Tensors and its transformations
Here, we introduce the definition of tensors and the visualization behind it. Then, we discuss its transformation to another basis with a worked example.
Definition
Consider a rotation of the \(\{e_i\}\) basis (frame \(S\)) to the \(\{e'_i\}\) basis (frame \(S'\)). This is called a passive rotation.
The components of two arbitrary vectors \(\mathbf{a}\) and \(\mathbf{b}\) in the two frames are related by:
\[a'_i = \ell_{ij} a_j\] \[b'_i = \ell_{ij} b_j\]\(T'_{ijk\cdots op} = \ell_{ir} \ell_{js} \ell_{kt} \cdots \ell_{ov} \ell_{pw} T_{rst\cdots vw}\) is the transformation law for a tensor.
In the two dimensional case, this is equivalent to:
\[T'_{ij} = \ell_{ip} \ell_{jq} T_{pq}\]The astute reader will realize this is equivalent to:
\[T'=LTL\]Another way to view Tensor Transformations
\(T\) maps \(v\) to \(Tv\),
\(T'\) maps \(v'\) to \(T'v'\).
We can map from \(v\) to \(Tv\) in the way below:
- Rotate \(L\) to new basis
- Apply \(T'\)
- Rotate back
This gives us: \(Tv=L^{-1}T'Lv\).
From which we can obtain \(T'=LTL^{-1}\).
Sample Problem
The moments of inertia along the principal axes \(\underline{e}_1, \underline{e}_2, \underline{e}_3\) of a rigid body are 1,2 and 3 respectively. An observer has a coordinate basis \(\underline{e}_1^{\prime}=\frac{\sqrt{2}}{2}\left(\underline{e}_1+ \underline{e}_2\right), \underline{e}_2^{\prime}=\frac{\sqrt{2}}{2}\left(-\underline{e}_1+ \underline{e}_2\right)\) and \(\underline{e}_3^{\prime}=\underline{e}_3\). What are the values of the components of the inertia tensor measured by the observer?
Solution
Using \(R_{ij}=\underline{e_i}' \cdot \underline{e_j}\), we get:
\[R=\left(\begin{array}{ccc} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{array}\right)\]And also:
\[I=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{array}\right)\]One might be tempted to use coordinate transformations, i.e. \(x_j=R_{ij} x_i\). However, this is not the case here. The key is using the definition below:
\[\begin{aligned} I_{i j}^{\prime} & =l_{i p} l_{j q} I_{p q} \\ & =(L)_{i p} I_{p q}\left(L^T\right)_{q j} \\ & =\left(L I L^T\right)_{i j} \end{aligned}\]We know that rotation matrices satisfy \(L^T=L^{-1}\), hence:
\[I_{i j}^{\prime} = \left(L I L^{-1}\right)_{i j}\]Hence, multiplying will get us:
\[\begin{aligned} I' &= \left(\begin{array}{ccc} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{array}\right) \left(\begin{array}{ccc} \frac{\sqrt{2}}{2} & \frac{-\sqrt{2}}{2} & 0 \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{array}\right) \\ &=\left(\begin{array}{ccc} \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & 0 \\ 0 & 0 & 3 \end{array}\right) \end{aligned}\]Credits
The intuition part is inspired by this post. The definition of tensors is from Dr BJ Pendelton’s Notes. The motivation part is inspired by Dr BJ Pendelton’s Problem Sheet.
Exercises for the readers
- What rank is a vector? What rank is a scalar? (Difficulty: University Year 2)
- Show: \(\det L = +1\) for rotations and \(\det L = -1\) for reflections and inversions of basis (Difficulty: University Year 1-2)
- Consider \(\mathbf{c} = \mathbf{a} \times \mathbf{b}\). Show that \(\underline{\mathbf{c'}} = \det L \, \underline{\mathbf{c}}\) by considering rotations, reflections and inversion. This is the definition of a psuedovector. (Difficulty: University Year 2)
- State an example of a psuedovector in physics (Difficulty: University Year 2)
- (*) Irrelevant but Interesting: Define the Levi-Civita symbol and show that it is a pseudotensor of rank 3 (Difficulty: University Year 2-3)
- (**) State an example of a pseudoscalar in physics (Difficulty: University Year 3-4). (Hint: Helicity)
Last Updated - 4/6/2024
