-m$

My goal is to derive the non-relativistic limit (Pauli-Schroedinger Equation) using the above. I consider all the fields/momentum to be nonzero now but small now. I am stuck on one single point:

It is standard to use the ansatz $\psi = (\tilde\phi_1,\tilde\phi_2,\tilde\chi_3,\tilde\chi_4)=(\tilde\psi,\tilde\chi)\equiv e^{-imt}(\phi,\chi)$. In the last statement, we are effectively factoring out the positive energy phase in our solution. From this, why is $\frac \partial {\partial t} \chi\sim 0$, in the limit that $P \ll m$ and $\Phi\ll m$?

My confusion is that clearly from (1) and (2), $\tilde\phi\sim e^{-imt}$ while $\tilde\chi\sim e^{imt}$. So doesn't this mean $\phi\sim 1$ while $\chi\sim e^{2imt}$? In this case, it seems that we do not get $\frac \partial {\partial t} \chi\sim 0$, but instead we get $\frac \partial {\partial t} \phi\sim 0$, which is the opposite of what I wanted, and is the opposite of what is in the literature (see "From Dirac Equation")

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2 Answers 2

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To get the non-relativistic limit, that is the Pauli equation, one can use the ansatz $\psi = e^{-imt}\begin{pmatrix}\phi\\ \chi\end{pmatrix}$, as you mentioned. In the ansatz, however, $\phi$ and $\chi$ are generic functions that you will find after plugging the ansatz into Dirac's equation and taking the non-relativistic limit, which is $|\partial\chi/\partial t|\ll|m\chi|$ and $|e\Phi\chi|\ll|m\chi|$ (i.e, kinetic and potential energy respectively much smaller then the rest energy).

It's not clear to me how you wish to use (1) and (2).

EDIT

Following your comment, it seems to me that you're plugging into the ansatz the form of the free-particle solutions (1) and (2). What I mean with

In the ansatz, however, $\phi$ and $\chi$ are generic functions[...]

Is that their form is not known when using the ansatz, so assuming they behave as in (1) and (2) would be incorrect. Indeed the aim is to retrieve the Pauli equation, whose solutions are not in the form of (1) and (2).

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  • $\begingroup$ I use (1) and (2) in the first sentences of the last paragraph: "My confusion is that clearly from (1) and (2), ϕ~$e^{−imt}$ while χ$∼e^{imt}$. So doesn't this mean ϕ∼1 while χ∼$e^{2imt}$?". Is this a correct statement? I might not have been as clear as possible following this statement, so I will be clearer: if this quoted argument is correct, then the energy of $\chi$ goes as $i \partial_t \chi=i \frac{\partial \chi }{\partial t}=i \cdot 2im = -2m\neq 0$. I wanted $i \partial_t \chi$ to be 0, but I get that is goes as $2m$. This is my confusion. $\endgroup$
    – Mondo Duke
    Commented Feb 22, 2022 at 8:58
  • $\begingroup$ @MondoDuke I have edited the answer. I hope it is clearer what I meant now. $\endgroup$ Commented Feb 22, 2022 at 10:14
  • $\begingroup$ thanks. A followup, with that logic why cant we also assume $\partial_t \phi << m \phi$ in the nonrelativistic limit? Clearly this would be invalid $\endgroup$
    – Mondo Duke
    Commented Feb 22, 2022 at 20:01
  • $\begingroup$ @MondoDuke Yes, you can assume that, but luckily the terms $m\phi$ cancel out in the equation for $\phi$, as expected. You can see this being done in the Wiki link in your post or in Greiner's "Relativistic Quantum Mechanics". $\endgroup$ Commented Feb 23, 2022 at 9:14
  • $\begingroup$ We know that each component of the Dirac bi-spinor verifies the Klein-Gordon equation*, you replace your 'ansatz' (1) in the K-G equation, possible that you find the equation that you seek to demonstrate and the same thing for the relation (2) with the consideration that one can restore (1) with exp(mt)=exp(-m(-t)), a particle (e) of negative energy which goes back in time is a particle (-e) of positive energy moving forward in time (Feynman-Stuckelberg interpretation). *en.wikipedia.org/wiki/Klein%E2%80%93Gordon_equation $\endgroup$
    – The Tiler
    Commented Feb 23, 2022 at 16:46
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To arrive at the non-relativistic form of the Dirac equation, we can reverse the process used, here, in this reply:

Is the Schrödinger equation incompatible with special relativity? If so, why?

starting from the Dirac equation, which you wrote in the following form: $$i \frac{∂ψ}{∂t} = (𝝰·𝐏 + βm + eΦ)ψ,$$ with the representation $$ 𝝰 = \left(\begin{matrix}𝟎&+𝞂\\+𝞂&𝟎\end{matrix}\right), \quad β = \left(\begin{matrix}+1&0\\0&-1\end{matrix}\right), \label{1}\tag{1} $$ where $$ 𝞂·𝐚 = \left(\begin{matrix}+a^3&a^1 - i a^2\\a^1 + i a^2&-a^3\end{matrix}\right), \\ 𝞂·𝐚 𝞂·𝐛 = 𝐚·𝐛 + i𝞂·𝐚×𝐛, $$ for $𝐚 = \left(a^0, a^1, a^2\right)$ and $𝞂 = \left(σ_1, σ_2, σ_3\right)$; which leads to: $$𝝰·𝐚 𝝰·𝐛 + 𝝰·𝐛 𝝰·𝐚 = 2 𝐚·𝐛, \quad 𝝰β + β𝝰 = 𝟎, \quad β^2 = 1 \label{2}\tag{2}.$$ This is more commonly seen with: $$ γ^0 = β = \left(\begin{matrix}+1&0\\0&-1\end{matrix}\right), \quad 𝝲 = \left(γ^1,γ^2,γ^3\right) = β𝝰 = \left(\begin{matrix}𝟎&+𝞂\\-𝞂&𝟎\end{matrix}\right). $$ I'll be reusing $𝝲$ for something different below.

The Dirac Equation - With Electromagnetic Potential
"Natural units", as you call them, are not appropriate since you're talking about $c → ∞$, so $c$ must be made explicit! So should $ħ = h/2π$ be. We will do more than you ask for: also adding in an $𝐀$ for magnetic (or "vector") potential, while treating $φ$ as the electric potential.

In the presence of the field, the energy $E$ of a body of charge $e$ has a potential energy $eφ$ added to it, and the momentum $𝐏$ a potential $e𝐀$. Thus, the operator correspondence is with: $$ E + eφ ⇔ iħ \frac{∂}{∂t}, \quad 𝐏 + e𝐀 ⇔ -iħ∇, $$ where $∇ = (∂/∂x,∂/∂y,∂/∂z)$, hence: $$ E ⇔ iħ \frac{∂}{∂t} - eφ, \quad 𝐏 ⇔ -iħ∇ - e𝐀, $$

In Relativity, the one-form $P = 𝐏·d𝐫 - Edt$ is an invariant (where $d𝐫 = (dx, dy, dz)$), as is the potential one-form: $A = 𝐀·d𝐫 - φdt$. Under the operator correspondence, this reads: $$P + eA ⇔ -iħ \left(d𝐫·∇ + dt \frac{∂}{∂t}\right) = -iħ d.$$

The Dirac equation is the operator form of the mass-shell condition $$E^2 - |𝐏c|^2 = \left(mc^2\right)^2,$$ rendered in linear form as: $$E = 𝝰·𝐏c + βmc^2,$$ which is: $$\left(iħ \frac{∂}{∂t} - eφ\right)ψ = \left(𝝰c·(-iħ∇ - e𝐀) + βmc^2\right)ψ,$$ or equivalently, in more standard form, as: $$γ^0\left(iħ \frac{∂}{∂t} - eφ\right)ψ + 𝝲c·(iħ∇ + e𝐀) = mc^2ψ. \label{3}\tag{3}$$ But - as I noted above - I won't be using that form of $𝝲$ for this (but for something else, instead); and am only stating it as a point of reference and comparison.

Instead, we will write it in this form: $$iħ\left(\frac{∂}{∂t} + 𝝰c·∇\right)ψ = e(φ - 𝝰c·𝐀)ψ + βmc^2ψ, \label{4}\tag{4}$$ for now.

The Langoliers / Premonition Transform (in reverse)
The reverse Langoliers Transform, introducing Steven King / Outer Limits Time $s$ (or, more appropriately: the reverse Premonition Transform, after the episode of The Outer Limits that he ripped off ... another discussion for another time), is now going to be introduced, alongside real time $t$. $$ ψ(𝐫,t,s) = e^{imc^2s/ħ} ψ(𝐫,t), \quad ψ(𝐫,t) ≡ ψ(𝐫,t,0). $$ With the extra argument, the extended function is the solution to the following: $$iħ\left(\frac{∂}{∂t} + 𝝰c·∇\right)ψ = e(φ - 𝝰c·𝐀)ψ + βmc^2ψ, \quad -iħ\frac{∂ψ}{∂s} = mc^2ψ,$$ or equivalently: $$iħ\left(\frac{∂}{∂t} + 𝝰c·∇ + β\frac{∂}{∂s}\right)ψ = e(φ - 𝝰c·𝐀)ψ, \quad -iħ\frac{∂ψ}{∂s} = mc^2ψ.$$

Now, we define: $$\check{ψ}(𝐫,t,u) ≡ ψ\left(𝐫,t,t+\frac{u}{c^2}\right) ⇔ ψ(𝐫,t,s) = \check{ψ}(𝐫,t,(s-t)c^2),$$ with $$\check{ψ}(𝐫,t) ≡ \check{ψ}(𝐫,t,0) = ψ(𝐫,t,t) = e^{imc^2t/ħ} ψ(𝐫,t).$$ Then: $$ \frac{∂ψ}{∂t} = \left(\frac{∂}{∂t} - c^2 \frac{∂}{∂u}\right)\check{ψ}, \quad \frac{∂ψ}{∂s} = c^2 \frac{∂\check{ψ}}{∂u}. $$ This leads to: $$iħ\left(\frac{∂}{∂t} + 𝝰c·∇ + (β-1)c^2 \frac{∂}{∂u}\right)\check{ψ} = e(φ - 𝝰c·𝐀)ψ, \quad -iħ \frac{∂\check{ψ}}{∂u} = m\check{ψ}.$$ With a suitably-defined matrix / Clifford algebra unit $δ$, this may be rewritten as: $$ iħ\left(δ \frac{∂}{∂t} + δ𝝰c·∇ + δ(β-1)c^2 \frac{∂}{∂u}\right)\check{ψ} = e(δφ - δ𝝰c·𝐀)\check{ψ}, \quad -iħ \frac{∂\check{ψ}}{∂u} = m\check{ψ}. $$ Defining $$𝝲 = δ𝝰c, \quad ε = δ(β-1)c^2, \label{5}\tag{5}$$ this can be written more succinctly as: $$iħ\left(δ \frac{∂}{∂t} + 𝝲·∇ + ε \frac{∂}{∂u}\right)\check{ψ} = e(δφ - 𝝲·𝐀)\check{ψ}, \quad -iħ \frac{∂\check{ψ}}{∂u} = m\check{ψ},$$ which is clearly equivalent to: $$iħ\left(δ \frac{∂}{∂t} + 𝝲·∇\right)\check{ψ} = εm\check{ψ} + e(δφ - 𝝲·𝐀)\check{ψ}, \quad -iħ \frac{∂\check{ψ}}{∂u} = m\check{ψ}. \label{6}\tag{6}$$ The general solution is: $$\check{ψ}(𝐫,t,u) = \check{ψ}(𝐫,t) e^{imu/ħ};$$ where $$\check{ψ}(𝐫,t) ≡ \check{ψ}(𝐫,t,0) = ψ(𝐫,t,t) = e^{imc^2t/ħ} ψ(𝐫,t).$$ The function $\check{ψ}(𝐫,t)$ is an adjustment of $ψ(𝐫,t)$ to where The Outer Limits Time $s$ is brought back into sync with Real Time $t$, just like at the end of that Outer Limits episode (except for the guy on permanent vacation in the $s$-zone).

The Outer Limits Time $s$ is a Lorentz invariant. It is the survival of the Absolute Time of non-relativistic theory into Relativity. It was there all along, but nobody saw it - except for those people in Premonition and in the Langoliers. The $u$ coordinate is time-dilation, $s - t$, ramped up by $c^2$, which means it survives in the non-relativistic limit - as a non-relativistic form of time dilation.

So, time is now both absolute and relative in both Relativity and non-relativistic theory - because they're different times. One is $s$ and the other is $t$. The one that goes with $mc^2$ is $s$, not $t$.

And the coordinate that goes with $m$ is $u$.

Bargmann Geometry
Ever wonder why Dirac called it $γ_5$ with the $5$?

The Dirac algebra with just $\left(γ^0, γ^1, γ^2, γ^3\right)$ - as a real Clifford algebra - is isomorphic to the algebra of $2×2$ quaternion matrices. The algebra that you normally use has an extra $i$ stuck into it and is called the "complexified" Dirac algebra. It is isomorphic to the $4×4$ complex matrices. As a real Clifford algebra, it has five units, not four! The fifth one is $i$. The same result applies if $γ_5$ is added in, instead, since it has an $i$ in it, too.

Dirac did it (apparently) unconsciously and (apparently) wasn't able to quite pin down what the $5$ was. But, that's what it actually is and where it's coming from.

Let $α$ denote the value $α = 1/c^2$ for Relativity and $α = 0$ in the non-relativistic case - not to be confused with the vector-valued matrix $𝝰$ above. The geometry with the coordinates $$𝐫 = (x,y,z) = \left(x^1, x^2, x^3\right), \quad t = x^4, \quad u = x^5,$$ is the Bargmann Geometry, when $α = 0$. The relativistic version of it, when $α = 1/c^2$, has no name that I am aware of. In the generic case, for arbitrary $α$, its defining invariants are: $$ dx^2 + dy^2 + dz^2 + 2 dt du + α du^2, \quad ds = dt + α du, \\ dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z} + dt \frac{∂}{∂t} + du \frac{∂}{∂u}, \\ \left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2 + 2 \left(\frac{∂}{∂t}\right) \left(\frac{∂}{∂u}\right) - α \left(\frac{∂}{∂t}\right)^2, \quad \frac{∂}{∂u} $$

The Total Energy $E$ of Relativity is reverted back to the Kinetic (plus Internal) Energy $H$, with the split-off $$E = H + mc^2.$$ The energy-momentum $4$-vector reverts to the energy-momentum-mass $5$-vector: $$𝐏 = \left(P_1,P_2,P_3\right), \quad P_4 = -H, \quad P_5 = m.$$ In place of the mass-shell condition for $(E, 𝐏)$ $$E^2 - |𝐏c|^2 = \left(mc^2\right)^2,$$ is now two invariants: $$|𝐏|^2 - 2mH - αH^2, \quad m.$$

With the expansion, the momentum invariant $P$ became $$P = 𝐏·d𝐫 - E dt + mc^2 ds = 𝐏·d𝐫 - H dt + m du.$$ The operator correspondence - with the contribution by the field intact - is now: $$P + eA ⇔ -iħ \left(d𝐫·∇ + dt \frac{∂}{∂t} + du \frac{∂}{∂u}\right) = -iħ d,$$ except the $d$ is now the 5-D $d$ of the Bargmann geometry and its unnamed relativistic version. Thus: $$H + eφ ⇔ iħ \frac{∂}{∂t}, \quad 𝐏 + e𝐀 ⇔ -iħ∇, \quad m[ + eb] ⇔ -iħ\frac{∂}{∂u}.$$ The extra scalar (denoted here as $b$) is one and the same as what is sometimes used with the "$B$-field formalism" in QED, and it would go with $m$ as $m + eb$. Correspondingly: $$H ⇔ iħ \frac{∂}{∂t} - eφ, \quad 𝐏 ⇔ -iħ∇ - e𝐀, \quad m ⇔ -iħ\frac{∂}{∂u}[ - eb].$$

A reference.

Nakanishi-Lautrup B-Field, Crossed Product &Duality
https://www.semanticscholar.org/paper/Nakanishi-Lautrup-B-Field%2C-Crossed-Product-B-Field/f60e8dd3bda029ed9f756fa8827873d2bd15cde3

This was mostly Nakanishi's thing. Most of the other references, related to the B-field formalism, that I find are in association with the "Stueckelberg trick" ... which also has connections with the geometric matters discussed here.

The Bargmann Clifford Algebra
The linearized form of the mass-shell condition is: $$δH - 𝝲·𝐏 - εm = 0.$$

First we focus on the relativistic case, $α = 1/c^2$.

To make this yield the quadratic form of the invariant requires: $$ 𝝲·𝐚 𝝲·𝐛 + 𝝲·𝐛 𝝲·𝐚 = 2 𝐚·𝐛, \quad 𝝲δ + δ𝝲 = 𝟎, \quad 𝝲ε + ε𝝲 = 𝟎, \\ δ^2 = -α = -\frac{1}{c^2}, \quad δε + εδ = 2, \quad ε^2 = 0. $$ In order to make this work, this requires, by ($\ref{2}$) and ($\ref{5}$): $$δ𝝰 + 𝝰δ = 𝟎, \quad δβ + βδ = 0,$$ which has the following as a solution, using ($\ref{1}$): $$ δ = \left(\begin{matrix}0&+1/c\\-1/c&0\end{matrix}\right), \quad 𝝲 = δ𝝰c = \left(\begin{matrix}+𝞂&𝟎\\𝟎&-𝞂\end{matrix}\right), \quad ε = δ(β-1)c^2 = \left(\begin{matrix}0&-2c\\0&0\end{matrix}\right). \label{7}\tag{7} $$

Steam-Punk Light Speed $V$ Versus The Absolute Speed $c$
In Relativity, where $α > 0$, the Absolute Speed $c = \sqrt{1/α}$ is one and the same as the speed $V$ of electromagnetic (and other) waves in a vacuum. In non-relativistic theory, $α = 0$, the Absolute Speed $c = \sqrt{1/α}$ is infinite: $c = ∞$. Hence, the need to distinguish $V$ from $c$. The letter $V$ is what was used in the older literature up to - and including - Einstein's 1905 paper.

Now, make a distinction between steam-punk light-speed, $V$ (e.g. $V = 1/\sqrt{εμ}$ in the 19th century "vacuum" medium) versus the Absolute Speed light-speed, $c$; and rewrite ($\ref{1}$) as: $$𝝰 = \left(\begin{matrix}𝟎&+V𝞂/c\\+c𝞂/V&𝟎\end{matrix}\right), \quad β = \left(\begin{matrix}+1&0\\0&-1\end{matrix}\right).$$

Then, in place of ($\ref{7}$) will be: $$δ = \left(\begin{matrix}0&+αV\\-1/V&0\end{matrix}\right), \quad 𝝲 = \left(\begin{matrix}+𝞂&𝟎\\𝟎&-𝞂\end{matrix}\right), \quad ε = \left(\begin{matrix}0&-2V\\0&0\end{matrix}\right). \tag{8}\label{8} $$ Then, ($\ref{6}$) is still equivalent to the Dirac equation with electromagnetic potentials - in either form as ($\ref{3}$) or ($\ref{4}$), when $α = 1/c^2 > 0$. So, forget about $𝝰$, $β$ and ($\ref{5}$). They're no longer needed and - with the modification made to $𝝰$ - will not be meaningful in the limit $α → 0$. We can also drop back down to the 4D version of ($\ref{6}$), by setting $u = 0$ and considering only $\check{ψ}(𝐫,t) = \check{ψ}(𝐫,t,0)$ and remove the extra eigenvalue equation for $u$ and $m$ to get: $$iħ\left(δ \frac{∂}{∂t} + 𝝲·∇\right)\check{ψ} = εm\check{ψ} + e(δφ - 𝝲·𝐀)\check{ψ}.\label{9}\tag{9}$$ This is still equivalent to the Dirac equation, when $α = 1/c^2$, with the definitions ($\ref{8}$) for the matrices.

The $α = 0$ version of ($\ref{9}$), is the non-relativistic Dirac equation.

The Pauli-Schrödinger Equation
First, we will rewrite ($\ref{9}$) as: $$\left(δ \left(iħ\frac{∂}{∂t} - eφ\right) + 𝝲·(iħ∇ + e𝐀)\right)\check{ψ} = εm\check{ψ}.$$

Carry out the following decompositions: $$ ψ = \left(\begin{matrix}ζ\\\bar{χ}\end{matrix}\right). $$ In spinor component form, $ζ = \left(ζ_A\right)$ and $\bar{χ} = \left(χ^{\dot{B}}\right)$. Similarly, write: $$ \check{ψ} = \left(\begin{matrix}\check{ζ}\\\check{\bar{χ}}\end{matrix}\right). $$ Then, setting $\check{\bar{ω}} = V\check{\bar{χ}}$, we obtain the following $2$-component spinor equations: $$ \left(2m + α \left(iħ \frac{∂}{∂t} - eφ\right)\right) \check{\bar{ω}} + 𝞂·(iħ∇ + e𝐀)\check{ζ} = 0,\\ \left(iħ\frac{∂}{∂t} - eφ\right)\check{ζ} + 𝞂·(iħ∇ + e𝐀)\check{\bar{ω}} = 0.$$ With the previously-mentioned expediencies, this is equivalent to the Dirac equation when $α = 1/c^2$.

When $α = 0$, it becomes the non-relativistic version: $$2m \check{\bar{ω}} + 𝞂·(iħ∇ + e𝐀)\check{ζ} = 0, \quad \left(iħ\frac{∂}{∂t} - eφ\right)\check{ζ} + 𝞂·(iħ∇ + e𝐀)\check{\bar{ω}} = 0,$$ or $$ 2m\check{\bar{ω}} = -𝞂·(iħ∇ + e𝐀)\check{ζ},\\ 2m\left(iħ\frac{∂}{∂t} - eφ\right)\check{ζ} = 𝞂·(iħ∇ + e𝐀)𝞂·(iħ∇ + e𝐀)\check{ζ}, $$ or just: $$iħ\frac{∂\check{ζ}}{∂t} = \left(eφ + \frac{(𝞂·(iħ∇ + e𝐀))^2}{2m}\right)\check{ζ}, $$ with $\check{\bar{ω}}$ being largely superfluous.

Hmm, where have we seen that before? Oh, I know! That guy who invented Vitamin C? Pauling! They had some kind of Exclusion Principle to keep him out of science, because of that, or something. So, this is called the Pauling Exclusion Principle Equation.

Even ChatGPT is going: "Dude! You can't say that!"

So, you thought the Schrödinger equation was first order in $t$ and second order in the spatial coordinates $(x,y,z)$ and that time and space were being treated differently? No! You just didn't see that extra coordinate $u$. It's hidden away under $m$. It's second order, plus an extra first order equation for $m$.

You can do that with all parabolic systems. They are hyperbolic in one higher dimension. That applies to the heat equation, too. That's why things ran down in the "s" dimension in Langoliers and Premonition. Oscillation in "t" time became exponentially decaying in "s" time. Does any of that sound familiar - even a tad bit Hawkingish or Bogoliubovish?

So, everything Dirac did was a total red herring! Every single element of his narrative: that "first-order-t versus second order spatial derivatives" talking point, that "this is relativistic" talking point, the negative energy sea. He just stumbled his way onto the right equation for all the wrong reasons. That's ok - for discovering stuff - but not when the wrong reasons and false narratives stick on the wall of folklore, motivation, visualization even in the peer-reviewed science literature!

By the way: where did that "negative energy" solution go, in all of this? Did you keep track of it to see what it turned into? And, is the vacuum still being polarized, when you turn this into a field theory? Be careful about that! The non-relativistic version of Maxwell's equations is not of the equations coming from the Maxwell-Lorentz Lagrangian density, but is the limit of the Maxwell-Minkowski relations for moving media, because in the Steam Punk Era, the vacuum is a medium with a wave speed $V = 1/\sqrt{εμ}$ and a frame of reference. So, when you make non-relativistic QED out of all of this, you have to handle that too. Context is important. And we haven't even gotten to gravity yet. That's got a line element, too; e.g., $$dx^2 + dy^2 + dz^2 + 2 dt du - 2U dt^2 = 0, \quad ds = dt, \quad U = -\frac{GM}{r}$$ for the one-body solution.

There's Steam Punk General Relativity underlying this.

I have no idea what back-reaction the electromagnetic field will have on the geometry, other than that the "$U$" part can be either $$U = -\frac{GM}{r} + \frac{μ_0}{8π} \frac{Ge^2}{r^2} \quad \text{or} \quad U = -\frac{GM}{r} + \frac{1}{8πε_0c^2} \frac{Ge^2}{r^2}.$$

So, is that a "$c$" $c$ or a steam-punk "$V$" $c$? If it's a "$c$" $c$ then, the limit for $U$ is $-GM/r$. However, if it's a "$V$" $c$, with $V = 1/√{εμ}$ being that for the Steam-Punk Vacuum Medium, then it's $$U = -\frac{GM}{r} + \frac{μ}{8π} \frac{Ge^2}{r^2}.$$ This is something Maxwell, Hertz, Lorentz and Heaviside could have surmised in their day. Maxwell, himself, was already thinking about the matter on how to combine gravity with electromagnetism and said a few things about the electromagnetic field energies in connection with the gravitation potential.

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    $\begingroup$ Incredibly helpful, thank you! $\endgroup$
    – Mondo Duke
    Commented Dec 2, 2023 at 4:13

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