Spacetime Notation

Spacetime notation packages equations so that their transformation laws can be read from their indices. Its practical value is diagnostic: every term in a tensor equation must have the same free indices and physical dimensions. This worked guide applies the conventions owned by Metric and Units; it does not introduce a second convention set.

Required background. Metric and Units fixes the signature, coordinates, and phase used in the component checks.

Free indices and summed indices

A free index labels components of an equation. A repeated upper–lower pair is summed and is called a dummy index. For example,

$$b_\mu=\eta_{\mu\nu}a^\nu$$

has one free index, $\mu$, on each side. It represents four equations. The index $\nu$ occurs twice in the term on the right and can be renamed without changing the result. In contrast,

$$a_\mu b^\mu=a_\nu b^\nu$$

has no free indices and represents a scalar equality. A label occurring three times in one term is not a valid Einstein contraction. Neither is $a^\mu=b_\nu$ an equation with consistent free indices.

Greek indices include the time component. Spatial Latin indices in ordinary three-vector expressions are contracted with the Euclidean metric. Keep that notation separate from lowering a spatial component of a four-vector: the symbols $p_i$ as a spacetime covector component and $(\mathbf p)_i$ as a Cartesian three-vector component need not denote the same signed number.

A complete contraction

Take $a^\mu=(2,1,0,0)$ and $b^\mu=(3,-2,0,0)$ in a common unit. Lowering one index gives $a_\mu=(2,-1,0,0)$ and therefore

$$a_\mu b^\mu=2(3)+(-1)(-2)=8.$$

The Euclidean dot product of the two displayed upper-index columns would be $4$ and is the wrong invariant. One need not lower both columns: contracting two lower-index columns with an implicit Euclidean sum would make the same mistake in a different form.

Derivatives and the quantum phase

The placement of the derivative index follows from what it differentiates: $\partial_\mu=\partial/\partial x^\mu$. On a coordinate function,

$$\partial_\mu x^\nu=\delta_\mu{}^\nu, \qquad \partial_\mu x_\nu=\eta_{\nu\mu}.$$

For a plane wave, expand the phase before differentiating:

$$\begin{aligned} f(x)&=e^{-i(Et-\mathbf p\cdot\mathbf x)/\hbar},\\ i\hbar\partial_0 f&=(E/c)f,\\ i\hbar\partial_i f&=-p^i f. \end{aligned}$$

Thus $i\hbar\partial_\mu f=p_\mu f$. The spatial quantum momentum operator is $-i\hbar\nabla$, consistent with the upper-index spatial momentum. The opposite signs are a consequence of the index placement, not conflicting quantization rules.

As a second check,

$$\Box f=-\frac{p_\mu p^\mu}{\hbar^2}f.$$

On the mass shell this becomes $\Box f=-m^2c^2f/\hbar^2$, which fixes the relative sign in the Klein–Gordon equation.

Worldlines and their parameters

A worldline $x^\mu(\lambda)$ is a curve; $x^\mu$ is its coordinate value and $dx^\mu$ its infinitesimal displacement. For a timelike curve the proper time satisfies

$$c^2d\tau^2=c^2dt^2-d\mathbf x^2, \qquad d\tau=dt\sqrt{1-\mathbf v^2/c^2}.$$

The tangent with proper-time parametrization is $U^\mu=dx^\mu/d\tau$. If a particle travels at constant $v=3c/5$ along $x$, then

$$U^\mu=(5c/4,3c/4,0,0), \qquad U_\mu U^\mu=c^2.$$

The tangent $dx^\mu/dt=(c,\mathbf v)$ does not have this fixed norm because coordinate time is not invariant. A null curve has $d\tau=0$; its tangent must be defined using another parameter. Dividing by its proper time is not a limiting prescription for a massless four-velocity. Four-Vectors develops the transformation laws and physical examples.

Checking a covariant expression

For any proposed identity, perform three checks. First match free indices term by term. Then check dimensions, remembering that $x^0$ has dimensions of length. Finally expand one time and one spatial component. For instance, $j^\mu=(c\rho,\mathbf j)$ has uniform component dimensions, and

$$\partial_\mu j^\mu=\partial_t\rho+\nabla\cdot\mathbf j.$$

Writing $(\rho,\mathbf j)$ while retaining $x^0=ct$ fails the dimensional check. An expression can pass all three checks and still require a physical derivation; these checks detect errors but do not prove dynamics.

Exercises

1. Evaluate $\partial_\mu(x_\nu x^\nu)$ and identify its free index.

Solution

The product rule gives $\eta_{\nu\mu}x^\nu+x_\nu\delta_\mu{}^\nu =2x_\mu$. The result is a covector with free index $\mu$.

2. Find $\Box(x_\mu x^\mu)$ in four spacetime dimensions.

Solution

Use a different dummy index for the operator: $\partial^\nu\partial_\nu (x_\mu x^\mu)=2\partial^\nu x_\nu=2\delta^\nu{}_{\nu}=8$. Directly differentiating $c^2t^2-x^2-y^2-z^2$ gives the same answer.

3. A timelike displacement has $c\Delta t=5\,\mathrm m$ and $\Delta x=3\,\mathrm m$. Find the elapsed proper time for the straight inertial path connecting the events.

Solution

$c\Delta\tau=\sqrt{25-9}\,\mathrm m=4\,\mathrm m$. The qualification about the path matters: a different timelike path between the same endpoints can have a different accumulated proper time.

References

• J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1998, chapter 11 — covariant notation and relativistic kinematics.
• W. Rindler, Introduction to Special Relativity, 2nd ed., Oxford University Press, 1991 — four-vectors, intervals, and proper time.
• D. Tong, Lectures on Quantum Field Theory, University of Cambridge, 2006, section 1 — covariant derivatives and relativistic field equations.