Four-Vectors

Write down the four numbers $(c, v_x, v_y, v_z)$ for a particle moving at $0.8c$ along the $x$ -axis. It looks like a perfectly reasonable object: a time slot, three space slots, all with dimensions of velocity. Now boost to a frame moving at $0.5c$ and ask what the same particle's four numbers are there. Two routes are available. You can transform the tuple with the Lorentz matrix, which gives $(0.693c,\ 0.346c,\ 0,\ 0)$ . Or you can compute the particle's new velocity by the velocity-addition law — it is exactly $0.5c$ — and write the tuple down again, getting $(c,\ 0.5c,\ 0,\ 0)$ . Those two answers are not the same. They are not even close: the first has lost its $c$ in the time slot. The tuple $(c,\vec{v}\,)$ , which cost nothing to write and looked like a four-vector, is not one. And that is the whole point of the subject. "Four-vector" is not a shape — four numbers in a column — but a constraint: an object is a four-vector if and only if its components transform under a change of frame with exactly the same matrix $\Lambda$ that acts on the coordinate differences $\Delta x^\mu = (c\Delta t, \Delta \vec{x}\,)$ . Most quadruples fail that test. The ones that pass are the ones from which invariant physics can be built. The repair, once you see it, is a single symbol. Multiply the failed tuple by $\gamma$ and it works: $U^\mu = \gamma(c,\vec{v}\,)$ transforms correctly, in every frame, at every speed. The reason is that $\vec{v} = d\vec{x}/dt$ was built by dividing a four-vector by $dt$ — and $dt$ is frame-dependent — whereas $U^\mu = dx^\mu/d\tau$ divides by the invariant proper time. Divide by a scalar, keep the transformation law. Divide by a coordinate, destroy it. Everything else in relativistic physics is downstream of that sentence.

Key takeaways

Signature $(+,-,-,-)$ throughout: $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , so $A\cdot B = \eta_{\mu\nu}A^\mu B^\nu = A^0B^0 - \vec{A}\cdot\vec{B}$ , and lowering an index flips the sign of the spatial part: $A_\mu = (A^0, -\vec{A}\,)$ .
A four-vector is defined by its transformation law: $A'^\mu = \Lambda^\mu{}_\nu A^\nu$ , the same $\Lambda$ that moves the coordinate displacement. Four numbers in a column are not enough; the law is the certificate.
Because $\Lambda^{\mathsf T}\eta\,\Lambda = \eta$ , any full contraction $A\cdot B$ is a Lorentz scalar — the same number in every frame. This one identity is why invariants exist at all.
Owned result: $(c,\vec{v}\,)$ is NOT a four-vector; $\gamma(c,\vec{v}\,)$ is. Boosting $(c, 0.8c)$ by $v = 0.5c$ gives $(0.693c,\ 0.346c)$ — the right direction (the ratio is the velocity-addition answer $0.5$ ) but the wrong normalisation, off by exactly $\kappa = \gamma_{u'}/\gamma_u = 0.693$ . Multiply by $\gamma_u$ and $\kappa$ cancels identically. The missing $\gamma$ is $dt/d\tau$ .
Key idea: build four-vectors by dividing a four-vector by a Lorentz SCALAR, never by a coordinate. The four-velocity $U^\mu = dx^\mu/d\tau = \gamma(c,\vec{v}\,)$ then has the same invariant length for every particle in the universe, $U\cdot U = +c^2$ — and differentiating that constant gives the four-acceleration's orthogonality $A\cdot U = 0$ for free.

The core insight

An equation between four-vectors is true in every inertial frame or in none, because both sides carry the same $\Lambda$ : if $A^\mu - B^\mu = 0$ in one frame, then $\Lambda^\mu{}_\nu(A^\nu - B^\nu) = 0$ in every other. That is what "manifestly covariant" means, and it is the reason physicists went to the trouble of learning index notation. The price of admission is that you must check an object is a four-vector rather than assume it — and the check is not "does it have four components" but "does it transform with $\Lambda$ ".

Where this lives today

Four-vector arithmetic is the daily language of experimental particle physics, and it is almost the only thing a collider event display is really showing you. Every reconstructed particle in an LHC event is stored as a four-momentum; every question asked of that event is a contraction. The invariant mass of a pair of photons is $\sqrt{(p_1 + p_2)\cdot(p_1 + p_2)}$ , and it is because that contraction is frame-independent that a bump in the diphoton spectrum means a particle rather than an artefact of the beam. When ATLAS and CMS announced the Higgs boson in 2012, the plot they showed was a histogram of exactly this scalar. The rule that only scalars may divide four-vectors is what forces detector conventions too. Transverse momentum is used rather than momentum because the longitudinal boost between the lab and the parton collision frame is unknown; rapidity is used rather than velocity because rapidity differences are invariant under that boost. Both choices are the transformation law asserting itself through the design of an experiment. Beyond particles, the same discipline organises relativistic electromagnetism: the charge and current densities combine into a four-current $J^\mu = (c\rho, \vec{J}\,)$ , the potentials into $A^\mu = (\phi/c, \vec{A}\,)$ , and Maxwell's equations collapse to a single covariant statement whose form is manifestly the same in every inertial frame. That reorganisation — the recognition that the fields were never separate objects — is the S5 band's subject, and it is the pay-off of everything above.

The physical picture

In three dimensions, nobody would call $(x, y, \text{temperature})$ a vector. It has three components; it is still not a vector, because rotating your axes mixes $x$ and $y$ but leaves the temperature alone, and a genuine vector's components must all mix together under the same rotation matrix. The test is the transformation law, and it always was. Relativity enlarges the group. Boosts mix time into space and space into time, so a relativistic vector must have four components that mix among themselves under $\Lambda$ . The archetype is handed to us for free: the coordinate displacement $\Delta x^\mu = (c\Delta t, \Delta x, \Delta y, \Delta z)$ transforms with $\Lambda$ by definition of $\Lambda$ . Everything else in the theory is manufactured from it, and there is exactly one legal manufacturing operation.

Divide a four-vector by a Lorentz scalar; the result is a four-vector. That is it. And the discipline it imposes is severe, because most of the "obvious" divisions are illegal. Ordinary velocity is $\vec{v} = d\vec{x}/dt$ , and $dt$ is a coordinate, not a scalar: different frames disagree about it. Proper time $d\tau$ is a scalar — the arc length of the worldline, the same number for everyone. Divide the displacement by $d\tau$ instead of $dt$ and you get an object that survives boosts: the four-velocity. The difference between the two is a factor $dt/d\tau = \gamma$ , and that lone $\gamma$ is the entire content of this article.

Figure

PUNCHLINE — every massive particle's four-velocity lies on the SAME hyperbola U·U = c². At β = 0.6 (rapidity ln 2) it sits at U = (1.25c, 0.75c) and 1.25² − 0.75² = 1 exactly. The naive tuple (c, v⃗) instead lies on the flat line U⁰ = c — and a boost DOES NOT keep it there: boosting (c, 0.8c) by v = 0.5c lands at (0.693c, 0.346c), off the line. The hyperbola is preserved; the line is not. That is the whole test

Look at what the boost did to each object. The blue points all sit on the hyperbola $\gamma^2 - (\gamma\beta)^2 = 1$ , and a boost slides them along it: a four-velocity boosted is still a four-velocity, still of Minkowski length $c$ . The red point started on the flat line $U^0 = c$ — where the naive tuple lives by construction — and the boost threw it off the line entirely, to $(0.693c,\ 0.346c)$ . There is no way to interpret that as $(c, \vec{v}'\,)$ for any $\vec{v}'$ , because its first slot is no longer $c$ . And yet the failure is almost a success, in a way that tells you exactly how to fix it. The ratio of the boosted components is $0.346/0.693 = 0.5$ — precisely the velocity-addition answer. The direction in spacetime is right; only the length is wrong, by the factor $\kappa = 0.693$ . Supply that factor and the object works.

Figure

The four-vector family tree. The coordinate displacement is the archetype; every other four-vector is manufactured from it by dividing by a Lorentz SCALAR (dτ, or nothing at all) and multiplying by scalars (m). Divide by the coordinate dt instead — the dashed red branch — and the transformation law breaks

The family tree is short because the rules are strict. Displacement, four-velocity, four-acceleration, four-momentum, the wave four-vector, the four-current and four-potential of electromagnetism — each one is the archetype with scalars applied. The dashed red branch is the one that looks like it should work and does not.

The core equations

A four-vector is defined by how it behaves, not by what it is made of:

A'^{\mu} = \Lambda^{\mu}{}_{\nu}\,A^{\nu}, \qquad \Lambda^{\mu}{}_{\nu} = \begin{pmatrix}\gamma & -\gamma\beta & 0 & 0\\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix} \qquad (1)

where: $\Lambda$ — the Lorentz transformation, here a boost of speed $\beta c$ along $x$ ; $A^\mu$ — a candidate four-vector. Physical reading: this is a test, not a formula. Given four numbers, compute them in a second frame two different ways — by physics, and by applying $\Lambda$ — and see whether the answers agree. If they do not, the object is not a four-vector, whatever it looks like. The metric lowers indices and supplies the inner product. In $(+,-,-,-)$ :

A_\mu = \eta_{\mu\nu}A^{\nu} = (A^0, -\vec{A}\,), \qquad A\cdot B \equiv \eta_{\mu\nu}A^{\mu}B^{\nu} = A_\mu B^\mu = A^0B^0 - \vec{A}\cdot\vec{B} \qquad (2)

where: $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ . Physical reading: a four-vector with $A\cdot A > 0$ is timelike, $A\cdot A = 0$ null, $A\cdot A < 0$ spacelike. (In $(-,+,+,+)$ every one of those signs flips. Read the metric before you read the equation.) The minus signs on the spatial part of $A_\mu$ are not decoration — they are what makes the contraction come out invariant. And the reason it is invariant is one line of matrix algebra, from the defining property of $\Lambda$ :

\Lambda^{\mathsf T}\eta\,\Lambda = \eta \qquad\Longrightarrow\qquad A'\cdot B' = \eta_{\mu\nu}\Lambda^{\mu}{}_{\alpha}\Lambda^{\nu}{}_{\beta}A^{\alpha}B^{\beta} = \eta_{\alpha\beta}A^{\alpha}B^{\beta} = A\cdot B \qquad (3)

where: the middle step uses $\eta_{\mu\nu}\Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta = \eta_{\alpha\beta}$ , which is $\Lambda^{\mathsf T}\eta\,\Lambda = \eta$ in index form. Physical reading: every Lorentz invariant in physics — the interval, the rest mass, the phase of a wave, the invariant mass of a collision — is an instance of this single identity. Lorentz Invariance and Invariants develops it as a group statement. Finally, the manufacturing rule, and its most important product:

U^{\mu} \equiv \frac{dx^{\mu}}{d\tau} = \frac{dt}{d\tau}\frac{dx^{\mu}}{dt} = \gamma\,(c,\vec{v}\,), \qquad U\cdot U = \gamma^2\!\left(c^2 - v^2\right) = +\,c^2 \qquad (4)

where: $\tau$ — proper time, a Lorentz scalar; $\gamma = dt/d\tau$ . Physical reading: $U\cdot U = +c^2$ in this signature, not $-c^2$ . Every massive particle in the universe carries a four-velocity of the same Minkowski length. A fast particle and a slow one differ only in where their arrow points in spacetime, never in how long it is — which is why the four-velocities of all particles lie on one hyperbola, and why a boost can only slide a particle along it.

Where four-vectors come from

Derivation AThe transformation law forces the inner product to be invariant

The claim to be established is the engine of the entire formalism: that any full contraction of four-vectors with $\eta$ is a number every inertial observer agrees on. It sounds like an accident of the algebra and it is not — it follows in one line from what a Lorentz transformation is. Note the order of logic carefully: we do not first find invariants and then notice they are preserved. We define $\Lambda$ as the linear map that preserves $\eta$ , and invariance is then a theorem with no content beyond that definition.

Step 1. Fix the definition. A Lorentz transformation is a linear map on spacetime satisfying $\Lambda^{\mathsf T}\eta\,\Lambda = \eta$ , equivalently $\eta_{\mu\nu}\Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta = \eta_{\alpha\beta}$ . Check it on the standard boost: the $00$ entry of $\Lambda^{\mathsf T}\eta\Lambda$ is $\gamma^2 - \gamma^2\beta^2 = \gamma^2(1-\beta^2) = 1$ , the $11$ entry is $\gamma^2\beta^2 - \gamma^2 = -1$ , and the off-diagonal is $-\gamma^2\beta + \gamma^2\beta = 0$ . Why: this is the condition that the interval is preserved, written as a matrix identity — and it is the only input the rest of the argument needs.

Step 2. Transform both factors of a contraction. If $A$ and $B$ are four-vectors, $A'^\mu = \Lambda^\mu{}_\alpha A^\alpha$ and $B'^\nu = \Lambda^\nu{}_\beta B^\beta$ , so $A'\cdot B' = \eta_{\mu\nu}\Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta\,A^\alpha B^\beta$ . Why: the whole trick is that both factors carry a $\Lambda$ , so the two $\Lambda$ 's sandwich the $\eta$ — exactly the combination the defining condition controls.

Step 3. Collapse the sandwich. By Step 1 the middle is just $\eta_{\alpha\beta}$ , so $A'\cdot B' = \eta_{\alpha\beta}A^\alpha B^\beta = A\cdot B$ . Why: no properties of $A$ or $B$ were used beyond their being four-vectors. The invariance is a property of the group, not of the particular objects.

Step 4. Read off the corollaries. Taking $B = A$ gives $A\cdot A$ invariant — hence the sign of $A\cdot A$ (timelike, null, spacelike) is an absolute, frame-independent classification. Taking $A = B = \Delta x$ gives the invariant interval. And if $A^\mu = B^\mu$ holds in one frame, then $\Lambda^\mu{}_\nu(A^\nu - B^\nu) = 0$ in every other, so the equation holds everywhere. Why: this last corollary is <strong>covariance</strong> — the practical reason to write physics in four-vector form at all. A law written as an equality of four-vectors cannot fail to respect the principle of relativity. *Verified symbolically in verify_aq-sr-four-vectors.py: $\Lambda^{\mathsf T}\eta\Lambda = \eta$ and $\det\Lambda = +1$ for the standard boost; $A'\cdot B' - A\cdot B = 0$ for arbitrary symbolic $A$ , $B$ ; $(2c,c,0,0)$ is timelike with $A\cdot A = +3c^2$ , $(c,2c,0,0)$ is spacelike with $-3c^2$ , $(c,c,0,0)$ is null.*

Go deeper · Derivation B · 2 minThe four-tuple with c in the time slot is not a four-vector — and one gamma repairs it

Derivation BThe four-tuple with c in the time slot is not a four-vector — and one gamma repairs it

This is the article's owned result and its teaching punch. The object $(c,\vec{v}\,)$ is the first thing anyone writes down when told to build a "four-velocity", and it is wrong. Showing that it is wrong takes one boost; showing why, and how the repair is forced, takes a little more — and the payoff is a rule you can apply forever after: divide by a scalar, never by a coordinate. We do the computation both ways, componentwise and by norms, because each exposes a different half of the failure.

Step 1. Set up. A particle moves at speed $u$ along $x$ in frame $S$ . The naive candidate is $V^\mu = (c, u, 0, 0)$ . Boost to $S'$ , which moves at $v = \beta c$ along $x$ . Why: pick the simplest possible non-trivial case. If the object fails here it fails, full stop.

Step 2. Apply $\Lambda$ and read the components. $V'^0 = \gamma_v\!\left(c - \dfrac{uv}{c}\right)$ and $V'^1 = \gamma_v(u - v)$ . The time slot is no longer $c$ . Why: this alone settles it. Whatever $(c,\vec{v}\,)$ means, a boosted version of it is not of the same form, so it cannot be a four-vector.

Step 3. Find out how badly it fails. Factor the result: $\Lambda\,V = \kappa\cdot(c, u', 0, 0)$ where $u' = \dfrac{u-v}{1 - uv/c^2}$ is exactly the velocity-addition answer, and $\kappa = \gamma_v\!\left(1 - \dfrac{uv}{c^2}\right)$ . So the boosted tuple points in the right direction — its space/time ratio is $u'/c$ , precisely what the velocity-composition law demands — but its length is off by $\kappa$ . Why: this is much more informative than a bare "it fails". The failure is a single multiplicative factor, which means a single multiplicative repair will fix it.

Step 4. Identify $\kappa$ . Using $1 - u'^2/c^2 = \dfrac{(1-u^2/c^2)(1-v^2/c^2)}{(1-uv/c^2)^2}$ , we get $\gamma_{u'} = \gamma_u\gamma_v\!\left(1-\dfrac{uv}{c^2}\right)$ , hence $\boxed{\kappa = \gamma_{u'}/\gamma_u}$ . Why: the stray factor is a <strong>ratio of gammas</strong> — which is a very loud hint about what the tuple is missing.

Step 5. Repair it. Multiply the candidate by $\gamma_u$ : $\;\Lambda\big[\gamma_u(c,u,0,0)\big] = \gamma_u\,\kappa\,(c,u',0,0) = \gamma_u\cdot\dfrac{\gamma_{u'}}{\gamma_u}\cdot(c,u',0,0) = \gamma_{u'}(c,u',0,0)$ . The $\gamma_u$ cancels the $\kappa$ identically, and the boosted object is exactly the same construction built from $u'$ . So $U^\mu = \gamma(c,\vec{v}\,)$ is a four-vector. Why: and the $\gamma$ was never arbitrary — $\gamma = dt/d\tau$ , so $\gamma(c,\vec{v}\,) = (dt/d\tau)(dx^\mu/dt) = dx^\mu/d\tau$ . Dividing by the invariant $d\tau$ instead of the coordinate $dt$ is the same repair, seen from the other side.

Step 6. Confirm by norms, independently. The naive tuple has $V\cdot V = c^2 - u^2$ , which is manifestly frame-dependent — at $u = 0.8c$ it is $0.36c^2$ , and in the boosted frame the same construction gives $c^2 - u'^2 = 0.75c^2$ . Not invariant, so not a four-vector. But $U\cdot U = \gamma^2(c^2 - u^2) = c^2$ for every $u$ . Why: two independent proofs of the same fact, from the transformation law and from the norm. The second one also hands you the hyperbola of the punchline figure. *Verified symbolically in verify_aq-sr-four-vectors.py: $\Lambda V = \kappa(c,u',0,0)$ with $\kappa = \gamma_v(1-uv/c^2)$ ; $\kappa = \gamma_{u'}/\gamma_u$ (proved on squares and spot-checked numerically at four $(u,v)$ pairs); in rapidity variables $\Lambda\big[c(\cosh a, \sinh a)\big] = c(\cosh(a-w), \sinh(a-w))$ identically while the naive tuple picks up $\cosh(a-w)/\cosh a$ ; and the numbers $u = 0.8c$ , $v = 0.5c \Rightarrow u' = 0.5c$ , $\kappa = 0.6928$ , $\Lambda V = (0.693c, 0.346c)$ , $0.36c^2 \ne 0.75c^2$ .*

Go deeper · Derivation C · 2 minThe four-velocity has fixed length, and the four-acceleration is orthogonal to it

Derivation CThe four-velocity has fixed length, and the four-acceleration is orthogonal to it

Two structural facts fall out of the construction with almost no work, and both are used constantly. The first is that $U\cdot U = c^2$ for every particle at every speed — the four-velocity's length is not a property of the particle but of spacetime. The second is that the four-acceleration is always Minkowski-perpendicular to the four-velocity, which is the relativistic replacement for the Newtonian statement that speeding up and turning are different things. The second follows from the first by differentiation, which is the kind of economy that makes the formalism worth learning.

Step 1. Compute the norm. $U^\mu = \gamma(c,\vec{v}\,)$ , so $U\cdot U = \gamma^2 c^2 - \gamma^2 v^2 = \gamma^2 c^2\big(1 - v^2/c^2\big) = c^2$ , using $\gamma^2(1-\beta^2) = 1$ . Why: the $\gamma^2$ that made the object a four-vector is exactly the $\gamma^2$ that normalises it. One factor, two jobs.

Step 2. Read the geometry. Since $U\cdot U = c^2$ is a constant independent of $v$ , all four-velocities lie on the hyperbola $(U^0)^2 - (U^1)^2 = c^2$ , parametrised by rapidity as $U = c(\cosh\varphi, \sinh\varphi)$ with $\beta = \tanh\varphi$ and $\gamma = \cosh\varphi$ . At $\beta = 0.6$ (rapidity $\ln 2$ ) this reads $U = (1.25c,\ 0.75c)$ , and $1.25^2 - 0.75^2 = 1$ exactly. Why: a boost is a hyperbolic rotation, so it must slide points along this curve. That is the geometric reason the norm cannot change.

Step 3. Differentiate the constant. Define the four-acceleration $A^\mu = dU^\mu/d\tau$ . Then $\dfrac{d}{d\tau}\big(U\cdot U\big) = 2\,A\cdot U = \dfrac{d}{d\tau}\big(c^2\big) = 0$ , so $\boxed{A\cdot U = 0}$ . Why: $\eta$ is constant, so it passes through the derivative, and the product rule gives the factor $2$ . No coordinates, no components, no special case — the result holds for arbitrary motion.

Step 4. Interpret it. Go to the instantaneously comoving frame, where $U^\mu = (c,\vec{0}\,)$ . Then $A\cdot U = A^0 c = 0$ forces $A^0 = 0$ : the four-acceleration is purely spatial in the particle's own frame, and equals the proper acceleration there. Why: this is what a relativistic accelerometer reads. The orthogonality is not a curiosity — it is the statement that a particle cannot accelerate "in the time direction", which is precisely why $|\vec{v}|$ can approach $c$ but never reach it. *Verified symbolically in verify_aq-sr-four-vectors.py: $U\cdot U = c^2$ for an arbitrary $v(\tau)$ ; $A\cdot U = 0$ identically by differentiation; $U = c(\cosh\varphi,\sinh\varphi)$ satisfies $U\cdot U = c^2$ ; and the figure's marks $\beta = 0.6, 0.8, 0.9$ give $(\gamma,\gamma\beta) = (1.25, 0.75), (1.667, 1.333), (2.294, 2.065)$ , each with $\gamma^2 - (\gamma\beta)^2 = 1$ .*

Limiting cases

Case	Result	Meaning
$\beta \to 0$	$U^\mu \to (c, \vec{v}\,)$	the naive tuple is the low-speed limit of the true one — which is why it looks so plausible
$\gamma \to 1$ , i.e. $c \to \infty$	$\kappa \to 1$	the stray factor disappears: Galilean physics never notices the error
Norm sign $A\cdot A > 0$	timelike	a frame exists in which $\vec{A} = \vec{0}$ (the rest frame)
Norm sign $A\cdot A < 0$	spacelike	a frame exists in which $A^0 = 0$
Norm sign $A\cdot A = 0$ , $A \ne 0$	null	the wave four-vector of light, $k\cdot k = 0$ ; no rest frame exists
$v \to c$	$U^0, U^1 \to \infty$ , but $U\cdot U = c^2$ still	the four-velocity runs off to infinity along the hyperbola, never leaving it
Instantaneous rest frame	$U = (c,\vec 0\,)$ , $A = (0, \vec{a}_{\text{proper}})$	$A\cdot U = 0$ makes the four-acceleration purely spatial

Common trapFour components is not the test. The transformation law is the test.

The single most common error in relativistic kinematics is to assume that any physically natural quadruple is a four-vector. It is not. $(c,\vec{v}\,)$ is not. $\big(E, c\vec{p}\,\big)$ is not — the four-momentum is $\big(E/c, \vec{p}\,\big)$ , and the misplaced factor of $c$ breaks the law just as surely as a missing $\gamma$ . $\big(\text{energy density}, \text{energy flux}\big)$ is not (it is part of a tensor). Before contracting anything with $\eta$ and calling the result invariant, ask the only question that matters: does this object transform with $\Lambda$ ? If it was built by dividing a four-vector by a Lorentz scalar and multiplying by Lorentz scalars, yes. Otherwise, prove it or drop it.

Common trapThe sign of the norm — and the other convention

In $(+,-,-,-)$ : timelike $A\cdot A > 0$ , spacelike $A\cdot A < 0$ , and $U\cdot U = +c^2$ , $p\cdot p = +m^2c^2$ . In $(-,+,+,+)$ every one of those flips: timelike is negative, and $U\cdot U = -c^2$ . Rindler and particle physics use ours; Taylor–Wheeler and Misner–Thorne–Wheeler use the other. No physics depends on the choice; every intermediate sign does. The AllQFT base for the neighbouring proper-time article prints $u^\mu u_\mu = -c^2$ , which is correct in $(-,+,+,+)$ and wrong here — a reminder that a formula copied without its metric is a formula copied without its meaning.

Worked examples

1 · Boost a particle two ways and watch the naive tuple break

A particle moves at $u = 0.8c$ along $x$ in the lab. Frame $S'$ moves at $v = 0.5c$ along $x$ . Compute the particle's four numbers in $S'$ (a) by applying the Lorentz matrix to $(c, u, 0, 0)$ , and (b) by first finding $u'$ and rewriting the tuple. Then do the same for $\gamma(c,u,0,0)$ .

Step 1 — identify. The test of a four-vector is whether these two routes agree. Route (a) is "transform the object". Route (b) is "recompute the object from the physics in the new frame". A four-vector makes them the same computation.

Step 2 — the physics route. Velocity addition: $u' = \dfrac{u - v}{1 - uv/c^2} = \dfrac{0.8c - 0.5c}{1 - 0.40} = \dfrac{0.3c}{0.6} = \mathbf{0.5c}$ exactly. So in $S'$ the particle moves at half the speed of light, and the naive tuple there would be $(c,\ 0.5c,\ 0,\ 0)$ .

Step 3 — the matrix route. With $\beta = 0.5$ , $\gamma_v = 1/\sqrt{1-0.25} = 1.1547$ . Then $V'^0 = \gamma_v(c - uv/c) = 1.1547 \times (1 - 0.40)c = \mathbf{0.693c}$ and $V'^1 = \gamma_v(u - v) = 1.1547 \times 0.3c = \mathbf{0.346c}$ .

Step 4 — compare. $(0.693c,\ 0.346c)$ versus $(c,\ 0.5c)$ . The routes disagree. They differ by the overall factor $\kappa = 0.693$ , and note that $0.346/0.693 = 0.5$ — the direction survived, the normalisation did not. So $(c,\vec{v}\,)$ is not a four-vector: proved, with numbers.

Step 5 — the repair, with numbers. $\gamma_u = 1/\sqrt{1-0.64} = 5/3 = 1.6667$ , so the four-velocity in the lab is $U = 1.6667\,(c,\ 0.8c) = (1.667c,\ 1.333c)$ . Boost it: $U'^0 = 1.1547\,(1.667 - 0.5\times1.333)c = 1.1547 \times 1.0c = \mathbf{1.1547c}$ and $U'^1 = 1.1547\,(1.333 - 0.5\times1.667)c = 1.1547 \times 0.5c = \mathbf{0.577c}$ . Now recompute from the physics: $\gamma_{u'} = 1/\sqrt{1-0.25} = 1.1547$ , so $\gamma_{u'}(c, u') = 1.1547\,(c,\ 0.5c) = (1.1547c,\ 0.577c)$ . Identical. The two routes agree, which is exactly what being a four-vector means.

Step 6 — the boundary number. The discrepancy factor is $\kappa = \gamma_{u'}/\gamma_u = 1.1547/1.6667 = \mathbf{0.693}$ , and it is $1$ only when there is no boost. That single number is the whole difference between an object that obeys the principle of relativity and one that does not.

2 · Two norms, one invariant

Check directly that the naive tuple's norm changes under the boost of Example 1 while the four-velocity's does not — and identify what the four-velocity's norm actually is.

Step 1 — the naive norm in the lab. $(c, 0.8c)\cdot(c, 0.8c) = c^2 - (0.8c)^2 = \mathbf{0.36\,c^2}$ .

Step 2 — the naive norm in $S'$ . Rebuilding the tuple from $u' = 0.5c$ : $(c, 0.5c)\cdot(c, 0.5c) = c^2 - 0.25c^2 = \mathbf{0.75\,c^2}$ . Two frames, two different numbers, from a construction that was supposed to describe the same particle. It cannot be an invariant, so it cannot be the norm of a four-vector.

Step 3 — the four-velocity's norm. $U\cdot U = \gamma_u^2\big(c^2 - u^2\big) = (5/3)^2 \times 0.36\,c^2 = \dfrac{25}{9}\times\dfrac{9}{25}c^2 = \mathbf{c^2}$ . In $S'$ : $\gamma_{u'}^2(c^2 - u'^2) = (4/3)\times(0.75\,c^2) = c^2$ . The same number, and it will be the same number in every frame, for every particle, at every speed.

Step 4 — the boundary number. $U\cdot U = c^2$ is not a fact about this particle. It is a fact about spacetime: it says that everything with mass moves through spacetime at "speed" $c$ , and that speeding up in space is nothing but rotating that fixed-length arrow toward the light cone. On the punchline figure, the whole hyperbola is one particle's possible states — and the boost simply chooses a point on it.

Physics of the portrait

Minkowski gave physics the geometry in 1908 and died four months later, leaving behind a matrix calculus that most working physicists found forbidding. The man who turned it into something usable — a vector algebra, with rules you could apply to a real electrodynamics problem without first learning a new branch of mathematics — was his former student.

Origins · 1 minArnold Sommerfeld and how this was discovered

Arnold Sommerfeld

1868–1951 · Königsberg · Aachen · Munich

Arnold Johannes Wilhelm Sommerfeld took his doctorate at Königsberg under Lindemann and served as Felix Klein's assistant at Göttingen, where he absorbed the invariant-theoretic view of geometry that would later make Minkowski's spacetime feel natural rather than exotic. In 1910 he published a pair of papers that did for four-dimensional physics what Gibbs and Heaviside had done for three: "Zur Relativitätstheorie I: Vierdimensionale Vektoralgebra" (Annalen der Physik 32, 749) and "II: Vierdimensionale Vektoranalysis" (Ann. Phys. 33, 649). Where Minkowski had written matrices, Sommerfeld wrote vectors and their products, extending the operations physicists already knew from electrodynamics into four dimensions so that the theory's simplifications became visible rather than merely provable. He also supplied the editorial notes to the 1913 reprint of Minkowski's "Raum und Zeit" in Blumenthal's anthology, using them to record what Poincaré had done first. Sommerfeld is remembered as much for what he grew as for what he wrote: his Munich institute produced Heisenberg, Pauli, Debye, Bethe and Peierls, and he was nominated for the Nobel Prize eighty-four times without ever receiving it — more than any other physicist. He died in Munich in 1951, struck by a car while out walking with his grandchildren.

History

1906

— Poincaré's Palermo memoir "Sur la dynamique de l'électron" (received July 1905) treats

(ct, x, y, z)

as coordinates in a four-dimensional space with the invariant quadratic form.

1908

— Minkowski's Cologne lecture "Raum und Zeit" makes spacetime geometry the language of the theory, using a matrix calculus.

1910

— Sommerfeld recasts that calculus as a four-dimensional vector algebra and analysis (Ann. Phys. 32, 749 and 33, 649), giving four-vectors the form physicists still use.

1911

— Max von Laue's Das Relativitätsprinzip, the first relativity monograph, adopts the four-dimensional formalism throughout and makes it standard pedagogy.

1913

— Sommerfeld's editorial notes accompany the reprint of "Raum und Zeit" in Blumenthal's collection of relativity papers.

1928

— Dirac's equation is written in manifestly four-vector form, with the four-momentum operator; covariance becomes the organising principle of relativistic quantum theory.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §5 · Taylor & Wheeler, Spacetime Physics, 2nd ed., ch. 5–7 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.1 · Hartle, Gravity, ch. 5 · Jackson, Classical Electrodynamics, 3rd ed., §11.6 · Sommerfeld, "Zur Relativitätstheorie I: Vierdimensionale Vektoralgebra", Ann. Phys. 32, 749 (1910); "II: Vierdimensionale Vektoranalysis", Ann. Phys. 33, 649 (1910).

Provenance. Signature $(+,-,-,-)$ , so $A\cdot B = A^0B^0 - \vec A\cdot\vec B$ and $U\cdot U = +c^2$ . Everything above is recomputed in verify_aq-sr-four-vectors.py: $\Lambda^{\mathsf T}\eta\Lambda = \eta$ and $\det\Lambda = 1$ ; invariance of $A\cdot B$ for symbolic $A$ , $B$ ; the owned result in two independent forms (componentwise, with $\kappa = \gamma_v(1-uv/c^2) = \gamma_{u'}/\gamma_u$ , and in rapidity variables where $\Lambda\big[c(\cosh a,\sinh a)\big] = c(\cosh(a-w),\sinh(a-w))$ exactly while the naive tuple picks up $\cosh(a-w)/\cosh a$ ); $U\cdot U = c^2$ and $A\cdot U = 0$ for arbitrary $v(\tau)$ ; and the numbers $u = 0.8c$ , $v = 0.5c \Rightarrow u' = 0.5c$ , $\kappa = 0.693$ , $\Lambda(c, 0.8c) = (0.693c, 0.346c)$ , naive norms $0.36c^2 \ne 0.75c^2$ , figure marks $(\gamma, \gamma\beta) = (1.25, 0.75)$ , $(1.667, 1.333)$ , $(2.294, 2.065)$ .

Scope. Four-momentum $p^\mu = mU^\mu$ appears here only as the canonical example of a four-vector. The dynamics — $E = \gamma mc^2$ , $\vec{p} = \gamma m\vec{v}$ , $E^2 = (pc)^2 + (mc^2)^2$ — belongs to the relativistic-dynamics band and is not derived here. Likewise the wave four-vector $k^\mu$ is stated and shown to be null for light; the relativistic Doppler effect that follows from boosting it is treated in its own article.

Builds on. The Minkowski Metric · The Spacetime Interval · Proper Time and Proper Length. Leads to. Lorentz Invariance and Invariants · Rapidity · Hyperbolic Geometry of Boosts · Relativistic Four-Momentum.

If (c, v⃗) is not a four-vector, why does it look so much like one?

Because it is the low-speed limit of one. $U^\mu = \gamma(c,\vec{v}\,)$ and $\gamma = 1 + \tfrac12\beta^2 + \cdots$ , so at ordinary speeds $U^\mu$ and $(c,\vec{v}\,)$ agree to a part in $10^{16}$ for a jet aircraft. The discrepancy factor $\kappa$ tends to $1$ as $c \to \infty$ . Newtonian physics therefore never notices the error — which is precisely why the error is so easy to make, and why it only bites when you go fast enough for the transformation law to matter.

Is (E, cp⃗) a four-vector?

No — but $\big(E/c,\ \vec{p}\,\big)$ is, and so is $\big(E,\ c\vec{p}\,\big)$ up to an overall constant, which is the subtlety. Multiplying a four-vector by the scalar $c$ gives a four-vector, so $\big(E, c\vec{p}\,\big) = c\cdot\big(E/c, \vec{p}\,\big)$ is fine. What is not fine is $\big(E, \vec{p}\,\big)$ with no factor of $c$ anywhere, which is what people usually write down in natural units and then misremember when they restore the units. The lesson is the same as the article's: the components must all carry the same dimensions and mix under one $\Lambda$ , and a stray $c$ is as fatal as a stray $\gamma$ .

Why is the four-velocity's norm the same for a photon and an electron?

It is not — and it cannot be, because a photon has no four-velocity at all. The four-velocity is defined as $dx^\mu/d\tau$ , and along a null worldline $d\tau = 0$ : the derivative does not exist. Light is described instead by a null four-vector, the wave four-vector $k^\mu = (\omega/c, \vec{k}\,)$ with $k\cdot k = 0$ , or by a null four-momentum with $p\cdot p = 0$ . The statement " $U\cdot U = c^2$ " applies to everything with mass, and the light cone is exactly the boundary it cannot reach.

What does it actually mean to say an equation is "manifestly covariant"?

It means the equation is written as an equality between objects that carry the same transformation law, so that its truth cannot depend on the frame. If $A^\mu = B^\mu$ in one frame, then applying $\Lambda$ to both sides gives $A'^\mu = B'^\mu$ in any other, automatically. "Manifestly" is doing real work in that phrase: a law can be Lorentz-invariant without looking it — Maxwell's equations in three-vector form are — and the point of the four-vector formalism is to make the invariance visible on the page, so that you cannot accidentally write down a law that violates the principle of relativity.

The four-acceleration is orthogonal to the four-velocity. Orthogonal in what sense?

In the Minkowski sense: $A\cdot U = \eta_{\mu\nu}A^\mu U^\nu = 0$ , with the metric's minus signs included. This does not mean the two arrows look perpendicular on a spacetime diagram drawn on Euclidean paper — they generally do not, and Minkowski-orthogonal vectors appear to make equal angles with the light cone rather than a right angle. The physical content is unambiguous, though: in the particle's instantaneous rest frame $U = (c, \vec{0}\,)$ and the orthogonality forces $A^0 = 0$ , so the four-acceleration is purely spatial there and equals the proper acceleration an accelerometer would read.

Do I have to check the transformation law every single time?

No, and that is the point of the manufacturing rule. If an object was built from four-vectors and Lorentz scalars only — by adding four-vectors, multiplying by scalars, or differentiating with respect to a scalar — then it is a four-vector, and you may skip the check. The cases where you must be careful are exactly the cases where a coordinate has crept in: dividing by $dt$ , or by an energy, or by a component of some other four-vector. Whenever the recipe mentions a quantity that is different in different frames, stop and prove it.