Lorentz Transformation - Derivation - From Group Postulates

From Group Postulates

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

Coordinate transformations as a group

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:

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′′, there exists a transformation, directly from an inertial frame K to inertial frame K′′.
Associativity: the result of and is the same, K → K′′′.
Identity element: there is an identity element, a transformation K → K.
Inverse element: for any transformation K → K′ there exists an inverse transformation K′ → K.

Group symmetry
Transformation matrices consistent with group axioms 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 z and z′ axes along the relative velocity vector and also that the events (t, z) = (0, 0) and (t′, z′) = (0, 0) coincide. Since the velocity boost is along the z (and z′) 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, z) into a linear motion in (t′, z′) coordinates. Therefore it must be a linear transformation. The general form of a linear transformation is $\begin{bmatrix} t' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma & \delta \\ \beta & \alpha \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix},$ 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′, z′ = 0), while in the K frame it has coordinates (t, z = 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 . 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 . Combining these two gives α = γ and the transformation matrix has simplified, $\begin{bmatrix} t' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma & \delta \\ -v\gamma & \gamma \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix},$ 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: $\begin{bmatrix} t \\ z \end{bmatrix} = \frac{1}{\gamma^2+v\delta\gamma} \begin{bmatrix} \gamma & -\delta \\ v\gamma & \gamma \end{bmatrix} \begin{bmatrix} t' \\ z' \end{bmatrix}.$ 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: $\begin{bmatrix} t \\ z \end{bmatrix} = \begin{bmatrix} \gamma(-v) & \delta(-v) \\ v\gamma(-v) & \gamma(-v) \end{bmatrix} \begin{bmatrix} t' \\ z' \end{bmatrix},$ 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′′: $\begin{align} \begin{bmatrix} t'' \\ z'' \end{bmatrix} & = \begin{bmatrix} \gamma(v') & \delta(v') \\ -v'\gamma(v') & \gamma(v') \end{bmatrix} \begin{bmatrix} \gamma(v) & \delta(v) \\ -v\gamma(v) & \gamma(v) \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix}\\ & = \begin{bmatrix} \gamma(v')\gamma(v)-v\delta(v')\gamma(v) & \gamma(v')\delta(v)+\delta(v')\gamma(v) \\ -(v'+v)\gamma(v')\gamma(v) & -v'\gamma(v')\delta(v)+\gamma(v')\gamma(v) \end{bmatrix} \begin{bmatrix} t\\z \end{bmatrix}. \end{align}$ 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; . 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 and thus the transformation matrix, consistent with the group axioms, is given by $\begin{bmatrix} t' \\ z' \end{bmatrix} = \frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix}.$ 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: with the universal constant κ = 0, and with κ < 0. Galilean transformations If κ = 0 then we get the Galilean-Newtonian kinematics with the Galilean transformation, $\begin{bmatrix} t' \\ z' \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix}\;,$ where time is absolute, t′ = t, and the relative velocity v of two inertial frames is not limited. Lorentz transformations 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 $\begin{bmatrix} t' \\ z' \end{bmatrix} = \frac{1}{\sqrt{1 - {v^2 \over c^2}}} \begin{bmatrix} 1 & {- v \over c^2} \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix}\;,$ 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.

Group symmetry

Transformation matrices consistent with group axioms

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 z and z′ axes along the relative velocity vector and also that the events (t, z) = (0, 0) and (t′, z′) = (0, 0) coincide. Since the velocity boost is along the z (and z′) 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, z) into a linear motion in (t′, z′) coordinates. Therefore it must be a linear transformation. The general form of a linear transformation is

$\begin{bmatrix} t' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma & \delta \\ \beta & \alpha \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix},$

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′, z′ = 0), while in the K frame it has coordinates (t, z = 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

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

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

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

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:

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

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:

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

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′′:

$\begin{align} \begin{bmatrix} t'' \\ z'' \end{bmatrix} & = \begin{bmatrix} \gamma(v') & \delta(v') \\ -v'\gamma(v') & \gamma(v') \end{bmatrix} \begin{bmatrix} \gamma(v) & \delta(v) \\ -v\gamma(v) & \gamma(v) \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix}\\ & = \begin{bmatrix} \gamma(v')\gamma(v)-v\delta(v')\gamma(v) & \gamma(v')\delta(v)+\delta(v')\gamma(v) \\ -(v'+v)\gamma(v')\gamma(v) & -v'\gamma(v')\delta(v)+\gamma(v')\gamma(v) \end{bmatrix} \begin{bmatrix} t\\z \end{bmatrix}. \end{align}$

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;

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

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

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

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:

with the universal constant κ = 0, and
with κ < 0.

Galilean transformations

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

$\begin{bmatrix} t' \\ z' \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix}\;,$

where time is absolute, t′ = t, and the relative velocity v of two inertial frames is not limited.

Lorentz transformations

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

$\begin{bmatrix} t' \\ z' \end{bmatrix} = \frac{1}{\sqrt{1 - {v^2 \over c^2}}} \begin{bmatrix} 1 & {- v \over c^2} \\ -v & 1 \end{bmatrix} \begin{bmatrix} t \\ z \end{bmatrix}\;,$

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.

Read more about this topic: Lorentz Transformation, Derivation

Famous quotes containing the words group and/or postulates:

“My routines come out of total unhappiness. My audiences are my group therapy.”
—Joan Rivers (b. 1935)

“The more reasonable a student was in mathematics, the more unreasonable she was in the affairs of real life, concerning which few trustworthy postulates have yet been ascertained.”
—George Bernard Shaw (1856–1950)

Related Subjects

Lorentz Transformations

Related Phrases

Coordinate Systems

Electromagnetic Field

Inertial Frames

Kennedy–thorndike Experiment

Light Beam

Light Cone

List of Relativistic Equations

Lorentz Group

Minkowski Space