Maxwell equations in 2+1 D

Question

I have a problem with the Maxwell equations in (2+1) dimensions using differential form. Following J. Baez "Gauge Fields, Knots and Gravity" page 93 (or any other book), the equations are

\begin{align}dF &=0\\ *d*F &=J \end{align}

with $J=-\rho dt+j$ which is a 1-form, and $F$ is the electromagnetic tensor field (2-form). These equations give the correct Maxwell equations in (3+1) dimension but in (2+1) there is a different equation with a minus sign

\begin{align}\text{div} \;E=-\rho. \end{align}

To detail the things \begin{align} F &=B dx\wedge dy-E_x dt\wedge dx-E_y dt \wedge dy\\ *F &=B dt+E_x dy-E_y dx\\ d*F &= -(\partial_t E_y-\partial_x B) dt\wedge dx+(\partial_t E_x-\partial_y B) dt\wedge dy+(\partial_x E_x+\partial_y E_y) dx\wedge dy\\ *d*F &= (\partial_x E_x+\partial_y E_y) dt+(\partial_t E_x-\partial_y B) dx+(\partial_t E_y-\partial_x B) dy \end{align}

which implies

\begin{align}\text{div} \;E=-\rho. \end{align}

WHY, there is a negative sign?

Answer 1

There's something wrong with your sign permutations in the Hodge star operator calculation. If

$F = B + E \wedge dt$,

then, in 2D,

$F = B dx \wedge dy + E_x dx \wedge dt + E_y dy \wedge dt$,

as you wrote yourself. Now, let us take our initial Hodge star as $\star dx \wedge dy = dt$. This means that $\star dt \wedge dx = dy$ and $\star dy \wedge dt = dx$, by cyclic permutation (right hand rule, and here's your mistake, I think). This means that

$\star F = B dt - E_x dy + E_y dx$,

which disagrees in sign to what you got. Indeed, it is easy to see that

$(\star d \star F)(\partial_t) = - (\partial_x E_x + \partial_y E_y) $ ,

which fixes the sign implying $\nabla \cdot E = \rho$.

Answer 2

You're right, and the book is wrong. Its $J$ has the wrong sign.

Bear in mind, this is a monograph, not a peer-reviewed article. Monographs of any level of complexity are extremely prone to and laden with errors. For example,

Exact Solutions Of Einstein's Field Equations
https://www.cambridge.org/core/books/exact-solutions-of-einsteins-field-equations/11CF6CFCC10CC62B9B299F08C32C37A6

writes down the wrong definition for the Riemann Curvature Tensor (!!) and totally mangles up the definition of Lie derivatives. In

Gravitation: Foundations and Frontiers
https://www.amazon.com/Gravitation-Foundations-Frontiers-T-Padmanabhan/dp/0521882230

most of the equations in Chapter 1 are mangled up and wrong; and the book you're citing

Gauge Fields, Knots and Gravity (Knots and Everything)
https://www.amazon.com/GAUGE-FIELDS-KNOTS-GRAVITY-Everything/dp/9810220340

is filled with errors! That's one of them.

That's only the tip of the iceberg, there are a whole lot of more examples like that.

In part, this is because books like these were written before the publishing technology, like TeX, became as accessible and easy to use as it is today; making writing as transparent and computer-powered as it is today. TeX, itself, was a lot harder to use, too.

So, basically you were stuck in the slow lane on paper or (worse) a chalkboard, and that alone creates enough of a distraction to causes errors to crop up. Nowadays, with a good editor and (even) automation, a lot of the grunge work - and the distraction that comes with it - can go away. Part of "computer-powered" also means that it's much easier to correct the errors when they are found and easier to do a run-through to find them. If you weren't around when people used typewriters, then you don't know how good you have it today.

So, let's do this carefully and rigorously.

The potential one form in the coordinates $\left(x^0, x^1, x^2\right) = (t, x, y)$ is: $$A_μ dx^μ = A_x dx + A_y dy - φ dt.$$ The field strength two-form, given by, component-wise, by $F_{μν} = ∂_μ A_ν - ∂_ν A_μ$ is: $$F = ½ F_{μν} dx^μ dx^ν = B dx dy + \left(E_x dx + E_y dy\right) dt.$$ In here and the following, I will be denoting wedge products by juxtaposition, and ordinary products of coordinate differentials will stand for wedge products - except for the metric line element, which I'll only mention and use twice below. The field strength components satisfy the equations: $$B = \frac{∂A_y}{∂x} - \frac{∂A_x}{∂y}, \hspace 1em E_x = -\frac{∂A_x}{∂t} - \frac{∂φ}{∂x}, \hspace 1em E_y = -\frac{∂A_y}{∂t} - \frac{∂φ}{∂y}, \hspace 1em \frac{∂B}{∂t} = \frac{∂E_x}{∂y} - \frac{∂E_y}{∂x}. $$

So, the first thing we note is that, unlike 4-dimensions, there is a scale-dependency on the metric in the formula for the Lagrangian density: $$𝔏 = -¼ k \sqrt{|g|} g^{μρ} g^{νσ} F_{μρ} F_{νσ}.$$ If you adopt a metric with the line element $$g_{μν} dx^μ dx^ν = β dt^2 - α \left(dx^2 + dy^2\right), \hspace 1em β = α c^2,$$ with coordinates $\left(x^0, x^1, x^2\right) = (t, x, y)$, then you will need to set $k = ε_0 c \sqrt{|α|}$ to get the correct Lagrangian density $$𝔏 = ½ \left(ε_0\left({E_x}^2 + {E_y}^2\right) - \frac{B^2}{μ_0}\right), \hspace 1em μ_0 = \frac{1}{ε_0c^2}.$$

So, the natural scaling for the metric is with $|α| = 1$. The sign of $α$ will have no bearing on anything, so we make this a line element for proper distance, with $α = -1$ and $β = -c^2$: $$g_{μν} dx^μ dx^ν = dx^2 + dy^2 - c^2 dt^2.$$

Second, we will then have $$d^3x = dt dx dy, \hspace 1em \sqrt{|g|} d^3x = c dt dx dy.$$

The Hodge Dual is defined in terms of the contraction operator $(\_)˩(\_)$ and metric by: $$\star{\left(dx^μ⋯dx^ν\right)} = ∂^ν ˩ ⋯ ∂^μ ˩ \sqrt{|g|} d^3 x$$ Thus $$ \star{(dx dy)} = ∂^y ˩ ∂^x ˩ c dt dx dy = c ∂_y ˩ ∂_x ˩ dt dx dy = -c ∂_y ˩ dt dy = c dt, \\ \star{(dx dt)} = ∂^t ˩ ∂^x ˩ c dt dx dy = -\frac{∂_t ˩ ∂_x ˩ dt dx dy}{c} = \frac{∂_t ˩ dt dy}{c} = \frac{dy}{c}, \\ \star{(dy dt)} = ∂^t ˩ ∂^y ˩ c dt dx dy = -\frac{∂_t ˩ ∂_y ˩ dt dx dy}{c} = -\frac{∂_t ˩ dt dx}{c} = -\frac{dx}{c}. \\ $$

The components of the response fields $𝔇^x$, $𝔇^y$, $ℌ$ and source $𝔍^x$, $𝔍^y$, $ρ$ are of tensor densities (hint: the names "charge density", "current density"), defined in terms of the Lagrangian density by $$ 𝔇^x = \frac{∂𝔏}{∂E_x}, \hspace 1em 𝔇^y = \frac{∂𝔏}{∂E_y}, \hspace 1em ℌ = -\frac{∂𝔏}{∂B}, \\ 𝔍^x = \frac{∂𝔏}{∂A_x}, \hspace 1em 𝔍^y = \frac{∂𝔏}{∂A_y}, \hspace 1em ρ = -\frac{∂𝔏}{∂φ}. $$ Then, from the total variational $$\begin{align} Δ𝔏 &= ΔE_x 𝔇^x + ΔE_y 𝔇^y - ΔB ℌ + ΔA_x 𝔍^x + ΔA_y 𝔍^y - Δφ ρ \\ &= -\frac{∂}{∂t} \left(ΔA_x 𝔇^x + ΔA_y 𝔇^y\right) - \frac{∂}{∂x} \left(ΔA_y ℌ + Δφ 𝔇^x\right) + \frac{∂}{∂y} \left(ΔA_x ℌ - Δφ 𝔇^y\right) \\ &+ Δφ \left(\frac{∂𝔇^x}{∂x} + \frac{∂𝔇^y}{∂y} - ρ\right) + ΔA_x \left(𝔍^x + \frac{∂𝔇^x}{∂t} - \frac{∂ℌ}{∂y}\right) + ΔA_y \left(𝔍^y + \frac{∂𝔇^y}{∂t} + \frac{∂ℌ}{∂x}\right), \end{align}$$ the corresponding Euler-Lagrange equations and (resulting) continuity equation can then be read off: $$ \frac{∂𝔇^x}{∂x} + \frac{∂𝔇^y}{∂y} = ρ, \hspace 1em \frac{∂ℌ}{∂y} - \frac{∂𝔇^x}{∂t} = 𝔍^x, \hspace 1em -\frac{∂ℌ}{∂x} - \frac{∂𝔇^y}{∂t} = 𝔍^y, \hspace 1em \frac{∂ρ}{∂t} + \frac{∂𝔍^x}{∂x} + \frac{∂𝔍^y}{∂y} = 0. $$

In terms of differential forms, the variational of the Lagrangian 3-form $L = 𝔏 d^3x$ can be written as: $$ΔL = (ΔA)J - (ΔF)G = d(-(ΔA)G) + (ΔA)(J - dG),$$ and the corresponding Euler-Lagrange equation and continuity equation as: $$dG = J, \hspace 1em dJ = 0,$$ provided we set: $$G = ℌ dt + 𝔇^x dy - 𝔇^y dx, \hspace 1em J = ρ dx dy + \left(𝔍^x dy - 𝔍^y dx\right) dt.$$

From the Lagrangian density, as originally stated, we can directly read off the constitutive relations between the field strengths and response fields: $$B = μ_0 ℌ, \hspace 1em 𝔇^x = ε_0 E_x, \hspace 1em 𝔇^y = ε_0 E_y.$$ Upon substitution into $G$, we get: $$\begin{align} G &= \frac{B}{μ_0} dt + ε_0 \left(E_x dy - E_y dx\right) \\ &= ε_0 c \left(B cdt + E_x \frac{dy}{c} - E_y \frac{dx}{c}\right) \\ &= ε_0 c \left(B \star{(dx dy)} + E_x \star{(dx dt)} + E_y \star{(dy dt)}\right) \\ &= ε_0 c \star{F}. \end{align}$$

By virtue of $J = dG$, my $J$ is your $ε_0 c d\star{F}$. Therefore, by virtue of your equation $\star{d\star{F}} = J$, my $\star{J}$ is your $ε_0 c J$. Taking the Hodge-dual of my $J$ results in: $$\begin{align} \star{J} &= ρ \star{(dx dy)} + 𝔍^x \star{(dy dt)} - 𝔍^y \star{(dx dt)} \\ &= ρ c dt - 𝔍^x \frac{dx}{c} - 𝔍^y \frac{dy}{c} \\ &= ρ c dt - \frac{𝔍^x dx + 𝔍^y dy}{c}. \end{align}$$ Therefore, your $J$, in component form, is: $$\frac{ρ}{ε_0} dt - μ_0 \left(𝔍^x dx + 𝔍^y dy\right).$$