# 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.

The 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.

The notations in this article are: lowercase bold for three-dimensional vectors, hats for three-dimensional unit vectors, capital bold for four dimensional vectors (except for the four-gradient), and tensor index notation.

## Four-vector algebra

### Four-vectors in a real-valued basis

A four-vector A is a vector with a "timelike" component and three "spacelike" components, and can be written in various equivalent notations:[1]

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

The upper indices indicate contravariant components. Here the standard convention that Latin indices take values for spatial components, so that i = 1, 2, 3, and Greek indices take values for space and time components, so α = 0, 1, 2, 3, used with the summation convention. 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 are given below), or raising and lowering indices.

In special relativity, the spacelike basis e1, e2, e3 and components A1, A2, A3 are often Cartesian basis and components:

\begin{align} \mathbf{A} & = (A_t, \, A_x, \, A_y, \, A_z) \\ & = A_t \mathbf{e}_t + A_x \mathbf{e}_x + A_y \mathbf{e}_y + A_z \mathbf{e}_z \\ \end{align}

although, of course, any other basis and components may be used, such as spherical polar coordinates

\begin{align} \mathbf{A} & = (A_t, \, A_r, \, A_\theta, \, A_\phi) \\ & = A_t \mathbf{e}_t + A_r \mathbf{e}_r + A_\theta \mathbf{e}_\theta + A_\phi \mathbf{e}_\phi \\ \end{align}
\begin{align} \mathbf{A} & = (A_t, \, A_r, \, A_\theta, \, A_z) \\ & = A_t \mathbf{e}_t + A_r \mathbf{e}_r + A_\theta \mathbf{e}_\theta + A_z \mathbf{e}_z \\ \end{align}

or any other orthogonal coordinates, or even general curvilinear coordinates. Note the coordinate labels are always subscripted as labels and are not indices taking numerical values. In general relativity, local curvilinear coordinates in a local basis must be used. Geometrically, a four-vector can still be interpreted as an arrow, but in spacetime - not just space. In relativity, the arrows are drawn as part of a spacetime diagram or Minkowski diagram. In this article, four-vectors will be referred to simply as vectors.

It is also customary to represent the bases by column vectors:

$\mathbf{e}_0 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \,,\quad \mathbf{e}_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \,,\quad \mathbf{e}_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} \,,\quad \mathbf{e}_3 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$

so that:

$\mathbf{A} = \begin{pmatrix} A^0 \\ A^1 \\ A^2 \\ A^3 \end{pmatrix}$

The relation between the covariant and contravariant coordinates is through the Minkowski metric tensor, η which raises and lowers indices as follows:

$A_{\mu} = \eta_{\mu \nu} A^{\nu} \,,$

and in various equivalent notations the covariant components are:

\begin{align} \mathbf{A} & = (A_0, \, A_1, \, A_2, \, A_3) \\ & = A_0\mathbf{e}^0 + A_1 \mathbf{e}^1 + A_2 \mathbf{e}^2 + A_3 \mathbf{e}^3 \\ & = A_0\mathbf{e}^0 + A_i \mathbf{e}^i \\ & = A_\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.

The bases can be represented by row vectors:

$\mathbf{e}^0 = \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \,,\quad \mathbf{e}^1 = \begin{pmatrix} 0 & 1 & 0 & 0 \end{pmatrix} \,,\quad \mathbf{e}^2 = \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix} \,,\quad \mathbf{e}^3 = \begin{pmatrix} 0 & 0 & 0 & 1 \end{pmatrix}$

so that:

$\mathbf{A} = \begin{pmatrix} A_0 & A_1 & A_2 & A_3 \end{pmatrix}$

The motivation for the above conventions are that the inner product is a scalar, see below for details.

### Lorentz transformation

Given two inertial or rotated frames of reference, a four-vector is defined as a quantity which transforms according to the Lorentz transformation matrix Λ:

$\mathbf{A}'=\boldsymbol{\Lambda}\mathbf{A}$

In index notation, the contravariant and covariant components transform according to, respectively:

${A'}^\mu = \Lambda^\mu {}_\nu A^\nu \,,\quad {A'}_\mu = \Lambda_\mu {}^\nu A_\nu$

in which the matrix Λ has components Λμν in row μ and column ν, and the inverse matrix Λ−1 has components Λμν in row μ and column ν.

For background on the nature of this transformation definition, see tensor. All four-vectors transform in the same way, and this can be generalized to four-dimensional relativistic tensors; see special relativity.

#### Pure rotations about an arbitrary axis

For two frames rotated by a fixed angle θ about an axis defined by the unit vector:

$\hat{\mathbf{n}} = (\hat{n}_1,\hat{n}_2,\hat{n}_3)\,,$

without any boosts, the matrix Λ has components given by:[2]

$\Lambda_{00} = 1$
$\Lambda_{0i} = \Lambda_{i0} = 0$
$\Lambda_{ij} = (\delta_{ij} - \hat{n}_i \hat{n}_j) \cos\theta - \varepsilon_{ijk} \hat{n}_k \sin\theta + \hat{n}_i \hat{n}_j$

where δij is the Kronecker delta, and εijk is the three-dimensional Levi-Civita symbol. The spacelike components of 4-vectors are rotated, while the time-like components remain unchanged.

For the case of rotations about the z-axis only, the spacelike part of the Lorentz matrix reduces to the rotation matrix about the z-axis:

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

#### Pure boosts in an arbitrary direction

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

For two frames moving at constant relative 3-velocity v (not 4-velocity, see below), it is convenient to denote and define the relative velocity in units of c by:

$\boldsymbol{\beta} = (\beta_1,\,\beta_2,\,\beta_3) = \frac{1}{c}(v_1,\,v_2,\,v_3) = \frac{1}{c}\mathbf{v} \,.$

Then without rotations, the matrix Λ has components given by:[3]

\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} \,\!

where the Lorentz factor is defined by:

$\gamma = \frac{1}{\sqrt{1- \boldsymbol{\beta}\cdot\boldsymbol{\beta}}} \,,$

and δij is the Kronecker delta. Contrary to the case for pure rotations, the spacelike and timelike components are mixed together under boosts.

For the case of a boost in the x-direction only, the matrix reduces to;[4][5]

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

Where the rapidity ϕ expression has been used, written in terms of the hyperbolic functions:

$\gamma = \cosh \varphi$

This Lorentz matrix illustrates the boost to be a hyperbolic rotation in four dimensional spacetime, analogous to the circular rotation above in three-dimensional space.

### Properties

#### Linearity

Four-vectors have the same linearity properties as Euclidean vectors in three dimensions. They can be added in the usual entrywise way:

$\mathbf{A}+\mathbf{B} = (A^0, A^1, A^2,A^3) + (B^0, B^1, B^2,B^3) = (A^0 + B^0, A^1 + B^1, A^2 + B^2, A^3 + B^3)$

and similarly scalar multiplication by a scalar λ is defined entrywise by:

$\lambda\mathbf{A} = \lambda(A^0, A^1, A^2,A^3) = (\lambda A^0, \lambda A^1, \lambda A^2, \lambda A^3)$

Then subtraction is the inverse operation of addition, defined entrywise by:

$\mathbf{A}+(-1)\mathbf{B} = (A^0, A^1, A^2,A^3) + (-1)(B^0, B^1, B^2,B^3) = (A^0 - B^0, A^1 - B^1, A^2 - B^2, A^3 - B^3)$

#### Inner product

The inner product (also called the scalar product) of two four-vectors A and B is defined, using Einstein notation, as

$\mathbf{A} \cdot \mathbf{B} = A^{\mu} \eta_{\mu \nu} B^{\nu}$

where η is the Minkowski metric. The inner product in this context is also called the Minkowski inner product. For visual clarity, it is convenient to rewrite the definition in matrix form:

$\mathbf{A \cdot B} = \begin{pmatrix} A^0 & A^1 & A^2 & A^3 \end{pmatrix} \begin{pmatrix} \eta_{00} & \eta_{01} & \eta_{02} & \eta_{03} \\ \eta_{10} & \eta_{11} & \eta_{12} & \eta_{13} \\ \eta_{20} & \eta_{21} & \eta_{22} & \eta_{23} \\ \eta_{30} & \eta_{31} & \eta_{32} & \eta_{33} \end{pmatrix} \begin{pmatrix} B^0 \\ B^1 \\ B^2 \\ B^3 \end{pmatrix}$

in which case ημν above is the entry in row μ and column ν of the Minkowski metric as a square matrix. The Minkowski metric is not a Euclidean metric, because it is indefinite (see metric signature). The inner product can be rewritten in a number of other ways because the metric tensor raises and lowers the components of A and B. For contra/co-variant components of A and co/contra-variant components of B, we have:

$\mathbf{A} \cdot \mathbf{B} = A_{\nu} B^{\nu} = A^{\mu} B_{\mu}$

so in the matrix notation:

$\mathbf{A \cdot B} = \begin{pmatrix} A_0 & A_1 & A_2 & A_3 \end{pmatrix} \begin{pmatrix} B^0 \\ B^1 \\ B^2 \\ B^3 \end{pmatrix} = \begin{pmatrix} B_0 & B_1 & B_2 & B_3 \end{pmatrix} \begin{pmatrix} A^0 \\ A^1 \\ A^2 \\ A^3 \end{pmatrix}$

while for A and B each in covariant components:

$\mathbf{A} \cdot \mathbf{B} = A_{\mu} \eta^{\mu \nu} B_{\nu}$

with a similar matrix expression to the above.

The inner product of a four-vector A with itself is the square of the norm of the vector, denoted and defined by:

$\|\mathbf{A}\|^2 = \mathbf{A \cdot A} = A^\mu \eta_{\mu\nu} A^\nu$

and intuitively represents (the square of) the length or magnitude of the vector. However, in general, four-vectors can have nonpositive length, contrary to three-dimensional vectors in Euclidean space.

Following are two common choices for the metric tensor in the standard basis (essentially Cartesian coordinates). If orthogonal coordinates are used, there would be scale factors along the diagonal part of the spacelike part of the metric, while for general curvilinear coordinates the entire spacelike part of the metric would have components dependent on the curvilinear basis used.

##### Standard basis, (+−−−) signature

In the (+−−−) metric signature, evaluating the summation over indices gives:

$\mathbf{A} \cdot \mathbf{B} = A^0 B^0 - A^1 B^1 - A^2 B^2 - A^3 B^3$

while in matrix form:

$\mathbf{A \cdot B} = \begin{pmatrix} A^0 & A^1 & A^2 & A^3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} B^0 \\ B^1 \\ B^2 \\ B^3 \end{pmatrix}$

It is a recurring theme in special relativity to take the expression

$\mathbf{A}\cdot\mathbf{B} = A^0 B^0 - A^1 B^1 - A^2 B^2 - A^3 B^3 = C$

in one reference frame, where C is the value of the inner product in this frame, and:

$\mathbf{A}'\cdot\mathbf{B}' = {A'}^0 {B'}^0 - {A'}^1 {B'}^1 - {A'}^2 {B'}^2 - {A'}^3 {B'}^3 = C'$

in another frame, in which C′ is the value of the inner product in this frame. Then since the inner product is an invariant, these must be equal:

$\mathbf{A}\cdot\mathbf{B} = \mathbf{A}'\cdot\mathbf{B}'$

that is:

$C = A^0 B^0 - A^1 B^1 - A^2 B^2 - A^3 B^3 = {A'}^0 {B'}^0 - {A'}^1 {B'}^1 - {A'}^2 {B'}^2 - {A'}^3{B'}^3$

Considering that physical quantities in relativity are four-vectors, this equation has the appearance of a "conservation law", but there is no "conservation" involved. 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. The components of the four-vectors change from one frame to another; A and A′ are connected by a Lorentz transformation, and similarly for B and B′, although the inner products are the same in all frames. Nevertheless, this type of expression is exploited in relativistic calculations on a par with conservation laws, since the magnitudes of components can be determined without explicitly performing any Lorentz transformations. A particular example is with energy and momentum in the energy-momentum relation derived from the four-momentum vector (see also below).

In this signature, the norm of the vector A is:

$\|\mathbf{A}\|^2 = (A^0)^2 - (A^1)^2 - (A^2)^2 - (A^3)^2$

With the signature (+−−−), four-vectors may be classified as either spacelike if ||A|| < 0, timelike if ||A|| > 0, and null vectors if ||A|| = 0.

##### Standard basis, (−+++) signature

Some authors define η with the opposite sign, in which case we have the (−+++) metric signature. Evaluating the summation with this signature:

$\mathbf{A \cdot B} = - A^0 B^0 + A^1 B^1 + A^2 B^2 + A^3 B^3$

while the matrix form is:

$\mathbf{A \cdot B} = \left( \begin{matrix}A^0 & A^1 & A^2 & A^3 \end{matrix} \right) \left( \begin{matrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix}B^0 \\ B^1 \\ B^2 \\ B^3 \end{matrix} \right)$

Note that in this case, in one frame:

$\mathbf{A}\cdot\mathbf{B} = - A^0 B^0 + A^1 B^1 + A^2 B^2 + A^3 B^3 = -C$

while in another:

$\mathbf{A}'\cdot\mathbf{B}' = - {A'}^0 {B'}^0 + {A'}^1 {B'}^1 + {A'}^2 {B'}^2 + {A'}^3 {B'}^3 = -C'$

so that:

$-C = - A^0 B^0 + A^1 B^1 + A^2 B^2 + A^3 B^3 = - {A'}^0 {B'}^0 + {A'}^1 {B'}^1 + {A'}^2 {B'}^2 + {A'}^3{B'}^3$

which is equivalent to the above expression for C in terms of A and B. Either convention will work. With the Minkowski metric defined in the two ways above, the only difference between covariant and contravariant four-vector components are signs, therefore the signs depend on which sign convention is used.

The square of the norm in this signature is:

$\|\mathbf{A}\|^2 = - (A^0)^2 + (A^1)^2 + (A^2)^2 + (A^3)^2$

With the signature (−+++), four-vectors may be classified as either spacelike if ||A|| > 0, timelike if ||A|| < 0, and null vectors if ||A|| = 0.

##### Dual vectors

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

$\mathbf{A \cdot B} = A^*(\mathbf{B}) = A{_\nu}B^{\nu}.$

Here the Aνs are the components of the dual vector A* of A in the dual basis and called the covariant coordinates of A, while the original Aν components are called the contravariant coordinates.

## Four-vector calculus

### Derivatives and differentials

In special relativity (but not general relativity), the derivative of a four-vector with respect to a scalar λ (invariant) is itself a four-vector. It is also useful to take the differential of the four-vector, dA and divide it by the differential of the scalar, :

$\underset{\text{differential}}{d\mathbf{A}} = \underset{\text{derivative}}{\frac{d\mathbf{A}}{d\lambda}} \underset{\text{differential}}{d\lambda}$

where the contravariant components are:

$d\mathbf{A} = (dA^0, dA^1, dA^2, dA^3)$

while the covariant components are:

$d\mathbf{A} = (dA_0, dA_1, dA_2, dA_3)$

In relativistic mechanics, one often takes the differential of a four-vector and divides by the differential in proper time (see below).

## Fundamental four-vectors

### Four-position

A point in Minkowski space is a time and spatial position, called an "event", or sometimes the position 4-vector or 4-position, described in some reference frame by a set of four coordinates:

$\mathbf{X}= \left(ct, \mathbf{r}\right)$

where r is the three-dimensional space position vector. If r is a function of coordinate time t in the same frame, i.e. r = r(t), this corresponds to a sequence of events as t varies. The definition X0 = ct ensures that all the coordinates have the same units (of distance).[6][7][8] 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:

$\Delta \mathbf{X} = \left(c\Delta t, \Delta \mathbf{r} \right)$

The scalar product of the 4-position with itself is;[9]

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

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:

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

defining the differential line element ds and differential proper time increment dτ, but this norm is also:

$\|d\mathbf{X}\|^2 = (cdt)^2 - d\mathbf{r}\cdot d\mathbf{r} \,,$

so that:

$(c d\tau)^2 = (cdt)^2 - d\mathbf{r}\cdot d\mathbf{r} \,.$

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 above differential invariant spacetime interval, then dividing by (cdt)2 to obtain:

$\left(\frac{cd\tau}{cdt}\right)^2 = 1 - \left(\frac{d\mathbf{r}}{cdt}\cdot \frac{d\mathbf{r}}{cdt}\right) = 1 - \frac{\mathbf{u}\cdot\mathbf{u}}{c^2} = \frac{1}{\gamma(\mathbf{u})^2} \,,$

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

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

is the Lorentz factor. This provides a useful relation between the differentials in coordinate time and proper time:

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

This relation can also be found from the time transformation in the Lorentz transformations. Important four-vectors in relativity theory can be defined by dividing by this differential.

Considering that partial derivatives are linear operators, one can form a four-gradient from the partial time derivative /t and the spatial gradient ∇. Using the standard basis, in index and abbreviated notations, the contravariant components are:

\begin{align} \boldsymbol{\partial} & = \left(\frac{\partial }{\partial x_0}, \, -\frac{\partial }{\partial x_1}, \, -\frac{\partial }{\partial x_2}, \, -\frac{\partial }{\partial x_3} \right) \\ & = (\partial^0, \, - \partial^1, \, - \partial^2, \, - \partial^3) \\ & = \mathbf{e}_0\partial^0 - \mathbf{e}_1\partial^1 - \mathbf{e}_2\partial^2 - \mathbf{e}_3\partial^3 \\ & = \mathbf{e}_0\partial^0 - \mathbf{e}_i\partial^i \\ & = \mathbf{e}_\alpha \partial^\alpha \\ & = \left(\frac{1}{c}\frac{\partial}{\partial t} , \, - \nabla \right) \\ & = \mathbf{e}_0\frac{1}{c}\frac{\partial}{\partial t} - \nabla \\ \end{align}

Note the basis vectors are placed in front of the components, to prevent confusion between taking the derivative of the basis vector, or simply indicating the partial derivative is a component of this four-vector. The covariant components are:

\begin{align} \boldsymbol{\partial} & = \left(\frac{\partial }{\partial x^0}, \, \frac{\partial }{\partial x^1}, \, \frac{\partial }{\partial x^2}, \, \frac{\partial }{\partial x^3} \right) \\ & = (\partial_0, \, \partial_1, \, \partial_2, \, \partial_3) \\ & = \mathbf{e}^0\partial_0 + \mathbf{e}^1\partial_1 + \mathbf{e}^2\partial_2 + \mathbf{e}^3\partial_3 \\ & = \mathbf{e}^0\partial_0 + \mathbf{e}^i\partial_i \\ & = \mathbf{e}^\alpha \partial_\alpha \\ & = \left(\frac{1}{c}\frac{\partial}{\partial t} , \, \nabla \right) \\ & = \mathbf{e}^0\frac{1}{c}\frac{\partial}{\partial t} + \nabla \\ \end{align}

Since this is an operator, it doesn't have a "length", but evaluating the inner product of the operator with itself gives another operator:

$\partial^\mu \partial_\mu = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2$

called the D'Alembert operator.

## Kinematics

### Four-velocity

The four-velocity of a particle is defined by:

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

Geometrically U is a tangent vector along the world line of the particle. Using the differential of the 4-position, the magnitude of the 4-velocity can be obtained:

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

in short, the magnitude of the 4-velocity for any all objects is always a fixed constant:

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

The norm is also:

$\|\mathbf{U}\|^2 = {\gamma(\mathbf{u})}^2 \left( c^2 - \mathbf{u}\cdot\mathbf{u} \right) \,,$

so that:

$c^2 = {\gamma(\mathbf{u})}^2 \left( c^2 - \mathbf{u}\cdot\mathbf{u} \right) \,,$

which reduces to the definition the Lorentz factor.

### Four-acceleration

The four-acceleration is given by:

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

where a = du/dt is the coordinate 3-acceleration. Since the magnitude of 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:

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

which is true for all world lines. The geometric meaning of 4-acceleration is the curvature vector of the world line in Minkowski space.

## Dynamics

### Four-momentum

For a massive particle of rest mass (or invariant mass) m, the four-momentum is given by:

$\mathbf{P} = m \mathbf{U} = m\gamma(\mathbf{u})(c, \mathbf{u}) = (E/c, \mathbf{p})$

where the total energy of the moving particle is:

$E = \gamma(\mathbf{u}) mc^2$

and the total relativistic momentum is:

$\mathbf{p} = \gamma(\mathbf{u}) m \mathbf{u}$

Taking the inner product of the four-momentum with itself:

$\|\mathbf{P}\|^2 = P^\mu P_\mu = m^2 U^\mu U_\mu = m^2 c^2$

and also:

$\|\mathbf{P}\|^2 = \frac{E^2}{c^2} - \mathbf{p}\cdot\mathbf{p}$

which leads to the energy–momentum relation:

$E^2 = c^2 \mathbf{p}\cdot\mathbf{p} + (mc^2)^2 \,.$

This last relation is useful relativistic mechanics, essential in relativistic quantum mechanics and relativistic quantum field theory, all with applications to particle physics.

### Four-force

The four-force acting on a particle is defined analogously to the 3-force as the time derivative of 3-momentum in Newton's second law:

$\mathbf{F} = \frac {d \mathbf{P}} {d \tau} = \gamma(\mathbf{u})\left(\frac{1}{c}\frac{dE}{dt},\frac{d\mathbf{p}}{dt}\right) = \gamma(\mathbf{u})(P/c,\mathbf{f})$

where P is the power transferred to move the particle, and f is the 3-force acting on the particle. For a particle of constant invariant mass m, this is equivalent to

$\mathbf{F} = m \mathbf{A} = m\gamma(\mathbf{u})\left( \frac{d{\gamma}(\mathbf{u})}{dt} c, \left(\frac{d{\gamma}(\mathbf{u})}{dt} \mathbf{u} + \gamma(\mathbf{u}) \mathbf{a}\right) \right)$

An invariant derived from the 4-force is:

$\mathbf{F}\cdot\mathbf{U} = F^\mu U_\mu = m A^\mu U_\mu = 0$

from the above result.

## Thermodynamics

### Four-heat flux

The 4-heat flux vector field, is essentially similar to the 3d heat flux vector field q, in the local frame of the fluid:[10]

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

where T is absolute temperature and k is thermal conductivity.

### Four-baryon number flux

The flux of baryons is:[11]

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

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:[12]

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

where s is the entropy per baryon, and T the absolute temperature, in the local rest frame of the fluid.[13]

## Electromagnetism

Examples of four-vectors in electromagnetism include the following.

### Four-current

The electromagnetic 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 ϕ. The four-potential is not uniquely determined, because it depends on a choice of gauge.

## Waves

### Four-frequency

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

$\mathbf{N} = \nu\left(1 , \hat{\mathbf{n}} \right)$

where ν is the frequency of the wave and $\hat{\mathbf{n}}$ is a unit vector in the travel direction of the wave. Now:

$\|\mathbf{N}\| = N^\mu N_\mu = \nu ^2 \left(1 - \hat{\mathbf{n}}\cdot\hat{\mathbf{n}}\right) = 0$

so the 4-frequency is always a null vector.

### Four-wavevector

The quantities reciprocal to time t and space r are the angular frequency ω and wave vector k, respectively. The form the components of the 4-wavevector or wave 4-vector:

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

A wave packet of nearly monochromatic light can be described by:

$\mathbf{K} = \frac{2\pi}{c}\mathbf{N} = \frac{2\pi}{c} \nu(1,\hat{\mathbf{n}}) = \frac{\omega}{c}\left( 1 , \hat{\mathbf{n}} \right) \,.$

For matter waves, the de Broglie relations become one equation:

$\mathbf{P} = \hbar \mathbf{K}\,.$

where ħ is the Planck constant divided by 2π. The square of the norm is:

$\| \mathbf{K} \|^2 = K^\mu K_\mu = \left(\frac{\omega}{c}\right)^2 - \mathbf{k}\cdot\mathbf{k}\,,$

and by the de Broglie relation:

$\| \mathbf{K} \|^2 = \frac{1}{\hbar^2} \| \mathbf{P} \|^2 = \left(\frac{mc}{\hbar}\right)^2 \,,$

we have the matter wave analogue of the energy-momentum relation:

$\left(\frac{\omega}{c}\right)^2 - \mathbf{k}\cdot\mathbf{k} = \left(\frac{mc}{\hbar}\right)^2 \,.$

Note that for massless particles, in which case m = 0, we have:

$\left(\frac{\omega}{c}\right)^2 = \mathbf{k}\cdot\mathbf{k} \,,$

or ||k|| = ω/c. Note this is consistent with the above case; for photons with a 3-wavevector of modulus ω/c, in the direction of wave propagation defined by the unit vector $\hat{\mathbf{n}}$.

## Quantum theory

In quantum mechanics, the 4-probability current or probability 4-current is analogous to the electromagnetic 4-current:[14]

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

where ρ is the probability density function corresponding to the time component, and j is the probability current vector. In non-relativistic quantum mechanics, this current is always well defined because the expressions for density and current are positive definite and can admit a probability interpretation. In relativistic quantum mechanics and quantum field theory, it is not always possible to find a current, particularly when interactions are involved.

Replacing the energy by the energy operator and the momentum by the momentum operator in the four-momentum, one obtains the four-momentum operator, used in relativistic wave equations.

## Other formulations

### Four-vectors in the algebra of physical space

A four-vector A can also be defined in using the Pauli matrices as a basis, again in various equivalent notations:[15]

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

or explicitly:

\begin{align} \mathbf{A} & = A^0\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + A^1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + A^2 \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} + A^3 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ & = \begin{pmatrix} A^0 + A^3 & A^1 -i A^2 \\ A^1 + i A^2 & A^0 - A^3 \end{pmatrix} \end{align}

and in this formulation, the four-vector is represented as a unitary matrix (the matrix transpose and complex conjugate of the matrix leaves it unchanged), rather than a real-valued column or row vector. The determinant of the matrix is the modulus of the four-vector, so the determinant is an invariant:

\begin{align} |\mathbf{A}| & = \begin{vmatrix} A^0 + A^3 & A^1 -i A^2 \\ A^1 + i A^2 & A^0 - A^3 \end{vmatrix} \\ & = (A^0 + A^3)(A^0 - A^3) - (A^1 -i A^2)(A^1 + i A^2) \\ & = (A^0)^2 - (A^1)^2 - (A^2)^2 - (A^3)^2 \end{align}

This idea of using the Pauli matrices as basis vectors is employed in the algebra of physical space, an example of a Clifford algebra.

### Four-vectors in spacetime algebra

In spacetime algebra, another example of Clifford algebra, the gamma matrices can also form a basis. (They are also called the Dirac matrices, owing to their appearance in the Dirac equation). There is more than one way to express the gamma matrices, detailed in that main article.

The Feynman slash notation is a shorthand for a four-vector A contracted with the gamma matrices:

$\mathbf{A}\!\!\!\!/ = A_\alpha \gamma^\alpha = A_0 \gamma^0 + A_1 \gamma^1 + A_2 \gamma^2 + A_3 \gamma^3$

The four-momentum contracted with the gamma matrices is an important case in relativistic quantum mechanics and relativistic quantum field theory. In the Dirac equation and other relativistic wave equations, terms of the form:

$\mathbf{P}\!\!\!\!/ = P_\alpha \gamma^\alpha = P_0 \gamma^0 + P_1 \gamma^1 + P_2 \gamma^2 + P_3 \gamma^3 = \dfrac{E}{c} \gamma^0 + p_x \gamma^1 + p_y \gamma^2 + p_z \gamma^3$

appear, in which the energy E and momentum components (px,py, pz) are replaced by their respective operators.

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