We know, the the vector form of Maxwell equations \begin{align} \vec\nabla\cdot\vec{E} &= 4\pi\rho \label{Diff I}\\ \vec\nabla\times\vec{B} &= \dfrac{4\pi}{c} \vec{j}+\dfrac{1}{c}\dfrac{\partial\vec{E}}{\partial t} \label{Diff IV}\\ \vec\nabla\times\vec{E} &= -\dfrac{1}{c}\dfrac{\partial\vec{B}}{\partial t} \label{Diff III}\\ \vec\nabla\cdot\vec{B} &= 0. \label{Diff II} \end{align}

The last two of them allow us to introduce the potentials: \begin{align} \vec{E} &= -\frac1c \frac{\partial \vec{A}}{\partial t} - \vec\nabla\phi\\ \vec{B} &= \vec\nabla\times\vec A \end{align} which tells us about gauge invariance of equations.

All four of Maxwell's equations can be written compactly as

\begin{align} \partial_{\mu}F^{\mu\nu} &= \frac{4\pi}{c}j^{\nu} \tag{1}\\ \partial_{[\mu}F_{\alpha\beta]} &= 0\;. \tag{2} \end{align}

And according to the last one equation, the first one we can rewrite (use preferred gauge) in form: \begin{equation} \Box A^{\mu} = -\frac{4\pi}{c} j^{\mu}. \end{equation}

Now we consider the Einstein GR equations: \begin{equation} R_{\mu\nu} = 8\pi G (T_{\mu\nu} - \frac12g_{\mu\nu}T). \end{equation}

Or in "$\Gamma-$field" form (indexes are omitted): \begin{equation} \partial \Gamma - \partial \Gamma + \Gamma\Gamma - \Gamma\Gamma = 8\pi G (T_{\mu\nu} - \frac12g_{\mu\nu}T). \end{equation}

We know, in weak field limit the equations get simply form externally similar to Maxwell ones (so called gravitomagnetism) in 3-vector form, or in covariant form: \begin{equation} \Box h_{\mu\nu} = -16\pi G (T_{\mu\nu} - \frac12\eta_{\mu\nu}T). \end{equation}

Thus, the question: Why for covariant form of Maxwell equations we need two different entities of equations, but for the GR the only one? Or another words, can we write the Einstein GR equations for weak field limit similar to Maxwell equations in field form, not via potentials?

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3 Answers 3


If you want to compare Maxwell's EM with GR, note they're respectively obtainable from extending a global $U(1)$ variance to a local one and Lorentz invariance to invariance under general coordinate transformations. Thus the equivalent of introducing $A_\mu$ in the gauge covariant derivative is introducing Christoffel symbols in the connection, while the commutator $F_{\mu\nu}$ of gauge covariant derivatives is analogous to the commutator $R_{\mu\nu\rho\sigma}$ of connections acting on a vector field. So the equivalent of (1) is Einstein's usual equations, while the equivalent of (2) is $R_{\mu[\nu\rho\sigma]}=0$. In neither case do we need the second equation; a Lagrangian formulation obtains the first, but the second is just a tautology.

  • $\begingroup$ Ok, as I understand, the answer, the Bianchi relation is the GR analog to the (2). $\endgroup$
    – Sergio
    Commented Dec 24, 2018 at 8:24
  • $\begingroup$ I believe it is the other Bianchi identity $R^{\alpha}_{\;\beta[\mu\nu ;\sigma ]}=0$ that is the analog of the EM Bianchi identity. $\endgroup$
    – fewfew4
    Commented Aug 14, 2021 at 0:07
  1. The counterpart to $(\text{metric }g,\text{curvature }R)$ in GR is $(\text{gauge potential }A,\text{field strength }F)$ in E&M.

  2. If you write $F=F(A)$ in E&M, then the Bianchi relation (2) is not needed/a tautology, cf. e.g. this Phys.SE post.

  3. For how GEM appears as a limit of GR, see e.g. this Phys.SE post.


The Analogy Is With Non-Abelian Gauge Fields
The relevant analogy is with non-Abelian gauge fields. Using the conventions I wrote out in my reply here

Yang-Mills vs Einstein-Hilbert Action

the relevant analogy is with the non-linear equations: $$dA + A A = F, \hspace 1em dF + AF - FA = 0.$$ For electromagnetism and gauge fields, one has $$A = A^a Y_a = A_μ dx^μ = A^a_μ Y_a dx^μ, \hspace 1em F = F^c Y_c = ½ F_{μν} dx^μ dx^ν = ½ F^c_{μν} Y_c dx^μ dx^ν,$$ where I'm using juxtaposition $dx^μ dx^ν$ for wedge products $dx^μ∧dx^ν$ and adopting the convention of allowing the gauge basis $Y_a$ to freely intermix with the form, e.g. $Y_a dx^μ = dx^μ Y_a$.

For the Cartan structure equations $$dθ^a + ω^a_c θ^c = Θ^a, \hspace 1em dω^a_b + ω^a_c ω^c_b = Ω^a_b,$$ one has $$A = θ^a Y_a + ω^a_b Y_a^b, \hspace 1em F = Θ^a Y_a + Ω^a_b Y_a^b,$$ with the frame one-forms $θ^a = h^a_μ dx^μ$, connection one-forms $ω^a_b = Γ^a_{μb} dx^μ$, torsion two-forms $Θ^a = ½ T^a_{μν} dx^μ dx^ν$ and curvature two-forms $Ω^a_b = ½ R^a_{bμν} dx^μ dx^ν$.

This plays the role of $dA + A A = F$, for a suitably-defined set of Lie brackets defined for the basis elements $Y_a$ and $Y_a^b$. (The Lie brackets are those for the Lie algebra of the gauge group $GA(4)$ - the 4D general affine group).

The role of the "Bianchi identity" equations $dA + A F - FA = 0$ is then played by the equations derived from the Cartan structure equations: $$dΘ^a + ω^a_c Θ^c = Ω^a_c θ^c, \hspace 1em dΩ^a_b + ω^a_c Ω^c_b = Ω^a_c ω^c_b.$$

Because this is subject to the constraint that the (constant) frame metric $g_{ab}$ has zero-covariant derivative, $$∇g_{ab} ≡ dg_{ab} - ω^c_a g_{cb} - ω^c_b g_{ac} = 0,$$ then along with the equations derived from it: $$d∇g_{ab} - ω^c_a ∇g_{cb} - ω^c_b ∇g_{ac} = Ω^c_a g_{cb} - Ω^c_b g_{ac},$$ this entails that the connection one-form and curvature two-form are anti-symmetric in their frame indices, when raised or lowered by the metric. The preferred position that dovetails into the above-mentioned reply is with them up: $$ω^a_c g^{cb} = ω^{ab} = -ω^{ba}, \hspace 1em Ω^a_c g^{cb} = Ω^{ab} = -Ω^{ba}.$$ In that case, the constraint can be removed, reducing the one-form $A$ and two-form $F$ to: $$A = θ^a P_a + ½ ω^{ab} S_{ab}, \hspace 1em F = Θ^a P_a + ½ Ω^{ab} S_{ab},$$ the metric $g_{ab}$ now being woven into the structure coefficients for the basis elements $P_a$ and $S_{ab}$. (The relevant gauge group is the inhomogeneous extension $ISO(3,1)$ of the Lorentz group $SO(3,1)$.)

The actual reduction - with more detail on the correspondence laid out here - is provided in my earlier reply here

What are the analogues of $F_{\mu\nu}$ in General Relativity?

The space-time metric, itself, is the frame metric converted to space-time indices: $g_{μν} = h^a_μ h^b_ν g_{ab}$.

In my later reply, cited on top above, which starts out with this reduced form, I made a minor change to the $[P,P]$ Lie brackets to produce this as the field strength $$F = Θ^aP_a + ½\left(Ω^{ab} + λθ^aθ^b\right)S_{ab}$$ so as to directly weave in the cosmological coefficient, in the process also enabling the action to be expressed equivalently as a quadratic function of the gauge field strengths. Depending on what the coefficient $λ$ used with the modification is, the Lie algebra can be that for either $SO(3,2)$ or $SO(4,1)$.

What Are The "Second Set" Of Fields?
First, you should identify what actually are the "second set" of fields; and that is obscured by the form you wrote the equations in. I will rewrite your equations, plus the continuity equation (which you left out), in the following form: $$ ∇·𝐞 = 4πr, \hspace 1em ∇×𝐛 = \frac{4π}{c}𝐣 + \frac{1}{c}\frac{∂𝐞}{∂t} \hspace 1em⇒\hspace 1em ∇·𝐣 + \frac{1}{c}\frac{∂r}{∂t} = 0, \\ ∇×𝐞 = -\frac{1}{c}\frac{∂𝐛}{∂t}, \hspace 1em ∇·𝐛 = 0 \hspace 1em⇐\hspace 1em ∇·𝐞 = -\frac{1}{c}\frac{∂𝐚}{∂t} - ∇f, \hspace 1em 𝐛 = ∇×𝐚. $$ The first set of equations arise from an action principle given by the Lagrangian density $$𝔏 = \frac{|𝐞|^2 - |𝐛|^2}{8π} + 𝔏_0(𝐚,f),$$ the additional "interaction" part $𝔏_0(𝐚,f)$ arises from all other fields that have non-trivial gauge transformations.

The extra fields arise as derivatives of the Lagrangian density, whose total variational can be written in terms of them as: $$δ𝔏 = δ𝐚·𝐣 - δf r + δ𝐞·𝐝 - δ𝐛·𝐡,$$ with the Lagrangian density, itself, serving as a device to generate a set of constitutive relations that connect those fields with the gauge fields: $$ 𝐝 = \frac{∂𝔏}{∂𝐞} = \frac{𝐞}{4π}, \hspace 1em 𝐡 = -\frac{∂𝔏}{∂𝐛} = \frac{𝐛}{4π}, \hspace 1em r = -\frac{∂𝔏}{∂f} = -\frac{∂𝔏_0}{∂f}, \hspace 1em 𝐣 = \frac{∂𝔏}{∂𝐚} = \frac{∂𝔏_0}{∂𝐚}. $$ From this, your first set of equations arise as Euler-Lagrange equations $$∇·𝐝 = r, \hspace 1em ∇×𝐡 = \frac{𝐣}{c} + \frac{1}{c}\frac{∂𝐝}{∂t},$$ for the response fields $(𝐝,𝐡)$ and sources $(r,𝐣)$.

This is best seen by restoring the constitutive coefficients, making the change: $$(f,𝐚,𝐛,𝐞) = \sqrt{4πε_0} (φ,c𝐀,c𝐁,𝐄), \hspace 1em (r,𝐣,𝐝,𝐡) = \frac{(ρ,𝐉,𝐃,𝐇/c)}{\sqrt{4πε_0}},$$ thereby giving us back the Maxwell (and SI) form of the equations: $$ F = dA \hspace 1em⇒\hspace 1em 𝐁 = ∇×𝐀, \hspace 1em 𝐄 = -∇φ - \frac{∂𝐀}{∂t}, \\ dF = 0 \hspace 1em⇒\hspace 1em ∇·𝐁 = 0, \hspace 1em ∇×𝐄 + \frac{∂𝐁}{∂t} = 𝟎, $$ with the one-form $A$ and two-form $F$ being written $$A = 𝐀·d𝐫 - φ dt, \hspace 1em F = 𝐁·d𝐒 + 𝐄·d𝐫 dt,$$ where $d𝐫 = (dx, dy, dz)$ and $d𝐒 = (dy dz, dz dx, dx dy)$. The non-Abelian analogue of this would be: $$ 𝐁 = ∇×𝐀 + 𝐀×𝐀, \hspace 1em 𝐄 = -∇φ - \frac{∂𝐀}{∂t} + φ𝐀 - 𝐀φ, \\ ∇·𝐁 + 𝐀·𝐁 - 𝐁·𝐀 = 0, \hspace 1em ∇×𝐄 + \frac{∂𝐁}{∂t} + 𝐀×𝐄 + 𝐄×𝐀 + 𝐁φ - φ𝐁 = 𝟎. $$

There are no $c$'s in any of this, because there is no connection with the Minkowski - or any - metric in these equations.

The second set of fields are the response fields and sources that are generated by the Lagrangian density, arising as its derivatives, with the variational being: $$δ𝔏 = δ𝐀·𝐉 - δφ ρ + δ𝐄·𝐃 - δ𝐁·𝐇.$$ With the Lagrangian density now expressed as $$𝔏 = \frac{ε_0|𝐄|^2}{2} - \frac{|𝐁|^2}{2μ_0} + 𝔏_0(φ,𝐀),$$ where $μ_0 = 1/(ε_0c^2)$ this yields the constitutive relations $$𝐃 = \frac{∂𝔏}{∂𝐄} = ε_0 𝐄, \hspace 1em 𝐇 = -\frac{∂𝔏}{∂𝐁} = \frac{𝐁}{μ_0}, \hspace 1em ρ = -\frac{∂𝔏}{∂φ} = -\frac{∂𝔏_0}{∂φ}, \hspace 1em 𝐉 = \frac{∂𝔏}{∂𝐀} = \frac{∂𝔏_0}{∂𝐀},$$ along with the following as the Euler-Lagrange equations and (derived from it) the continuity equation $$∇·𝐃 = ρ, \hspace 1em ∇×𝐇 - \frac{∂𝐃}{∂t} = 𝐉, \hspace 1em ∇·𝐉 + \frac{∂ρ}{∂t} = 0.$$

In terms of differential forms, the response fields and sources are: $$G = ½ 𝔊^{μν} ∂_ν ˩ ∂_μ ˩ d^4x = 𝐃·d𝐒 - 𝐇·d𝐫 dt, \hspace 1em J = 𝔍^μ ∂_μ ˩ d^4x = ρ dV - 𝐉·d𝐒 dt,$$ where $$dV = dx dy dz, \hspace 1em d^4 x = dt dV = dt dx dy dz.$$ The contraction operator $∂_μ ˩ α$ is defined recursively on differential forms by $∂_μ ˩ (dx^ν ∧ α) = δ_μ^ν α - dx^ν ∧ (∂_μ ˩ α)$ and $∂_μ ˩ f = 0$ for scalars (a.k.a. 0-forms) $f$. For instance, $$ \frac{∂}{∂t} ˩ d^4 x = \frac{∂}{∂t} ˩ dt dV = dV, \\ \frac{∂}{∂x} ˩ d^4 x = \frac{∂}{∂x} ˩ dt dx dy dz = -dt \left(\frac{∂}{∂x} ˩ dx dy dz\right) = -dt dy dz = -dy dz dt. $$

The non-Abelian version of these equations are: $$ ∇·𝐃 + 𝐀·𝐃 - 𝐃·𝐀 = ρ, \hspace 1em ∇×𝐇 - \frac{∂𝐃}{∂t} + 𝐀×𝐇 + 𝐇×𝐀 + φ𝐃 - 𝐃φ = 𝐉, \\ ∇·𝐉 + \frac{∂ρ}{∂t} + 𝐀·𝐉 - 𝐉·𝐀 + ρφ - φρ = 𝐃·𝐄 - 𝐄·𝐃 + 𝐁·𝐇 - 𝐇·𝐁. $$ The response fields and sources are now: $$G = G_c y^c = ½ 𝔊_c^{μν} y^c ∂_ν ˩ ∂_μ ˩ d^4x, \hspace 1em J = J_c y^c = 𝔍^μ ∂_μ ˩ d^4x = 𝔍_c^μ y^c ∂_μ ˩ d^4x.$$ They are referenced to the dual Lie basis $y^a$ and are not even the same type of objects as the gauge fields. They Lie co-vectors, while the gauge fields are Lie vectors.

That's what the "second set" of fields actually are.

In terms of the Lagrangian 4-form $L = 𝔏 d^4x$, the variational may be written: $$ΔL = \left(ΔA^a\right) J_a - \left(ΔF^a\right) G_a.$$

For Yang-Mills actions and Lagrangians, $FG - GF = 0$ and the terms $𝐃·𝐄 - 𝐄·𝐃 = 0$ and $𝐁·𝐇 - 𝐇·𝐁 = 0$ drop out from the continuity equation.

All of this is obscured by the form your equations were written in, which essentially equated $𝐝$ with $𝐞$ and $𝐛$ with $𝐡$.

Where Are The Second Set Of Fields?
The Einstein-Hilbert Lagrangian with a cosmological coefficient - as a 4-form - can be written, along with the corresponding action $S$, as: $$S = \int L, \hspace 1em L = ε_{abcd} \sqrt{|g|} \left(k Ω^{ab} + l θ^aθ^b\right)θ^cθ^d,$$ for suitably-defined coefficients $k$ and $l$.

Technically, if the torsion is not constrained to be 0, it is the Einstein-Cartan action (with cosmological term), while it is the Einstein-Hilbert action (with cosmological term) expressed on a Riemann-Cartan geometry as a Palatini action, if the torsion is constrained to be 0. But with matter-free Lagrangians, or outside of matter of a matter term is included, the torsion reduces to 0 as a result of the field equations.

If $l ≠ 0$, the action can be equivalently written as $$S = \int L', \hspace 1em L' = m ε_{abcd} \sqrt{|g|} \left(Ω^{ab} + λθ^aθ^b\right)\left(Ω^{cd} + λθ^cθ^d\right),$$ where $$m = \frac{k^2}{4l}, \hspace 1em λ = \frac{2l}{k},$$ because $ε_{abcd} \sqrt{|g|} Ω^{ab} Ω^{cd}$ drops out from the action as a boundary term.

Up to proportionality $ε_{abcd} \sqrt{|g|} Ω^{ab} θ^cθ^d$ is just $R \sqrt{|g|} d^4x$, itself, where $R$ is the curvature scalar.

In both cases, you can write down the variational in terms of a set of response fields and sources: $$ΔL = \left(Δθ^a\right) J_a + ½ \left(Δω^{ab}\right) J_{ab} - \left(ΔΘ^a\right) G_a - ½ \left(ΔΩ^{ab}\right) G_{ab}.$$ But the $G$'s and $J$'s are the bona fide response and source fields only for the field strength $F$ without the $λ$ adjustment.

With the $λ$ adjustment, the fields would be: $$ ΔL = \left(Δθ^a\right) \bar{J}_a + ½ \left(Δω^{ab}\right) J_{ab} - \left(ΔΘ^a\right) G_a - ½ \left(Δ\bar{Ω}^{ab}\right) G_{ab},\\ \bar{Ω}^{ab} = Ω^{ab} + λθ^aθ^b, \hspace 1em \bar{J}_a = J_a + λG_{ab}θ^b. $$

The 3-currents associated with the source fields $$J_a = 𝔍^μ_a ∂_μ ˩ d^4 x, \hspace 1em J_{ab} = 𝔍^μ_{ab} ∂_μ ˩ d^4 x,$$ are naturally identified as the 3-currents, respectively, for momentum $p_a$ and internal angular momentum $s_{ab}$.

The corresponding Euler-Lagrange equations are $$ dG_a - G_c ω^c_a = J_a, \\ dG_{ab} - G_{cb} ω^c_a - G_{ac} ω^c_b = J_{ab} + θ_a G_b - θ_b G_a, $$ the resulting "continuity" equations being: $$ dJ_a + J_c ω^c_a = -G_c Ω^c_a, \\ dJ_{ab} + J_{cb} ω^c_a + J_{ac} ω^c_b - θ_b J_a + θ_a J_b = -G_{ac} Ω^c_b - G_{cb} Ω^c_a - Θ_a G_b + Θ_b G_a. $$

So ... what are the extra fields? With the unadjusted Lagrangian, they are: $$ G_a = 0, \hspace 1em G_{ab} = 2k \sqrt{|g|} ε_{abcd} θ^c θ^d, \hspace 1em J_c = \sqrt{|g|} ε_{abcd} \left(2k Ω^{ab} + 4 l θ^a θ^b\right) θ^d, \hspace 1em J_{cd} = 0. $$ Up to a constant factor the curvature term $ε_{abcd} Ω^{ab} θ^d$ in the source $J_c$ is the Einstein tensor, itself, expressed as a 3-current density: $\sqrt{|g|} e^ν_c \left(R^μ_ν - ½ δ^μ_ν R\right) ∂_μ ˩ d^4 x$, where $e^ν_c$ is the inverse to the frame matrix $h^μ_a$.

The $λ$-adjusted response fields and sources, in terms of the $λ$-adjusted action and Lagrangian, take on a much simpler form: $$G_a = 0, \hspace 1em G_{ab} = 2m \sqrt{|g|} ε_{abcd} \bar{Ω}^{cd}, \hspace 1em \bar{J}_a = 0, \hspace 1em J_{ab} = 0.$$ The response field $G_{ab}$ picks up an extra curvature term $2m \sqrt{|g|} ε_{abcd} Ω^{cd}$, compared to the unadjusted response field. It arises from the boundary term $ε_{abcd} \sqrt{|g|} Ω^{ab} Ω^{cd}$ and drops out from the Euler-Lagrange equations (as a result of the "Bianchi identities" that came from $dF + AF - FA = 0$). The unadjusted source term $J_a$ picks up a contribution proportional to $λ G_{ab} θ^b$ that yields the 3-current density associated with the Einstein tensor.


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