Why symmetrization of energy-momentum tensor doesn't add additional term to Lagrangian density? [duplicate]

Question

I am self-studying the book “James H. Luscombe, Core Principles of Special and General Relativity”. In “CHAPTER 9 : Energy-momentum of fields” of the book, it starts by introducing Noether’s theorem and then shows that invariance under spacetime translations implies conservation of field energy and momentum: $$T^{μν}≡g^{μν} L-(∂^μ ϕ_α)\frac{∂L}{∂(∂_ν ϕ_α)}$$ $$∂_ν T^{μν}=0$$ in which $T^{μν}$ is canonical energy-momentum tensor, and then shows that invariance under Lorentz transformations implies conservation of angular momentum current density tensor: $$J^{ρτμ}≡x^ρ T^{τμ}-x^t T^{ρμ}+ϕ^τ \frac{∂L}{∂(∂_μ ϕ_ρ )} -ϕ^ρ \frac{∂L}{(∂(∂_μ ϕ_τ))}$$ $$∂_μ J^{ρτμ}=0.$$ Then the book groups terms of $J^{ρτμ}$ in two part

1. orbital momentum density $L^{ρτμ}=x^ρ T^{τμ}-x^t T^{ρμ}$ and

2. spin density $S^{ρτμ}=\frac{∂L}{∂(∂_μ ϕ_ρ )} -ϕ^ρ\frac{∂L}{(∂(∂_μ ϕ_τ))}$.

Then the book define an equivalent symmetric energy-momentum tensor $$T^{μν}→θ^{μν}≡T^{μν}+\frac{1}{2} ∂_λ (S^{μλν}+S^{νλμ}-S^{μνλ})$$ to write $J^{ρτμ}$ as $$J^{ρτμ}→J_θ^{ρτμ}≡x^ρ θ^{τμ}-x^t θ^{ρμ}=J^{ρτμ}+∂_λ (x^ρ ψ^{τλμ}-x^τ ψ^{ρλμ}).$$

My question is "Does symmetrization of energy-momentum tensor add additional term to Lagrangian density? Why?"

Simple answer is that it doesn’t change Lagrangian density because conservation laws remain unchanged. But I think this is not enough!