Physlib.Relativity.Tensors.RealTensor.Vector.Tensorial

Tensorial nature of Lorentz vectors

We define the tensorial instance on `Lorentz.Vector`, and show prove properties related to the Lorentz group action and the basis.

Tensorial

Basis

The action of the Lorentz group

B. The continuous action of the Lorentz group

2 declarations

theorem

Evaluation of $F \in V \otimes V$ at $\mu, \nu$ equals its basis coefficient $F^{\mu\nu}$

#toField_eval_eval_eq_tensorProduct_repr

Let $d$ be the number of spatial dimensions and $V_d$ be the space of real Lorentz vectors. For any rank-2 tensor $F \in V_d \otimes_{\mathbb{R}} V_d$ and spacetime indices $\mu, \nu \in \text{Fin } 1 \oplus \text{Fin } d$ , let $\{F| [\mu] [\nu]\}^T$ denote the evaluation of the tensor $F$ at the indices $\mu$ and $\nu$ . The scalar value obtained by this evaluation (mapped to the field $\mathbb{R}$ via $\text{toField}$ ) is equal to the $(\mu, \nu)$ -th coefficient of $F$ with respect to the tensor product of the standard basis for $V_d$ .

theorem

Tensor Basis Coefficients of $F \in V_d \otimes V_d$ Equal Tensor Product Basis Coefficients

#tensor_basis_repr_toTensor_prod_apply

For a given spatial dimension $d$ , let $V_d$ be the space of Lorentz vectors. For any element $F \in V_d \otimes_{\mathbb{R}} V_d$ in the tensor product space and any multi-index $b = (b_0, b_1)$ representing two contravariant indices, the component of the tensor representation $\text{toTensor}(F)$ at index $b$ in the canonical tensor basis is equal to the component of $F$ at indices $(b_0, b_1)$ with respect to the tensor product of the standard bases of $V_d$ .