# The Question

Given Maxwell's equations of the form \begin{align} \bar{\nabla}\times \bar{B} = \dfrac{4\pi}{c} \bar{J} + \partial_0 \bar{E} \\ \bar{\nabla}\times \bar{E} = -\partial_0 \bar{B} \\ \bar{\nabla} \cdot \bar{B}=0 \\ \bar{\nabla} \cdot \bar{E} = 4 \pi \rho, \end{align} what is the physical significance of the following transformation of Maxwell's equations: \begin{align} -i \bar{\nabla} \times \bar{G} + \bar{\nabla} G_0 = \dfrac{4\pi}{c} \bar{R} + \partial_{0}\bar{G} \tag{Amp-Far Dipole} \\ \bar{\nabla} \cdot \bar{G} - \partial_{0}G_0 = 4\pi R_0 , \tag{Gauss Dipole} \end{align} where $\bar{G} =(i\bar{r} \times\bar{F}-x_0 \bar{F})$, $G_0 = (\bar{r} \cdot \bar{F})$, $\bar{R} = (\rho c \bar{r}-x_0\bar{J}+i \bar{r} \times\bar{J})$, $R_0 = (\bar{r}\cdot\bar{J} - x_0\rho c)/c$ and $\bar{F} = \bar{E} - i \bar{B}$.

# The Definitions

In this post, I refer to (Amp-Far Dipole) and (Gauss Dipole) as the dipole equations, because I do not know what there actual names are or who first published these equations. I just stumbled upon them by accident on pencil and paper.

$x_0 = ct$ is the time variable multiplied by the speed of light for condensed notation purposes.

$\bar{R}$ is the complex combination of the electric dipole field density $\rho c \bar{r}-x_0\bar{J}$ and the magnetic dipole field density $\bar{r} \times\bar{J}$.

$\bar{R}$ is interpreted as a fictitious current in the dipole equations (Amp-Far Dipole). The corresponding fictitious charge density of $\bar{R}$ is $R_0$, which is equal to the Minkowski inner product of the four-position and the four-current.

# Physical Consequences

Though $R_0$ and $\bar{R}$ are fictitious charge and current, they are conserved as a current when $G_0 = 0$. This implies that $G_0$ breaks charge conservation of the fictitious charge and current $R_0$ and $\bar{R}$.

An interesting consequence of the dipole equations is they are identical to Maxwell's equations when $G_0 = 0$.

# Complex Formulation of Maxwell's Equations

I first write Ampere's law, Faraday's law and Gauss' law in complex form \begin{align} -i\bar{\nabla} \times \bar{F} = \dfrac{4\pi}{c} \bar{J} + \partial_{0} \bar{F} \tag{Amp-Far} \\ \bar{\nabla} \cdot \bar{F} = 4\pi \rho \tag{Gauss} , \end{align} where $\bar{F} = \bar{E} + i \bar{B}$.

# Formulation of Ampere-Faraday Law in Dipole form

I use the following differential vector calculus identity \begin{align} \bar{r} \times (\bar{\nabla} \times) + \bar{r} (\bar{\nabla} \cdot) + x_0(\partial_{0}) = \bar{\nabla} \times (\bar{r} \times) + \bar{\nabla} (\bar{r} \cdot) + \partial_{0}(x_0) \end{align} to transform (Amp-Far) into the following: \begin{align} \bar{r} \times (\bar{\nabla} \times \bar{F}) + \bar{r} (\bar{\nabla} \cdot \bar{F}) + x_0(\partial_{0} \bar{F}) =\\ \bar{r} \times \left( i\dfrac{4\pi}{c} \bar{J} + i\partial_{0} \bar{F} \right) + \bar{r} \left(4\pi \rho\right) + x_0\left(-i\bar{\nabla} \times \bar{F} - \dfrac{4\pi}{c} \bar{J}\right) =\\ \dfrac{4\pi}{c} (i\bar{r} \times\bar{J}) + \partial_{0} (i\bar{r} \times\bar{F}) + 4\pi \left( \rho \bar{r} \right) - i\bar{\nabla} \times (x_0\bar{F}) - \dfrac{4\pi}{c} (x_0\bar{J}) =\\ \bar{\nabla} \times (\bar{r} \times \bar{F}) + \bar{\nabla} (\bar{r} \cdot \bar{F}) + \partial_{0}(x_0 \bar{F}) , \end{align} which reduces to the following expression \begin{align} -i \bar{\nabla} \times \left( i\bar{r} \times \bar{F} - x_0\bar{F} \right) + \bar{\nabla} (\bar{r} \cdot \bar{F}) =\\ \dfrac{4\pi}{c} \left( \rho c \bar{r} - x_0\bar{J} + i \bar{r} \times\bar{J} \right) + \partial_{0} \left( i\bar{r} \times\bar{F} - x_0 \bar{F} \right) . \end{align} One can perform the following substitutions $\bar{G} =(i\bar{r} \times\bar{F}-x_0 \bar{F})$, $G_0 = (\bar{r} \cdot \bar{F})$, and $\bar{R} = (\rho c \bar{r}-x_0\bar{J}+i \bar{r} \times\bar{J})$ to obtain \begin{align} -i \bar{\nabla} \times \bar{G} + \bar{\nabla} G_0 = \dfrac{4\pi}{c} \bar{R} + \partial_{0}\bar{G} . \tag{Amp-Far Dipole} \end{align}

# Formulation of Gauss' Law in Dipole form

I use the following differential vector calculus identity \begin{align} -x_0 (\nabla\cdot) + \bar{r}\cdot (-i\bar{\nabla}\times) = \bar{\nabla} \cdot (i(\bar{r}\times)- x_0) \end{align} to transform (Gauss) into the following: \begin{align} -x_0 (\nabla\cdot\bar{F}) + \bar{r}\cdot (-i\bar{\nabla}\times\bar{F}) =\\ -x_0 (4\pi \rho) + \bar{r}\cdot \left(\dfrac{4\pi}{c} \bar{J} + \partial_{0} \bar{F}\right) =\\ \dfrac{4\pi}{c} (\bar{r}\cdot\bar{J} - x_0\rho c) + \partial_{0} (\bar{r}\cdot\bar{F}) =\\ \bar{\nabla} \cdot (i\bar{r}\times\bar{F} - x_0\bar{F}) , \end{align} which reduces to the following expression \begin{align} \bar{\nabla} \cdot (i\bar{r}\times\bar{F} - x_0\bar{F}) - \partial_{0} (\bar{r}\cdot\bar{F}) = \dfrac{4\pi}{c} (\bar{r}\cdot\bar{J} - x_0\rho c) . \end{align} One can perform the following substitutions $R_0 = (\bar{r}\cdot\bar{J} - x_0\rho c)/c$ to obtain \begin{align} \bar{\nabla} \cdot \bar{G} - \partial_{0}G_0 = 4\pi R_0 . \tag{Gauss Dipole} \end{align}

• (a) What is $x_0$? (b) What is your question? (c) How am I to think of these terms physically. – Kyle Kanos Mar 18 '14 at 16:27
• You didn't address (2) (which, in terms of this site, is the most important one to address). And the terms I was referring to are your "dipole" terms, $F$ and $G$; how am I to think of these. I believe your math, I just don't see the use of the above. – Kyle Kanos Mar 18 '14 at 16:51
• @KyleKanos Does this change address issue(2)? Thanks. – linuxfreebird Mar 18 '14 at 17:01

No explicit complexification is needed to derive this breakdown of Maxwell's equations. This can be understood wholly through the real vector space of special relativity.

Let's start with Maxwell's equations for the EM field, in the clifford algebra language called STA: the spacetime algebra. Maxwell's equations take the form

$$\nabla F = -J$$

where $\nabla F = \nabla \cdot F + \nabla \wedge F$, $F = e_0 E + B \epsilon_3$, in the $(-, +, +, +)$ sign convention.

Let $x$ be the spacetime position vector. It's generally true that, for a vector $v$ and a constant bivector $C$,

$$\nabla (C \cdot x) = -2 C, \quad \nabla (C \wedge x) = 2C \implies \nabla (Cx) = 0$$

One can then evaluate the expression

$$\nabla (Fx) = (\nabla F)x + \dot \nabla (F \dot x)$$

where the overdot means that only $x$ is differentiated in the second term; using the product rule, $F$ is "held constant" and so the above formulas apply. We just argued that the second term is zero, so we get $\nabla (Fx) = (\nabla F) x$. Thus, we arrive at the following transformation of Maxwell's equations:

$$\nabla (Fx) = -Jx$$

Now, we could always write $F$ as a "complex bivector" in the sense that, using $\epsilon = e_0 \epsilon_3$, and $\epsilon \epsilon = -1$, we have

$$F = e_0 E - B \epsilon_3 e_0 \epsilon_3 \epsilon = e_0 (E + \epsilon B)$$

It's crucial to note that $\epsilon$ does not commute with any vector.

What are the components of $Fx$? Write $x = t e_0 + r$ and we can write them as

$$Fx = e_0 (Ex + \epsilon Bx) = e_0 (E \cdot r + E \wedge r - e_0 Et + \epsilon B \cdot r - e_0 B \times r + \epsilon B t e_0)$$

This too can be written in a "complex" form:

$$Fx = (e_0 E \cdot r + Et + B \times r) + \epsilon (E \times r + e_0 B\cdot r + Bt)$$

We seem to differ on some signs, but this is recognizably the same quantity you have called $G$.

Now, to talk about how these equations break down, let's write $G = G_1 + G_3$, where $G_1 = (e_0 E \cdot r + \ldots)$ and $G_3 = \epsilon (E \times r + \ldots)$. Let's also write for $R = Jx = R_0 + R_2$.

Maxwell's equations then become

$$\nabla \cdot G_1= R_0, \quad \nabla \wedge G_1 + \nabla \cdot G_3 = R_2, \quad \nabla \wedge G_3 = 0$$

The first and third equations are the components of the Gauss dipole; the second equation is the Ampere-Faraday dipole equation.

Now, what does it all mean? The expression for $G = Fx$ includes both rotational moments of the EM field as well as some dot products, so it measures both how much the spacetime position is in the same plane as the EM field as well as how much the spacetime position is out of the plane.

It's probably more instructive to look at the source term $-Jx$. This tells us both about the moments of the four-current as well as how it goes toward or away from the coordinate origin. The description for the moments is wholly in the Ampere-Faraday dipole equation. What kinds of moments would this describe? A pair of two opposite point charges at rest, separated by a spatial vector $2 \hat v$ and centered on the origin, each with current at rest $j_0$, would create a $R = Jx = + j_0 e_t \hat v - j_0 e_t (-\hat v) = 2 j_0 e_t \hat v$, so this would be described wholly by the A-F dipole equation.

That's at time zero, however. At later times, $R$ will pick up these weird time terms. Say we're at time $\tau$. Then $R = 2 j_0 e_t \hat v + j_0 e_t (\tau e_t) - j_0 e_t (\tau e_t)$. So for this case, there's no problem: the extra stuff will just cancel. A single charge, however, would start picking up this term.

In a few words, these equations are weird.

• I think your answer makes sense. I liked the use of the word "weird." If you have time could you take a stab at physics.stackexchange.com/questions/103535/… ? I think this question implies the possibility of a fourth component in the Faraday field, maybe? $G_0$ is the fourth component of $\bar{G}$. – linuxfreebird Mar 18 '14 at 21:18
• You should not be deceived because $G$ has eight real components. An arbitrary field $P$ with the same nonzero components as $G$ could have would have 8 degrees of freedom, yes. When you multiply $P$ by $x$ on the right, you get 8 numbers: the 6 bivector components, a scalar component, and a pseudoscalar component. This is not true of $G$. When you multiply $Gx$, you get that the scalar and pseudoscalar components are zero. Though $G$ has eight components, only six of them represent real degrees of freedom, just as the Faraday bivector $F$ has only six components. – Muphrid Mar 18 '14 at 23:11
• I cannot remember if you answered another question of mine. I think what your saying is that since G is related to F, F only has 6 degrees of freedom, so G only has 6 degrees of freedom. What sparked my interest was G express a new flavor of Maxwell's equations with a mysterious fourth field component. If there existed a fourth component of F in the same form as G, what would it be? Could it be expressed in terms of the four-potential? If F did have 8 independent components, then G would have 8 independent components. – linuxfreebird Mar 19 '14 at 0:29
• I'll answer this question on your actual question with this question. – Muphrid Mar 19 '14 at 3:04

I was browsing this and it looks like you can improve your notation a little bit. You can define a product on four vectors given by $(a^0,\vec{a}) * (b^0,\vec{b}) = (\vec{a} \cdot \vec{b} - a^0 b^0, b^0 \vec{a} - a^0 \vec{b} +i \vec{a} \times \vec{b})$. Also you can define a related product $\bar{*}$ by $(a^0,\vec{a}) \bar{*} (b^0,\vec{b}) = (\vec{a} \cdot \vec{b} - a^0 b^0, b^0 \vec{a} - a^0 \vec{b} -i \vec{a} \times \vec{b})$. Then you can define a four vector $F$ by $F=(0,\vec{F})$.

Then maxwell's equations becomes $\partial \bar{*} F = 4\pi j$, you can define a 4 vector $R=r * j$, and $G=r * F$. These are the four vector analogues of your $R$ and $G$. Having framed everything in terms of these products your new equations become $\partial \bar{*} G =4\pi R$. I haven't explained the physical meaning, but hopefully this makes the problem easier to think about. Also hopefully I didn't get any minus signs wrong. Please comment or fix the answer if I did.

# Edit

I guess I can give an attempt and giving some interpretation, but I don't think it will be too insightful. Basically you can think of $G$ as a moment of the field, and $R$ as a moment of the current. You have found that if the fields and currents satisfy maxwell's equations, then the moments must as well. This reminds me of how if the four-vector potential satisfies the field equation then so must the fields, becuase the process of differentiating the vector potential commutes with applying the field operator. So here you seem to be saying that if the fields and currents satisfy maxwell's equations, then the moments must as well, because the process of taking the moments commutes with applying $\partial \bar{*}$.

• I think your notation is using bi-quaternion? I think you are missing a sign, but the idea is the same. By the way you are representing F and G as four-component vectors? Please take a look at this posted question: physics.stackexchange.com/questions/103535/… – linuxfreebird Mar 18 '14 at 18:14
• The answer in the linked to question is the way I am used to seeing it, but I think the way you did it here looks interesting. I was thinking biquaternions would be relevant, but I didn't intend for my four-vectors to be interpreted as a kind of quaternion, though they do look similar. $F$ is a four-vector, though its time-like component is zero. $G$ is also a four-vector and it in general has a non-zero time component. Also where was my sign error? – Brian Moths Mar 18 '14 at 18:19
• I attempted to add physical explanation. I don't think it is so helpful, but it is something. – Brian Moths Mar 18 '14 at 18:28
• It will take me sometime to fix the missing sign problem. It comes down to choice of sign convention. To perform a quick fix just write out * and -* as equal expressions and then change the sign of a_0 for one and that will fix the sign problem, but again choosing any of the vector component combinations will work. I trying to figure out what is the best choice of convention. – linuxfreebird Mar 18 '14 at 21:14