# Formula for the Maxwell Stress tensor in arbitrary coordinates

This question is nearly identical to my last, except this time its the Maxwell stress tensor, not the Cauchy stress tensor. I often see its components written as $$\sigma_{ij}=\varepsilon_0E_iE_j+\frac{1}{\mu_0}B_iB_j+\frac{\delta_{ij}}{2}\left(\varepsilon_0|\boldsymbol E|^2+\frac{1}{\mu_0}|\boldsymbol{B}|^2\right)$$ With $$E_i$$ being understood as the components of the vector $$\boldsymbol E$$. But I thought, "hang on", $$\boldsymbol E$$ is a vector, and thus is contravariant, and its components should be written $$E^i$$, and similarly for $$\boldsymbol B$$. But, I know that the stress tensor should be fully covariant, since it measures force (a vector) per unit area (which can be represented as a normal vector). I.e, it takes in two vector inputs and outputs a scalar, so it should be second order covariant. So I thought, we should replace $$E_i$$ with $$g_{ki}E^k$$. Similarly, the Kronecker delta bugs me as well - it is defined as a $$(1,1)$$ tensor, with components $$\delta^i_j=1 \text{ if }i=j, ~\delta^i_j=0\text{ if }i\neq j$$ So $$\delta_{ij}=g_{ki}\delta^k_j=g_{ij}$$ So, the "correct" formula, written out in all its glory, should really be $$\sigma_{ij}=\varepsilon_0(g_{ki}E^k)(g_{lj}E^j)+\frac{1}{\mu_0}(g_{ki}B^k)(g_{lj}B^j)+\frac{g_{ij}}{2}\left(\varepsilon_0|\boldsymbol E|^2+\frac{1}{\mu_0}|\boldsymbol{B}|^2\right)$$ Or of course, in shorter form $$\sigma_{ij}=\varepsilon_0E_iE_j+\frac{1}{\mu_0}B_iB_j+\frac{g_{ij}}{2}\left(\varepsilon_0|\boldsymbol E|^2+\frac{1}{\mu_0}|\boldsymbol{B}|^2\right)$$ Where $$E_i$$ are recognized not as the components of $$\boldsymbol E$$, but rather as the components of $$\boldsymbol E^{\flat}$$, its dual. And of course $$|\boldsymbol E|^2=g_{ab}E^aE^b$$. Am I right?

It is actually better if we define the Lorentz invariant Maxwell stress tensor. It is a second order contravariant tensor: $$T^{\alpha\beta}=\frac{1}{\mu_0}\left(F^{\alpha\gamma}F^\beta{}_{\gamma}-\frac{1}{4}\eta^{\alpha\beta}F^{\gamma\delta}F_{\gamma\delta}\right)$$

Here $$F^{\mu\nu}$$ is the $$\mu,\nu$$ component of the field strength tensor and $$\eta^{\mu\nu}$$ is the $$\mu,\nu$$ component of the inverse spacetime metric. This formula works no matter which coordinate system we use (but, the components of $$\boldsymbol{\eta}$$, $$\mathbf{F}$$ might be very complicated). We use the well known Lorentz scalar $$F^{\mu\nu}F_{\mu\nu}=2|\boldsymbol B|^2-2|\boldsymbol E^2|/c^2$$ Let $$i,j,k,l,m\in\{1,2,3\}$$. Let's compute the spacial components of the MST:

$$T^{ij}=\frac{1}{\mu_0}\left(F^{i\gamma}F^j{}_{\gamma}-\frac{\eta^{ij}}{2}(|\boldsymbol B|^2-|\boldsymbol E^2|/c^2)\right)$$

Working for example in Cartesian spacial coordinates (i.e $$\boldsymbol\eta=\operatorname{diag}(-1,1,1,1)$$), we have the following identities: $$F^{i~0}=\frac{1}{c}E^i~~;~~F^{ij}=\varepsilon_{ijk}B^k \\ F^j{}_\gamma=\eta_{\gamma\lambda}F^{j\lambda} \\ \implies F^j{}_0=-F^{j~0}=\frac{-1}{c}E^j \\ \text{and}~~F^j{}_k=\eta_{k\lambda}F^{j\lambda}=\eta_{kk}F^{jk}=\varepsilon_{jkl}B^l$$ Hence $$T^{ij}=\frac{1}{\mu_0}\left(F^{i~0}F^j{}_0+F^{ik}F^j{}_k-\frac{\delta^i_j}{2}(|\boldsymbol B|^2-|\boldsymbol E^2|/c^2)\right) \\ =\frac{1}{\mu_0}\left(-\frac{1}{c^2}E^iE^j+\varepsilon_{ikl}B^l\varepsilon_{jkm}B^m-\frac{\delta^i_j}{2}(|\boldsymbol B|^2-|\boldsymbol E^2|/c^2)\right)$$ Now, $$\varepsilon_{ikl}\varepsilon_{jkm}=\varepsilon_{ilk}\varepsilon_{jmk}=\delta^i_j\delta^l_m-\delta^i_m\delta^l_j$$ So $$\varepsilon_{ikl}B^l\varepsilon_{jkm}B^m=(\delta^i_j\delta^l_m-\delta^i_m\delta^l_j)B^lB^m=\delta^i_jB^lB^l-B^iB^j$$ Hence, in Cartesian coords, $$\boxed{T^{ij}=\frac{1}{\mu_0}\left(-\frac{1}{c^2}E^iE^j-B^iB^j+\frac{\delta^i_j}{2}\left(|\boldsymbol B|^2+\frac{1}{c^2}|\boldsymbol E|^2\right)\right)}$$

The canonical stress tensor density is given in component form by $$𝔓^ρ_ν = \frac{∂𝔏}{∂\left(∂_ρ A_μ\right)} ∂_ν A_μ - δ^ρ_ν 𝔏,$$ where $$𝔏$$ is the Lagrangian density that generates the field theory. In terms of the response field (the $$𝐃$$ and $$𝐇$$ fields), which is the tensor density given by $$𝔊^{μν} = -\frac{∂𝔏}{∂F_{μν}}$$ where $$F_{μν} = ∂_μ A_ν - ∂_ν A_μ$$ is the field strength (the $$𝐄$$ and $$𝐁$$ fields), it is: $$𝔓^ρ_ν = -𝔊^{ρμ} ∂_ν A_μ - δ^ρ_ν 𝔏.$$

The stress tensor density $$𝔗$$ is an adjustment of $$𝔓$$, given by: $$𝔗^ρ_ν = 𝔓^ρ_ν + ∂_μ 𝔭^{ρμ}_ν,$$ where the tensor density $$𝔭$$ arises from the Belinfante correction and is given by: $$𝔭^{ρμ}_ν = 𝔊^{ρμ} A_ν.$$

The response field is subject to the Euler-Lagrange equations $$∂_ν 𝔊^{μν} = 𝔍^μ,$$ where the current density (also a tensor density, as the name indicates) is given by $$𝔍^μ = \frac{∂𝔏}{∂A_μ}.$$

Assuming we're talking about the free field, i.e. where the current density vanishes, then: $$∂_ν 𝔊^{μν} = 𝔍^μ = 0,$$ and the divergence of the Belinfante correction term yields: $$∂_μ 𝔭^{ρμ}_ν = ∂_μ𝔊^{ρμ} A_ν + 𝔊^{ρμ} ∂_μA_ν = 𝔊^{ρμ} ∂_μA_ν.$$ Therefore, \begin{align} 𝔗^ρ_ν &= -𝔊^{ρμ} ∂_ν A_μ - δ^ρ_ν 𝔏 + 𝔊^{ρμ} ∂_μA_ν\\ &= 𝔊^{ρμ} \left(∂_μA_ν - ∂_ν A_μ\right) - δ^ρ_ν 𝔏\\ &= 𝔊^{ρμ} F_{μν} - δ^ρ_ν 𝔏. \end{align} This only applies for free fields! If there is a non-zero current density, there is the additional term that arises from it.

For the Maxwell-Lorentz Lagrangian density for the free field: $$𝔏 = -\frac{ε_0c}{4} \sqrt{|g|} g^{μρ}g^{νσ}F_{μν}F_{ρσ} = -\frac{ε_0c}{4} \sqrt{|g|} F_{μν}F^{μν},$$ the response field and current density are $$𝔊^{μν} = ε_0c \sqrt{|g|} g^{μρ}g^{νσ}F_{ρσ} = ε_0c \sqrt{|g|} F^{μν},$$ using the metric to raise indices, and the stress tensor density is: $$𝔗^ρ_ν = ε_0c \sqrt{|g|} \left(F^{ρμ} F_{μν} + \frac{1}{4}δ^ρ_ν F^{νσ}F_{νσ}\right).$$

Generically, one can write the following component forms adapted to the coordinates $$x^0 = t, \quad \left(x^1, x^2, x^3\right) = (x,y,z)$$ as: $$φ = -A_0, \quad 𝐀 = \left(A_1, A_2, A_3\right), \quad 𝐁 = \left(F_{23}, F_{32}, F_{12}\right), \quad 𝐄 = \left(F_{10}, F_{20}, F_{30}\right),\\ ρ = 𝔍^0, \quad 𝐉 = \left(𝔍^1, 𝔍^2, 𝔍^3\right), \quad 𝐃 = \left(𝔊^{01},𝔊^{02}, 𝔊^{03}\right), \quad 𝐇 = \left(𝔊^{23}, 𝔊^{31}, 𝔊^{12}\right),$$ with $$ρ = -\frac{∂𝔏}{∂φ}, \quad 𝐉 = +\frac{∂𝔏}{∂𝐀}, \quad 𝐃 = +\frac{∂𝔏}{∂𝐄}, \quad 𝐇 = -\frac{∂𝔏}{∂𝐁},$$ and (in tensor-dyad form): $$𝔗^0_0 = 𝐃·𝐄 - 𝔏, \quad \left(𝔗^1_0, 𝔗^2_0, 𝔗^3_0\right) = 𝐄×𝐇, \quad \left(𝔗^0_1, 𝔗^0_2, 𝔗^0_3\right) = 𝐁×𝐃,\\ 𝔗^i_j = D^iE_j + B^iH_j - δ^i_j(𝐁·𝐇 + 𝔏) = \left(𝐃𝐄 + 𝐁𝐇 - 𝕀(𝐁·𝐇 + 𝔏)\right)^i_j\quad (i,j = 1, 2, 3).$$

The trace is: $$𝔗^μ_μ = 2(𝐃·𝐄 - 𝐁·𝐇) - 4𝔏 = 2\left(𝐄·\frac{∂𝔏}{∂𝐄} + 𝐁·\frac{∂𝔏}{∂𝐁} - 2𝔏\right).$$ This shows that a free field with a trace-free stress tensor can only arise from a free field Lagrangian that is homogeneous to the second degree in the field strengths. If it is also Lorentz invariant, then it is a function that is homogeneous to the first degree in the field invariants: $$ℑ_0 ≡ \frac{|𝐄|^2 - |𝐁|^2c^2}{2}, \quad ℑ_1 ≡ 𝐄·𝐁,$$ expressible as $$𝔏 = ε ℑ_0 + θ ℑ_1 = ε \frac{|𝐄|^2 - |𝐁|^2c^2}{2} + θ 𝐄·𝐁,$$ where the derivatives of the Lagrangian density $$ε ≡ \frac{∂𝔏}{∂ℑ_0}, \quad θ ≡ \frac{∂𝔏}{∂ℑ_1},$$ are constant: $$ε = ε_0$$ and $$θ = θ_0$$. The term $$θ 𝐄·𝐁$$ is a boundary term and can be eliminated, which reduces the Lagrangian density to the form: $$𝔏 = ε_0 \frac{|𝐄|^2 - |𝐁|^2c^2}{2},$$ which is exactly what the Maxwell-Lorentz Lagrangian density is. The response fields, for this Lagrangian density are: $$𝐃 = ε_0 𝐄, \quad 𝐁 = ε_0c^2 𝐇,$$ and the components of the stress tensor density reduce to: $$𝔗^0_0 = ε_0\frac{|𝐄|^2 + |𝐁|^2c^2}{2}, \quad \left(𝔗^1_0, 𝔗^2_0, 𝔗^3_0\right) = ε_0𝐄×𝐁c^2, \quad \left(𝔗^0_1, 𝔗^0_2, 𝔗^0_3\right) = ε_0𝐁×𝐄,\\ 𝔗^i_j = ε_0\left(𝐄𝐄 + 𝐁𝐁c^2 - 𝕀\frac{|𝐄|^2 + |𝐁|^2c^2}{2}\right)^i_j\quad (i,j = 1, 2, 3).$$

This generalizes to gauge fields, except that the potentials $$A^a_μ$$, field strengths $$F^c_{μν}$$, response fields $$𝔊_c^{μν}$$ and current density $$𝔍_a^μ$$ now all have an extra index, referring to a Lie algebra basis (for $$A$$ and $$F$$) or dual basis (for $$𝔊$$ and $$𝔍$$).

The expression of $$F$$ in terms of $$A$$ and the Euler-Lagrange equations that express $$𝔍$$ in terms of $$𝔊$$ may be non-linear - as they are if the underlying Lie algebra is non-Abelian - but the expressions for the stress tensor density are of the same form, e.g. $$𝔗^0_0 = 𝐃_a·𝐄^a - 𝔏$$.

The role played by $$ε_0c$$ is then played by the Lie algebra metric $$k_{ab}$$, whose specification is also part of the specification of the gauge field. The analogue of the Maxwell-Lorentz Lagrangian density, then, is precisely what the Yang-Mills Lagrangian density is, $$𝔏 = -\frac{k_{ab}}{4} \sqrt{|g|} g^{μρ}g^{νσ}F^a_{μν}F^b_{ρσ}.$$

For both Maxwell-Lorentz and Yang-Mills, the term $$\sqrt{|g|} g^{μρ}g^{νσ}$$ is independent of which scaling convention (and sign) you use for the metric $$g_{μν}$$ and its inverse $$g^{μν}$$ and all the expressions are convention-independent, as is the stress tensor density.

If, on the other hand, you take out the $$\sqrt{|g|}$$ to get the stress tensor $$T^ρ_ν = \frac{𝔗^ρ_ν}{\sqrt{|g|}}, \quad T_{μν} = g_{μρ}T^ρ_ν,$$ then it will depend on what convention you use for the metric, as well as for the coordinates.

So, there's really no consistent form for the stress tensor - it's convention-dependent and author-dependent. Only the stress tensor density transcends all that and only it is directly connected to the various quantities involved in the integrals that densities are integrands for.