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
$${_1}^1 = \frac{∂}{∂(∂/∂x)}·\frac{∂}{∂x} - .$$
The field-potential relations are $ = -∇φ - ∂⁄∂t = -∂⁄∂t$, $ = ∇×$, for your gauge $φ = 0$ and the Lagrangian density is
$$ = \frac 1 2 \left(ε_0 E^2 - \frac{B^2}{μ_0}\right).$$
It depends on the gradients of the potentials only through its dependence on the field strengths $$ and $$, with respective derivatives:
$$ ≡ \frac{∂}{∂} = ε_0, \hspace 1em ≡ -\frac{∂}{∂} = \frac{}{μ_0}.$$
The dependence on $∂/∂x$ occurs only through the dependence on $$, with the respective derivatives:
$$\frac{∂}{∂(∂/∂x)} = \left(\frac{∂}{∂\left(∂A_1/∂x\right)}, \frac{∂}{∂\left(∂A_2/∂x\right)}, \frac{∂}{∂\left(∂A_3/∂x\right)}\right) = \left(0, -\frac{∂}{∂B^3}, \frac{∂}{∂B^2}\right) = (0, H_3, -H_2).$$
Consequently,
$${_1}^1 = (0, H_3, -H_2)·\left(\frac{∂A_1}{∂x}, \frac{∂A_2}{∂x}, \frac{∂A_3}{∂x}\right) - = -\left(×\frac{∂}{∂x}\right)^1 - $$
In general
$${_j}^i = -\left(×\frac{∂}{∂x^j}\right)^i - δ_j^i $$
In contrast, the stress tensor density for the field (in the absence of sources) is given by
$${_1}^1 = D^1 E_1 + B^1 H_1 - (· + ) = D^1 E_1 - B^2 H_2 - B^3 H_3 - $$
More generally,
$${_j}^i = D^i E_j + B^i H_j - δ^i_j (· + )$$
This generalizes further, by including the 0 indexes to the following (and I'll be using the summation convention here and below, in which repeated indexes in a monomial term are understood to be summed over)
$${_μ}^ρ = ^{ρν} F_{νμ} - δ^ρ_μ ,$$
where the potentials and fields are given by:
$$F_{μν} = ∂_μ A_ν - ∂_ν A_μ = -F_{νμ} ⇒ = (F_{23},F_{31},F_{12}), = (F_{10},F_{20},F_{30}),$$
and the response fields are tensor densities given by
$$^{μν} = -\frac{∂}{∂F_{μν}} = -^{νμ} ⇒ = (^{01},^{02},^{03}), = (^{23},^{31},^{12}).$$
The potentials, themselves are given by
$$φ = -A_0, \hspace 1em = (A_1,A_2,A_3).$$
Taking the difference, we have
$$\begin{align}
{_1}^1 - {_1}^1 & = D^1 E_1 - B^2 H_2 - B^3 H_3 + \left(×\frac{∂}{∂x}\right)^1 \\
& = D^1\left(-\frac{∂A_1}{∂t}\right) - \left(B^2 + \frac{∂A_3}{∂x}\right) H_2 - \left(B^3 - \frac{∂A_2}{∂x}\right) H_3
\end{align}$$
This works out to:
$$\begin{align}
{_1}^1 - {_1}^1 & = -D^1 \frac{∂A_1}{∂t} - H_2 \frac{∂A_1}{∂z} + H_3 \frac{∂A_1}{∂y} \\
& = \frac{∂(-D^1A_1)}{∂t} + \frac{∂(H_3A_1)}{∂y} + \frac{∂(-H_2A_1)}{∂z} + A_1 \left(\frac{∂}{∂t} - ∇× \right)^1.
\end{align}$$
"On shell" (that is "upon application of the source-free field equation: $∇× - ∂⁄∂t = = $), this reduces to a total divergence:
$${_1}^1 - {_1}^1 = \frac{∂(-D^1A_1)}{∂t} + \frac{∂(H_3A_1)}{∂y} + \frac{∂(-H_2A_1)}{∂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}) = A_1 (-D^1, 0, H_3, -H_2) = A_1 (^{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, \hspace 1em {_{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 ˩ d^4x, \hspace 1em J(Δω) = ½ {_{μν}}^ρ Δω^{μν} ∂_ρ ˩ d^4x$$
where the total angular momentum current density is given by
$${_{μν}}^ρ = x_μ {_ν}^ρ - x_ν {_μ}^ρ + {_{μν}}^ρ.$$
Each of these satisfies continuity equations:
$$∂_ρ {_μ}^ρ = 0, \hspace 1em ∂_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}, \hspace 1em {_{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
$${_μ}^ρ → {_μ}^ρ = {_μ}^ρ + ∂_σ {_μ}^{ρσ}, \hspace 1em {_{μν}}^ρ → {_{μν}}^ρ = {_{μν}}^ρ + ∂_σ {_{μν}}^{ρσ}$$
is the one which reduces $$ to 0, so that one has the following
$${_{μν}}^ρ = x_μ {_ν}^ρ - x_ν {_μ}^ρ.$$
The actual expression for $$ and $$ in terms of $$ can be found by substituting for $(,)$ 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
$${_{μν}}^ρ = {{_μ}^ρ}_ν - {{_ν}^ρ}_μ, \hspace 1em {_{μν}}^{ρσ} = x_μ {_ν}^{ρσ} - x_ν {_μ}^{ρσ}.$$
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}).$$
For a field theory given by an action principle with a Lagrangian density $(q,v)$ that is a function of field variables $q = (q^A)$ and its gradients $v = ({v^A}_μ) = (∂_μ q^A)$; if we have a symmetry, given infinitesimally by ${λ^μ}_ν$, that is metric-preserving:
$$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{∂}{∂{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{∂}{∂(∂_ρ A_σ)} (Σ^{μν} A)_σ = \frac{∂}{∂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.$$
If the metric is Minkowski, then it has the form (up to a scale factor)
$$g_{00} = β, g_{11} = g_{22} = g_{33} = -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, $$ an infinitesimal boost, and
$$\begin{align}
({λ^3}_2,{λ^1}_3,{λ^2}_1) &= = -({λ^2}_3,{λ^3}_1,{λ^1}_2), \\
({λ^0}_1,{λ^0}_2,{λ^0}_3) &= -α, \\
({λ^1}_0,{λ^2}_0,{λ^3}_0) &= -β, \\
({λ^0}_0,{λ^1}_1,{λ^2}_2,{λ^3}_3) &= (0,0,0,0).
\end{align}$$
This also induces a transform on the differential operators
$$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(∂_μ A_ν - ∂_ν A_μ)$$
and work this out in detail to show that
$$ΔF_{μν} = -{λ^ρ}_μ F_{ρν} - {λ^ρ}_ν F_{μρ}.$$
In component form, this transform is given by:
$$\begin{align}
Δ &= × - αφ, & Δφ &= -β·, \\
Δ &= × - α×, & Δ &= × + β×.
\end{align}$$
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 $(α,β) = (1/c^2,1)$ 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 $-c^2$ to make it a metric for distance.
