Four-Vectors

A four-vector is not merely a list of four numbers. It is an object whose components transform with the Lorentz matrix between inertial frames. If $a^\mu$ is contravariant, then

$$a'^\mu=\Lambda^\mu{}_{\nu}a^\nu,$$

and the Lorentz condition

$$\Lambda^{\mathsf T}\eta\Lambda=\eta$$

guarantees that $a_\mu b^\mu$ is frame independent. Coordinates, four- momentum, derivatives, electromagnetic potentials, and conserved currents fit into this structure, but ordinary velocity and Dirac spinors transform in different representations.

Required background. Metric and Units supplies the mostly-minus metric and all index, phase, and unit conventions. This page also assumes familiarity with Lorentz matrices and active-versus- passive transformations.

Lorentz transformations preserve the metric

In inertial Cartesian coordinates, a Lorentz transformation is a real linear map satisfying

$$\eta_{\rho\sigma} \Lambda^\rho{}_{\mu} \Lambda^\sigma{}_{\nu} = \eta_{\mu\nu}.$$

For two contravariant four-vectors,

$$\begin{aligned} a'\cdot b' &= \eta_{\mu\nu}a'^\mu b'^\nu \\ &= \eta_{\mu\nu} \Lambda^\mu{}_{\rho} \Lambda^\nu{}_{\sigma} a^\rho b^\sigma \\ &= \eta_{\rho\sigma}a^\rho b^\sigma =a\cdot b. \end{aligned}$$

The inverse transformation follows from the metric:

$$(\Lambda^{-1})^\mu{}_{\nu} = \eta^{\mu\rho}\Lambda^\sigma{}_{\rho}\eta_{\sigma\nu}.$$

A covector transforms with that inverse,

$$b'_\mu=(\Lambda^{-1})^\nu{}_{\mu}b_\nu,$$

so the contraction $b_\mu a^\mu$ is a scalar. Raising or lowering an index converts between these transformation laws; it is not decorative typography.

A standard boost

Consider a passive change to a frame moving with speed $v$ along $+x$. Define

$$\beta=\frac{v}{c}, \qquad \gamma=\frac{1}{\sqrt{1-\beta^2}}.$$

Then

$$\begin{pmatrix} x'^0\\x'^1\\x'^2\\x'^3 \end{pmatrix} = \begin{pmatrix} \gamma&-\beta\gamma&0&0\\ -\beta\gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix} \begin{pmatrix} x^0\\x^1\\x^2\\x^3 \end{pmatrix}.$$

The same matrix acts on every contravariant four-vector. For $p^\mu=(E/c,\mathbf p)$,

$$E'=\gamma(E-vp_x), \qquad p_x'=\gamma\left(p_x-\frac{vE}{c^2}\right),$$

with $p_y'=p_y$ and $p_z'=p_z$. These equations mix energy and momentum in the same way that a boost mixes time and position.

As a worked check, take a massive particle initially at rest, $p^\mu=(mc,0,0,0)$. The boosted frame measures

$$E'=\gamma mc^2, \qquad p_x'=-\gamma mv.$$

The minus sign is correct for this passive convention: in a frame moving along $+x$, the formerly stationary particle moves along $-x$. An active boost of the particle while holding the coordinate frame fixed would use the inverse matrix and reverse this sign.

Position, proper time, and four-velocity

The position four-vector is $x^\mu=(ct,\mathbf x)$. Along a timelike worldline, proper time is defined by

$$c^2d\tau^2=dx_\mu dx^\mu.$$

The four-velocity is

$$U^\mu = \frac{dx^\mu}{d\tau} = \gamma_v(c,\mathbf v),$$

where $\gamma_v=(1-\mathbf v^2/c^2)^{-1/2}$. Its invariant norm is

$$U_\mu U^\mu=c^2.$$

The ordinary three-velocity $\mathbf v=d\mathbf x/dt$ is not the spatial part of a four-vector: the spatial part is $\gamma_v\mathbf v$. For a particle of invariant mass $m>0$,

$$p^\mu=mU^\mu = \left(\frac{E}{c},\mathbf p\right),$$

so $E=\gamma_vmc^2$ and $\mathbf p=\gamma_vm\mathbf v$.

A massless particle follows a null worldline and has no proper-time parameter or rest frame. Its four-momentum is well defined, but the formula $p^\mu=mU^\mu$ is not: both $m$ and the proper-time construction fail in that limit.

Four-momentum and the invariant phase

With the shared conventions,

$$p\cdot x = p_\mu x^\mu = Et-\mathbf p\cdot\mathbf x.$$

Under the simultaneous transformations $p'=\Lambda p$ and $x'=\Lambda x$,

$$p'\cdot x'=p\cdot x.$$

Therefore the plane-wave phase

$$e^{-ip\cdot x/\hbar}$$

is a Lorentz scalar. The components $E$ and $\mathbf p$ change between frames, but the complete phase does not. This is the bridge between relativistic kinematics and the Fourier modes used in wave equations.

The invariant $p^2=p_\mu p^\mu$ classifies four-momentum. For an on-shell massive particle $p^2=m^2c^2>0$; for an on-shell massless particle $p^2=0$. Spacelike four-momenta can appear as momentum transfers or off-shell variables, but not as the four-momentum of a free physical massive particle.

Derivatives are covectors

For a scalar function $f(x)$, the chain rule gives

$$\partial'_\mu f'(x') = (\Lambda^{-1})^\nu{}_{\mu}\partial_\nu f(x),$$

where $f'(x')=f(x)$. Thus $\partial_\mu$ is a covector operator. Raising its index produces a contravariant operator:

$$\partial'^\mu = \Lambda^\mu{}_{\nu}\partial^\nu.$$

Consequently

$$\partial_\mu\partial^\mu=\Box$$

is a Lorentz scalar operator. This observation is the structural reason the Klein–Gordon equation can be written covariantly.

Four-current and local conservation

Suppose $\rho$ is a density and $\mathbf j$ its spatial flux in a chosen frame. The associated four-current is

$$j^\mu=(c\rho,\mathbf j).$$

Its four-divergence is

$$\partial_\mu j^\mu = \frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf j.$$

If $j^\mu$ transforms as a four-vector, the continuity equation

$$\partial_\mu j^\mu=0$$

is a Lorentz-scalar statement. Different observers split the same current into density and three-flux differently. Covariance alone does not guarantee that $\rho$ is positive: the Klein–Gordon charge current supplies the crucial counterexample. The Dirac current is future-directed causal (or zero), so $j^0\geq0$ in every proper orthochronous inertial frame.

Four-potential and gauge dependence

The electromagnetic potential transforms as

$$A^\mu=\left(\frac{\Phi}{c},\mathbf A\right).$$

Its Lorentz transformation law makes it a four-vector in a fixed gauge, but its components are not directly gauge invariant. The replacement

$$A_\mu\longmapsto A_\mu-\partial_\mu\chi$$

changes the potential without changing the electromagnetic field strength. “Transforms as a four-vector” and “is an observable” are therefore different claims.

Dirac spinors provide the complementary warning. A four-component spinor is not a four-vector. It transforms with a spinor matrix $S(\Lambda)$ satisfying an intertwining relation with the gamma matrices, not with $\Lambda^\mu{}_{\nu}$ acting directly on its four entries.

Common pitfalls

Calling any four-component column a four-vector. Transformation law, not component count, defines the object. Spinors and collections of four scalars are different representations.

Mixing an active boost with a passive formula. The two use inverse Lorentz matrices. State what changes—the frame or the physical system—before interpreting a momentum sign.

Using proper time for a null worldline. Proper time vanishes along a massless trajectory, so four-velocity is not defined there. Use a different affine parameter or work directly with four-momentum.

Assuming covariance implies gauge invariance. $A^\mu$ has a Lorentz transformation law and remains gauge dependent. The field strength and gauge-covariant observables carry the physical information.

Exercises

1. Four-velocity norm

Show that $U^\mu=\gamma_v(c,\mathbf v)$ satisfies $U^2=c^2$.

Solution

Using the mostly-minus metric,

$$U^2 = \gamma_v^2(c^2-\mathbf v^2) = \frac{c^2(1-\mathbf v^2/c^2)}{1-\mathbf v^2/c^2} =c^2.$$

2. Boosted photon

A photon moves along $+x$ with $p^\mu=(E/c,E/c,0,0)$. Apply the passive boost above and verify that it remains null.

Solution

The transformed components are

$$E'=\gamma E(1-\beta), \qquad p_x'=\frac{E'}{c}.$$

Hence

$$p'^2=\frac{E'^2}{c^2}-p_x'^2=0.$$

The energy changes by the longitudinal Doppler factor, but the null character is invariant.

3. Invariant plane-wave phase

For the one-dimensional boost, verify directly that $E't'-p_x'x'=Et-p_xx$.

Solution

Insert

$$ct'=\gamma(ct-\beta x), \qquad x'=\gamma(x-\beta ct),$$

and the corresponding formulas for $E'$ and $p_x'$. Expanding gives terms proportional to $\gamma^2(1-\beta^2)=1$; the mixed terms cancel, leaving $Et-p_xx$.

4. Covariant continuity equation

Show that if $j'^\mu=\Lambda^\mu{}_{\nu}j^\nu$ for a constant Lorentz matrix, then $\partial'_\mu j'^\mu=\partial_\mu j^\mu$.

Solution

The derivative transforms with the inverse matrix, so

$$\begin{aligned} \partial'_\mu j'^\mu &= (\Lambda^{-1})^\rho{}_{\mu}\partial_\rho \Lambda^\mu{}_{\sigma}j^\sigma \\ &= \delta^\rho{}_{\sigma}\partial_\rho j^\sigma = \partial_\rho j^\rho. \end{aligned}$$

Constancy of $\Lambda$ is appropriate for global inertial-frame changes.

References

• J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1998.
• S. Navas et al. (Particle Data Group), “Kinematics,” in Review of Particle Physics, Physical Review D 110, 030001, 2024, doi:10.1103/PhysRevD.110.030001.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, doi:10.1017/9781139540940.
• S. Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995.