Conservation of angular momentum from Noether's theorem and Lorentz invariance

Noether's theorem has been stated in the form that if the assignment $$\phi \mapsto \phi + \delta \phi$$, $$x^{\mu} \mapsto x^{\mu} + \delta x^{\mu}$$ leaves the action invariant, then the quantity $$f^{\mu} := \dfrac{\partial \mathcal L}{\partial ( \partial_{\mu} \phi ) } \delta \phi - \dfrac{\partial \mathcal L}{\partial ( \partial_{\mu} \phi )} \partial_{\nu} \phi \delta x^{\nu} + \mathcal L \delta x^{\mu}$$ obeys the continuity equation $$\partial _{\mu} f ^{\mu} = 0$$.

This also leads to a corresponding conserved quantity of the form $$P := \displaystyle\int \text{d} ^3 x f^{0}$$

In the case of a Lorentz transformation, the mapping is given by $$\delta \phi = 0, \delta x^{\mu} = \omega^{\mu \nu} x_{\nu}$$ where $$\omega^{\mu \nu}$$ is an antisymmetric tensor.

Let the expression for the stress energy tensor be given as $$T^{\mu \nu} = \dfrac{ \partial \mathcal L}{\partial ( \partial_{\mu} \phi )} \partial ^{\nu} \phi - \mathcal L g^{\mu \nu}$$

The sources I've seen online simply state that the conserved quantities are then of the form $$L^{\mu \nu} = \displaystyle \int \text{d} ^3 x \; T^{0 \mu} x^{\nu} - T^{0 \nu} x^{\mu}$$ without giving an explanation of how the stress energy tensor arises in this expression.

What I've done so far is to substitute the given Lorentz transformation into Noether's theorem and switch the indices slightly leading to a function $$f^{\mu} = \left( - \dfrac{\partial \mathcal L}{\partial ( \partial_{\mu} \phi )} \partial_{\nu} \phi \omega^{\nu \rho} + \mathcal L \omega^{\mu \rho} \right) x^{\rho}$$

In particular, the zeroth component of this expression is given by $$f^{0} = \left( - \dfrac{\partial \mathcal L}{\partial \dot{ \phi} } \partial_{\nu} \phi \omega^{\nu \rho} + \mathcal L \omega^{0 \rho} \right) x^{\rho}$$

Since $$\omega$$ is antisymmetric, its diagonal components must vanish which allows us to get rid of the latter $$\omega^{0 \rho}$$ term for $$\rho = 0$$ and similarly for the diagonal terms of the $$\omega^{\nu \rho}$$ sum but I can't see how to manipulate the above summation in such a way that the energy-momentum tensor comes into the picture (the first term on the left looks like part of the expression for $$T^{0 \nu}$$ but what happens with the metric term...? )