Derivation of Maxwell stress tensor from EM Lagrangian From Noether's theorem applied to fields we can get the general expression for the stress-energy-momentum tensor for some fields:
$$T^{\mu}_{\;\nu} = \sum_{i} \left(\frac{\partial \mathcal{L}}{\partial \partial_{\mu}\phi_{i}}\partial_{\nu}\phi_{i}\right)-\delta^{\mu}_{\;\nu}\mathcal{L}$$
The EM Lagrangian, in the Weyl gauge, is:
$$\mathcal{L} = \frac{1}{2}\epsilon_{0}\left(\frac{\partial \vec{A}}{\partial t}\right)^{2}-\frac{1}{2\mu_{0}}\left(\vec{\nabla}\times \vec{A}\right)^{2}$$
Applying the above, all I manage to get for the pressure along x, which I believe corresponds to the first diagonal element of the Maxwell stress tensor, is:
$$p_{x} = \sigma_{xx} = -T^{xx} = \frac{-1}{\mu_{0}}\left(\left(\partial_{x}A_{z}\right)^{2}-\partial_{x}A_{z}\partial_{z}A_{x}-\left(\partial_{x}A_{y}\right)^{2}+\partial_{x}A_{y}\partial_{y}A_{x}\right)+\mathcal{L}$$
But I can't see how this can be equal to what is given in Wikipedia.Why is this?
 A: Hint: The canonical stress-energy tensor from Noether's theorem is not necessarily symmetric, and often needs to be improved with appropriate improvements terms. This is e.g. the case for EM. See also e.g. this Phys.SE post and links therein.
References:


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*Landau and Lifshitz, Vol.2, The Classical Theory of Fields, $\S$33.

A: First, let's denote the canonical stress tensor density (and it is a tensor density, by the way, not a tensor) by $$, rather than by $$ or $T$. This is a rank (1,1) tensor density and the diagonal component is
$${_1}^1 = \frac{∂ℒ}{∂/∂x}·\frac{∂}{∂x} - ℒ.$$
The field-potential relations are $ = -∇φ - ∂⁄∂t = -∂⁄∂t$, $ = ∇×$, for your gauge $φ = 0$ and the Lagrangian density is
$$ℒ = \frac 1 2 \left(ε₀ E² - \frac{B²}{μ₀}\right).$$
It depends on the gradients of the potentials only through its dependence on the field strengths $$ and $$, with respective derivatives:
$$ ≡ \frac{∂ℒ}{∂} = ε₀,$$
$$ ≡ -\frac{∂ℒ}{∂} = \frac{}{μ₀}.$$
The dependence on $∂/∂x$ occurs only through the dependence on $$, with the respective derivatives:
$$\frac{∂ℒ}{∂/∂x} = \left(\frac{∂ℒ}{∂A₁/∂x}, \frac{∂ℒ}{∂A₂/∂x}, \frac{∂ℒ}{∂A₃/∂x}\right) = \left(0, -\frac{∂ℒ}{∂B³}, \frac{∂ℒ}{∂B²}\right) = (0, H₃, -H₂).$$
Consequently,
$${_1}^1 = (0, H₃, -H₂)·\left(\frac{∂A₁}{∂x}, \frac{∂A₂}{∂x}, \frac{∂A₃}{∂x}\right) - ℒ = -\left(×\frac{∂}{∂x}\right)^1 - ℒ$$
In general
$${_j}^i = -\left(×{\frac{∂}{∂x^j}}\right)^i - δ_j^i ℒ$$
In contrast, the stress tensor density for the field (in the absence of sources) is given by
$${_1}^1 = D¹ E₁ + B¹ H₁ - (· + ℒ) = D¹ E₁ - B² H₂ - B³ H₃ - ℒ$$
More generally,
$${_j}^i = D^i E_j + B^i H_j - δ^i_j (· + ℒ)$$
This generalizes further, by including the 0 indexes to the following (and I'll be using the summation convention here and below, in which repeated indexes in a monomial term are understood to be summed over)
$${_μ}^ρ = ^{ρν} F_{νμ} - δ^ρ_μ ℒ,$$
where the potentials and fields are given by:
$$F_{μν} = ∂_μ A_ν - ∂_ν A_μ = -F_{νμ} ⇒  = (F₂₃,F₃₁,F₁₂),  = (F₁₀,F₂₀,F₃₀),$$
and the response fields are tensor densities given by
$$^{μν} = -\frac{∂ℒ}{∂F_{μν}} = -^{νμ} ⇒  = (⁰¹,⁰²,⁰³),  = (²³,³¹,¹²).$$
The potentials, themselves are given by
$$φ = -A₀,  = (A₁,A₂,A₃).$$
Taking the difference, we have
$${_1}^1 - {_1}^1 = D¹ E₁ - B² H₂ - B³ H₃ + \left(×\frac{∂}{∂x}\right)^1$$
$$= D¹\left(-\frac{∂A₁}{∂t}\right) - \left(B² + \frac{∂A₃}{∂x}\right) H₂ - \left(B³ - \frac{∂A₂}{∂x}\right) H₃$$
This works out to:
$${_1}^1 - {_1}^1 = -D¹ \frac{∂A₁}{∂t} - H₂ \frac{∂A₁}{∂z} + H₃ \frac{∂A₁}{∂y}$$
$$= \frac{∂(-D¹A₁)}{∂t} + \frac{∂(H₃A₁)}{∂y} + \frac{∂(-H₂A₁)}{∂z} + A₁ \left(\frac{∂}{∂t} - ∇× \right)^1.$$
"On shell" (that is "upon application of the source-free field equation: $∇× - ∂⁄∂t =  = $), this reduces to a total divergence:
$${_1}^1 - {_1}^1 = \frac{∂(-D¹A₁)}{∂t} + \frac{∂(H₃A₁)}{∂y} + \frac{∂(-H₂A₁)}{∂z} = \frac{∂{_1}^{10}}{∂t} + \frac{∂{_1}^{11}}{∂x} + \frac{∂{_1}^{12}}{∂y} + \frac{∂{_1}^{13}}{∂z}$$
where
$$({_1}^{10}, {_1}^{11}, {_1}^{12}, {_1}^{13}) = A₁ (-D¹, 0, H₃, -H₂) = A₁ (¹⁰,¹¹,¹²,¹³)$$
This generalizes to:
$${_μ}^{ρσ} = A_μ ^{ρσ}.$$
This extra contribution can be accounted for by the Belinfante relocation of the canonical stress tensor. In general, given a stress tensor and spin tensor
$${_μ}^ρ, {_{μν}}^ρ = -{_{νμ}}^ρ$$
the momentum/energy and angular momentum/mass moment 3-current densities corresponding to infinitesimal translations $x^μ → x^μ + Δx^μ$ and rotations $Δω^{μν}$ are
$$P(Δx) = {_μ}^ρ Δx^μ ∂_ρ ˩ d⁴x$$
$$J(Δω) = ½ {_{μν}}^ρ Δω^{μν} ∂_ρ ˩ d⁴x$$
where the total angular momentum current density is given by
$${_{μν}}^ρ = x_μ {_ν}^ρ - x_ν {_μ}^ρ + {_{μν}}^ρ.$$
Each of these satisfies continuity equations:
$$∂_ρ {_μ}^ρ = 0$$
$$∂_ρ {_{μν}}^ρ = 0$$
which, in the latter case, produces the symmetrization condition
$$0 = ∂_ρ {_{μν}}^ρ = _{νμ} - _{μν} + ∂_ρ {_{μν}}^ρ ⇒ _{μν} - _{νμ} = ∂_ρ {_{μν}}^ρ$$
A relocation of $(, )$ is an adjustment by 2-current densities $(,)$
$${_μ}^ρ → {_μ}^ρ + ∂_σ {_μ}^{ρσ}$$
$${_{μν}}^ρ → {_{μν}}^ρ + ∂_σ {_{μν}}^{ρσ}$$
that preserves the continuity equations and the relation between $(,)$ and $$ (which, thus, also undergoes adjustment).
The anti-symmetry of $(,)$ in $(ρ,σ)$ ensures that the continuity equations are preserved, and the anti-symmetry of $$ in $(μ,ν)$ ensures that $$ likewise retain its anti-symmetry in $(μ,ν)$.
The Belinfante relocation
$${_μ}^ρ → {_μ}^ρ = {_μ}^ρ + ∂_σ {_μ}^{ρσ}$$
$${_{μν}}^ρ → {ℳ_{μν}}^ρ = {_{μν}}^ρ + ∂_σ {_{μν}}^{ρσ}$$
is the one which reduces $$ to 0, so that one has the following
$${ℳ_{μν}}^ρ = x_μ {_ν}^ρ - x_ν {_μ}^ρ.$$
The actual expression for $$ and $$ in terms of $$ can be found by substituting for $(ℳ,)$ in their relation and applying the anti-symmetry conditions on $(μ,ν)$ and $(ρ,σ)$.
The expression for, $$, in turn, comes out of the Lorentz transform law for the fields $A$ and $F$. Finally, the fact that this $$ will actually yield the correction $$ that we previously derived is a matter of a routine verification. In fact, we can use this point and work backwards, inverting the relation between $$ and $$, solving for $$ (since we already have $$) and determine what $$ ought to be and - from this - what the transforms on $A$ and $F$ ought to be.
Substitute the adjustments into the target relation for the Belinfante relocation:
$${_{μν}}^ρ + ∂_σ {_{μν}}^{ρσ} = x_μ ({_ν}^ρ + ∂_σ {_ν}^{ρσ}) - x_ν ({_μ}^ρ + ∂_σ {_μ}^{ρσ}),$$
integrate by parts and substitute for , to get
$$x_μ {_ν}^ρ - x_ν {_μ}^ρ + {_{μν}}^ρ + ∂_σ {_{μν}}^{ρσ} = x_μ {_ν}^ρ - x_ν {_μ}^ρ + ∂_σ (x_μ {_ν}^{ρσ} - x_ν {_μ}^{ρσ}) + {{_μ}^ρ}_ν - {{_ν}^ρ}_μ.$$
or
$${_{μν}}^ρ = ∂_σ (x_μ {_ν}^{ρσ} - x_ν {_μ}^{ρσ} - {_{μν}}^{ρσ}) + {{_μ}^ρ}_ν - {{_ν}^ρ}_μ,$$
which can be solved with
$${_{μν}}^ρ = {{_μ}^ρ}_ν - {{_ν}^ρ}_μ,$$
$${_{μν}}^{ρσ} = x_μ {_ν}^{ρσ} - x_ν {_μ}^{ρσ}.$$
The Belinfante correction $$, itself, would be obtained by inverting the $(,)$ relation, which (after lowering the index $ρ$) would be:
$$_{μνρ} = ½ (_{μρν} + _{νμρ} + _{νρμ}).$$
Since we already found that
$${_μ}^{ρσ} = A_μ ^{ρσ}$$
then it follows that we must have
$${_{μν}}^ρ = A_μ {^ρ}_ν - A_ν {^ρ}_μ = ^{ρσ} (A_μ g_{σν} - A_ν g_{σμ}).$$
For a field theory given by an action principle with a Lagrangian density $ℒ(q,v)$ that is a function of field variables $q = (q^A)$ and its gradients $v = ({v^A}_μ) = (∂_μ q^A)$; if we have a symmetry, given infinitesimally by ${λ^μ}_ν$, that is metric-preserving:
$$0 = Δg_{μν} = -{λ^ρ}_μ g_{ρν} - {λ^ρ}_ν g_{μρ} = -λ_{νμ} - λ_{μν}$$
and if the field variables transform under it as
$${Δq}^A = ½ λ_{μν} (Σ^{μν} q)^A = ½ λ_{μν} {{Σ^{μν}}^A}_B q^B$$
then the "spin" tensor for the $λ$-symmetry is given by
$$^{μνρ} = \frac{∂ℒ}{∂{v^A}_ρ} (Σ^{μν} q)^A.$$
When this is applied to the electromagnetic field $q = (A_ν)$, $v = (∂_μ A_ν)$, this yields the spin tensor:
$$^{μνρ} = \frac{∂ℒ}{∂_ρ A_σ} (Σ^{μν} A)_σ = \frac{∂ℒ}{F_{ρσ}} (Σ^{μν} A)_σ = -^{ρσ} (Σ^{μν} A)_σ.$$
Thus, we find that the transform ought to be given by
$$(Σ_{μν} A)_σ = A_ν g_{σμ} - A_μ g_{σν}$$
and
$$ΔA_σ = ½ λ^{μν} (Σ_{μν} q)_σ = ½ λ^{μν} (A_ν g_{σμ} - A_μ g_{σν}) = -{λ^μ}_σ A_μ.$$
If the metric is Minkowski, then it has the form (up to a scale factor)
$$g₀₀ = β, g₁₁ = g₂₂ = g₃₃ = -α,$$
and $g_{μν} = 0$ otherwise, where $αβ > 0$ and $c = \sqrt{β/α}$ and respects Lorentz symmetry,
$$Δg_{μν} = -{λ^ρ}_μ g_{ρν} - {λ^ρ}_ν g_{μρ}$$
given in infinitesimal form by:
$$Δx^ρ = {λ^ρ}_μ x^μ ⇒ Δt = -β·, Δ =  ×  - α  t ⇒ Δx^μ = {λ^μ}_ρ x^ρ$$
where $ = (x,y,z)$, $$ is an infinitesimal rotation, $$ an infinitesimal boost, and
$$(λ³₂,λ¹₃,λ²₁) =  = -(λ²₃,λ³₁,λ¹₂),$$
$$(λ⁰₁,λ⁰₂,λ⁰₃) = -α,$$
$$(λ¹₀,λ²₀,λ³₀) = -β,$$
$$(λ⁰₀,λ¹₁,λ²₂,λ³₃) = (0,0,0,0).$$
This also induces a transform on the differential operators
$$Δ∂_μ = -{λ^ρ}_μ ∂_ρ ⇒ Δ∇ = ×∇ + α\frac{∂}{∂t}, Δ\left(\frac{∂}{∂t}\right) = β·∇$$
which is obtained from the condition that the 1-form operator
$$d·∇ + dt \frac{∂}{∂t}$$
be an invariant.
As a consequence of the transforms on the operators, we may also write
$$ΔF_{μν} = Δ(∂_μA_ν - ∂_νA_μ)$$
and work this out in detail to show that
$$ΔF_{μν} = -{λ^ρ}_μ F_{ρν} - {λ^ρ}_ν F_{μρ}.$$
In component form, this transform is given by:
$$Δ = × - αφ,$$
$$Δφ = -β·,$$
$$Δ = × - α×,$$
$$Δ = × + β×.$$
The parameters $(α,β)$ can be normalized in a variety of ways, depending on how the coordinates and field components are scaled. The normalization that best accords with the correspondence limit of non-relativistic theory is $(α,β) = (1/c²,1)$ for relativity and $(α,β) = (0,1)$ as the non-relativistic limit; in which case the metric becomes a metric for proper time and can be rescaled by $-c²$ to make it a metric for distance.
