Four-vector

In the theory of relativity, a four-vector or 4-vector is a vector in a four-dimensional real vector space, called Minkowski space. It differs from a Euclidean vector in that four-vectors transform by the Lorentz transformations. The term four-vector tacitly assumes that its components refer to a vector basis. In a standard basis, the components transform between these bases as the space and time coordinate differences, (cΔt, Δx, Δy, Δz) under spatial translations, spatial rotations, spatial and time inversions and boosts (a change by a constant velocity to another inertial reference frame). The set of all such translations, rotations, inversions and boosts (called Poincaré transformations) forms the Poincaré group. The set of rotations, inversions and boosts (Lorentz transformations, described by 4×4 matrices) forms the Lorentz group.

This article considers four-vectors in the context of special relativity. Although the concept of four-vectors also extends to general relativity, some of the results stated in this article require modification in general relativity.

Mathematics of four-vectors

General Four-vectors

A four-vector V is defined as, in various equivalent notations:

$\begin{align} \mathbf{V} & = (V^0, \, V^1, \, V^2, \, V^3) \\ & = V^0\mathbf{e}_0 + V^1 \mathbf{e}_1 + V^2 \mathbf{e}_2 + V^3 \mathbf{e}_3 \\ & = V^0\mathbf{e}_0 + V^i \mathbf{e}_i \\ & = V^\alpha\mathbf{e}_\alpha\\ \end{align}$

where the upper index denotes it to be contravariant. Here the standard convention of Latin indices take values for spatial components (so i = 1, 2, 3), and Greek indices take values for space and time components (so α =0, 1, 2, 3). The split between the time component and the spatial components is a useful one to make when determining contractions of one four vector with other tensor quantities, such as for calculating Lorentz invariants in inner products (examples given throughout below), or raising and lowering indices.

The covariant form can be obtained using the metric tensor g, in various equivalent notations:

$\begin{align} V_\mu & = g_{\mu\nu}V^\nu \\ \mathbf{V} & = (V_0, \, V_1, \, V_2, \, V_3) \\ & = V_0\mathbf{e}^0 + V_1 \mathbf{e}^1 + V_2 \mathbf{e}^2 + V_3 \mathbf{e}^3 \\ & = V_0\mathbf{e}^0 + V_i \mathbf{e}^i \\ & = V_\alpha\mathbf{e}^\alpha\\ \end{align}$

where the lowered index indicates it to be covariant. Often the metric is diagonal, as is the case for orthogonal coordinates (see line element), but not in general curvilinear coordinates.

Scalar product

The scalar product of two four-vectors U and V is defined (using Einstein notation) as

$\begin{align} \mathbf{U \cdot V} & = U^{\mu} \eta_{\mu \nu} V^{\nu} \\ & = \left( \begin{matrix}U^0 & U^1 & U^2 & U^3 \end{matrix} \right) \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix} \right) \left( \begin{matrix}V^0 \\ V^1 \\ V^2 \\ V^3 \end{matrix} \right) \\ & = U^0 V^0 - U^1 V^1 - U^2 V^2 - U^3 V^3 = L \\ \end{align}$

where η_μν is the entry in the μth row and νth column of the Minkowski metric η, here in the (+−−−) metric signature, and L is the value of the inner product. Sometimes the inner product in this context is called the Minkowski inner product. It is a recurring theme in special relativity to take the expression;

$L = U^0 V^0 - U^1 V^1 - U^2 V^2 - U^3 V^3$

and use this for calculations in different reference frames. While all components of U and V change from one frame to another; L is independent of the frame, so for a different frame where the quantities representing U and V are measured to be U′ and V′ (as given by a Lorentz transformation, see below) we have;

$L = U^0 V^0 - U^1 V^1 - U^2 V^2 - U^3 V^3 = {U'}^0 {V'}^0 - {U'}^1 {V'}^1 - {U'}^2 {V'}^2 - {U'}^3 {V'}^3$

for any two frames. Notice, that the Minkowski metric is not a Euclidean metric because it is indefinite (see metric signature), and vector can have, in general, nonpositive length. Some authors define η with the opposite sign, in which case we have the (−+++) metric signature;

$\begin{align} \mathbf{U \cdot V} & = U^{\mu} \eta_{\mu \nu} V^{\nu} \\ & = \left( \begin{matrix}U^0 & U^1 & U^2 & U^3 \end{matrix} \right) \left( \begin{matrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix}V^0 \\ V^1 \\ V^2 \\ V^3 \end{matrix} \right) \\ & = - U^0 V^0 + U^1 V^1 + U^2 V^2 + U^3 V^3 = -L \\ \end{align}$

so that

$-L= - U^0 V^0 + U^1 V^1 + U^2 V^2 + U^3 V^3$

which is equivalent to the above expression for L in terms of U and V. Either convention will work, since the primary significance of the Minkowski inner product is that for any two four-vectors, its value is invariant for all observers; a change of coordinates does not result in a change in value of the inner product.

With the Minkowski metric as defined above, the only difference between covariant and contravariant four-vector components are signs, though the signs depend on which sign convention is used.

The inner product is often expressed as the effect of the dual vector of one vector on the other:

$\mathbf{U \cdot V} = U^*(\mathbf{V}) = U{_\nu}V^{\nu}.$

Here the U_νs are the components of the dual vector U* of U in the dual basis and called the covariant coordinates of U, while the original U^ν components are called the contravariant coordinates. Lower and upper indices indicate always covariant and contravariant coordinates, respectively.

The relation between the covariant and contravariant coordinates is:

$U_{\mu} = \eta_{\mu \nu} U^{\nu}. \,$

The four-vectors are arrows on the spacetime diagram or Minkowski diagram. In this article, four-vectors will be referred to simply as vectors.

Four-vectors may be classified as either spacelike, timelike or null. Spacelike, timelike, and null vectors are ones whose inner product with themselves is less than, greater than, and equal to zero respectively (assuming Minkowski metric with signature (+−−−)).

In special relativity (but not general relativity), the derivative of a four-vector with respect to a scalar (invariant) is itself a four-vector.

Lorentz transformation

Main article: Lorentz transformation

For two inertial frames of reference, in which one frame F' moves with 3-velocity v (not 4-velocity) relative to F, i..e. "F' is boosted with velocity v", all four-vectors transform in the same way according to the lorentz transformation matrix Λ. For relative motion in an arbitrary direction without rotations, a four-vector U as observed in F transforms to U' observed in F' according to

$\mathbf{U}'=\boldsymbol{\Lambda}\mathbf{U}\,\!$

where the matrix has components

$\begin{align} \Lambda_{00} & = \gamma, \\ \Lambda_{0i} & = \Lambda_{i0} = - \gamma \beta_{i}, \\ \Lambda_{ij} & = \Lambda_{ji} = ( \gamma - 1 )\dfrac{\beta_{i}\beta_{j}}{\beta^{2}} + \delta_{ij}= ( \gamma - 1 )\dfrac{v_i v_j}{v^2} + \delta_{ij}, \\ \end{align} \,\!$

in turn

$\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \,\!$

is the Lorentz factor,

$\beta = \frac{1}{c}(v_1,\,v_2,\,v_3)\,\!$

is the relative velocity in units of c, and δ_ij is the Kronecker delta.

Standard configuration of coordinate systems; for a Lorentz boost in the x-direction.

For the case of a boost in the x-direction only, the matrix can be reduced to;

$\begin{pmatrix} U'^0 \\ U'^1 \\ U'^2 \\ U'^3 \end{pmatrix} =\begin{pmatrix} \cosh\phi &-\sinh\phi & 0 & 0 \\ -\sinh\phi & \cosh\phi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} U^0 \\ U^1 \\ U^2 \\ U^3 \end{pmatrix}$

and from this; an interesting point is that this Lorentz matrix corresponds to a hyperbolic rotation. So just as 3-vectors are preserved under circular rotations in three-dimensions:

$\begin{pmatrix} U'^1 \\ U'^2 \\ U'^3 \end{pmatrix} = \begin{pmatrix} \cos\phi &-\sin\phi & 0 \\ \sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} U^1 \\ U^2 \\ U^3 \end{pmatrix}\ .$

which is the rotation matrix about the z-axis in three dimensions, 4-vectors are preserved under hyperbolic rotations in four-dimensions.

Four-position

A point in Minkowski space is called an "event" and is described in a standard basis by a set of four coordinates such as

$\mathbf{X} = X^{\mu} := \left(X^0, X^1, X^2, X^3 \right) = \left(ct, x, y, z \right)$

where $\mu$ = 0, 1, 2, 3, labels the spacetime dimensions and where c is the speed of light. The definition $X^0 = ct$ ensures that all the coordinates have the same units (of distance). These coordinates are the components of the position four-vector for the event. The displacement four-vector is defined to be an "arrow" linking two events:

$\Delta X^{\mu} := \left(c\Delta t, \Delta x, \Delta y, \Delta z \right)$

The position vector is the displacement vector when one of the two events is the origin of the coordinate system. Position vectors are relatively trivial; the general theory of four-vectors is concerned with displacement vectors.

The scalar product of the 4-position with itself is;

$\mathbf{X}\cdot\mathbf{X} = \|\mathbf{X}\|^2 = X^\mu X_\mu=(c\tau)^2=s^2 \,\!$

which defines the spacetime interval s and proper time τ in Minkowski spacetime, which are invariant. The scalar product of the differential 4-position with itself is:

$d\mathbf{X}\cdot d\mathbf{X} = \|d\mathbf{X}\|^2 = dX^\mu dX_\mu=c^2d\tau^2=ds^2 \,\!$

defining the differential line element ds and differential proper time increment dτ.

Dynamics

When considering physical phenomena, differential equations arise naturally; however, when considering space and time derivatives of functions, it is unclear which reference frame these derivatives are taken with respect to. It is agreed that time derivatives are taken with respect to the proper time τ. As proper time is an invariant, this guarantees that the proper-time-derivative of any four-vector is itself a four-vector. It is then important to find a relation between this proper-time-derivative and another time derivative (using the coordinate time t of an inertial reference frame). This relation is provided by taking the differential invariant spacetime interval;

$(cd\tau)^2 = (cdt)^2 - dr^2 \,,$

where

$dr^2 = dx^2 + dy^2 + dz^2$

is the distance differential of Cartesian coordinates x, y, z measured in some frame. Then dividing by (cdt)² gives:

$\left(\frac{cd\tau}{cdt}\right)^2 = 1 - \left(\frac{1}{c}\frac{dr}{dt}\right)^2 = 1 - \left(\frac{u}{c}\right)^2 = \frac{1}{\gamma(u)^2} \,,$

where u = dr/dt is the 3-velocity of an object measured in the same frame as the coordinates x, y, z, and coordinate time t, and

$\gamma(u) = \frac{1}{\sqrt{1-(u/c)^2}}$

is the Lorentz factor, so

$dt = \gamma(u)d\tau \,.$

It can also be found from the time transformation in the Lorentz transformations. Important four-vectors in relativity theory can now be defined.

Four-velocity

The four-velocity of an $\mathbf{X}(\tau)$ world line is defined by:

$\mathbf{U} := \frac{d\mathbf{X}}{d \tau}= \frac{d\mathbf{X}}{dt}\frac{dt}{d \tau}= \left(\gamma c, \gamma \mathbf{u} \right),$

where, using suffix notation,

$\mathbf{u} = u^i = \frac{dX^i}{dt}$

for i = 1, 2, 3. Using the differential of the 4-position, the magnitude of the 4-velocity can be obtained;

$\frac{dX^\mu dX_\mu}{d\tau^2} = \frac{dX^\mu dX_\mu}{d\tau d\tau}=U^\mu U_\mu = \|\mathbf{U}\|^2 = \mathbf{U}\cdot\mathbf{U} =c^2 \,\!$

in short

$\| \mathbf{U} \|^2 = U^{\mu} U_{\mu} = c^2 \,$

The geometric meaning of 4-velocity is the unit vector tangent to the world line in Minkowski space.

Four-acceleration

The four-acceleration is given by:

$\mathbf{A} =\frac{d\mathbf{U} }{d \tau} = \left(\gamma \dot{\gamma} c, \gamma \dot{\gamma} \mathbf{u} + \gamma^2 \dot{\mathbf{u}} \right).$

Since the magnitude $\sqrt{ | U_\mu U^\mu | }$ of $\mathbf{U}$ is a constant, the four acceleration is (pseudo-)orthogonal to the four velocity, i.e. the Minkowski inner product of the four-acceleration and the four-velocity is zero:

$A^\mu U_\mu = \frac{dU^\mu}{d\tau} U_\mu = \frac{1}{2} \, \frac{d}{d\tau} (U^\mu U_\mu) = 0 \,$

which is true for all world lines.

The geometric meaning of 4-acceleration is the curvature vector of the world line in Minkowski space.

Four-momentum

The four-momentum for a massive particle is given by:

$\mathbf{P} :=m \mathbf{U} = (\gamma mc, \mathbf{p}) = (\gamma mc, \gamma m\mathbf{u})$

where m is the invariant mass of the particle and p is the relativistic momentum.

Four-force

The four-force is defined by:

$\mathbf{F} := \frac {d \mathbf{P}} {d \tau}.$

For a particle of constant mass, this is equivalent to

$\mathbf{F} = m \mathbf{A} = \left(\gamma \dot{\gamma} mc, \gamma \mathbf{f} \right)$

where

$\mathbf{f} = \frac{d\mathbf{p}}{dt} = \dot{\gamma} m\mathbf{u} + \gamma m\dot{\mathbf{u}}.$

Thermodynamics

See also: Relativistic heat conduction

Four-heat flux

The 4-heat flux vector field, is essentially similar to the 3d heat flux vector field q:

$\mathbf{Q} = - k\left( \frac{1}{c}\frac{\partial T}{\partial t}, \nabla T\right)$

where T is absolute temperature and k is thermal conductivity.

The components are:

$Q_\alpha = - k \frac {\partial T}{\partial x^\alpha}$

in the local frame of the fluid, where ∂_α = ∂/∂x^α is the four-gradient.

Four-baryon number flux

The flux of baryons is:

$\mathbf{S}= n\mathbf{U}$

where n is the number density of baryons in the local rest frame of the baryon fluid (positive values for baryons, negative for antibaryons), and U the 4-velocity field (of the fluid) as above.

Four-entropy

The 4-entropy vector is defined by:

$\mathbf{s}= s\mathbf{S} + \frac{\mathbf{Q}}{T}$

where s is the entropy per baryon, and T the absolute temperature, in the local rest frame of the fluid.

Electromagnetism

Examples of four-vectors in electromagnetism include the following.

Four-current

The four-current is defined by

$\mathbf{J} := \left( \rho c, \mathbf{j} \right)$

formed from the current density j and charge density ρ.

Four-potential

The electromagnetic four-potential defined by

$\mathbf{A} := \left( \phi /c, \mathbf{a} \right)$

formed from the vector potential a and the scalar potential $\phi \,$. The four-potential is not uniquely determined, because it depends on a choice of gauge.

Four-frequency

A plane electromagnetic wave can be described by the four-frequency defined as

$\mathbf{N} :=\left(\nu, \nu \mathbf{n} \right)$

where $\nu$ is the frequency of the wave and n is a unit vector in the travel direction of the wave. Notice that

$N^\mu N_\mu = \nu ^2 \left(n^2 - 1 \right) = 0$

so that the four-frequency is always a null vector.

Four-wavevector

See also: De Broglie relation

A wave packet of nearly monochromatic light can be characterized by the wave vector, or four-wavevector

$\mathbf{K} = \left(\frac{\omega}{c}, \mathbf{k} \right) = 2\pi\frac{\mathbf{N}}{c}.$

Quantum theory

In relativistic quantum mechanics, the 4-probability current is:

$\mathbf{J} = (\rho c, \mathbf{j})= (c|\Psi|^2, \mathbf{j})$

where ρ is the probability density function corresponding to the time component, in turn Ψ is the wavefunction, and j is the probability current vector.

Physics of four-vectors

One advantage of the four-vector formalism, over that of three-vectors, is illuminated by showing that known relations between energy and matter are contained in the four-momentum vector.

Energy of massive particles

Here, an expression for the total energy of a particle

$E = \gamma m c^2\,$

will be derived. The kinetic energy (K) of a particle is defined analogously to the classical definition, namely as

$\frac{dK}{dt}= \mathbf{f} \cdot \mathbf{u}$

with f as above. Noticing that $F^\mu U_\mu = 0$ and expanding this out we get

$\gamma^2 \left(\mathbf{f} \cdot \mathbf{u} - \dot{\gamma} mc^2 \right) = 0$

Hence

$\frac{dK}{dt} = c^2 \frac{d\gamma m}{dt}$

which yields

$K = \gamma m c^2 + S \,$

for some constant S. When the particle is at rest (u = 0), we take its kinetic energy to be zero (K = 0). This gives

$S = -m c^2. \,$

Thus, we interpret the total energy E of the particle as composed of its kinetic energy K and its rest energy m c². Thus, we have

$E = \gamma m c^2. \,$

Total energy and invariant mass

We can also derive the energy–momentum relation:

$E^2 = (pc)^2 + (mc^2)^2\,\!$

using the four-vector formalism. Using the relation

$E = \gamma mc^2$

we can write the four-momentum as

$\mathbf{P} = \left(\frac{E}{c}, \mathbf{p} \right).$

Taking the inner product of the four-momentum with itself in two different ways, we obtain the relation

$\frac{E^2}{c^2} - p^2 = P^\mu P_\mu = m^2 U^\mu U_\mu = m^2 c^2$

reducing to

$\frac{E^2}{c^2} - p^2 = m^2 c^2.$

Hence

$E^2 = p^2 c^2 + m^2 c^4. \,$

This last relation is useful in many areas of physics.

See also

• Relativistic mechanics
• paravector
• wave vector
• Dust (relativity) for the number-flux four-vector
• Basic introduction to the mathematics of curved spacetime
• Minkowski space

References

• Rindler, W. Introduction to Special Relativity (2nd edn.) (1991) Clarendon Press Oxford ISBN 0-19-853952-5