User:Mpatel/sandbox/Derivations of the Lorentz transformations

Derivations of the Lorentz transformations are ways of obtaining the Lorentz transformations, a set of equations that describe how space and time measurements in two inertial reference frames change.

In the fundamental branches of modern physics, namely general relativity and its widely applicable subset special relativity, as well as relativistic quantum mechanics and relativistic quantum field theory, the Lorentz transformation is the transformation rule under which all four-vectors and tensors containing physical quantities transform.

The prime examples of such four-vectors are the four-position and four-momentum of a particle, and for fields the electromagnetic tensor and stress–energy tensor. The fact that these objects transform according to the Lorentz transformation is what mathematically defines them as vectors and tensors, see tensor.

Given the components of the four vectors or tensors in some frame, the "transformation rule" allows one to determine the altered components of the same four-vectors or tensors in another frame, which could be boosted or accelerated, relative to the original frame. A "boost" should not be conflated with spatial translation, rather it's characterized by the relative velocity between frames. The transformation rule itself depends on the relative motion of the frames. In the simplest case of two inertial frames the relative velocity between enters the transformation rule. For rotating reference frames or general non-inertial reference frames, more parameters are needed, including the relative velocity (magnitude and direction), the rotation axis and angle turned through. There are many ways to derive the Lorentz transformations utilizing a variety of mathematical tools, spanning from elementary algebra and hyperbolic functions, to linear algebra and group theory.

This article provides a few of the easier ones to follow in the context of special relativity, for the simplest case of a Lorentz boost in standard configuration, i.e. two inertial frames moving relative to each other at constant (uniform) relative velocity less than the speed of light, and using Cartesian coordinates so that the x and x&prime; axes are collinear.

Different derivations allow for comparison and highlight certain assumptions that are stronger than others.

Forms of the Lorentz Transformations
The Lorentz transformations take on various forms when the frames are moving along coordinate axes (boost along a coordinate axis), in a general direction (boost in any direction), or when the frames are rotated (not rotating) relative to each other.

Boost in the x, y or z directions
For a boost in the x-direction (frames in standard configuration),


 * $$\begin{align}

t' &= \gamma \left( t - \frac{vx}{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,
 * $$\ \gamma = \frac{1}{ \sqrt{1 - { \beta^2}}}$$ is the Lorentz factor (Greek lowercase gamma),
 * $$\ \beta = \frac{v}{c}$$ is the velocity coefficient (Greek lowercase beta), again for the x-direction.

For the y-direction, where v is now the relative velocity between frames in the y-direction,


 * $$\begin{align}

t' &= \gamma \left( t - \frac{vy}{c^2} \right)  \\ x' &= x \\ y' &= \gamma \left( y - vt \right)\\ z' &= z \end{align}$$

For the z-direction, where v is now the relative velocity between frames in the z-direction,


 * $$\begin{align}

t' &= \gamma \left( t - \frac{vz}{c^2} \right)  \\ x' &= x \\ y' &= y \\ z' &= \gamma \left( z - vt \right) \end{align}$$

Boost in any direction
For a boost in an arbitrary direction with velocity $v$, that is, $O$ observes $O&prime;$ to move in direction $v$ in the $F$ coordinate frame, while $O&prime;$ observes $O$ to move in direction $−v$ in the $F&prime;$ coordinate frame,


 * $$\begin{align}t' & = \gamma \left(t - \frac{\mathbf{r} \cdot \mathbf{v}}{c^{2}} \right) \end{align}$$


 * $$\mathbf{r}' = \mathbf{r} + \left(\frac{\gamma-1}{v^2}\mathbf{r}\cdot\mathbf{v} - \gamma t \right)\mathbf{v}\,. $$

where


 * $$\gamma(\mathbf{v}) = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}}$$.

Historical background
The usual treatment (e.g., Einstein's original work) is based on the invariance of the speed of light. However, this is not necessarily the starting point: indeed (as is exposed, for example, in the second volume of the Course of Theoretical Physics by Landau and Lifshitz), what is really at stake is the locality of interactions: one supposes that the influence that one particle, say, exerts on another can not be transmitted instantaneously. Hence, there exists a theoretical maximal speed of information transmission which must be invariant, and it turns out that this speed coincides with the speed of light in vacuum. The need for locality in physical theories was already noted by Newton (see Koestler's The Sleepwalkers), who considered the notion of action at a distance "philosophically absurd" and believed that gravity must be transmitted by an agent (such as an interstellar aether) which obeys certain physical laws.

Michelson and Morley in 1887 designed an experiment, employing an interferometer and a half-silvered mirror, that was accurate enough to detect aether flow. The mirror system reflected the light back into the interferometer. If there were an aether drift, it would produce a phase shift and a change in the interference that would be detected. However, no phase shift was ever found. The null result of the Michelson–Morley experiment left the concept of aether (or its drift) undermined. There was consequent perplexity as to why light evidently behaves like a wave, without any detectable medium through which wave activity might propagate.

In a 1964 paper, Erik Christopher Zeeman showed that the causality preserving property, a condition that is weaker in a mathematical sense than the invariance of the speed of light, is enough to assure that the coordinate transformations are the Lorentz transformations.

Assumptions
Some derivations are of a more physically intuitive nature whereas others are more mathematically rigorous. However, both approaches involve some standard physical and mathematical starting points. The fewer the assumptions used in derivations, the more philosophically elegant and powerful the derivation is deemed to be.

Physical assumptions
All derivations of the Lorentz transformations involve at least one of the following assumptions:


 * Constancy of the speed of light.
 * Special principle of relativity.
 * Homogeneity and isotropy of space and time.
 * Causality.
 * Correspondence principle.

Mathematical assumptions
Mathematical assumptions are necessary for writing equations and they indicate allowable mathematical procedures (for example, differentiating functions). Common assumptions are:


 * Spacetime is a four-dimensional manifold.
 * The transformations are linear functions of the spacetime coordinates.
 * The transformations form a group.
 * Invariance of the spacetime interval.

Proofs of linearity
Linearity of the transformations can be proven in various ways.

Frames in standard configuration


The problem is usually restricted to two spacetime dimensions by using a velocity along the x axis such that the y and z coordinates do not intervene. The following is similar to that of Einstein. As in the Galilean transformation, the Lorentz transformation is linear since the relative velocity of the reference frames is constant as a vector; otherwise, inertial forces would appear. They are called inertial or Galilean reference frames. According to relativity no Galilean reference frame is privileged. Another condition is that the speed of light must be independent of the reference frame, in practice of the velocity of the light source.

Spherical wavefronts of light
Consider two inertial frames of reference O and O′, assuming O to be at rest while O′ is moving with a velocity v with respect to O in the positive x-direction. The origins of O and O′ initially coincide with each other. A light signal is emitted from the common origin and travels as a spherical wave front. Consider a point P on a spherical wavefront at a distance r and r′ from the origins of O and O′ respectively. According to the second postulate of the special theory of relativity the speed of light is the same in both frames, so r and r′ will be different only if t and t′ are different:
 * $$\begin{align}

r &= ct \\ r' &= ct'. \end{align}$$

The equation of the spherical wavefront in frame O will be
 * $$x^2 + y^2 + z^2 = r^2$$

or
 * $$x^2 + y^2 + z^2 = c^2t^2.$$

Similarly, the equation of the spherical wavefront in frame O′ will be
 * $$x'^2 + y'^2 + z'^2 = r'^2$$

or
 * $$x'^2 + y'^2 + z'^2 = c^2t'^2.$$

The origin O′ is moving along x-axis. Therefore,
 * $$\begin{align}

y' &= y \\ z' &= z. \end{align}$$

The relation between x and x′ should be in linear form and be such that it reduces to the Galilean transformation at v ≪ c. Therefore, such a relation can be written in the form:
 * $$x' = \gamma \left( x - v t \right)$$

where γ is to be determined. At this point γ is not necessarily a constant and independent of the coordinates t, x, t' , x' , but is required to reduce to 1 for v ≪ c.

The inverse is:
 * $$x = \gamma \left( x' + v t' \right).$$

The above two equations give the relation between t and t′ as:
 * $$x = \gamma \left[ \gamma \left( x - v t \right) + v t' \right]$$

or
 * $$t' = \gamma t + \frac{ \left( 1 - { \gamma^2} \right)x}{ \gamma v}.$$

Substituting the expressions of x′, y′, z′ and t′ in terms of x, y, z and t in spherical wavefront equation of O′ frame,
 * $$x'^2 + y'^2 + z'^2 = c^2t'^2,$$

produces
 * $$ {\gamma^2} \left( x - v t \right)^2 + y^2 + z^2 = c^2 \left[ \gamma t + \frac{ \left( 1 - { \gamma^2} \right)x}{ \gamma v} \right]^2$$

and therefore,
 * $$ \gamma^2 x^2 + \gamma^2 v^2 t^2 - 2 \gamma^2 v t x + y^2 + z^2 = c^2 {\gamma^2} t^2 + \frac{ \left( 1 - {\gamma^2} \right)^2 c^2 x^2}{ {\gamma^2} v^2} + 2 \frac{ \left( 1 - {\gamma^2} \right) t x c^2}{ v}$$

which implies,
 * $$ \left[ {\gamma^2} - \frac{ \left( 1 - {\gamma^2} \right)^2 c^2}{ {\gamma^2} v^2} \right] x^2 - 2 {\gamma^2} v t x + y^2 + z^2 = \left( c^2 {\gamma^2} - v^2 {\gamma^2} \right) t^2 + 2 \frac{ \left( 1 - {\gamma^2} \right) t x c^2}{ v}$$

comparing the coefficients of t2 from above equation with the spherical wavefront equation of O frame produces
 * $$c^2 {\gamma^2} - v^2 {\gamma^2} = c^2$$

or
 * $${\gamma^2} = \frac{1}{1 - \frac{v^2}{c^2}}$$

or, choosing the positive root to ensure that the x and x' axes and the time axes point in the same direction,
 * $${\gamma} = \frac{1}{ \sqrt{1 - \frac{v^2}{c^2}}}$$

which is called the Lorentz factor. This produces the Lorentz transformation from the above expression. It is given by
 * $$\begin{align}

x' &= \gamma \left( x - v t \right)\\ t' &= \gamma \left( t - \frac{vx}{c^2} \right) \\ y' &= y \\ z' &= z \end{align}$$

The Lorentz transformation is not the only transformation leaving invariant the shape of spherical waves, as there is a wider set of spherical wave transformations in the context of conformal geometry, leaving invariant the expression $$\lambda\left(\delta x^{2}+\delta y^{2}+\delta z^{2}-c^{2}\delta t^{2}\right)$$. However, scale changing conformal transformations cannot be used to symmetrically describe all laws of nature including mechanics, whereas the Lorentz transformations (the only one implying $$\lambda=1$$) represent a symmetry of all laws of nature and reduce to Galilean transformations at $$v\ll c$$.

Galilean and Einstein's relativity
In classical kinematics, the total displacement x in the R frame is the sum of the relative displacement x′ in frame R′ and of the distance between the two origins x − x′. If v is the relative velocity of R′ relative to R, the transformation is: x = x′ + vt, or x′ = x − vt. This relationship is linear for a constant v, that is when R and R′ are Galilean frames of reference.
 * Galilean reference frames

In Einstein's relativity, the main difference from Galilean relativity is that space and time coordinates are intertwined, and in different inertial frames t ≠ t′.

Since space is assumed to be homogeneous, the transformation must be linear. The most general linear relationship is obtained with four constant coefficients, A, B, γ, and b:
 * $$x'=\gamma x + b t \,$$
 * $$t'=A x + B t. \,$$

The Lorentz transformation becomes the Galilean transformation when γ = B = 1, b = −v and A = 0.

An object at rest in the R′ frame at position x′ = 0 moves with constant velocity v in the R frame. Hence the transformation must yield x′ = 0 if x = vt. Therefore, b = −γv and the first equation is written as
 * $$x'=\gamma (x - v t) \,.$$


 * Principle of relativity

According to the principle of relativity, there is no privileged Galilean frame of reference: therefore the inverse transformation for the position from frame R′ to frame R should have the same form as the original but with the velocity in the opposite direction, i.o.w. replacing v with -v:
 * $$x=\gamma\left(x' - (-v)t'\right),$$

and thus
 * $$x=\gamma\left(x' + vt'\right) .$$

Since the speed of light is the same in all frames of reference, for the case of a light signal, the transformation must guarantee that t = x/c and t′ = x′/c.
 * The speed of light is constant

Substituting for t and t′ in the preceding equations gives:
 * $$x'= \gamma\left(1 - v/c\right) x, $$
 * $$x= \gamma\left(1 + v/c\right) x'. $$

Multiplying these two equations together gives,
 * $$xx' = \gamma^2 \left(1 - v^2/c^2\right) xx'. $$

At any time after t = t′ = 0, xx′ is not zero, so dividing both sides of the equation by xx′ results in
 * $$\gamma=\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}, $$

which is called the "Lorentz factor".

When the transformation equations are required to satisfy the light signal equations in the form x = ct and x′ = ct′, by substituting the x and x'-values, the same technique produces the same expression for the Lorentz factor.

The transformation equation for time can be easily obtained by considering the special case of a light signal, satisfying
 * Transformation of time
 * $$\begin{cases}

x' = ct' \\ x = ct. \end{cases}$$ Substituting term by term into the earlier obtained equation for the spatial coordinate
 * $$x'=\gamma (x - v t), \,$$

gives
 * $$ct'=\gamma \left(ct - \frac{v}{c} x\right), $$

so that
 * $$t'=\gamma \left(t - \frac{v}{c^2} x\right), $$

which determines the transformation coefficients A and B as
 * $$A= -\gamma v/c^2,\,$$
 * $$B=\gamma. \,$$

So A and B are the unique coefficients necessary to preserve the constancy of the speed of light in the primed system of coordinates.

Einstein's popular derivation
In his popular book Einstein derived the Lorentz transformation by arguing that there must be two non-zero coupling constants λ and μ such that
 * $$\begin{cases}

x' - ct' = \lambda \left( x - ct \right) \\ x' + ct' = \mu \left( x + ct \right) \, \end{cases}$$ that correspond to light traveling along the positive and negative x-axis, respectively. For light x = ct if and only if x′ = ct′. Adding and subtracting the two equations and defining
 * $$\begin{cases}

\gamma = \left( \lambda + \mu \right) /2 \\ b = \left( \lambda - \mu \right) /2 ,\, \end{cases}$$ gives
 * $$\begin{cases}

x' = \gamma x - bct \\ ct' = \gamma ct - bx. \, \end{cases}$$ Substituting x′ = 0 corresponding to x = vt and noting that the relative velocity is v = bc/&gamma;, this gives
 * $$\begin{cases}

x' = \gamma \left( x - vt \right) \\ t' = \gamma \left( t - \frac{v}{c^2} x \right) \, \end{cases}$$ The constant γ can be evaluated as was previously shown above.

The Lorentz transformations can also be derived by simple application of the special relativity postulates and using hyperbolic identities. It is sufficient to derive the result for a boost in  one direction, since for an arbitrary direction the decomposition of the position vector into parallel and perpendicular components can be done after, and generalizations therefrom follow, as outlined above.

Hyperbolic geometry



 * Relativity postulates

Start from the equations of the spherical wave front of a light pulse, centred at the origin:


 * $$(ct)^2 - (x^2+y^2+z^2) = (ct')^2 - (x'^2+y'^2+z'^2) =0$$

which take the same form in both frames because of the special relativity postulates. Next, consider relative motion along the x-axes of each frame, in standard configuration above, so that y = y′, z = z′, which simplifies to


 * $$(ct)^2 - x^2 = (ct')^2 - x'^2 $$


 * Linearity

Now assume that the transformations take the linear form:


 * $$ \begin{align}

x' & = Ax + Bct \\ ct' & = Cx + Dct \end{align}$$

where A, B, C, D are to be found. If they were non-linear, they would not take the same form for all observers, since fictitious forces (hence accelerations) would occur in one frame even if the velocity was constant in another, which is inconsistent with inertial frame transformations.

Substituting into the previous result:


 * $$(ct)^2 - x^2 = [(Cx)^2 + (Dct)^2 + 2CDcxt] - [(Ax)^2 + (Bct)^2 + 2ABcxt] $$

and comparing coefficients of x2, t2, xt:


 * $$ \begin{align}

- 1 = C^2 - A^2 & \Rightarrow & A^2 - C^2 = 1 \\ c^2 = (Dc)^2 - (Bc)^2 & \Rightarrow & D^2 - B^2 = 1 \\ 2CDc - 2ABc = 0 & \Rightarrow & AB = CD \end{align}$$


 * Hyperbolic rotation

The formulae resemble the hyperbolic identity


 * $$\cosh^2\phi-\sinh^2\phi=1$$

Introducing the rapidity parameter ϕ as a parametric hyperbolic angle allows the self-consistent identifications


 * $$A=D=\cosh\phi\,,\quad C = B = -\sinh\phi $$

where the signs after the square roots are chosen so that x and t increase. The hyperbolic transformations have been solved for:


 * $$\begin{align}

x' & = \cosh\phi x - \sinh\phi ct \\ ct' & = -\sinh\phi x + \cosh\phi ct \end{align}$$

If the signs were chosen differently the position and time coordinates would need to be replaced by −x and/or −t so that x and t increase not decrease.

To find what ϕ actually is, from the standard configuration the origin of the primed frame x′ = 0 is measured in the unprimed frame to be x = vt (or the equivalent and opposite way round; the origin of the unprimed frame is x = 0 and in the primed frame it is at x′ = −vt):


 * $$ 0 = \cosh\phi vt - \sinh\phi ct \, \Rightarrow \, \tanh\phi = \frac{v}{c} = \beta$$

and manipulation of hyperbolic identities leads to


 * $$ \cosh\phi = \gamma,\,\sinh\phi=\beta\gamma $$

so the transformations are also:


 * $$\begin{align}

x' = \gamma x - \frac{\gamma v}{c}ct & \Rightarrow & x' = \gamma(x - vt) \\ ct' = -\frac{\gamma v}{c} x + \gamma ct & \Rightarrow & t' = \gamma\left(t-\frac{vx}{c^2}\right) \end{align}$$

From group postulates
Following is a classical derivation (see, e.g., and references therein) based on group postulates and isotropy of the space.

The coordinate transformations between inertial frames form a group (called the proper Lorentz group) with the group operation being the composition of transformations (performing one transformation after another). Indeed the four group axioms are satisfied:
 * Coordinate transformations as a group
 * 1) Closure: the composition of two transformations is a transformation: consider a composition of transformations from the inertial frame K to inertial frame K′, (denoted as K → K′), and then from K′ to inertial frame K′′, [K′ → K′′], there exists a transformation, [K → K′][K′ → K′′], directly from an inertial frame K to inertial frame K′′.
 * 2) Associativity: the result of ([K → K′][K′ → K′′])[K′′ → K′′′] and [K → K′]([K′ → K′′][K′′ → K′′′]) is the same, K → K′′′.
 * 3) Identity element: there is an identity element, a transformation K → K.
 * 4) Inverse element: for any transformation K → K′ there exists an inverse transformation K′ → K.

Let us consider two inertial frames, K and K′, the latter moving with velocity v with respect to the former. By rotations and shifts we can choose the x and x′ axes along the relative velocity vector and also that the events (t, x) = (0, 0) and (t′, x′) = (0, 0) coincide. Since the velocity boost is along the x (and x′) axes nothing happens to the perpendicular coordinates and we can just omit them for brevity. Now since the transformation we are looking after connects two inertial frames, it has to transform a linear motion in (t, x) into a linear motion in (t′, x′) coordinates. Therefore it must be a linear transformation. The general form of a linear transformation is
 * Transformation matrices consistent with group axioms

\begin{bmatrix} t' \\ x' \end{bmatrix} = \begin{bmatrix} \gamma & \delta \\ \beta & \alpha \end{bmatrix} \begin{bmatrix} t \\ x \end{bmatrix}, $$ where α, β, γ, and δ are some yet unknown functions of the relative velocity v.

Let us now consider the motion of the origin of the frame K′. In the K′ frame it has coordinates (t′, x′ = 0), while in the K frame it has coordinates (t, x = vt). These two points are connected by the transformation

\begin{bmatrix} t' \\ 0 \end{bmatrix} = \begin{bmatrix} \gamma & \delta \\ \beta & \alpha \end{bmatrix} \begin{bmatrix} t \\ vt \end{bmatrix}, $$ from which we get
 * $$\beta=-v\alpha \,$$.

Analogously, considering the motion of the origin of the frame K, we get

\begin{bmatrix} t' \\ -vt' \end{bmatrix} = \begin{bmatrix} \gamma & \delta \\ \beta & \alpha \end{bmatrix} \begin{bmatrix} t \\ 0 \end{bmatrix}, $$ from which we get
 * $$\beta=-v\gamma \,$$.

Combining these two gives α = γ and the transformation matrix has simplified,

\begin{bmatrix} t' \\ x' \end{bmatrix} = \begin{bmatrix} \gamma & \delta \\ -v\gamma & \gamma \end{bmatrix} \begin{bmatrix} t \\ x \end{bmatrix}. $$

Now let us consider the group postulate inverse element. There are two ways we can go from the K′ coordinate system to the K coordinate system. The first is to apply the inverse of the transform matrix to the K′ coordinates:



\begin{bmatrix} t \\ x \end{bmatrix} = \frac{1}{\gamma^2+v\delta\gamma} \begin{bmatrix} \gamma & -\delta \\ v\gamma & \gamma \end{bmatrix} \begin{bmatrix} t' \\ x' \end{bmatrix}. $$

The second is, considering that the K′ coordinate system is moving at a velocity v relative to the K coordinate system, the K coordinate system must be moving at a velocity −v relative to the K′ coordinate system. Replacing v with −v in the transformation matrix gives:



\begin{bmatrix} t \\ x \end{bmatrix} = \begin{bmatrix} \gamma(-v) & \delta(-v) \\ v\gamma(-v) & \gamma(-v) \end{bmatrix} \begin{bmatrix} t' \\ x' \end{bmatrix}, $$

Now the function γ can not depend upon the direction of v because it is apparently the factor which defines the relativistic contraction and time dilation. These two (in an isotropic world of ours) cannot depend upon the direction of v. Thus, γ(−v) = γ(v) and comparing the two matrices, we get

\gamma^2+v\delta\gamma=1. \, $$

According to the closure group postulate a composition of two coordinate transformations is also a coordinate transformation, thus the product of two of our matrices should also be a matrix of the same form. Transforming K to K′ and from K′ to K′′ gives the following transformation matrix to go from K to K′′:



\begin{align} \begin{bmatrix} t \\ x \end{bmatrix} & = \begin{bmatrix} \gamma(v') & \delta(v') \\ -v'\gamma(v') & \gamma(v') \end{bmatrix}

\begin{bmatrix} \gamma(v) & \delta(v) \\ -v\gamma(v) & \gamma(v) \end{bmatrix}

\begin{bmatrix} t \\ x \end{bmatrix}\\

& = \begin{bmatrix} \gamma(v')\gamma(v)-v\delta(v')\gamma(v) & \gamma(v')\delta(v)+\delta(v')\gamma(v) \\ -(v'+v)\gamma(v')\gamma(v) & -v'\gamma(v')\delta(v)+\gamma(v')\gamma(v) \end{bmatrix}

\begin{bmatrix} t\\z \end{bmatrix}. \end{align} $$

In the original transform matrix, the main diagonal elements are both equal to γ, hence, for the combined transform matrix above to be of the same form as the original transform matrix, the main diagonal elements must also be equal. Equating these elements and rearranging gives:



\gamma(v')\gamma(v)-v\delta(v')\gamma(v)=-v'\gamma(v')\delta(v)+\gamma(v')\gamma(v)\, $$



v\delta(v')\gamma(v)=v'\gamma(v')\delta(v)\, $$



\frac{\delta(v)}{v\gamma(v)}=\frac{\delta(v')}{v'\gamma(v')}.\, $$

The denominator will be nonzero for nonzero v, because γ(v) is always nonzero;


 * $$\gamma^2 + v \delta \gamma = 1$$.

If v = 0 we have the identity matrix which coincides with putting v = 0 in the matrix we get at the end of this derivation for the other values of v, making the final matrix valid for all nonnegative v.

For the nonzero v, this combination of function must be a universal constant, one and the same for all inertial frames. Define this constant as δ(v)/vγ(v) = κ where κ has the dimension of 1/v2. Solving

1 = \gamma^2 + v\delta\gamma = \gamma^2 (1 + \kappa v^2) \,$$ we finally get
 * $$\gamma=1/\sqrt{1 + \kappa v^2}$$

and thus the transformation matrix, consistent with the group axioms, is given by



\begin{bmatrix} t' \\ x' \end{bmatrix} = \frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ x \end{bmatrix}. $$

If κ > 0, then there would be transformations (with κv2 ≫ 1) which transform time into a spatial coordinate and vice versa. We exclude this on physical grounds, because time can only run in the positive direction. Thus two types of transformation matrices are consistent with group postulates:

If κ = 0 then we get the Galilean-Newtonian kinematics with the Galilean transformation,
 * Galilean transformations

\begin{bmatrix} t' \\ x' \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ x \end{bmatrix}\;, $$ where time is absolute, t′ = t, and the relative velocity v of two inertial frames is not limited.

If κ < 0, then we set c = 1/√(−κ) which becomes the invariant speed, the speed of light in vacuum. This yields κ = −1/c2 and thus we get special relativity with Lorentz transformation
 * Lorentz transformations

\begin{bmatrix} t' \\ x' \end{bmatrix} = \frac{1}{\sqrt{1 - {v^2 \over c^2}}} \begin{bmatrix} 1 & {- v \over c^2} \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ x \end{bmatrix}\;, $$ where the speed of light is a finite universal constant determining the highest possible relative velocity between inertial frames.

If v ≪ c the Galilean transformation is a good approximation to the Lorentz transformation.

Only experiment can answer the question which of the two possibilities, κ = 0 or κ < 0, is realised in our world. The experiments measuring the speed of light, first performed by a Danish physicist Ole Rømer, show that it is finite, and the Michelson–Morley experiment showed that it is an absolute speed, and thus that κ < 0.

From experiments
Howard Percy Robertson and others showed that the Lorentz transformation can also be derived empirically. In order to achieve this, it's necessary to write down coordinate transformations that include experimentally testable parameters. For instance, let there be given a single "preferred" inertial frame $$X, Y, Z, T$$ in which the speed of light is constant, isotropic, and independent of the velocity of the source. It is also assumed that Einstein synchronization and synchronization by slow clock transport are equivalent in this frame. Then assume another frame $$x, y, z, t$$ in relative motion, in which clocks and rods have the same internal constitution as in the preferred frame. The following relations, however, are left undefined:


 * $$a(v)$$ differences in time measurements,
 * $$b(v)$$ differences in measured longitudinal lengths,
 * $$d(v)$$ differences in measured transverse lengths,
 * $$\varepsilon(v)$$ depends on the clock synchronization procedure in the moving frame,

then the transformation formulas (assumed to be linear) between those frames are given by:


 * $$\begin{align}

t & =a(v)T+\varepsilon(v) x\\ x & =b(v)(X-vT)\\ y & =d(v)Y\\ z & =d(v)Z \end{align}$$

$$\varepsilon (v)$$ depends on the synchronization convention and is not determined experimentally, it obtains the value $$-v/c^{2}$$ by using Einstein synchronization in both frames. The ratio between $$b(v)$$ and $$d(v)$$ is determined by the Michelson–Morley experiment, the ratio between $$a(v)$$ and $$b(v)$$ is determined by the Kennedy–Thorndike experiment, and $$a(v)$$ alone is determined by the Ives–Stilwell experiment. In this way, they have been determined with great precision to $$1/a(v)=b(v)=\gamma$$ and $$d(v)=1$$, which converts the above transformation into the Lorentz transformation.

Boost in any direction
The Lorentz transformations obtained above for frames in standard configuration are often known as boosts in the x-direction. To obtain the transformations for a boost in any direction, begin as follows.