Introduction to Tensors

The word "tensor" can sound intimidating, yet you have been using tensors for years without the name: every scalar and every vector is one. Tensors are the language that engineers use for stress in a bridge, that physicists use for Einstein's curved space-time, and that machine-learning libraries use to store data. This introduction builds the idea step by step and works through practical examples.

Why We Need Tensors

Some physical quantities need just one number (a scalar), such as temperature or mass. Others need a magnitude and a direction (a vector), such as force or velocity. But some quantities link one direction to another.

Think of the forces inside a loaded steel beam. If you slice the beam with an imaginary cut, the force per unit area acting on that cut depends on which way the cut faces. A vertical cut and a slanted cut experience different forces. The quantity that turns "a direction of cut" into "a force per area on that cut" is a rank-2 tensor, the stress tensor.

Rank: Counting Directions

The rank (or order) of a tensor is the number of indices needed to label its components. In three dimensions a rank-\( n \) tensor has \( 3^n \) components.

A rank-2 tensor can be displayed as a \( 3 \times 3 \) matrix, but a tensor is more than an array of numbers, as we will see.

Index Notation and the Summation Convention

Tensor work becomes much shorter with index notation. Indices \( i, j, k \) run over 1, 2, 3 (meaning \( x, y, z \)). Albert Einstein introduced a convenient rule in 1916: a repeated index means "sum over it". So

$$ a_i b_i = a_1 b_1 + a_2 b_2 + a_3 b_3 $$

is simply the dot product, and

$$ t_i = \sigma_{ij} n_j $$

means \( t_1 = \sigma_{11}n_1 + \sigma_{12}n_2 + \sigma_{13}n_3 \), and similarly for \( t_2 \) and \( t_3 \): matrix-times-vector in one line.

Two special tensors appear everywhere:

• The Kronecker delta \( \delta_{ij} \), equal to 1 if \( i = j \) and 0 otherwise. It is the identity matrix, and \( \delta_{ii} = 1 + 1 + 1 = 3 \).
• The Levi-Civita symbol \( \varepsilon_{ijk} \), equal to \( +1 \) for even orderings of (1, 2, 3), \( -1 \) for odd ones and 0 if any index repeats. It gives the cross product: \( (\mathbf{a} \times \mathbf{b})_i = \varepsilon_{ijk} a_j b_k \).

What Makes a Tensor a Tensor

The key property is how a tensor behaves when you rotate your coordinate axes. The physical object does not change, but its components do, in a precise way. If \( R \) is the rotation matrix taking old axes to new ones:

$$ v'_i = R_{ij} v_j \qquad \text{(rank 1)} $$

$$ \sigma'_{ij} = R_{ik} R_{jl}\,\sigma_{kl} \qquad \text{(rank 2)} $$

In matrix form, \( \sigma' = R\,\sigma\,R^{\mathsf{T}} \). One rotation matrix appears per index. Any quantity that transforms this way is a tensor, and this guarantees that physical laws written with tensors are true in every coordinate system.

Worked Example 1: Traction on an Inclined Plane

Problem: At a point in a loaded steel bracket, the stress tensor (in MPa) is

$$ \sigma = \begin{pmatrix} 50 & 20 & 0 \\ 20 & 30 & 0 \\ 0 & 0 & 10 \end{pmatrix} \text{ MPa} $$

Find the traction vector (force per unit area) on a plane whose unit normal is \( \mathbf{n} = \left(\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}, 0\right) \), and the normal and shear stress on that plane.

Step 1 (Cauchy's formula). \( t_i = \sigma_{ij} n_j \):

$$ t_1 = \frac{50 + 20}{\sqrt{2}} = \frac{70}{\sqrt{2}} \approx 49.50 \text{ MPa} $$

$$ t_2 = \frac{20 + 30}{\sqrt{2}} = \frac{50}{\sqrt{2}} \approx 35.36 \text{ MPa} $$

$$ t_3 = \frac{0 + 0}{\sqrt{2}} = 0 \text{ MPa} $$

Step 2 (normal stress). Project the traction onto the normal:

$$ \sigma_n = \mathbf{t} \cdot \mathbf{n} = \frac{70}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{50}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{70 + 50}{2} = 60 \text{ MPa} $$

Step 3 (shear stress). The magnitude of the traction is

$$ |\mathbf{t}| = \sqrt{49.50^2 + 35.36^2} = \sqrt{2450 + 1250} = \sqrt{3700} \approx 60.83 \text{ MPa} $$

so the shear part is

$$ \tau = \sqrt{|\mathbf{t}|^2 - \sigma_n^2} = \sqrt{3700 - 3600} = \sqrt{100} = 10 \text{ MPa} $$

Step 4 (sanity check). The principal (extreme) normal stresses in the \( xy \)-plane are \( 40 \pm \sqrt{10^2 + 20^2} \approx 62.4 \) MPa and \( 17.6 \) MPa, and our 60 MPa lies between them, as any normal stress must. Also \( \sigma_n^2 + \tau^2 = |\mathbf{t}|^2 \) holds by construction, and the shear stress of 10 MPa is far below the shear yield strength of mild structural steel (roughly 140 MPa), so the bracket is safe at this point.

Worked Example 2: Rotating the Axes

Problem: Using the same stress state, rotate the axes by \( \theta = 45^\circ \) about the \( z \)-axis and find the new component \( \sigma'_{xx} \). Confirm it matches Example 1.

Step 1. For a rotation about \( z \), the rank-2 transformation rule reduces to the familiar formula

$$ \sigma'_{xx} = \sigma_{xx}\cos^2\theta + \sigma_{yy}\sin^2\theta + 2\sigma_{xy}\sin\theta\cos\theta $$

Step 2. With \( \cos 45^\circ = \sin 45^\circ = \tfrac{1}{\sqrt{2}} \), we have \( \cos^2\theta = \sin^2\theta = \sin\theta\cos\theta = \tfrac{1}{2} \):

$$ \sigma'_{xx} = 50\left(\tfrac{1}{2}\right) + 30\left(\tfrac{1}{2}\right) + 2(20)\left(\tfrac{1}{2}\right) = 25 + 15 + 20 = 60 \text{ MPa} $$

Step 3 (check). The new \( x' \)-axis points along \( \mathbf{n} \) from Example 1, and \( \sigma'_{xx} \) is exactly the normal stress on that plane, 60 MPa. Two completely different routes give the same answer, which is the whole point of tensors: the physics is independent of the coordinates.

Step 4 (invariant check). The trace \( \sigma_{ii} \) is the same in every coordinate system. Originally it is \( 50 + 30 + 10 = 90 \) MPa. In the rotated frame, \( \sigma'_{yy} = 50(\tfrac12) + 30(\tfrac12) - 20 = 20 \) MPa, and \( \sigma'_{zz} = 10 \) MPa, so the trace is \( 60 + 20 + 10 = 90 \) MPa. Unchanged, as required.

A Short History

Tensor calculus was developed by the Italian mathematician Gregorio Ricci-Curbastro from the 1880s and set out with his student Tullio Levi-Civita in a 1900 paper on the "absolute differential calculus". The word "tensor" in its modern sense was introduced by the German physicist Woldemar Voigt in 1898, while studying stresses in crystals (from Latin tendere, "to stretch"). Around 1912, Albert Einstein learned the subject with help from his friend, the mathematician Marcel Grossmann, and used it to write general relativity in 1915. The Einstein field equations \( G_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu} \) are a tensor equation linking the curvature of space-time to matter and energy.

Where Tensors Are Used

• Civil and mechanical engineering: stress and strain tensors in beams, bridges, pressure vessels and aircraft.
• Rigid-body dynamics: the inertia tensor relates angular velocity to angular momentum, \( L_i = I_{ij}\omega_j \).
• Materials science: conductivity and permeability of crystals depend on direction.
• Machine learning: libraries such as TensorFlow and PyTorch store images and data as multi-dimensional arrays called tensors (a looser use of the word).

Did You Know?

• In 3D the stress tensor is symmetric, so only 6 of its 9 components are independent.
• The 81 components of the elasticity tensor reduce to just 2 independent constants for an isotropic material such as steel: for example, Young's modulus and Poisson's ratio.
• Levi-Civita corresponded with Einstein and pointed out errors in Einstein's early tensor calculations.

Key Takeaways

• Tensors generalise scalars (rank 0) and vectors (rank 1); a rank-\( n \) tensor in 3D has \( 3^n \) components.
• A tensor is defined by how its components transform under a change of axes: \( \sigma' = R\sigma R^{\mathsf{T}} \) for rank 2.
• Index notation with Einstein's summation convention makes tensor equations short and clear.
• Cauchy's formula \( t_i = \sigma_{ij}n_j \) gives the traction on any plane from the stress tensor.
• Invariants like the trace stay the same in every coordinate system, a powerful check on calculations.

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