In the frame of special relativity theory, the history of an inertial particle is described by a geodesic straight line in the four dimensional Minkowski space, endowed with the metric. This geodesic is a timelike straight line and its orthogonal complement is the physical space of the particle formed by all its simultaneous events. The passage from one inertial particle to another one is done through a special Lorentz matrix, which is called the boost, and this process is the Lorentz-Poincar?? transformation. But for a non inertial particle, all this is lost since its worldline is no more a straight line and there is no Lorentz transformation and boost associated to it. In order to fill this gap we suggest a deeper insight into the action of the Lie group of Lorentz matrices (and its Lie algebra) on the Minkowski space. This leads us to a new definition of a tangent boost along a worldline. This notion may be used in both situations of special or general relativity theories. Therefore we introduce a matrix belonging to the Lie algebra, which, together with the tangent boost, describes completely the dynamical system: acceleration and instantaneous Thomas rotation.
In a first part, we present properties of Lie matrices and of their reduced forms and we show that the Lie group of special and orthochronous Lorentz matrices has four one-parameter subgroups. These tools permit to introduce the Thomas rotation in a quite general way. Then, we give some applications of these tools: we first consider the case of an uniformly accelerated system and the one of an electron rotating on a circular orbit around the atom nucleus. We then present the case of the so-called “Langevin’s twins” and we show that, when the twin who made a journey into space returns home, he is not only younger than the twin who stayed on Earth but he is also disorientated with respect to the terrestial frame because his gyroscope has turned with respect to the earth referential frame  .
Let us underline that this formalism can be used both in Special and in General Relativity.
2. The Lie Algebra of a Lie Group
A Lie group is a smooth manifold with a compatible group structure, which means that the product and inverse operations are smooth. The Lie algebra of this Lie group can be seen as the tangent space to the manifold at the unit element e of the group multiplication. This tangent space is a vector space endowed with the Lie bracket of two tangent vectors.
Example: The Lie Group and Its Lie Algebra
Let’s start with the group of -matrices having determinant. As a smooth manifold, it can be regarded as a 3-dimensional submanifold of defined by the 6 equations resulting from the orthogonal matrix definition:. Let's denote it, as usual, by.
Its Lie algebra is the 3-dimensional vector space of skew-symmetric matrices endowed with the bracket
This manifold is obviously isomorphic to the euclidean space endowed with the cross product. The vectors of our Lie algebra should be regarded as
tangent vectors of smooth paths on the manifold.
The left translation on the group by shall be. Of course if,. The linear mapping which by definition is
equal to its differential maps the tangent vector from the tangent space to the tangent space:
Derivating the relation gives, which means that is skew-symmetric. Of course it would be possible to obtain the same through right translation by, using.
Application to kinematics of rotation of a rigid body around a fixed point.
Keeping in mind later comparisons, we shall apply the results of the previous section to the study of the motion of a rigid body. We want to show the interest of looking at the action of on when this group is regarded as a subgroup of isometries of the euclidean space, and the meaning of its Lie algebra. Let us associate to a three-dimensional point O two coordinate systems and defined by two orthonormal basis e and E respec- tively. And let A denote the matrix mapping e to E. A smooth path on the manifold corresponds to a rotation movement of around O with respect to the coordinate system.
Let us denote by and the coordinates of a point M in the neighborhood of O in the reference systems and respectively. Looking at the movement of M with respect to, we then have
composing by the left translation and using (1) we get
This relation expresses the derivation rule of the movement of a point X in the moving coordinate system. The absolute derivative of X with respect to t is equal to the sum of the relative derivative and of the training derivative defined by
is a skew-symmetric covariant tensor whose adjoint gives the components of a vector in and permits us to express the training velocity in the well known vector form:
An analogous process starting from the right translation would lead us to the same derivation rule as given in (2) but using the components in. Note that
There is another interesting application of the identification: The exponential map
is a diffeomorphism of the open ball of radius in into an open subset of, and we thus obtain an interesting parametrization of: writing the norm of and, we have
Such a formula has an obvious geometrical meaning: is a rotation through the angle about the axis having the direction of the vector. Note that, being independant of t, the function defines a one-parameter subgroup of, and that the matrix product is the solution of the linear differential equation
with the initial condition. This equation is nothing but (2) when written in the coordinate system. Its solution defines a uniform rotation.
3. The Lie Group of Lorentz Matrices Application to Special Relativity
In special relativity the motion of an inertial particle with respect to an inertial observer is described by a Lorentz-Poincar transformation. This transformation is associated to a -matrix belonging to the subgroup of orthochronous Lorentz matrices of determinant: they map the subset of timelike future oriented vectors into itself. They transform a h-orthonormal basis (associated to the inertial observer) into another h-orthonormal basis (associated to the particle).
The columns of such a matrix have a clear physical and geometrical interpre- tation: the first column is the 4-velocity of the particle (a unitary timelike 4-vector tangent to the worldline), and the three other columns define an orthonormal basis of the physical space of the particle. We turn now to the more general situation of a non inertial particle: the relative motion between two non inertial particles, or between a non inertial particle and another (inertial or non inertial) observer will be described by a time-dependent function with values in the group of Lorentz transformations. We thus naturally come to the notion of tangent boost along a worldline, we shall now study its main properties.
3.2. The Lie Group of Lorentz Matrices and Its Associated Lie Algebra
The shall denote by the subgroup of the Lie group of Lorentz matrices consisting of all orthochronous (Lorentz) matrices with determinant. It is a 6-dimensional submanifold of as defined by the 10 equations involving the 16 coordinates of the matrix S, obtained from the relation.
This group acts as a group of isometries on the Minkowski space. We shall now be interested in the tangent space of taken at the identity matrix I.
Let be a smooth curve on the manifold, and its tangent vector belonging to, the element of shall simply be:
From the relation we deduce that and that the covariant tensor of type, is skew-symmetric (this is obtained by derivating the relation), and so the element of the Lie algebra can be rewritten:
As a conclusion to this subsection, the Lie algebra is the linear space of matrices, where is a skew-symmetric covariant tensor of type.
The skew-symmetric tensor associated to the Lie bracket is
The exponential mapping from the Lie algebra to the group
defines a diffeomorphism from into (recall is the open ball of radius in).
3.3. Properties of the Lie Algebra Matrices
Every matrix belonging to the Lie algebra can be written
where and are spacelike vectors.
Note the relation
We then have the following proposition about the reduced forms of the matrices: Given any matrix of such that (usual inner product), there exists an h-orthonormal basis with respect to which can be written:
where and are two real numbers and where is a timelike 4-vector, the three other 4-vectors are spacelike.
If, setting and, the three reduced matrix forms, according to the three conditions, and respectively, shall be
We shall use following notations:
n1) Let be a h-orthonormal basis consisting of 4-vectors, where is a timelike vector and where is the basis of the orthogonal complement of the straight line.
n2) To every 3-vector a 4-vector can be associated, which obviously is space like. Note that this process leeds us to the definition of a linear mapping q.
Any 4-vector can be written. The inner product of two 4-vectors and shall be written, and we have the formula:
n3) With the aim of more elegant computations we shall write C the cross product, and the euclidean norms of the 3-vectors and C respectively. The following formulas will be useful:
n4) With the aim of studying the action of on 4-vectors, we write as:
Let be any 4-vector, we get the following formulas for the matrix products:
To obtain the reduced form of lets start with the study of. Indeed its characteristic polynomial simply factorizes where
is the minimal polynomial of the matrix, which means that we have the matrix relation
where and are the two zeros of. We wrote I and O for the identity and the zero -matrices.
Note by the way the formulas linking the roots of the polynomial (11)):
The first columns of the matrices and are the 4-vectors obtained by computing the product respectively, using (10):
The relation (12) means that the columns of the matrix generate the eigenspace of associated to the eigenvalue and that the columns of the matrix generate the eigenspace associated to. Let us write and these two 2-dimensional eigenspaces.
The eigenspace IIα associated to the eigenvalue α2:
Writing the vector defined by the first column of and, is an h-orthogonal basis of. Indeed, on the one hand:
shows that belongs to and on the other hand, using (n3) to compute we get:
Apart from the relation we also have:
which means that is timelike and that is spacelike. Also note the relation.
Writing now, , defines an orthonormal basis of the space like plane with:
The eigenspace IIω associated to the eigenvalue −ω2:
Writing the vector defined by the first column of and, is an h-orthogonal basis of. Moreover and are spacelike and is the orthogonal complement of. Here is an outline of the computations:
Apart from the relation we also have:
The plane is obviously spacelike, and writing:
constitutes an orthonormal basis of with the relations:
These two relations (19), with the former relations (18) linking and, show that the matrix gets the reduced form (7) in the h-orthonormal basis. Let us recall that is timelike and that the three other are spacelike; they define an orthonormal basis of, the orthogonal complement of the line.
The situation where A and B are orthogonal: In the case we need to discuss according to the sign of since the two roots of (12) are and and matrices and are of rank 2. The minimal polynomial (11) can be simplified and the relation (12) becomes:
1) Assume. is defined by and (16) with:
is the kernel of. It is generated by (first column of) and
As above, normalizing the four vectors and writing them respectively, the reduced form of the matrix in this new basis shall be the first matrix of (8)
2) Assume: Noting we have
is timelike since it is generated by the two vectors belonging to the kernel of:
The plane is space like. It is generated the first column of and:
As above, normalizing the four vectors (the first one being timelike) and writing them, the reduced form of in this new basis shall be the second matrix of (8)
3) Assume: when the minimal polynomial of is simply that is to say:
form an h-orthonormal basis and generate an eigenspace with the relations:
We thus obtain the third reduced form in (8).
The Lie group of special and orthochronous Lorentz matrices has four one- parameter subgroups which can be obtained by integrating the linear differential equation
where is one of the four reduced forms obtained above.
The solution of this equation is that is to say
4. Inertial Particles in Special Relativity
Let O and M be two inertial particles in the Minkowski space. Their worldlines and are two geodesic straight lines of generated by the timelike future oriented unitary 4-vectors t and by (which define the 4- velocities of O and M respectively).
Let be the h-orthonormal basis associated to O along and let us recall that is a basis of the hyperplane of passing through O and orthogonal to the worldline of O. This hyperplane is the physical space of O. We note t and the proper times of O and M respectively. We also denote the coordinates of M in the referential frame of O and
the 3-velocity of M. Using these notations, the 4-velocity can be written
with t he relations:
where is the Lorentz factor. All these quantities are constants.
In order to define the Lorentz-Poincar transform we may apply the orthonormalization Gram-Schmidt process to the basis. We thus obtain an h-orthonormal basis, where and where the three other vectors generate the basis of the physical space of M. This orthonormalization process directly gives the boost characterizing the relation between the two inertial particles:
In this result, is the column matrix of its components and is the unit matrix of size 3.
being a constant matrix, its associated matrix in the Lie algebra is the zero matrix. All this corresponds to the classical case of Special Relativity and can be summarized as follows: Any constant matrix defines a Lorentz transform relating two inertial particles.
1. The relation between O and M can be characterized by an infinity of Lorentz matrices. Each of them can be deduced from L by a left or a right multiplication of L with a pure rotation (a Lorentz matrix) R
where A is an orthogonal matrix of size 3. A left and a right multiplication correspond to a change of basis in the rest space of O and of M respectively.
2. The writing of the boost (20) can be simplified by choosing an appropriate basis of (recall that is the 4-velocity of O and let us note the 4-vector associated to the 3-velocity of M in). and are two orthogonal vectors in the Lorentz-Poincar?? transform plane. In fact, noting
We can define an h-orthonormal basis of this timelike plane by taking and. We thus obtain:
We also know that the two dimensional orthogonal complement is L- invariant. This can be seen by noting that the two 4-vectors and are orthogonal to and to and that they are linear- ly independant so that they form a basis. We can then construct an orthonormal basis of the spacelike plane which remains unchanged when orthonormalization process is applied:
These two vectors are eigenvectors of L associated to the double eigenvalue 1. We thus obtain a new h-orthonormal basis the transfert matrix beeing the Lorentz matrix Q:
is a pure rotation matrix () which only depends on the velocity direction.
the above expression can also be written
To summarize: there is a basis deduced from e through a space rotation of e for which the boost L can be written in the following canonical form:
With respect to, is the plane of the Lorentz transformation and is the invariant plane of that transformation.
5. Non Inertial Particles in Special Relativity. Tangent Boost along a Worldline
Let us now consider the case where O is an inertial particle and where M is not. Then, the wordline of M is no more a straight line and its 4-velocity is a vector field along. This leeds us to define the tangent boost along as being the boost of the inertial particle which coincides with M and the worldine of which is the tangent straight line at M. We thus obtain a field along where L is defined by (20). L being no more a constant
matrix, its associated matrix in the Lie algebra is no more the zero matrix. Before computing the 3-vectors A and B of the matrix, let us give some examples of using this latter.
5.1. Derivation Rule of a Vector X Defined by Its Components in the Referential Frame of M
Let us consider the two basis and, E being defined by the columns of L. Let and be the components of the vector in e and E respectively (). Let us derivate that relation with respect to t (or with respect to the proper time of M). Using then the left translation, we get:
where the subscripts e and E correspond to the basis e and E respectively. The above relation gives the derivative rule by its E-components that is the intrinsic vectorial relation:
Let us now apply that law to the 4-velocity of M the components of which are in E.
Equation (23) shows that the first column of is the 4-acceleration (notation n2 in paragraph 3.3) of M in E.
Now, let be a 4-vector defined by its components in and let us recall that is the 4-velocity of M. Noting the 3-vector, and the 4-accele- ration of M with, the 3-vector appears to be an instantaneous rotation defined by its components in E:
Changing into this last equation can also be written:
Note that there is a minor abuse of notation in the last line: B and W must be understood here as 3-vectors and no more as components in E as in previous lines. The term shows that B is an instantaneous rotation in the (physical) space of 3-vectors. It corresponds to Thomas rotation.
The matrix thus contains the 4-acceleration of M and the Thomas rotation. It therefore undoubtedly constitutes a valuable tool to describe the motion of any physical system.
5.2. Example of an Uniformly Accelerated Particle
In the referential frame of an inertial observer O, an uniform acceleration of M does not correspond to a constant 4-acceleration. In fact, the worldline of M is not a straigth line since it is not a geodesic. At two different points and, and are not parallel. In the case of an uniformly accelerated particule, we consequently only know that the norm of is a constant a. Moreover, in what follows, we will also consider that, for the inertial observer O, remains in a given plane. This plane is necessarily a timelike plane. The parametric equation of motion for M and its 4-velocity
The mere knowledge of permits to calcule the tangent boost L. Inserting into Equation (20) we get:
Let us then calculate its associated matrix in the Lie algebra
Using (25) in computing, the above equation gives:
where is the constant defined above (when, a is the norm of the 4-acceleration). Using the derivation rule, we obtain the components of the 4-acceleration in
Its components in are then obtained by a change of basis
Calculating we get the following conclusions: any uniformly accelerated particle is defined by a one-parameter subgroup of the Lie group,
and the 4-acceleration is uniform in the rest frame of M (note again that the basis E is defined by the columns of L). In an uniformly accelerated system, there is no Thomas rotation.
Let us now consider two nearby particles N and M, N being at rest with respect to M and their coordinates in being where. Let us calculate.
Knowing that X does not depend on, the derivation rule gives:
The components of in are then
where is a velocity (using we would get instead of). It is important to note that is not the 4-velocity of N and that
the proper time of N is not the same as the one of M. In fact, the norm of the 4-velocity of N, defined with its proper time s being 1, we obtain the following relation between and s:
This shows that in the case of a non inertial motion of M, it is impossible to synchronize the clocks in the rest frame of M.
Let us add that N has not the same acceleration as M. In fact, knowing that and consequently that, we get
5.3. Tangent Boost of a Worldline and Its Associated Matrix in the Lie Algebra in Special Relativity
In the referential frame of O, the parametric equations of the worldline are defined by cartesian coordinates where the parameter is the proper time of M:
Noting and the 3-velocity and the 3-accele- ration in the reference frame of O (with its propertime t) the 4-velocity and the 4-acceleration (first and second derivative of coordinates with respect to t) are:
The tangent boost (20) is:
and its associated matrix in the Lie algebra is
To summarize: using notations (6) we see that gives the complete dyna- mics of M. In:
• the 3-vector A is the acceleration of M in its rest frame:
• the 3-vector B gives the instantaneous Thomas rotation by its components in:
5.3.1. Writing the Tangent Boost and Its Associated Matrix in the Lie Algebra in a Rotating Frame. A first Insight on Thomas Rotation
The rotating basis is defined in the remark (2) of paragraph 4 but, in the present case, the rotation matrix Q now depends on the proper time of M. The tangent boost L in has the remarkable form (22). Our aim is to calculate the components of the matrix of the Lie algebra in the rotating frame in two ways:
1. Using the definition of in the moving referential frame
and applying the derivation rule to the tangent boost in
where is the antisymmetric matrix which defines the instantane- ous rotation of. Inserting this result in the previous equation gives the matrix of the Lie algebra of the boost as seen by the rotating observer
We thus obtain and
where and are the derivatives with respect to t of the three parameter defining (let us recall that).
2. Using (29) and (30) which give the 3-vectors A and B from and. we get:
Using, Equation (29) gives:
and Equation (30) gives the Thomas rotation:
To conclude: from the observer point of view, the L boost written in the rotating basis defines the rest frame with respect to. From the point of view, the two dimensional space of the Lorentz- Poincar transform, as well as its invariant space are not moving. The matrix Q defines the rotation of the rest frame of M with respect to.
These calculations show that we must clearly distinguish between the instantaneous rotation of (which is defined from the antisymmetric matrix) and the instantaneous Thomas rotation.
In order to get a better insight on Thomas rotation, let us consider the infinitesimal Lorentz matrix relating to:
A left-multiplication by of this result gives the Lorentz matrix
At first order, thus appears to be the product of an infinitesimal boost with an infinitesimal pure rotation (Thomas rotation):
We will see later that the Thomas rotation is a rotation of the rest frame of M with respect to the referential frame which is defined by the tangent boost.
5.3.2. Application to a Particle in Circular Motion at Constant Velocity
With respect to the frame of O, the parametric equations of the particule worldline are those of a circular helix with axis. Using cylindrical coordinates
where and are constant and where t is a function of.The 4-velocity is:
Noting that the Lorentz factor is constant, the 4-acceleration is:
where is the covariant derivative in the direction of expressed in cylindrical coordinate. In order to calculate the tangent boost we have to express in the h-orthonormal system obtained by applying the Gram-Schmidt orthonormalisation process to the natural basis and starting with the 4-vector. In the present case, calculations are very simple since the basis is already h-orthogonal. We get. The tangent boost is then defined by using (20):
Let us recall that the -columns give the referential frame of M, and that the 4-vectors are defined from their components in (e). Let us also note that is the plane of the Poincar-Lorentz transform, being the invariant orthogonal supplementary plane of the transformation.
The matrix of the Lie algebra is
It directly gives the 3-acceleration and the instantaneous Thomas rotation (which both are in the physical space of M). Let us note that it is also possible to obtain the 3-vectors A and B of from Equations (29) and (30): using and we in fact obtain the 3-acceleration and the instantaneous Thomas rotation in E (E is defined by the column vectors of L):
In order to understand the meaning of Thomas rotation, let us consider a gyroscope and let us recall the definition of a gyroscopic torque along a worldline as given in  and in   :
A gyroscopic torque along a worldline the 4-velocity and the proper time of which are V and respectively is a 4-vector G defined along, orthogonal to V and such that its derivative with respect to is proportional to V, that is to say:
These relations permit to calculate k. Noting the 4-acceleration, we in fact get:
The proportionality condition implies that the 4-vector (which belongs to the physical space of M along the worldline) rotates in that space. In fact, let us write the differential Equation (32) with respect to the components of in. Noting in, the components of in the inertial frame are then defined by:
In the inertial referential frame, Equation (32) thus becomes:
Using the covariant derivative in cylindrical coordinates and noting derivatives with respect to by accentuated characters we get:
Identifying this result with gives and the three differential equations (note that):
Taking initial conditions, the solutions of these differential equations are
Figure 1 shows the rotation of a gyroscope initially oriented following the x axis (), in when varies from 0 to that is to
say when M goes a 180 degree turn. It shows that in that case, the gyroscope indicates a half turn plus a rotation which corresponds to the Thomas rotation (in the clockwise direction). The gyroscope rotation in the plane is also shown when M goes a complete rotation (360-degree) in Figure 2. In that case, the gyroscope indicates a complete turn plus a part. For the sake of clarity, we only show this supplementary part.
Figure 1. The gyroscope rotation in the plane when M goes a 180 degree turn. Numerical values are and.
Figure 2. Gyroscope rotation in the plane when M goes a complete rotation (360-degree). In that case, the gyroscope indicates a complete turn plus a part. For the sake of clarity, we only show this supplementary part. We used here for R and the values.
It is also possible to highlight the Thomas rotation by applying the derivation rule (23) to (which is defined by its components in). Noting
and using (23) in that moving frame:
The left hand side of this equation is the Fermi-Walker derivative of in the direction. Using this last equation becomes
Consequently, the gyroscope rotates with respect to in the opposite direction to the instantaneous Thomas rotation. taking again its initial orientation after a complete period, this gap shows that the gyroscopes also rotate with respect to the inertial referential frame
It can be noted that the solution of (33) (with the initial condition also is the Fermi-Walker parallel transport of along    .
7. Conclusion: Langevin’s Twins and Thomas Precession
The main results of every dynamical system are contained in the tangent boost L (which gives its 4-velocity and the basis vectors of its rest frame), and its associated matrix of the Lie algebra which gives its acceleration and the instantaneous Thomas rotation.
The age of the electron with respect to the atom nucleus is then obtained by integrating over one period T. In the case of a uniform circular motion its value is
The gyroscope rotation in the physical space can be obtained by integrating over one period between and. We get
Figure 3. Illustration of “Langevin’s twins” in the case of an electron rotating on a circular orbit around the atom nucleus. The twins are denoted and M respectively. The straight line parallel to the axis of the cylinder is the worldline of; the helix is that of M.
We thus see that in the case of Langevin’s twins, (here, in the case of a uniform circular motion), when the twin who made a journey into space returns home he is not only younger than the twin who stayed on Earth but he is also disorientated with respect to the terrestrial frame because his gyroscope has turned with respect to earth referential frame. This effect is illustrated in Figure 3 in the case of an electron rotating on a circular orbit around the atom nucleus.