Tensor Transformation Rule

Relevant parts to questions...

• Using $T_{ij}^{\prime }=L_{im}{L}_{jn}{T}_{mn}$ (rank 2), $T_{ijk}^{\prime }=L_{im}{L}_{jn}{L}_{kp}{T}_{mnp}$ (rank 3), etc.: one $L$ per free index.
• Verifying tensor character by applying the rule and cancelling $L$s via orthogonality $L_{ij}{L}_{kj}={\delta }_{ik}$.
• Recognising dummy vs free indices on both sides: primed indices are free, original indices are dummy.

The Tensor Transformation Rule generalises the vector rule $v_i^{\prime }=L_{ij}{v}_j$ to tensors of any rank. A tensor of rank $r$ picks up exactly $r$ factors of the Coordinate Transformation Matrix $L$, one per free index:

$$\boxed{T_{i_1{i}_2\dots i_r}^{\prime }=L_{i_1{m}_1}{L}_{i_2{m}_2}\dots L_{i_r{m}_r}{T}_{m_1{m}_2\dots m_r}}$$

For the most common cases:

• Rank 0 (scalar)::$s^{\prime }=s$ - no $L$ factors, invariant.
• Rank 1 (vector)::$v_i^{\prime }=L_{ij}{v}_j$ - one factor of $L$.
• Rank 2::$T_{ij}^{\prime }=L_{im}{L}_{jn}{T}_{mn}$ - two factors of $L$, one per index. In matrix form: $T^{\prime }=LT{L}^T$.
• Rank 3::$T_{ijk}^{\prime }=L_{im}{L}_{jn}{L}_{kp}{T}_{mnp}$ - three factors.

This rule is the definition of a tensor: anything that transforms this way is a tensor, and anything that doesn’t, isn’t.

Properties

• Free vs dummy indices::the primed indices on the left are free, while the unprimed indices on the right are dummy (summed). Both sides carry the same free indices.
• Dimensional count::in 3D, a rank-$r$ tensor has $3^r$ components.
• Symmetry is frame-independent::if $T_{ij}=T_{ji}$ in one frame, then $T_{ij}^{\prime }=T_{ji}^{\prime }$ in every frame. The rule preserves Symmetry and Antisymmetry.
• Cartesian-specific::this rule assumes $L$ is a constant rotation matrix. In generalised coordinates it is replaced by the more general rule with separate covariant/contravariant factors (see Mixed Components).

Applications

1. Verifying a quantity is a tensor by direct application::e.g. the Kronecker Delta ${\delta }_{ij}$ is rank-2 since $L_{ik}{L}_{jm}{\delta }_{km}=L_{ik}{L}_{jk}={\delta }_{ij}={\delta }_{ij}^{\prime }$.
2. Proving the Gradient of a vector field is a rank-2 tensor, via product + chain rule::$\frac{\partial u_i^{\prime }}{\partial x_j^{\prime }}=L_{ik}{L}_{j\ell }\frac{\partial u_k}{\partial x_{\ell }}$.
3. Matrix form for rank-2::$T^{\prime }=LT{L}^T$ is the quickest way to compute $T_{ij}^{\prime }$ numerically when $L$ and $T$ are given as matrices.
4. Shortcut via the Quotient Rule::if contracting $T$ with an arbitrary tensor produces a tensor, then $T$ itself is a tensor - no need to verify the transformation rule directly.

Kronecker delta is a rank-2 tensor

We need ${\delta }_{ij}^{\prime }=L_{ik}{L}_{jm}{\delta }_{km}$. Applying the rule:

$L_{ik}{L}_{jm}{\delta }_{km}=L_{ik}{L}_{jk}={\delta }_{ij}$, using orthogonality.

Since ${\delta }_{ij}$ is defined identically in every frame, ${\delta }_{ij}^{\prime }={\delta }_{ij}=\boxed{L_{ik}{L}_{jm}{\delta }_{km}}$. ✓

One $L$ per free index; dummies sum.