# Lorentz transformation

(Redirected from Lorentz transformations)

In physics, the Lorentz transformation (or transformations) is named after the Dutch physicist Hendrik Lorentz. It was the result of attempts by Lorentz and others to explain how the speed of light was observed to be independent of the reference frame, and to understand the symmetries of the laws of electromagnetism. The Lorentz transformation is in accordance with special relativity, but was derived before special relativity.

The transformations describe how measurements related to events in space and time by two observers, in inertial frames moving at constant velocity with respect to each other, are related. They reflect the fact that observers moving at different velocities may measure different distances, elapsed times, and even different orderings of events. They supersede the Galilean transformation of Newtonian physics, which assumes an absolute space and time (see Galilean relativity). The Galilean transformation is a good approximation only at relative speeds much smaller than the speed of light.

The Lorentz transformation is a linear transformation. It may include a rotation of space; a rotation-free Lorentz transformation is called a Lorentz boost.

In Minkowski space, the Lorentz transformations preserve the spacetime interval between any two events. They describe only the transformations in which the spacetime event at the origin is left fixed, so they can be considered as a hyperbolic rotation of Minkowski space. The more general set of transformations that also includes translations is known as the Poincaré group.

## History

Many physicists, including Woldemar Voigt, George FitzGerald, Joseph Larmor, and Hendrik Lorentz himself had been discussing the physics implied by these equations since 1887.[1] Early in 1889, Oliver Heaviside had shown from Maxwell's equations that the electric field surrounding a spherical distribution of charge should cease to have spherical symmetry once the charge is in motion relative to the ether. FitzGerald then conjectured that Heaviside’s distortion result might be applied to a theory of intermolecular forces. Some months later, FitzGerald published the conjecture that bodies in motion are being contracted, in order to explain the baffling outcome of the 1887 ether-wind experiment of Michelson and Morley. In 1892, Lorentz independently presented the same idea in a more detailed manner, which was subsequently called FitzGerald–Lorentz contraction hypothesis.[2] Their explanation was widely known before 1905.[3]

Lorentz (1892–1904) and Larmor (1897–1900), who believed the luminiferous ether hypothesis, were also seeking the transformation under which Maxwell's equations are invariant when transformed from the ether to a moving frame. They extended the FitzGerald–Lorentz contraction hypothesis and found out that the time coordinate has to be modified as well ("local time"). Henri Poincaré gave a physical interpretation to local time (to first order in v/c) as the consequence of clock synchronization, under the assumption that the speed of light is constant in moving frames.[4] Larmor is credited to have been the first to understand the crucial time dilation property inherent in his equations.[5]

In 1905, Poincaré was the first to recognize that the transformation has the properties of a mathematical group, and named it after Lorentz.[6] Later in the same year Albert Einstein published what is now called special relativity, by deriving the Lorentz transformation under the assumptions of the principle of relativity and the constancy of the speed of light in any inertial reference frame, and by abandoning the mechanical aether.[7]

## Derivation

From Einstein's second postulate of relativity follows immediately

$c^2(t_2 - t_1)^2 - (x_2 - x_1)^2 - (y_2 - y_1)^2 - (z_2 - z_1)^2 = 0$

in all reference frames for events connected by light signals. An event is something which happens at a certain place and certain time, and in any inertial frame can be defined by a time coordinate t and some set of position coordinates, here Cartesian coordinates x, y, z are used. The quantity on the left is called the spacetime interval. The interval between any two reference frames is in fact invariant, as is shown here (where one can also find several more explicit derivations than presently given). The transformation sought after thus must possess the property that

$c^2(t_2 - t_1)^2 - (x_2 - x_1)^2 - (y_2 - y_1)^2 - (z_2 - z_1)^2 = c^2(t_2' - t_1')^2 - (x_2' - x_1')^2 - (y_2' - y_1')^2 - (z_2' - z_1')^2.$

where t, x, y, z are the spacetime coordinates used to define events in one frame, and t′, x′, y′, z are the coordinates in another frame. Notice immediately in general that

$t_2' - t_1' \neq t_2 - t_1 \,,$

and the possibility of time progressing at different rates, depending on which frame one measures from, is allowed. This is in stark contrast to Newtonian mechanics and Galilean relativity in which time is absolute and progresses at the same rate for all observers. Now one observes that a linear solution to the simpler problem

$c^2t^2 - x^2 - y^2 - z^2 = c^2t'^2 - x'^2 - y'^2 - z'^2$

will solve the general problem too. This is just a matter of look-up in the theory of classical groups that preserve bilinear forms of various signature. The Lorentz transformation is thus an element of the group O(3, 1) or, for those that prefer the other metric signature, O(1, 3).

The relations between the primed and unprimed spacetime coordinates are the Lorentz transformations, each coordinate in one frame is a function of all the coordinates in the other frame,

$t' = t'(t,x,y,z) \,,$
$x' = x'(t,x,y,z) \,,$
$y' = y'(t,x,y,z) \,,$
$z' = z'(t,x,y,z) \,,$

and the inverse functions are the inverse transformation,

$t = t(t',x',y',z') \,,$
$x = x(t',x',y',z') \,,$
$y = y(t',x',y',z') \,,$
$z = z(t',x',y',z') \,.$

In each case, depending on how the frames move relative to each other (e.g. moving in straight lines at constant velocity, accelerating in curved paths, rotating, etc.) other parameters will enter the transformation equations. For relative motion with constant velocity (called a "boost") and rotations through an angle about an axis (but not rotational motion with constant angular velocity), the transformations are linear transformations, and the transformations of the coordinates can be expressed in vector-matrix form using a transformation matrix. The connections between the matrix elements and physical quantities are made in the next sections.

## Rotations of frames (static)

Consider the simpler case where the square of the magnitude of the position vector r = (x, y, z) as measured in frame F, and r′ = (x′, y′, z′) as measured in frame F, are invariant,

$|\mathbf{r}|^2 = |\mathbf{r}'|^2 \,,$

or

$x^2 +y^2 + z^2 = {x'}^2 + {z'}^2 + {z'}^2\,,$

and the times are equal,

$t'= t \,.$

The transformation between the coordinates are related by rotations, so one frame is simply tilted relative to the other, and there is no relative motion.

### Rotation about the z axis (in the xy plane)

If a frame F with coordinates x′, y′, z′, t has its z axis coincident with the z axis of another frame F with coordinates x, y, z, t, each aligned in the same direction, and is rotated about the common zz axis (or equivalently in the coincident xy and xy planes) through an angle θ anticlockwise, the transformations of coordinates are

 Lorentz rotation (z axis) $t' = t \,,$ $z' = z \,,$ $x' = x\cos\theta - y\sin\theta \,,$ $y' = x\sin\theta + y\cos\theta \,,$

or in matrix form (see matrix product for how to multiply matrices)

$\begin{bmatrix}ct' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix}ct \\ x \\ y \\ z \end{bmatrix}$

(The factor of the speed of light c is included for consistency with other Lorentz transformations, it does not need to be included here).

The transformation matrix is a function of the angle of rotation, which parametrizes the rotation. It can be partitioned into block matrix form

$\begin{bmatrix} 1 & 0 \\ 0 & \mathbf{R} \end{bmatrix}\,, \quad \mathbf{R} = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\,,$

making apparent which parts of the transformation matrix act on the time coordinate and spatial coordinates. The 3d z-axis rotation matrix is a submatrix of the 4d rotation matrix.

The inverse transformations are easy to obtain, since frame F′ will measure the rotation to be through the same angle but in the opposite sense, simply negating the angle θ → −θ while leaving the direction of the rotation axis unchanged, and exchanging unprimed and primed coordinates, gives the inverse transformation

 Inverse Lorentz rotation (z axis) $t = t' \,,$ $z = z' \,,$ $x = x'\cos\theta + y'\sin\theta \,,$ $y = -x'\sin\theta + y'\cos\theta \,,$

in matrix form

$\begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & \sin \theta & 0 \\ 0 & -\sin \theta & \cos \theta & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix}ct' \\ x' \\ y' \\ z' \end{bmatrix}$

where the transformation matrix here is the matrix inverse of the original one.

### Rotations about the x and y axes

For a similar setup but with rotations about the common xx axes, we have

$\begin{bmatrix}ct' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos \theta & -\sin \theta \\ 0 & 0 & \sin \theta & \cos \theta \\ \end{bmatrix} \begin{bmatrix}ct \\ x \\ y \\ z \end{bmatrix} \,,$

as for the common yy axes

$\begin{bmatrix}ct' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & 0 & \sin \theta \\ 0 & 0 & 1 & 0 \\ 0 & -\sin \theta & 0 & \cos \theta \\ \end{bmatrix} \begin{bmatrix}ct \\ x \\ y \\ z \end{bmatrix} \,.$

### Rotations about an axis in any direction

Decomposition of r into parallel and perpendicular components, using geometric properties of the dot and cross products. Expanding the vector triple product verifies the link between the orthogonal components. Rotation of r anticlockwise through angle θ about a unit vector a defining an axis of rotation. The angle of rotation and axis vector can be combined into an "angular vector" θ = θa.

For a rotation about a unit vector a which defines the rotation axes, through an angle θ anticlockwise, the full transformation of position is less obvious. Decompose the position vectors r and its rotated counterpart r into components parallel and perpendicular to a, so in the original and rotated frames respectively

$\mathbf{r}=\mathbf{r}_\perp+\mathbf{r}_\|\,,\quad \mathbf{r}' = \mathbf{r}_\perp' + \mathbf{r}_\|' \,,$

where ‖ means "parallel" to a and ⊥ means "perpendicular" to a. The parallel component will not change magnitude or direction,

$\mathbf{r}_\parallel ' = \mathbf{r}_\parallel \,,$

only the component perpendicular to the axis will rotate according to (see diagram)

$\mathbf{r}_\perp ' = \cos\theta \mathbf{r}_\perp + \sin\theta \mathbf{a}\times\mathbf{r}_\perp \,.$

Inserting the parallel and rotated perpendicular components into r and simplifying, the full set of transformation equations are therefore

$t' = t$
$\mathbf{r}' = \mathbf{r} + \mathbf{a}\times(\mathbf{a}\times\mathbf{r})(1-\cos\theta) + \mathbf{a}\times\mathbf{r}\sin\theta \,.$

To translate this into matrix language, notice the cross product term is equivalent to the matrix product

$\mathbf{a}\times\mathbf{r} \leftrightarrow \begin{bmatrix} a_y z - a_z y \\ a_z x - a_x z \\ a_x y - a_y x \end{bmatrix} = \begin{bmatrix} 0 & -a_z & a_y \\ a_z & 0 & -a_x \\ -a_y & a_x & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \,,$

so that

$\mathbf{a}\times(\mathbf{a}\times\mathbf{r}) \leftrightarrow \begin{bmatrix} 0 & -a_z & a_y \\ a_z & 0 & -a_x \\ -a_y & a_x & 0 \end{bmatrix}^2 \begin{bmatrix} x \\ y \\ z \end{bmatrix} \,.$

Continuing to organize the components of r into column vectors, denoting the square antisymmetric cross product matrix by A, and I for the 3×3 identity matrix, the rotation of the position vector is Rodrigues' rotation formula

$\mathbf{r}' = \mathbf{R}\mathbf{r} \,,\quad \mathbf{R} = \mathbf{I} + \mathbf{A}^2(1-\cos\theta) + \mathbf{A}\sin\theta \,.$

The matrix A has an origin from group generators, as shown later. In block matrix form

$\begin{bmatrix} ct' \\ \mathbf{r}' \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \mathbf{R} \end{bmatrix} \begin{bmatrix} 1 \\ \mathbf{r}\end{bmatrix}\,,$

in full,

$\begin{bmatrix}ct' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta +a_x^2 \left(1-\cos \theta\right) & a_x a_y \left(1-\cos \theta\right) - a_z \sin \theta & a_x a_z \left(1-\cos \theta\right) + a_y \sin \theta \\ 0 & a_y a_x \left(1-\cos \theta\right) + a_z \sin \theta & \cos \theta + a_y^2\left(1-\cos \theta\right) & a_y a_z \left(1-\cos \theta\right) - a_x \sin \theta \\ 0 & a_z a_x \left(1-\cos \theta\right) - a_y \sin \theta & a_z a_y \left(1-\cos \theta\right) + a_x \sin \theta & \cos \theta + a_z^2\left(1-\cos \theta\right) \end{bmatrix} \begin{bmatrix}ct \\ x \\ y \\ z \end{bmatrix} \,.$

Notice the matrix is neither symmetric nor antisymmetric. The inverse transformations are as always found by negating the angle and switching unprimed and primed quantities,

$t' = t$
$\mathbf{r} = \mathbf{r}' + \mathbf{a}\times(\mathbf{a}\times\mathbf{r}')(1-\cos\theta) - \mathbf{a}\times\mathbf{r}'\sin\theta \,.$

## Boosts (constant relative velocity)

Consider two observers O and O, each using their own Cartesian coordinate system to measure space and time intervals. O uses (t, x, y, z) and O uses (t′, x′, y′, z′). Assume further that the coordinate systems are oriented so that, in 3 dimensions, the x-axis and the x-axis are collinear, the y-axis is parallel to the y-axis, and the z-axis parallel to the z-axis. The relative velocity between the two observers is v along the common x-axis; O measures O′ to move at velocity v along the coincident xx′ axes, while O′ measures O to move at velocity v along the coincident xx′ axes. Also assume that the origins of both coordinate systems are the same, that is, coincident times and positions. If all these hold, then the coordinate systems are said to be in standard configuration. The formulae below give the Lorentz transformations (boosts) for this configuration.

The inverse of a Lorentz transformation relates the coordinates the other way round; from the coordinates O measures (t′, x′, y′, z′) to the coordinates O measures (t, x, y, z), so t, x, y, z are in terms of t′, x′, y′, z. The mathematical form is nearly identical to the original transformation; the only difference is the negation of the uniform relative velocity (from v to v), and exchange of primed and unprimed quantities, because O moves at velocity v relative to O, and equivalently, O moves at velocity v relative to O. This symmetry makes it effortless to find the inverse transformation (carrying out the exchange and negation saves a lot of rote algebra), although more fundamentally; it highlights that all physical laws should remain unchanged under a Lorentz transformation.

Below, the Lorentz transformations are called "boosts" in the stated directions.

### Boost in the x-direction

The spacetime coordinates of an event, as measured by each observer in their inertial reference frame (in standard configuration) are shown in the speech bubbles.
Top: frame F moves at velocity v along the x-axis of frame F.
Bottom: frame F moves at velocity −v along the x-axis of frame F.[8]

These are the simplest forms. The Lorentz transformation for frames in standard configuration can be shown to be[9]

 Lorentz boost (x direction) \begin{align} t' &= \gamma \left( t - \frac{v x}{c^2} \right) \\ x' &= \gamma \left( x - v t \right)\\ y' &= y \\ z' &= z \end{align}

where v is the relative velocity between frames in the x-direction, c is the speed of light,

$\gamma = \frac{1}{ \sqrt{1 - { \beta^2}}}$

(lowercase gamma) is the Lorentz factor, and β = v/c (lowercase beta) is the velocity parameter. The use of β and γ is standard throughout the literature. The magnitude of relative velocity v cannot equal or exceed c, so that 0 ≤ v < c. Correspondingly β cannot equal or exceed 1, so that 0 ≤ β < 1. If v = c, the transformations are undefined because γ is infinite. For v > c, the Lorentz factor is a complex number, and the transformations make no sense because they are complex-valued. The space and time coordinates are measurable quantities and numerically must be real numbers, not complex.

Since the above is set of coupled linear equations, in other words a linear transformation, they can be written in a single matrix equation (see matrix product for how to multiply matrices)

$\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma & -\beta \gamma & 0 & 0\\ -\beta \gamma & \gamma & 0 & 0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} ,$

where the coordinates are separated off into column vectors. The transformation matrix depends on the relative velocity between the frames; v is the parameter of the transformation.

According to the principle of relativity, there is no privileged frame of reference, so the inverse transformations frame F to frame F must be given by simply negating v and exchanging primed and unprimed variables

 Inverse Lorentz boost (x direction) \begin{align} t &= \gamma \left( t' + \frac{v x'}{c^2} \right) \\ x &= \gamma \left( x' + v t' \right)\\ y &= y' \\ z &= z', \end{align}

where the value of γ remains unchanged. These equations can also be obtained by algebraically solving the original set of equations for the variables t, x, y, z. Arranging into matrix form,

$\begin{bmatrix} c t \\ x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \gamma & \beta \gamma & 0 & 0\\ \beta \gamma & \gamma & 0 & 0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} \,,$

the matrix in this equation is the inverse matrix of the original transformation matrix. So instead of solving two or more equations algebraically, the original matrix equation can be inverted to find the inverse transformation.

Orthogonality and rotation of coordinate systems compared between left: Euclidean space through circular angle ϕ, right: in Minkowski spacetime through hyperbolic angle ϕ (red lines labelled c denote the worldlines of a light signal, a vector is orthogonal to itself if it lies on this line).[10]
The momentarily co-moving inertial frames along the world line of a rapidly accelerating observer (center). The vertical direction indicates time, while the horizontal indicates distance, the dashed line is the spacetime trajectory ("world line") of the observer. The small dots are specific events in spacetime. If one imagines these events to be the flashing of a light, then the events that pass the two diagonal lines in the bottom half of the image (the past light cone of the observer in the origin) are the events visible to the observer. The slope of the world line (deviation from being vertical) gives the relative velocity to the observer. Note how the momentarily co-moving inertial frame changes when the observer accelerates.

The Lorentz transformations can be derived in a way that resembles circular rotations in 3d space using the hyperbolic functions. For the boost in the x direction, the results are

 Lorentz boost (x direction with rapidity ϕ) \begin{align} ct' &= ct \cosh\phi - x \sinh\phi \\ x' &= x \cosh\phi - ct \sinh\phi \\ y' &= y \\ z' &= z \end{align}

where ϕ is a parameter called rapidity. In matrix form

$\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \cosh\phi &-\sinh\phi & 0 & 0 \\ -\sinh\phi & \cosh\phi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} c t \\ x \\ y \\ z \end{bmatrix}\,.$

Given the strong resemblance to the rotation matrices about Cartesian axes, the Lorentz transformation can be thought of as a hyperbolic rotation of coordinates in Minkowski space, where the parameter ϕ represents the hyperbolic angle of rotation. This transformation can be illustrated with a Minkowski diagram as displayed above.

By comparison between the Lorentz transformations in terms of the relative velocity and rapidity, or otherwise shown more rigorously, the connections between β and ϕ are

$\gamma = \cosh\phi \,,$
$\beta \gamma = \sinh\phi \,.$
$\beta = \tanh\phi \,,$

The hyperbolic expression for β can be intuitively seen from the Minkowski spacetime diagrams above, since the velocity of the "moving" frame is related to the slope of the world line as measured from the other "stationary" frame. Taking the inverse hyperbolic tangent gives the rapidity

$\phi = \tanh^{-1}\beta \,.$

Notice the ranges 0 ≤ β < 1 and 0 ≤ ϕ < ∞.

Finding the inverse transformation is easy with the rapidity; as well as exchanging primed and unprimed quantities, simply negating rapidity ϕ → −ϕ is equivalent to negating the relative velocity, which follows from the relation between ϕ and β. Therefore

 Inverse Lorentz boost (x direction with rapidity ϕ) \begin{align} ct & = ct' \cosh\phi + x' \sinh\phi \\ x &= x' \cosh\phi + ct' \sinh\phi \\ y &= y' \\ z &= z' \end{align}

which in matrix form is

$\begin{bmatrix} c t \\ x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \cosh\phi & \sinh\phi & 0 & 0 \\ \sinh\phi & \cosh\phi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix}\,.$

As always, the matrix in the inverse transformation is the inverse of the original transformation matrix.

Above, the transformations have been applied to the position four-vector X = (ct, x, y, z). A Lorentz transformation changes this to X′ = (ct′, x′, y′, z′). The components of each are different and related by the transformation equations, but their norms or magnitudes are equal,

$ct^2 - x^2 - y^2 - z^2 = c{t'}^2 - {x'}^2 - {y'}^2 - {z'}^2\,.$

### Boost in the y or z directions

The above collection of equations apply only for a boost in the x-direction. The standard configuration works equally well in the y or z directions instead of x, and so the results are similar.

For a boost in the y-direction with velocity v

$\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma&0&-\beta \gamma&0\\ 0&1&0&0\\ -\beta \gamma&0&\gamma&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} ,$

likewise for the z-direction

$\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma&0&0&-\beta \gamma\\ 0&1&0&0\\ 0&0&1&0\\ -\beta \gamma&0&0&\gamma\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \,.$

### Boost in any direction

Boost in an arbitrary direction.

For a boost in an arbitrary direction with relative velocity vector v = cβ which has magnitude v = , O observes O′ to move in direction v in the F coordinate frame, while O′ observes O to move in direction v in the F′ coordinate frame. The coordinate axes of each frame are still parallel and orthogonal. The magnitude of relative velocity |v| = v cannot equal or exceed c, so that 0 ≤ v < c. Correspondingly |β| = β cannot equal or exceed 1, so that 0 ≤ β < 1.

#### Orthogonal decomposition and reconstruction

For the transformations in the x, y, and z directions, the coordinates perpendicular to the relative motion remain unchanged, while those parallel to the relative motion do change along with the time coordinate. For this reason, it is convenient to decompose the spatial position vector r = (x, y, z) as measured in F, and r′ = (x′, y′, z′) as measured in F′, each into components perpendicular and parallel to v = (vx, vy, vz),

$\mathbf{r}=\mathbf{r}_\perp+\mathbf{r}_\|\,,\quad \mathbf{r}' = \mathbf{r}_\perp' + \mathbf{r}_\|' \,,$

where ‖ means "parallel" to v and ⊥ means "perpendicular" to v. The transition from the boost in the x direction to a boost in any direction follows by making the identifications

$\mathbf{v} = v\mathbf{e}_x \,,\quad \mathbf{r}_\parallel = x\mathbf{e}_x \,,\quad \mathbf{r}_\perp = y\mathbf{e}_y + z\mathbf{e}_z \,,$

where ex, ey, ez are the Cartesian basis vectors, a set of mutually perpendicular unit vectors along their indicated directions. Then the Lorentz transformations take the form

$t' = \gamma \left(t - \frac{\mathbf{r}_\parallel \cdot \mathbf{v}}{c^{2}} \right)$
$\mathbf{r}_\|' = \gamma (\mathbf{r}_\| - \mathbf{v} t)$
$\mathbf{r}_\perp' = \mathbf{r}_\perp$

where • indicates the dot product. (Starting from the x-boost is for concreteness, the boosts in the y or z directions could be used to obtain the same result). These equations are vector equations and therefore true in any direction. However, they are in terms of the parallel and perpendicular components of r as measured in F, instead of the entire position vector itself.

Diagram of the unit vector n, relative velocity v, and parallel and perpendicular components of r. The procedure for r is identical.

To find the parallel and perpendicular components, it is useful to introduce a unit vector in the direction of v, which can be expressed as n = v/v = β/β. This has the advantage of simplifying equations and makes alternative parametrizations easier. Then the relative velocity is v = vn with magnitude v and direction n.

Substituting r, r, and v = vn into the entire position vector in F′ gives the desired transformation of the full position vector,

$\mathbf{r}' = \mathbf{r}_\perp + \gamma (\mathbf{r}_\| - \mathbf{v} t) = \mathbf{r} + (\gamma-1)(\mathbf{r}\cdot\mathbf{n})\mathbf{n} - \gamma t v\mathbf{n} \,.$

Using the equivalence between rv and rvn, the transformation of time immediately takes the form

$t' = \gamma \left(t - \frac{\mathbf{r} \cdot v\mathbf{n}}{c^{2}} \right) \,.$

The transformation of time is one equation, and the transformation of position is three equations for the three components of the position vectors. There are four equations for the linear transformation of the four spacetime coordinates. There are three numbers which define the Lorentz boost in any direction, one for the magnitude v and two for the direction n,[nb 1] or in Cartesian components the three components of the relative velocity vector v = (vx, vy, vz). In summary

 Lorentz boost (in direction n) $t' = \gamma \left(t - \frac{\mathbf{r} \cdot v\mathbf{n}}{c^2} \right) \,,$ $\mathbf{r}' = \mathbf{r} + (\gamma-1)(\mathbf{r}\cdot\mathbf{n})\mathbf{n} - \gamma t v\mathbf{n} \,.$

In the position transformation, the position vector can be neatly factorized using the dot and dyadic products of vectors. Although dyadic tensors are an archaic formalism not often used in this context, the notation is well-suited to the equations here, which are

$t' = \gamma t - \gamma\frac{v\mathbf{n} \cdot \mathbf{r}}{c^2} \,,$
$\mathbf{r}' = - \gamma t v\mathbf{n} + ( \mathbf{I} + (\gamma-1)\mathbf{n}\mathbf{n})\cdot\mathbf{r} \,.$

where I is the unit dyadic and nn is the dyadic product of n with itself.[nb 2] The identities I · r = r and nn · r = n(n · r) follow from the definitions of the dot and dyadic products.

The full set of equations are

$ct' = \gamma c t - \gamma \frac{v n_x}{c}x - \gamma \frac{v n_y}{c}y - \gamma \frac{v n_z}{c} z \,.$
$x' = - \gamma \frac{v n_x}{c} c t + (1 + (\gamma-1) n_x^2 )x + (\gamma-1) n_x n_y y + (\gamma-1) n_x n_z z \,,$
$y' = - \gamma \frac{v n_y}{c} ct + (\gamma-1) n_y n_x x + (1 + (\gamma-1) n_y^2 ) y + (\gamma-1) n_y n_z z \,,$
$z' = - \gamma \frac{v n_z}{c} ct + (\gamma-1) n_z n_x x + (\gamma-1) n_z n_y y + (1 + (\gamma-1) n_z^2 ) z \,.$

Collecting these four transformed coordinates into a single matrix equation, the Lorentz boost reads

$\begin{bmatrix} c\,t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma&-\gamma\,\beta n_x&-\gamma\,\beta n_y&-\gamma\,\beta n_z\\ -\gamma\,\beta n_x&1+(\gamma-1)n_x^2&(\gamma-1)n_x n_y&(\gamma-1)n_x n_z\\ -\gamma\,\beta n_y&(\gamma-1)n_y n_x&1+(\gamma-1)n_y^2&(\gamma-1)n_y n_z\\ -\gamma\,\beta n_z&(\gamma-1)n_z n_x&(\gamma-1)n_z n_y&1+(\gamma-1)n_z^2\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix}\,.$

This can be compressed into block matrix form as

$\begin{bmatrix} c t' \\ \mathbf{r}' \end{bmatrix} = \begin{bmatrix} \gamma & - \gamma \beta\mathbf{n}^\mathrm{T} \\ -\gamma\beta\mathbf{n} & \mathbf{I} + (\gamma-1) \mathbf{n}\mathbf{n}^\mathrm{T} \\ \end{bmatrix} \begin{bmatrix} c t \\ \mathbf{r} \end{bmatrix}\,$

where I is the 3×3 identity matrix, and here the components of n = β/β are arranged into a column vector, the transpose (indicated by T) of which is a row vector,

$\mathbf{n} = \begin{bmatrix} n_x \\ n_y \\ n_z \\ \end{bmatrix} \,, \quad \mathbf{n}^\mathrm{T} = \begin{bmatrix} n_x & n_y & n_z \end{bmatrix} \,.$

The position vectors r and r are also column vectors.

This transformation is only a "boost," i.e., a transformation between two frames whose x, y, and z axis are parallel and whose spacetime origins coincide. The most general proper Lorentz transformation also contains a rotation of the three axes, because the composition of two boosts is not a pure boost but is a boost followed by a rotation. The rotation gives rise to Thomas precession. The boost is given by a symmetric matrix, but the general Lorentz transformation matrix need not be symmetric.

The inverse transformations are easy to obtain, as always exchange primed for unprimed indices and negate the relative velocity (which is relative motion in the opposite direction), v → −v, which also amounts to simply negating the unit vector n → −n since the magnitude v is always positive,

 Inverse Lorentz boost (in direction n) $t = \gamma \left(t' + \frac{\mathbf{r}' \cdot v\mathbf{n}}{c^{2}} \right) \,,$ $\mathbf{r} = \mathbf{r}' + (\gamma-1)(\mathbf{r}'\cdot\mathbf{n})\mathbf{n} + \gamma t' v\mathbf{n} \,.$

Using the same unit vector n = v/v = β/β, the vector form of the rapidity relations are[11]

$\gamma = \cosh\phi \,,$
$\boldsymbol{\beta} = \beta \mathbf{n} = \mathbf{n} \tanh\phi \, ,$
$\gamma\boldsymbol{\beta} = \gamma\beta\mathbf{n} = \mathbf{n} \sinh\phi \, ,$
$\boldsymbol{\phi} = \phi\mathbf{n} = \mathbf{n}\tanh^{-1}\beta \,.$

The magnitude of the "rapidity vector" ϕ is the rapidity scalar |ϕ| = ϕ. The vector ϕ will be used more later on.

Substituting these into the vector transformation of r and t gives the transformation parametrized by the direction of relative motion (two independent numbers) and the rapidity ϕ, in all three numbers,

 Lorentz boost (in direction n with rapidity ϕ) $ct' = ct\cosh \phi - \mathbf{r}\cdot\mathbf{n}\sinh\phi$ $\mathbf{r}' = \mathbf{r} + \mathbf{n}(\mathbf{r}\cdot\mathbf{n})(\cosh\phi - 1) - ct\mathbf{n}\sinh \phi$

The transformation of position has a circular rotation counterpart in ordinary 3d space.

The full matrix transformation is

$\begin{bmatrix} c\,t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \cosh\phi & -n_x\sinh\phi & -n_y\sinh\phi & -n_z\sinh\phi\\ -n_x\sinh\phi & 1+(\cosh\phi-1)n_x^2&(\cosh\phi-1)n_x n_y&(\cosh\phi-1)n_x n_z\\ -n_y\sinh\phi & (\cosh\phi-1)n_y n_x&1+(\cosh\phi-1)n_y^2&(\cosh\phi-1)n_y n_z\\ -n_z\sinh\phi & (\cosh\phi-1)n_z n_x&(\cosh\phi-1)n_z n_y&1+(\cosh\phi-1)n_z^2\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix}\,.$

which can be systematically and compactly arranged into the block matrix form

$\begin{bmatrix} c t' \\ \mathbf{r}' \end{bmatrix} = \begin{bmatrix} \cosh\phi & - \mathbf{n}^\mathrm{T} \sinh\phi \\ -\mathbf{n}\sinh\phi & \mathbf{I} + (\cosh\phi-1)\mathbf{n}\mathbf{n}^\mathrm{T} \\ \end{bmatrix} \begin{bmatrix} c t \\ \mathbf{r} \end{bmatrix} \,,$

The inverse transformations are

 Inverse Lorentz boost (in direction n with rapidity ϕ) $ct' = ct\cosh \phi + \mathbf{r}\cdot\mathbf{n}\sinh\phi$ $\mathbf{r}' = \mathbf{r} + \mathbf{n}(\mathbf{r}\cdot\mathbf{n})(\cosh\phi - 1) + ct\mathbf{n}\sinh \phi$

#### Rotation matrices

Another way to obtain the boost in an arbitrary direction is to rotate the coordinates into a boost along a direction for which the Lorentz transformation is simple and known, then perform that Lorentz transformation, then rotate back; summarized by[12]

$B' = R_1 B R_2$

where

$R_1 = \begin{bmatrix} 1 & 0 \\ 0 & \mathbf{R}_1 \end{bmatrix}\,,\quad R_2 = \begin{bmatrix} 1 & 0 \\ 0 & \mathbf{R}_2 \end{bmatrix} \,,$

in which R1 and R2 are 3d rotation matrices on the spatial coordinates only, leaving the time components unchanged. For example, B could be the Lorentz boost along any of the x, y, or z directions.

## Boost and rotation matrices

In the matrix form of the Lorentz transformations, the components of the four positions are arranged into column vectors, and the Lorentz transformation denoted Λ (Greek capital Lambda) can always be compactly written as a single matrix equation of the form

$X' = \Lambda X \,.$

All "pure" Lorentz transformations take this form, those which do not include additional displacement in spacetime (this more general case is detailed later). The six matrices for rotations or boosts about or along the Cartesian axes are of particular importance. In summary they all are

Cartesian axis Rotation Boost
x $R (\mathbf{e}_x,\theta) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos \theta & -\sin \theta \\ 0 & 0 & \sin \theta & \cos \theta \\ \end{bmatrix}$ $B (\mathbf{e}_x,\phi) = \begin{bmatrix} \cosh\phi &-\sinh\phi & 0 & 0 \\ -\sinh\phi & \cosh\phi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$
y $R (\mathbf{e}_y,\theta) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & 0 & \sin \theta \\ 0 & 0 & 1 & 0 \\ 0 & -\sin \theta & 0 & \cos \theta \\ \end{bmatrix}$ $B (\mathbf{e}_y,\phi) = \begin{bmatrix} \cosh\phi & 0 & -\sinh\phi & 0 \\ 0 & 1 & 0 & 0 \\ -\sinh\phi & 0 & \cosh\phi & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$
z $R (\mathbf{e}_z,\theta) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$ $B (\mathbf{e}_z,\phi) = \begin{bmatrix} \cosh\phi & 0 & 0 & -\sinh\phi \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\sinh\phi & 0 & 0 & \cosh\phi \\ \end{bmatrix}$

The general rotation matrix is (in block form for compactness)

$R(\mathbf{a},\theta) = \begin{bmatrix} 1 & 0 \\ 0 & \mathbf{I} + \mathbf{A}^2(1-\cos\theta) + \mathbf{A}\sin\theta \end{bmatrix} \,,$

and general boost matrix is

$B(\mathbf{n},\phi) = \begin{bmatrix} \cosh\phi & - \mathbf{n}^\mathrm{T} \sinh\phi \\ -\mathbf{n}\sinh\phi & \mathbf{I} + (\cosh\phi-1)\mathbf{n}\mathbf{n}^\mathrm{T} \\ \end{bmatrix} \,.$

Successive transformations are applied on the left. Two rotations are another rotation

$R_2(\mathbf{a}_2,\theta_2) R_1(\mathbf{a}_1,\theta_1) = R_3(\mathbf{a}_3,\theta_3)$

but they are not commutative unless the rotations share the same axis,

$R_2(\mathbf{a},\theta_2) R_1(\mathbf{a},\theta_1) = R_1(\mathbf{a},\theta_1) R_2(\mathbf{a},\theta_2) = R(\mathbf{a},\theta_1+\theta_2) \,.$

Two boosts along different directions (not collinear with each other) form a boost followed by a rotation, instead of a another single boost,

$B_2(\mathbf{n}_2,\phi_2) B_1(\mathbf{n}_1,\phi_1) = B(\mathbf{n},\phi)R(\mathbf{a},\theta) \,,$

however two boosts along the same direction form a boost along that direction without rotation, and are commutative,

$B_2(\mathbf{n},\phi_2) B_1(\mathbf{n},\phi_1) = B_1(\mathbf{n},\phi_1) B_2(\mathbf{n},\phi_2) = B(\mathbf{n},\phi_1+\phi_2) \,,$

The general inverse rotation matrix is

$R (\mathbf{a} ,\theta)^{-1} = R (\mathbf{a} ,-\theta) = R (-\mathbf{a} , \theta) \,.$

similarly the general inverse boost matrix is

$B (\mathbf{n} ,\phi)^{-1} = R (\mathbf{n} , -\phi) = R (-\mathbf{n} , \phi) \,.$

The most general Lorentz transformation Λ is a boost and rotation, either can be performed before the other, but the results are different since the boost and rotation matrices do not commute. To show this explicitly, the boost followed by a rotation is

\begin{align}\Lambda_{RB} & = R(\mathbf{a},\theta)B(\mathbf{n},\phi) \\ & = \begin{bmatrix} 1 & 0 \\ 0 & \mathbf{R} \end{bmatrix} \begin{bmatrix} \cosh\phi & - \mathbf{n}^\mathrm{T} \sinh\phi \\ -\mathbf{n}\sinh\phi & \mathbf{I} + (\cosh\phi-1)\mathbf{n}\mathbf{n}^\mathrm{T} \\ \end{bmatrix} \\ & = \begin{bmatrix}\cosh\phi & -\mathbf{n}^{\mathrm{T}}\sinh\phi\\ -\mathbf{R}\mathbf{n}\sinh\phi & \mathbf{R}(\mathbf{I}+(\cosh\phi-1)\mathbf{n}\mathbf{n}^{\mathrm{T}}) \end{bmatrix} \end{align}

while the rotation followed by a boost is

\begin{align} \Lambda_{BR} & = B(\mathbf{n},\phi)R(\mathbf{a},\theta) \\ & = \begin{bmatrix} \cosh\phi & - \mathbf{n}^\mathrm{T} \sinh\phi \\ -\mathbf{n}\sinh\phi & \mathbf{I} + (\cosh\phi-1)\mathbf{n}\mathbf{n}^\mathrm{T} \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & \mathbf{R} \end{bmatrix} \\ & =\begin{bmatrix} \cosh\phi & -\mathbf{n}^\mathrm{T}\mathbf{R}\sinh\phi\\ -\mathbf{n}\sinh\phi & (\mathbf{I}+(\cosh\phi-1)\mathbf{n}\mathbf{n}^\mathrm{T})\mathbf{R} \end{bmatrix} \end{align}

where again R = I + A2(1 − cosθ) + Asinθ is the general rotation matrix. It follows in general

$\Lambda_{RB} \neq \Lambda_{BR} \,.$

One important aspect of the boost and rotation matrices is they form a group $\mathcal{L}$, since

• the operation of composition can be defined (here matrix multiplication),
• boosts and rotations are Lorentz transformations, the product of any two (two rotations, two boosts, one rotation and boost) is also a Lorentz transformation, so the set of these matrices is closed under this operation of composition,
$R(\mathbf{a},\theta) \in \mathcal{L} \,, \quad B(\mathbf{n},\phi) \in \mathcal{L} \,,$
$R(\mathbf{a},\theta)B(\mathbf{n},\phi) \in \mathcal{L} \,, \quad B(\mathbf{n},\phi)R(\mathbf{a},\theta) \in \mathcal{L} \,,$
$R(\mathbf{a},0) = B(\mathbf{n},0) = \mathbf{I} \in \mathcal{L} \,,$
• there are inverse elements (those which correspond to the inverse Lorentz transformations, i.e. a rotation which "undoes" another rotation, likewise for boosts),
$R(\mathbf{a},-\theta)R(\mathbf{a},\theta) = \mathbf{I} \,, \quad R(\mathbf{a},-\theta) \in \mathcal{L} \,,$
$B(\mathbf{n},-\phi)B(\mathbf{n},\phi) = \mathbf{I} \,, \quad B(\mathbf{n},-\phi) \in \mathcal{L} \,,$
• given three transformations, each of which may be a rotation and/or boost, the composition is associative, for example
$R_2(B R_1) = (R_2 B)R_1 = R_2 B R_1 \,.$

The R and B matrices are the elements of the Lorentz group, an example of a Lie group. The parameters are continuous variables. The number of parameters in the group is six, since three are for the boost and three for the rotation, therefore the Lorentz group is six-dimensional.

## Generators of the Lorentz group

All the group elements can be derived from the generators and parameters of the group. In this context, the generators of the Lorentz group are operators which correspond to important symmetries in spacetime: the rotation generators are physically angular momentum, and the boost generators correspond to the motion of the system in spacetime.

### Rotation generators

The Lie-algebraic rotation generators are defined by the partial derivatives (denoted ∂/∂) of the rotation matrices with respect to the angle of rotation, then the angle is set to zero,

$J_x = \left. \frac{\partial R(\mathbf{e}_x,\theta)}{\partial \theta} \right|_{\theta=0} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}$
$J_y = \left. \frac{\partial R(\mathbf{e}_y,\theta)}{\partial \theta} \right|_{\theta=0} = \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \end{bmatrix}$
$J_z = \left. \frac{\partial R(\mathbf{e}_z,\theta)}{\partial \theta} \right|_{\theta=0} = \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$

For example, a rotation about the z axis is the matrix exponential,

$R(\mathbf{e}_z,\theta) \equiv e^{\theta J_z} \equiv \exp\left(\theta \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \right)$

and similarly for the other directions.

For a rotation about an axis defined defined from a unit vector a = (ax, ay, az), the full rotation generator, a vector J = (Jx, Jy, Jz), must be projected into the axis

$\mathbf{a}\cdot\mathbf{J} = a_x J_x + a_y J_y + a_z J_z = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -a_z & a_y \\ 0 & a_z & 0 & -a_x \\ 0 & -a_y & a_x & 0 \end{bmatrix}$

The square is

$(\mathbf{a}\cdot\mathbf{J})^2=\begin{bmatrix}0 & 0 & 0 & 0\\ 0 & -(a_y^2+a_z^2) & a_xa_y & a_za_x\\ 0 & a_xa_y & -(a_z^2+a_x^2) & a_ya_z\\ 0 & a_za_x & a_ya_z & -(a_x^2+a_y^2) \end{bmatrix}$

and the cube and higher powers follow from the first and second powers

$(\mathbf{a}\cdot\mathbf{J})^3=-(\mathbf{a}\cdot\mathbf{J})$
$(\mathbf{a}\cdot\mathbf{J})^4=-(\mathbf{a}\cdot\mathbf{J})^2$

while the zeroth power is the 4×4 identity, (a · J)0 = I. These first four powers are enough to show the general pattern.

 In general for k = 1, 2, 3, ..., the odd powers are $(\mathbf{a}\cdot\mathbf{J})^{2k+1} = (-1)^k(\mathbf{a}\cdot\mathbf{J})$ then multiplying by (a · J) obtains the even powers $(\mathbf{a}\cdot\mathbf{J})^{2k+2} = (-1)^k(\mathbf{a}\cdot\mathbf{J})^2$ To put these in a more useful form, for the odd powers n = 1, 3, 5, ... set n = 2k + 1 so that k = (n − 1)/2 $(\mathbf{a}\cdot\mathbf{J})^n = (-1)^{(n-1)/2}(\mathbf{a}\cdot\mathbf{J})$ while for the even powers n = 2, 4, 6, ... set n = 2k + 2 so that k = (n − 2)/2 $(\mathbf{a}\cdot\mathbf{J})^n = (-1)^{(n-2)/2}(\mathbf{a}\cdot\mathbf{J})^2 \,.$

Expanding the matrix exponential of θ(a · J) in its power series,

$e^{\theta\mathbf{a}\cdot\mathbf{J}}=\sum_{n=0}^{\infty}\frac{1}{n!}\theta^n(\mathbf{a}\cdot\mathbf{J})^n= \sum_{n=1,3,5\ldots}^{\infty}\frac{1}{n!}\theta^n(\mathbf{a}\cdot\mathbf{J})^n+\sum_{n=0,2,4\ldots}^{\infty}\frac{1}{n!}\theta^n(\mathbf{a}\cdot\mathbf{J})^n$

for the odd powers,

\begin{align}\sum_{n=1,3,5\ldots}^{\infty}\frac{1}{n!}\theta^n(\mathbf{a}\cdot\mathbf{J})^n & =\left[\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}+\cdots\right](\mathbf{a}\cdot\mathbf{J}) \\ &=\sin\theta(\mathbf{a}\cdot\mathbf{J})\end{align}

for the even powers,

\begin{align}\sum_{n=0,2,4\ldots}^{\infty}\frac{1}{n!}\theta^n(\mathbf{a}\cdot\mathbf{J})^n & =\mathbf{I}+\left[\frac{1}{2!}\theta^2-\frac{1}{4!}\theta^4+\frac{1}{6!}\theta^6+\cdots\right](\mathbf{a}\cdot\mathbf{J})^2\\ & =\mathbf{I}+\left[1-1+\frac{1}{2!}\theta^2-\frac{1}{4!}\theta^4+\frac{1}{6!}\theta^6+\cdots\right](\mathbf{a}\cdot\mathbf{J})^2\\ & =\mathbf{I}+(1-\cos\theta)(\mathbf{a}\cdot\mathbf{J})^2 \end{align}

and substituting the summations into the exponential series gives the full transformation, the same as Rodrigues' rotation formula

$R(\mathbf{a},\theta) = e^{\theta\mathbf{a}\cdot\mathbf{J}}= \sin\theta(\mathbf{a}\cdot\mathbf{J})+\mathbf{I}+(1-\cos\theta)(\mathbf{a}\cdot\mathbf{J})^2 \,.$

Negating the angle gives the inverse matrix,

$R(\mathbf{a},-\theta) = e^{-\theta\mathbf{a}\cdot\mathbf{J}} = -\sin\theta(\mathbf{a}\cdot\mathbf{J})+\mathbf{I}+(1-\cos\theta)(\mathbf{a}\cdot\mathbf{J})^2 \,.$

### Boost generators

The Lie-algebraic hyperbolic rotation generators, called boost generators, are defined similarly to the rotational case. They are the partial derivative of the boost matrices with respect to the rapidity, then the rapidity is set to zero,

$K_x = \left. \frac{\partial B(\mathbf{e}_x,\phi)}{\partial \phi} \right|_{\phi=0} = \begin{bmatrix} 0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$
$K_y = \left. \frac{\partial B(\mathbf{e}_y,\phi)}{\partial \phi} \right|_{\phi=0} = \begin{bmatrix}0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$
$K_z = \left. \frac{\partial B(\mathbf{e}_z,\phi)}{\partial \phi} \right|_{\phi=0} = \begin{bmatrix}0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 \end{bmatrix}$

For example, the 4×4 matrix for a boost in the x direction can be written compactly as a matrix exponential,

$B(\mathbf{e}_x,\phi) \equiv e^{-\phi K_x} \equiv \exp \left( - \phi \begin{bmatrix} 0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}\right)\,.$

and similarly for the other directions.

For a boost in any direction, the full boost generator, which is a vector K = (Kx, Ky, Kz), must be projected into the direction of the boost as follows

$\mathbf{n}\cdot\mathbf{K} = n_x K_x + n_y K_y + n_z K_z = \begin{bmatrix}0 & n_x & n_y & n_z\\ n_x & 0 & 0 & 0\\ n_y & 0 & 0 & 0\\ n_z & 0 & 0 & 0 \end{bmatrix}$

The square is

$(\mathbf{n}\cdot\mathbf{K})^2=\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & n_x^2 & n_xn_y & n_xn_z\\ 0 & n_yn_x & n_y^2 & n_yn_z\\ 0 & n_zn_x & n_zn_y & n_z^2 \end{bmatrix}$

but the cube (n · K)3 returns to (n · K), and as always the zeroth power is the 4×4 identity, (n · K)0 = I.

 In general the odd powers n = 1, 3, 5, ... are $(\mathbf{n}\cdot\mathbf{K})^n = (\mathbf{n}\cdot\mathbf{K})$ while the even powers n = 2, 4, 6, ... are $(\mathbf{n}\cdot\mathbf{K})^n = (\mathbf{n}\cdot\mathbf{K})^2$

Expanding the matrix exponential of ϕ(n · K) in its power series

$e^{-\phi\mathbf{n}\cdot\mathbf{K}}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\phi^n(\mathbf{n}\cdot\mathbf{K})^n=-\sum_{n=1,3,5\ldots}^{\infty}\frac{1}{n!}\phi^n(\mathbf{n}\cdot\mathbf{K})^n+\sum_{n=0,2,4\ldots}^{\infty}\frac{1}{n!}\phi^n(\mathbf{n}\cdot\mathbf{K})^n$

for the odd powers,

\begin{align}\sum_{n=1,3,5\ldots}^{\infty}\frac{1}{n!}\phi^n(\mathbf{n}\cdot\mathbf{K})^n & =\left[\phi+\frac{\phi^3}{3!}+\frac{\phi^5}{5!} +\cdots \right] (\mathbf{n}\cdot\mathbf{K}) \\ &=\sinh\phi(\mathbf{n}\cdot\mathbf{K})\end{align}

for the even powers (except zero),

\begin{align}\sum_{n=0,2,4\ldots}^{\infty}\frac{1}{n!}\phi^n(\mathbf{n}\cdot\mathbf{K})^n & =\mathbf{I}+\left[\frac{1}{2!}\phi^2+\frac{1}{4!}\phi^4 +\frac{1}{6!}\phi^6 +\cdots\right](\mathbf{n}\cdot\mathbf{K})^2\\ &=\mathbf{I} +\left[ -1 + 1 + \frac{1}{2!}\phi^2+\frac{1}{4!}\phi^4 +\frac{1}{6!}\phi^6 +\cdots\right](\mathbf{n}\cdot\mathbf{K})^2\\ & =\mathbf{I}+(-1+\cosh\phi)(\mathbf{n}\cdot\mathbf{K})^2 \end{align}

then substituting these summations into the exponential formula gives the full transformation similar to Rodrigues' rotation formula

$B(\mathbf{n},\phi) = e^{-\phi\mathbf{n}\cdot\mathbf{K}} = -\sinh\phi(\mathbf{n}\cdot\mathbf{K})+\mathbf{I}+(\cosh\phi-1)(\mathbf{n}\cdot\mathbf{K})^2 \,,$

Again, negating the rapidity in the exponential gives the inverse transformation matrix,

$B(\mathbf{n},-\phi) = e^{\phi\mathbf{n}\cdot\mathbf{K}} = \sinh\phi(\mathbf{n}\cdot\mathbf{K})+\mathbf{I}+(\cosh\phi-1)(\mathbf{n}\cdot\mathbf{K})^2 \,.$

### Commutation relations

$[J_a ,J_b ] = \varepsilon_{abc} J_c$
$[K_a ,K_b ] = - \varepsilon_{abc} J_c$
$[J_a ,K_b ] = \varepsilon_{abc} K_c$

where the bracket

$[X,Y] = XY-YX$

is the commutator, εabc is the Levi-Civita symbol, a, b, c are indices each taking values 1, 2, 3, and the correspondence between Cartesian and numerical components are made as always, Jx = J1, Jy = J2, Jz = J3, etc.

## Transformation of other physical quantities

For the notation used, see Ricci calculus.

Writing the general matrix transformation

$\begin{bmatrix} {X'}^0 \\ {X'}^1 \\ {X'}^2 \\ {X'}^3 \end{bmatrix} = \begin{bmatrix} \Lambda^0{}_0 & \Lambda^0{}_1 & \Lambda^0{}_2 & \Lambda^0{}_3 \\ \Lambda^1{}_0 & \Lambda^1{}_1 & \Lambda^1{}_2 & \Lambda^1{}_3 \\ \Lambda^2{}_0 & \Lambda^2{}_1 & \Lambda^2{}_2 & \Lambda^2{}_3 \\ \Lambda^3{}_0 & \Lambda^3{}_1 & \Lambda^3{}_2 & \Lambda^3{}_3 \\ \end{bmatrix} \begin{bmatrix} X^0 \\ X^1 \\ X^2 \\ X^3 \end{bmatrix}$

in tensor index notation allows the transformation of other physical quantities which cannot be expressed as four-vectors, e.g. tensors or spinors in 4d spacetime, to be defined,

${X'}^\alpha = \Lambda^\alpha {}_\beta X^\beta \,,$

where upper and lower indices label covariant and contravariant components respectively, and the summation convention is applied. It is a standard convention to use Greek indices which take the value 0 for time components, and 1, 2, 3 for space components, while Latin indices simply take the values 1, 2, 3, for spatial components.

For rotations, the rotation matrix entries Λαβ = Rαβ can be found by comparing the full matrix equation with the Lorentz rotation about any axis to obtain[13]

$R^0{}_0 = 1$
$R^0{}_i = R^i{}_0 = 0$
$R^i{}_j = a_i a_j + (\delta_{ij} - a_i a_j) \cos\theta - \varepsilon_{ijk} a_k \sin\theta$

where δij is the Kronecker delta, εijk is the three-dimensional Levi-Civita symbol, and the correspondence between Cartesian components and spatial indices is made as follows, a1 = ax, a2 = ay, a3 = az, and similarly for other 3d vectors.

Similarly the boost matrix entries Λαβ = Bαβ can be found by comparing the full matrix equation with the Lorentz boost in any direction to obtain[14]

$B^0{}_0 = \gamma \,,$
$B^0{}_i = B^i{}_0 = - \gamma \beta n_i \,,$
$B^i{}_j = B^j{}_i = \delta_{ij} + ( \gamma - 1 ) n_i n_j \,,$

or in terms of rapidity

$B^0{}_0 = \cosh\phi \,,$
$B^0{}_i = B^i{}_0 = - n_i \sinh\phi \,,$
$B^i{}_j = B^j{}_i = \delta_{ij} + ( \cosh\phi - 1 ) n_i n_j \,.$

The transformation matrix is universal for all four-vectors, not just 4-dimensional spacetime coordinates. If A is any four-vector, then in tensor index notation

$A^{\alpha'} = \Lambda^{\alpha'}{}_\alpha A^\alpha \,.$

in which the primed indices denote the indices of A in the primed frame.

More generally, the transformation of any tensor quantity T is given by:[15]

$T^{\alpha' \beta' \cdots \zeta'}_{\theta' \iota' \cdots \kappa'} = \Lambda^{\alpha'}{}_{\mu} \Lambda^{\beta'}{}_{\nu} \cdots \Lambda^{\zeta'}{}_{\rho} \Lambda_{\theta'}{}^{\sigma} \Lambda_{\iota'}{}^{\upsilon} \cdots \Lambda_{\kappa'}{}^{\phi} T^{\mu \nu \cdots \rho}_{\sigma \upsilon \cdots \phi}$

where Λχ′ψ is the inverse matrix of Λχ′ψ.

### Transformation of the electromagnetic field

For the transformation rules, see classical electromagnetism and special relativity.

Lorentz transformations can also be used to illustrate that magnetic and electric fields are simply different aspects of the same force — the electromagnetic force, as a consequence of relative motion between electric charges and observers.[16] The fact that the electromagnetic field shows relativistic effects becomes clear by carrying out a simple thought experiment.[17]

• Consider an observer measuring a charge at rest in a reference frame F. The observer will detect a static electric field. As the charge is stationary in this frame, there is no electric current, so the observer will not observe any magnetic field.
• Consider another observer in frame F′ moving at relative velocity v (relative to F and the charge). This observer will see a different electric field because the charge is moving at velocity −v in their rest frame. Further, in frame F′ the moving charge constitutes an electric current, and thus the observer in frame F′ will also see a magnetic field.

This shows that the Lorentz transformation also applies to electromagnetic field quantities when changing the frame of reference, given below in vector form.

## The correspondence principle

For relative speeds much less than the speed of light, the Lorentz transformations reduce to the Galilean transformation in accordance with the correspondence principle.

The correspondence limit is usually stated mathematically as: as v → 0, c → ∞. In words: as velocity approaches 0, the speed of light (seems to) approach infinity. Hence, it is sometimes said that nonrelativistic physics is a physics of "instantaneous action at a distance".[18]

## Spacetime interval

In a given coordinate system xμ, if two events 1 and 2 are separated by

$(\Delta t, \Delta x, \Delta y, \Delta z) = (t_2-t_1, x_2-x_1, y_2-y_1, z_2-z_1)\ ,$

the spacetime interval between them is given by

$s^2 = - c^2(\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2\ .$

This can be written in another form using the Minkowski metric. In this coordinate system,

$\eta_{\mu\nu} = \begin{bmatrix} -1&0&0&0\\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}\ .$

Then, we can write

$s^2 = \begin{bmatrix}c \Delta t & \Delta x & \Delta y & \Delta z \end{bmatrix} \begin{bmatrix} -1&0&0&0\\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{bmatrix}$

or, using the Einstein summation convention,

$s^2= \eta_{\mu\nu} x^\mu x^\nu\ .$

Now suppose that we make a coordinate transformation xμxμ. Then, the interval in this coordinate system is given by

$s'^2 = \begin{bmatrix}c \Delta t' & \Delta x' & \Delta y' & \Delta z' \end{bmatrix} \begin{bmatrix} -1&0&0&0\\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} c \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{bmatrix}$

or

$s'^2= \eta_{\mu\nu} x'^\mu x'^\nu\ .$

It is a result of special relativity that the interval is an invariant. That is, s2 = s2, see invariance of interval. For this to hold, it can be shown[19] that it is necessary and sufficient for the coordinate transformation to be of the form

$x'^\mu = x^\nu \Lambda^\mu_\nu + C^\mu\ ,$

where Cμ is a constant vector and Λμν a constant matrix, where we require that

$\eta_{\mu\nu}\Lambda^\mu_\alpha \Lambda^\nu_\beta = \eta_{\alpha\beta}\ .$

Such a transformation is called a Poincaré transformation or an inhomogeneous Lorentz transformation.[20][21] The Ca represents a spacetime translation. When Ca = 0, the transformation is called an homogeneous Lorentz transformation, or simply a Lorentz transformation.

Taking the determinant of

$\eta_{\mu\nu}{\Lambda^\mu}_\alpha{\Lambda^\nu}_\beta = \eta_{\alpha\beta}$

gives us

$\det (\Lambda^a_b) = \pm 1\ .$

The cases are:

• Proper Lorentz transformations have det(Λμν) = +1, and form a subgroup called the special orthogonal group SO(1,3).
• Improper Lorentz transformations are det(Λμν) = −1, which do not form a subgroup, as the product of any two improper Lorentz transformations will be a proper Lorentz transformation.

From the above definition of Λ it can be shown that (Λ00)2 ≥ 1, so either Λ00 ≥ 1 or Λ00 ≤ −1, called orthochronous and non-orthochronous respectively. An important subgroup of the proper Lorentz transformations are the proper orthochronous Lorentz transformations which consist purely of boosts and rotations. Any Lorentz transform can be written as a proper orthochronous, together with one or both of the two discrete transformations; space inversion P and time reversal T, whose non-zero elements are:

$P^0_0=1, P^1_1=P^2_2=P^3_3=-1$
$T^0_0=-1, T^1_1=T^2_2=T^3_3=1$

The set of Poincaré transformations satisfies the properties of a group and is called the Poincaré group. Under the Erlangen program, Minkowski space can be viewed as the geometry defined by the Poincaré group, which combines Lorentz transformations with translations. In a similar way, the set of all Lorentz transformations forms a group, called the Lorentz group.

A quantity invariant under Lorentz transformations is known as a Lorentz scalar.

## Alternative formalisms

The usual way to formulate the Lorentz transformation is tensor analysis and group theory as shown above. Other formalisms are shown below.

### Gyrovector formalism

The composition of two Lorentz boosts B(u) and B(v) of velocities u and v is given by[22][23]

$B(\mathbf{u})B(\mathbf{v})=B\left ( \mathbf{u}\oplus\mathbf{v} \right )\mathrm{Gyr}\left [ \mathbf{u},\mathbf{v}\right ]=\mathrm{Gyr}\left [\mathbf{u},\mathbf{v} \right ]B \left ( \mathbf{v}\oplus\mathbf{u} \right )$,

where

• B(v) is the 4 × 4 matrix that uses the components of v, i.e. v1, v2, v3 in the entries of the matrix, or rather the components of v/c in the representation that is used above,
• $\mathbf{u}\oplus\mathbf{v}$ is the velocity-addition,
• Gyr[u,v] (capital G) is the rotation arising from the composition. If the 3 × 3 matrix form of the rotation applied to spatial coordinates is given by gyr[u,v], then the 4 × 4 matrix rotation applied to 4-coordinates is given by[24]
$\mathrm{Gyr}[\mathbf{u},\mathbf{v}]= \begin{pmatrix} 1 & 0 \\ 0 & \mathrm{gyr}[\mathbf{u},\mathbf{v}] \end{pmatrix}\,$
• gyr (lower case g) is the gyrovector space abstraction of the gyroscopic Thomas precession, defined as an operator on a velocity w in terms of velocity addition:
$\text{gyr}[\mathbf{u},\mathbf{v}]\mathbf{w}=\ominus(\mathbf{u} \oplus \mathbf{v}) \oplus (\mathbf{u} \oplus (\mathbf{v} \oplus \mathbf{w}))$
for all w.

The composition of two Lorentz transformations L(u, U) and L(v, V) which include rotations U and V is given by:[25]

$L(\mathbf{u},U)L(\mathbf{v},V)=L(\mathbf{u}\oplus U\mathbf{v}, \mathrm{gyr}[\mathbf{u},U\mathbf{v}]UV)$

## Rotations of frames (dynamic)

A more complicated case is where the square of the magnitude of the position vector r = (x, y, z) as measured in frame F, and r′ = (x′, y′, z′) as measured in frame F, are invariant but change with time,

$|\mathbf{r}(t)|^2 = |\mathbf{r}'(t')|^2 \,,$

or

$x(t)^2 +y(t)^2 + z(t)^2 = {x'(t')}^2 + {z'(t')}^2 + {z'(t')}^2\,,$

and the times are equal,

$t'= t \,.$

Here the transformation between the coordinates are related by rotations with constant angular velocity about some axis, so one frame is spinning relative to the other. Since these frames accelerate relative to one another, they are non-inertial frames.

### Rotation about the z axis (in the xy plane) with angular velocity ω

If a frame F with coordinates x′, y′, z′, t has its z axis coincident with the z axis of another frame F with coordinates x, y, z, t, each aligned in the same direction, and rotates about the common zz axis (or equivalently in the coincident xy and xy planes) with constant angular speed ω anticlockwise, the angle turned in time t is θ = ωt, so the transformations of coordinates are[26][27]

$t' = t \,,$
$z' = z \,,$
$x' = x\cos(\omega t) - y\sin(\omega t) \,,$
$y' = x\sin(\omega t) + y\cos(\omega t) \,,$

Unlike the case of a "static" rotation, where there was no time-dependence on the coordinates, this transformation is nonlinear. It can still be written in matrix form

$\begin{bmatrix}ct' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos (\omega t) & -\sin(\omega t) & 0 \\ 0 & \sin (\omega t) & \cos (\omega t) & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix}ct \\ x \\ y \\ z \end{bmatrix}$

but the time coordinate cannot be separated out.

The transformation matrix is parametrized by the angular velocity. Here it will be denoted by R(ez, ω). The quantity ω is the magnitude of the angular velocity, the full angular velocity vector points in the same direction as the rotation axis, so that ω = ωez.

The other directions can easily be found by inserting θ = ωt into the previous 4d rotation matrices.

## Footnotes

1. ^ Since n is a unit vector, only two components of the vector are independent, the absolute value of the third is given by its unit magnitude
$|\mathbf{n}|^2 = n_x^2 + n_y^2 + n_z^2 = 1$
(the sign must be chosen appropriately after taking the root to point in the correct direction). To see this another way, express the unit vector in the spherical polar angles in the Cartesian basis
$\mathbf{n} = \sin\theta(\cos\varphi\mathbf{e}_x + \sin\varphi\mathbf{e}_y) + \cos\theta\mathbf{e}_z$
which clearly shows two numbers uniquely define the direction.
2. ^ The unit dyadic is
$\mathbf{I} = \mathbf{e}_x\mathbf{e}_x + \mathbf{e}_y\mathbf{e}_y + \mathbf{e}_z\mathbf{e}_z$
and taking the dot product of I with r gives
\begin{align} \mathbf{I} \cdot \mathbf{r} & = (\mathbf{e}_x\mathbf{e}_x + \mathbf{e}_y\mathbf{e}_y + \mathbf{e}_z\mathbf{e}_z)\cdot \mathbf{r} \\ & = \mathbf{e}_x(\mathbf{e}_x \cdot \mathbf{r}) + \mathbf{e}_y(\mathbf{e}_y \cdot \mathbf{r}) + \mathbf{e}_z (\mathbf{e}_z \cdot \mathbf{r}) \\ & = \mathbf{e}_x x + \mathbf{e}_y y + \mathbf{e}_z z \\ & = \mathbf{r} \end{align}
The dyadic product of n with itself is
\begin{align}\mathbf{nn} & = (n_x \mathbf{e}_x + n_y \mathbf{e}_y + n_z \mathbf{e}_z)(n_x \mathbf{e}_x + n_y \mathbf{e}_y + n_z \mathbf{e}_z) \\ & = n_x^2\mathbf{e}_x\mathbf{e}_x + n_y n_x\mathbf{e}_y\mathbf{e}_x + n_z n_x\mathbf{e}_z\mathbf{e}_x \\ & + n_x n_y\mathbf{e}_x\mathbf{e}_y + n_y^2 \mathbf{e}_y\mathbf{e}_y + n_z n_y\mathbf{e}_z\mathbf{e}_y \\ & + n_x n_z\mathbf{e}_x\mathbf{e}_z + n_y n_z\mathbf{e}_y\mathbf{e}_z + n_z^2 \mathbf{e}_z\mathbf{e}_z \end{align}
and taking the dot product of nn with r gives
\begin{align}\mathbf{nn}\cdot\mathbf{r} & = (\mathbf{nn})\cdot\mathbf{r} = \mathbf{n}(\mathbf{n}\cdot\mathbf{r}) \\ & = n_x (n_x x + n_y y + n_z z) \mathbf{e}_x + n_y (n_x x + n_y y + n_z z) \mathbf{e}_y + n_z (n_x x + n_y y + n_z z) \mathbf{e}_z \end{align}
In matrix notation, I corresponds to the 3×3 unit matrix, and nn to the matrix product nnT if n is a column vector.

## Notes

1. ^ John & O'Connor 1996
2. ^ Brown 2003
3. ^ Rothman 2006, pp. 112f.
4. ^ Darrigol 2005, pp. 1–22
5. ^ Macrossan 1986, pp. 232–34
6. ^ The reference is within the following paper:Poincaré 1905, pp. 1504–1508
7. ^ Einstein 1905, pp. 891–921
8. ^ Young & Freedman 2008
9. ^ Forshaw & Smith 2009
10. ^ J.A. Wheeler, C. Misner, K.S. Thorne (1973). Gravitation. W.H. Freeman & Co. p. 58. ISBN 0-7167-0344-0.
11. ^ Barut 1964, p. 18–19
12. ^ Barut 1964, p. 18–19
13. ^ C.B. Parker (1994). McGraw Hill Encyclopaedia of Physics (2nd ed.). McGraw Hill. p. 1333. ISBN 0-07-051400-3.
14. ^ Misner, Thorne & Wheeler 1973 These authors give the results in terms of βi, not the unit vector used here. The substitution ni = βi obtains their result.
15. ^ Carroll 2004, p. 22
16. ^ Grant & Phillips 2008
17. ^ Griffiths 2007
18. ^ Einstein 1916
19. ^ Weinberg 1972
20. ^ Weinberg 2005, pp. 55–58
21. ^ Ohlsson 2011, p. 3–9
22. ^ Ungar 1989, pp. 1385–1396
23. ^ Ungar 2000, pp. 331–342
24. ^ Ungar 1989, pp. 1385–1396
25. ^ Ungar 1988
26. ^ Foster & Nightingale 1995, p. 93
27. ^

## References

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• Halpern, A. (1988). 3000 Solved Problems in Physics. Schaum Series. Mc Graw Hill. p. 688. ISBN 978-0-07-025734-4.
• Forshaw, J. R.; Smith, A. G. (2009). Dynamics and Relativity. Manchester Physics Series. John Wiley & Sons Ltd. p. 124–126. ISBN 978-0-470-01460-8.
• Grant, I. S.; Phillips, W. R. (2008). "14". Electromagnetism. Manchester Physics (2nd ed.). John Wiley & Sons. ISBN 0-471-92712-0.
• Sard, R. D. (1970). Relativistic Mechanics - Special Relativity and Classical Particle Dynamics. New York: W. A. Benjamin. ISBN 978-0805384918.