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Dedicated to those who never had the opportunity to receive a good education

Lorentz transformations

In the present article, the Lorentz transformations of the space-time coordinates, velocities, energy, momentum, accelerations and forces, are presented in a condensed form. It is explained how the Lorentz transformation for a boost in an arbitrary direction is obtained, and the relation between boosts in arbitrary directions and spatial rotations is discussed. The case when the respective coordinates axis of one of the inertial systems are not parallel to those of the other inertial system (This case is rarely discussed in the literature) is also analyzed. Besides, the derivation of the Lorentz transformations of the components of the spatial velocity is discussed, and the four-velocity is defined. The manner of introducing here the four-velocity is guided by mathematical intuition rather than by a physical motivation. It was decided to do it that way because it is easier, faster, more clear, and because the more physically motivated form of introducing it, it can be better appreciated after understanding the presentation given here. Nevertheless,  the traditional way of introducing the four-velocity will be studied in another article.

After introducing the Lorentz transformations of the velocity, it is advantageous to deduce immediately the Lorentz transformation for the energy, momentum, and acceleration. Indeed, the Lorentz transformation of energy and momentum is an immediate consequence of the transformation law for the four-velocity, and the transformation of the acceleration follows the same procedure as the derivation of the transformation of the spatial velocity.

It was decided to include here the Lorentz transformation of the force because it follows immediately from the transformation law of the four-momentum. The four-vector associated with the force is constructed similarly as the four-moment was, starting from the transformation law of the velocity.

Relativistic transformation of the space-time coordinates

Let’s consider two inertial reference systems S and S^{\prime}. Suppose that these systems are in relative motion at constant velocity one with respect to the other. If S^{\prime} travels in straight line with velocity \vec{v} respect to S, then, clearly S travels with respect to S^{\prime} in straight line with velocity -\vec{v}. Let’s suppose that there is a Cartesian system of coordinates in both, S and S^{\prime}, that they are equally oriented, and that their respective coordinate axes are parallel. Besides, let’s suppose that at t=0 in S, we have t^{\prime}=0 in S^{\prime}, and that the origin of both coordinates systems coincides at that time.

Lorentz transformations
The origin of the inertial coordinate system S^{\prime} travels with constant velocity \vec{v} along the x axis of S. Each of the coordinate axis in S^{\prime} are parallel to the respective axis in S.

Under these conditions, the relativistic transformation of the coordinates of space and time from S to S^{\prime}, are:

(1)   \begin{equation*} \begin{split} x^{\prime}&=\gamma_0\left(x-vt\right)\\ y^{\prime}&=y\\ z^{\prime}&=z\\ t^{\prime}&=\gamma_0\left(t-vx/c^2\right) \end{split} \end{equation*}

The derivation of this transformation law from the postulates of special relativity will be addressed in another article. Notice that the coordinates perpendicular to the direction of the relative motion between the systems S and S^{\prime},  in this case the coordinates of points in the yz-plane, are not affected by the velocity \vec{v}, i.e., they remain identical in both systems. This is a general feature of Lorenz transformations. The \gamma_0 factor appearing in these transformations is defined as

(2)   \begin{equation*} \begin{split} \gamma_0&=\frac{1}{\sqrt{1-\beta_0^2}}\\ \beta_0&=v/c \end{split} \end{equation*}

The sub-index zero in \gamma_0, and \beta_0, is to distinguish the gamma calculated with the relative velocity v between the two inertial systems S, and S^{\prime}, from the gamma which is calculated with the velocity u of a particle moving in S. The latter is simply called \gamma=1/\sqrt{1-\beta^2}, where \beta=u/c, i.e., without sub index.

The inverse of the transformation Eq.(1) is calculated by exchanging the role of the coordinates (t,x,y,z), and (t^{\prime},x^{\prime},y^{\prime},z^{\prime}), and changing the sign of v:

(3)   \begin{equation*} \begin{split} x&=\gamma_0\left(x^{\prime}+vt^{\prime}\right)\\ y&=y^{\prime}\\ z&=z^{\prime}\\ t&=\gamma_0\left(t^{\prime}+vx^{\prime}/c^2\right) \end{split} \end{equation*}

For convenience, usually the time t is replaced by the new variable x^0=ct, and the spatial coordinates are denoted x^1\equiv x, x^2\equiv y, and x^3\equiv z. The position of the indexes of the space-time coordinates as superscript, is a universal convention without exception. The super indexes are called contravariant. There are coordinates with sub-indexes also. Coordinates with sub-indexes refer to dual spaces and are called covariant. The meaning of the terms covariant and contravariant will be explained in another article. With the super-index notation, the transformations given in Eq.(1) can be written in matrix form:

(4)   \begin{equation*} \begin{pmatrix}x^{\prime 0}\\x^{\prime 1}\\x^{\prime 2}\\x^{\prime 3}\end{pmatrix}= \begin{pmatrix} \gamma_0 & -\gamma_0\beta_0 & 0 & 0 \\ -\gamma_0\beta_0 & \gamma_0 & 0 & 0 \\ 0&0&1&0 \\ 0&0&0&1 \end{pmatrix} \begin{pmatrix} x^0\\x^1\\x^2\\x^3\end{pmatrix} \end{equation*}

    \[\centerline{Boost in the $x$ direction}\]


Strictly speaking, Eq.(4) is what is called boost in the positive x direction. Boost in the directions y and z will be given by respective matrices which look like similar:

(5)   \begin{equation*} \begin{pmatrix}x^{\prime 0}\\x^{\prime 1}\\x^{\prime 2}\\x^{\prime 3}\end{pmatrix}= \begin{pmatrix} \gamma_0& 0 & -\gamma_0\beta_0 & 0  \\ 0&1&0&0\\ -\gamma_0\beta_0 &0& \gamma_0 & 0  \\ 0&0&0&1 \end{pmatrix} \begin{pmatrix} x^0\\x^1\\x^2\\x^3\end{pmatrix}\\ \end{equation*}

    \[\centerline{Boost in the $y$ direction}\]



(6)   \begin{equation*} \begin{pmatrix}x^{\prime 0}\\x^{\prime 1}\\x^{\prime 2}\\x^{\prime 3}\end{pmatrix}= \begin{pmatrix} \gamma_0& 0 & 0 & -\gamma_0\beta_0\\ 0&1&0&0\\ 0&0&1&0\\ -\beta_0\gamma_0&0&0&\gamma_0 \end{pmatrix} \begin{pmatrix} x^0\\x^1\\x^2\\x^3\end{pmatrix}\\ \end{equation*}

    \[\centerline{Boost in the $z$ direction}\]


The Boosts are usually called Lorentz transformations. Nevertheless, it has to be clear that, strictly speaking,

any transformation of the space-time coordinates, that leaves invariant the value I of the quadratic form I=(x^0)^2-(x^1)^2-(x^2)^2-(x^3)^2, is a Lorentz transformation.

Therefore, rotations of the spacial coordinates, time reversal t\to-t, parity (x,y,z)\to(-x,-y,-z), and any combination of them, are also Lorentz transformations. In matrix form they look as follows:

(7)   \begin{equation*} \begin{pmatrix}x^{\prime 0}\\x^{\prime 1}\\x^{\prime 2}\\x^{\prime 3}\end{pmatrix}= \begin{pmatrix} -1& 0 & 0 & 0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix} \begin{pmatrix} x^0\\x^1\\x^2\\x^3\end{pmatrix}\\ \end{equation*}

    \[\centerline{Time reversal transformation}\]


(8)   \begin{equation*} \begin{pmatrix}x^{\prime 0}\\x^{\prime 1}\\x^{\prime 2}\\x^{\prime 3}\end{pmatrix}= \begin{pmatrix} 1& 0 & 0 & 0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix} \begin{pmatrix} x^0\\x^1\\x^2\\x^3\end{pmatrix}\\ \end{equation*}

    \[\centerline{Parity transformation}\]


(9)   \begin{equation*} \begin{pmatrix}x^{\prime 0}\\x^{\prime 1}\\x^{\prime 2}\\x^{\prime 3}\end{pmatrix}= \begin{pmatrix} 1& 0 & 0 & 0\\ 0&\cos\theta&\sin\theta&0\\ 0&-\sin\theta&\cos\theta&0\\ 0&0&0&1 \end{pmatrix} \begin{pmatrix} x^0\\x^1\\x^2\\x^3\end{pmatrix}\\ \end{equation*}

    \[\centerline{Rotation by an angle $\theta$ about the $z$ axis}\]


The quadratic form I(x^0,x^1,x^2,x^3)=(x^0)^2-(x^1)^2-(x^2)^2-(x^3)^2 and its meaning will be discussed in another article. For the time being, we just indicate that I, stands for invariant, or more precisely, relativistic invariant. This means that I takes the same numerical value without regard to being calculated by using the coordinates in S or in S^{\prime}.

Rotation of the spatial coordinate axis about an arbitrary axis can be parametrized in different forms and needs to be studied more extensively. We will study rotations in another article. In Eq.(9) we have just given the example of the matrix representing the rotation of the spatial coordinates about the z axis. Spatial rotations do not change the time coordinate, and therefore, the first row and the first column, in the matrix of a spatial rotation, are the same as those in the matrix in Eq.(9).

Uniform translations of the space-time coordinates leave the value of I invariant also, but space-time translations are not called Lorentz transformations. The set of transformations, namely, Lorentz and translations, when considered together, are called Poincare Transformations. Some authors, call homogeneous to Lorentz transformations that do not include translations and call non-homogeneous to those Lorentz transformations that do include translations.

The group property, its associated Lie algebras, their representations, and several applications, will be studied in other articles.

Boost in an arbitrary direction

When the inertial system S^{\prime} travels at constant velocity \vec{v} in an arbitrary direction, the Lorentz transformation of the space time coordinates looks more complicated. To find the corresponding transformation, we first express the vector position \vec{r}=x\hat{x}+y\hat{y}+z\hat{x}, as the sum of its component \vec{r}_{||} parallel to \vec{v}, and its component \vec{r}_{\bot} perpendicular to \vec{v}:

(10)   \begin{equation*} \begin{split} \vec{r}&=\vec{r}_{||}+\vec{r}_{\bot}\\ \vec{r}_{||}&=\frac{\left(\vec{r}\cdot\vec{v}\right)}{|\vec{v}|^2}\ \vec{v}\\ &=\frac{\left(\vec{r}\cdot\vec{\beta}\right)}{\beta^2}\ \vec{\beta}\\ \vec{r}_{\bot}&=\vec{r}-\frac{\left(\vec{r}\cdot\vec{\beta}\right)}{\beta^2}\ \vec{\beta} \end{split} \end{equation*}

Our task now is to find the transformation rule of each component separately.

Lorentz transformations

The reason for separating \vec{r} in this form, is that the component \vec{r}_{\bot}, perpendicular to \vec{v} doesn’t change when we transform the spatial coordinates from S to S^{\prime}, while the parallel component \vec{r}_{||}, transforms just as a boost along the x axis:

(11)   \begin{equation*} \begin{split} \vec{r}^{\ \prime}_{\bot}&=\vec{r}_{\bot}\\ \vec{r}^{\ \prime}_{||}&=\gamma_0\left(\vec{r}_{||}-\vec{v}t\right)\\ \vec{r}^{\ \prime}&=\vec{r}^{\ \prime}_{||}+\vec{r}^{\ \prime}_{\bot}\\ \end{split} \end{equation*}

If we substitute Eq.(10) in Eq.(11), then, after some algebra, we arrives to:

(12)   \begin{equation*} \vec{r}^{\ \prime}=\vec{r}+\left\{(\gamma_0-1)\frac{\vec{r}\cdot\vec{\beta}_0}{\beta^2}-\gamma_0ct\right\}\vec{\beta}_0 \end{equation*}

The transformation of the time coordinate is the same as in a boost in the x direction, but with |\vec{v}||\vec{r}_{||}|=\vec{v}\cdot\vec{r} in place of vx:

(13)   \begin{equation*} ct^{\prime}=\gamma_0\left(ct-\vec{\beta}_0\cdot\vec{r}\right) \end{equation*}

Remembering that x^0=ct, x^1=x, x^2=y, and x^3=z, we can rewrite Eq.(13) and Eq.(12) (in that order), in matrix form as follows:

    \[ \begin{pmatrix}x^{\prime\ 0}\\x^{\prime\ 1}\\x^{\prime\ 2}\\x^{\prime\ 3}\end{pmatrix}=\Lambda\begin{pmatrix}x^{0}\\x^{1}\\x^{2}\\x^{3}\end{pmatrix} \]

Where the matrix \Lambda of the transformation is

(14)   \begin{equation*} \Lambda=\begin{pmatrix} \gamma_0 & -\gamma_0\beta_{0,x} & -\gamma_0\beta_{0,y} & -\gamma_0\beta_{0,z}\\ -\gamma_0\beta_{0,x} & 1+\frac{(\gamma_0-1)\beta^2_{0,x}}{\beta_0^2} & \frac{(\gamma_0-1)\beta_{0,x}\beta_{0,y}}{\beta_0^2} & \frac{(\gamma_0-1)\beta_{0,x}\beta_{0,z}}{\beta_0^2}\\ -\gamma_0\beta_{0,y} & \frac{(\gamma_0-1)\beta_{0,y}\beta_{0,x}}{\beta_0^2} & 1+\frac{(\gamma_0-1)\beta_{0,y}^2}{\beta_0^2} & \frac{(\gamma_0-1)\beta_{0,y}\beta_{0,z}}{\beta_0^2}\\ -\gamma_0\beta_{0,z} & \frac{(\gamma_0-1)\beta_{0,z}\beta_{0,x}}{\beta_0^2} & \frac{(\gamma_0-1)\beta_{0,z}\beta_{0,y}}{\beta_0^2} & 1+\frac{(\gamma_0-1)\beta_{0,z}^2}{\beta_0^2} \end{pmatrix} \end{equation*}

    \[\centerline{Boost in an arbitrary direction parallel to $\vec{\beta}_0=\vec{v}/c$}\]

(14)   \begin{align*}\Lambda=&\left(\begin{matrix} \gamma_0 & -\gamma_0\beta_{0,x} \\  -\gamma_0\beta_{0,x} & 1+\frac{(\gamma_0-1)\beta_{0,x}^2}{\beta_0^2}\\-\gamma_0\beta_{0,y} & \frac{(\gamma_0-1)\beta_{0,y}\beta_{0,x}}{\beta_0^2} \\ -\gamma_0\beta_{0,z} & \frac{(\gamma_0-1)\beta_{0,z}\beta_{0,x}}{\beta_0^2}\end{matrix}\right. \\&\left. \begin{matrix} -\gamma_0\beta_{0,y} & -\gamma_0\beta_{0,z}\\ \frac{(\gamma_0-1)\beta_{0,x}\beta_{0,y}}{\beta_0^2} & \frac{(\gamma_0-1)\beta_{0,x}\beta_{0,z}}{\beta_0^2}\\ 1+\frac{(\gamma_0-1)\beta_{0,y}^2}{\beta_0^2}& \frac{(\gamma_0-1)\beta_{0,y}\beta_{0,z}}{\beta_0^2} \\\frac{(\gamma_0-1)\beta_{0,z}\beta_{0,y}}{\beta_0^2} &  1+\frac{(\gamma_0-1)\beta_{0,z}^2}{\beta_0^2} \end{matrix}\right) \end{align*}

    \[\centerline{Boost in an arbitrary direction parallel to $\vec{\beta}_0=\vec{v}/c$}\]

Lorentz transformations
Boost in an arbitrary direction. The coordinate axis in S^{\prime} are parallel to the respective axes in S.


One could think that the expression for the boost in an arbitrary direction is more general than that for a boost along the x axis. In reality, this is not the case. Indeed, one can express the boost in an arbitrary direction as the product of rotations and the boost along the x axis. To this end, Consider the matrices R, and R^{\prime} that represent the rotations of the spatial coordinate axis, necessary to re-orientate the x, and x^{\prime} axes in the direction of the boost in S and S^{\prime}, respectively. The new coordinates in S are R(x^0,x^1,x^2,x^3). Similarly R^{\prime}(x^{0\ \prime},x^{1\ \prime},x^{2\ \prime},x^{3\ \prime}) , are the new coordinate axis in S^{\prime} (Remember that spatial rotations do not alter the time coordinate).

Lorentz transformations
Boost along an arbitrary direction when the respective coordinate systems in S and S^{\prime}, are not parallel between them. The reference system in blue has its axes parallel to those in S. They are shown to indicate that the axes in S^{\prime} are not parallel to those in S.

The relation between these new coordinates is just a boost along the new (i.e., rotated) x axis.

(15)   \begin{equation*} R^{\prime}\begin{pmatrix}x^{\prime 0}\\x^{\prime 1}\\x^{\prime 2}\\x^{\prime 3}\end{pmatrix}= \begin{pmatrix} \gamma_0 & -\gamma_0\beta_0 & 0 & 0 \\ -\gamma_0\beta_0 & \gamma_0 & 0 & 0 \\ 0&0&1&0 \\ 0&0&0&1 \end{pmatrix} R\begin{pmatrix} x^0\\x^1\\x^2\\x^3\end{pmatrix} \end{equation*}

R brings the x axis in S to the direction of \vec{v}, while R^{\prime} does the same thing with the x^{\prime} axis in S^{\prime}. After these rotations, the respective coordinate axis in S and S^{\prime} are parallel one the other, and the new x^{\prime} axis travels along the new x axis, and therefore, the transformation between the new axes is a boost along the x axis.

Let’s represent the boost along the x axis by \Lambda_x, and by \Lambda_{v} the boost along the direction of \vec{v}. Notice that \Lambda_v is the Lorentz transformation between X=(x^0, x^1, x^2,x^3) and X^{\prime}=(x^{\prime\ 0}, x^{\prime\ 1}, x^{\prime\ 2}, x^{\prime\ 3}), i.e., we have X^{\prime}=\Lambda_vX. We see from Eq.(15) that X^{\prime}=R^{\prime\ -1}\Lambda_xRX. This implies that \Lambda_v=R^{\prime\ -1}\Lambda_xR. This is the expression for the boost when the original axis in S and S^{\prime} are not parallel one to the other. If the original axis are parallel, then we will have R^{\prime}=R, and in such a case, we will have \Lambda_v=R^{-1}\Lambda_xR. As an example, let’s obtain the boost along the y axis, by performing first a rotation by \theta=\pi/2 about the z axis, to bring the x axis to the y axis. The rotation matrix is obtained from Eq.(9):

(16)   \begin{equation*} R= \begin{pmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&-1&0&0\\ 0&0&0&1 \end{pmatrix} \end{equation*}

The inverse is obtained from Eq.(9) with \theta=-\pi/2:

(17)   \begin{equation*} R^{-1}= \begin{pmatrix} 1&0&0&0\\ 0&0&-1&0\\ 0&1&0&0\\ 0&0&0&1 \end{pmatrix} \end{equation*}

One can check that R^{-1}\Lambda_xR=\Lambda_y, where \Lambda_x, and \Lambda_y are given in Eq.(4) and Eq.(5) respectively.

Relativistic transformation of the velocity

Let’s consider the transformation of the components of the velocity for a boost in the positive direction of the x axis. The components of the velocity in the reference system S are u_x=dx/dt, u_y=dy/dt, and u_z=dz/dt. In the reference system S^{\prime} the components of the velocity are calculated with the same derivatives but using the primed coordinates, i.e., u^{\prime}_{x}=dx^{\prime}/dt^{\prime}, etc.

In order to find the transformation rules for the velocity, let’s make an intermediate calculation first, namely, derive both sides of Eq.(1) with respect to t:

(18)   \begin{equation*} \begin{align} \frac{dx^{\prime}}{dt}&=\gamma_0\left(u_x-v\right)\\ \frac{dy^{\prime}}{dt}&=u_y\\ \frac{dz^{\prime}}{dt}&=u_z\\ \frac{dt^{\prime}}{dt}&=\gamma_0\left(1-vu_x/c^2\right) \end{align} \end{equation*}

Now use the chain rule as in the following example:

(19)   \begin{equation*} \begin{split} u^{\prime}_x&=\frac{dx^{\prime}}{dt^{\prime}}\\ &=\frac{dx^{\prime}}{dt}\frac{dt}{dt^{\prime}}\\ &=\frac{dx^{\prime}}{dt}\left(\frac{dt^{\prime}}{dt}\right)^{-1}\\ \end{split} \end{equation*}

Applying the preceding procedure, we can calculate the components u^{\prime}_x, u^{\prime}_y, and u^{\prime}_z, of the velocity in S^{\prime}. The resulting transformation rules are

(20)   \begin{equation*} \begin{split} u^{\prime}_x&=\frac{u_x-v}{1-\beta_0 u_x}\\ u^{\prime}_y&=\frac{u_y}{\gamma_0\left(1-\beta_0 u_x\right)}\\ u^{\prime}_z&=\frac{u_z}{\gamma_0\left(1-\beta_0 u_x\right)}\\ \end{split} \end{equation*}

Often it is convenient to normalize the velocities dividing by the speed of light c. when this is done, one defines \vec{\beta}=\vec{u}/c, \beta_x=u_x/c, etc. In terms of normalized velocities, the transformation rule becomes

(21)   \begin{equation*} \begin{split} \beta^{\prime}_x&=\frac{\beta_x-\beta_0}{1-\beta_0 \beta_x}\\ \beta^{\prime}_y&=\frac{\beta_y}{\gamma_0\left(1-\beta_0 \beta_x\right)}\\ \beta^{\prime}_z&=\frac{\beta_z}{\gamma_0\left(1-\beta_0 \beta_x\right)}\\ \end{split} \end{equation*}

The relativistic transformation of velocities is related to the phenomenon of aberration, which consists in the change of direction of motion of  either a particle or a light ray, when seen from different inertial systems in relative motion.


In the preceding paragraph we found the rule for transforming the spatial velocity from S to S^{\prime}. We can use the expressions given in Eq.(21) to calculate \gamma^{\prime}=1/\sqrt{1-\beta^{\prime 2}}. After some algebra one arrives to

(22)   \begin{equation*} \gamma^{\prime}=\gamma_0\gamma\left(1-\beta_0\beta_x\right) \end{equation*}

If we multiply both sides of Eq.(21) by Eq.(22), we obtain the following relations:

(23)   \begin{equation*} \begin{split} \gamma^{\prime}\beta^{\prime}_x&=\gamma_0\gamma\left(\beta_x-\beta_0\right)\\ \gamma^{\prime}\beta^{\prime}_y&=\gamma\beta_y\\ \gamma^{\prime}\beta^{\prime}_z&=\gamma\beta_z\\ \end{split} \end{equation*}

The set of equations (22) and (23) can be rewritten in matrix form:

(24)   \begin{equation*}\begin{pmatrix}\gamma^{\prime}\\\gamma^{\prime}\beta^{\prime}_x\\\gamma^{\prime}\beta^{\prime}_y\\\gamma^{\prime}\beta^{\prime}_z\end{pmatrix}= \begin{pmatrix} \gamma_0 & -\gamma_0\beta_0 & 0 & 0 \\ -\gamma_0\beta_0 & \gamma_0 & 0 & 0 \\ 0&0&1&0 \\ 0&0&0&1 \end{pmatrix} \begin{pmatrix} \gamma\\ \gamma\beta_x\\ \gamma\beta_y\\ \gamma\beta_z\end{pmatrix} \end{equation*}


This means that the vector of four components \left(\gamma,\gamma\vec{\beta}\right), which is calculated in S, and the corresponding one \left(\gamma^{\prime},\gamma^{\prime}\vec{\beta}^{\prime}\right), calculated in S^{\prime}, are related by a Lorentz transformation in exactly the same way as the space-time coordinates.

When a set of four components in S is related to another set of four components in S^{\prime} in the same way as the coordinates of space-time are, the set of four components is called four-vector. In special relativity, the components of four vectors are denoted with a super-index of the Greek alphabet, which runs from 0 to 3. Thus, the space-time coordinates are written x^{\mu}, \mu=0,1,2,3. The range of values of the super index is usually left understood and is not explicitly indicated. The index has to be up and is called contravariant index. The corresponding vector is called contravariant. The convention is that any set of four quantities defined in S, written in the form A^{\mu}, is called four-vector whenever the expression A^{\prime\mu} of these quantities in S^{\prime}, are obtained through a Lorentz transformation.

We see that \left(\gamma,\gamma\vec{\beta}\right) is a four vector. If we multiply this four-vector by the speed of light c, we obtain

(25)   \begin{equation*} u^{\mu}:=\gamma\left(c,\vec{u}\right), \end{equation*}

which is called four-velocity.

Explicitly, Eq.(25) reads:

(26)   \begin{equation*} \begin{split} u^{0}&=c/\sqrt{1-u^2/c^2}\\ u^{1}&=u_x/\sqrt{1-u^2/c^2}\\ u^{2}&=u_y/\sqrt{1-u^2/c^2}\\ u^{3}&=u_z/\sqrt{1-u^2/c^2}\\ \end{split} \end{equation*}

In the same form that we calculate the invariant length of the space-time four-vector x^{\mu}:=(x^0,x^1,x^2,x^3), namely, (x^0)^2-(x^1)^2-(x^2)^2-(x^3)^2, we can calculate the invariant length of the four-velocity u^{\mu}:=(u^0,u^1,u^2,u^3), which is (u^0)^2-(u^1)^2-(u^2)^2-(u^3)^2. If we use the expressions given in Eq.(26) to calculate explicitly the value of this ‘four-norm’ of the four-vector velocity, we obtain

(27)   \begin{equation*}(u^0)^2-(u^1)^2-(u^2)^2-(u^3)^2=c^2 \end{equation*}

This means that we get a constant that has the same value in all inertial reference system. We conclude that independent of the magnitude of the three dimensional velocity, the four-velocity has always the same magnitude. In the particular case of a body at rest in S, we have \gamma=1, and thus its four-velocity is (c,0,0,0). The invariant I associated to this four-velocity is c^2. From the point of view of S^{\prime}, the body is traveling with velocity -v. therefore the four-vector in S^{\prime}, has components \gamma_0(c,-v,0,0), and its magnitude is c^2 also.

Relativistic transformation of Energy and momentum

The relativistic expression for the total energy E, and three-dimensional momentum \vec{p}, of a free particle whose rest mass is m_0, are:

(28)   \begin{align*} E&=\gamma m_0c^2\\ \vec{p}&=\gamma m_0\vec{u}=\gamma m_0c\vec{\beta} \end{align*}

Where as usual \gamma=1/\sqrt{1-\beta^2}, with \vec{\beta}=\vec{u}/c, and \beta^2=\vec{\beta}\cdot\vec{\beta}. The corresponding expressions in S^{\prime} are

(29)   \begin{align*} E^{\prime}&=\gamma^{\prime} m_0c^2\\ \vec{p}^{\ \prime}&=\gamma^{\prime} m_0c\vec{\beta}^{\prime} \end{align*}

Let’s consider the transformation law for the energy first. We just have to substitute the expression for \gamma^{\prime} in the expression for E^{\prime}:

(30)   \begin{equation*} \begin{split} E^{\prime}&=\gamma^{\prime}m_0c^2\\ &=\gamma_0\gamma\left(1-\beta_0\beta_x\right)m_0c^2 \end{split} \end{equation*}

With help of Eq.(28), the latter equation can be rewritten in the form

(31)   \begin{equation*} \frac{1}{c}E^{\prime}=\gamma_0\frac{1}{c}E-\beta_0\gamma_0p_x \end{equation*}

To find the transformation law of the components of the linear momentum we just have to multiply both sides of Eq.(23) by m_0c, and use Eq.(28). This leads to

(32)   \begin{equation*} \begin{split} p_x^{\prime}&=\gamma_0p_x-\beta_0\gamma_0\frac{1}{c}E\\ p_y^{\prime}&=p_y\\ p_z^{\prime}&=p_z \end{split} \end{equation*}

If we put together the equations for the transformation law of the energy and momentum, we find that they form a four-vector (E/c,\vec{p}), i.e., the relation of this vector with that one calculated in S^{\prime}, is trough a Lorentz transformation, just as the space-time coordinates:

(33)   \begin{equation*} \begin{pmatrix}E^{\prime}/c\\p_x^{\prime}\\p_y^{\prime}\\p_z^{\prime}\end{pmatrix}= \begin{pmatrix} \gamma_0 & -\gamma_0\beta_0 & 0 & 0 \\ -\gamma_0\beta_0 & \gamma_0 & 0 & 0 \\ 0&0&1&0 \\ 0&0&0&1 \end{pmatrix} \begin{pmatrix} E/c\\p_x\\p_y\\p_z\end{pmatrix} \end{equation*}

The four-vector energy-momentum is thus defined as p^{\mu}:=(E/c,\vec{p}), where E, and \vec{p} are defined in Eq.(28).

The energy-momentum relativistic invariant

The quadratic form I can be applied to any four vector A^{\mu}. Thus, by I(A^{\mu}) we mean I(A^0,A^1,A^2,A^3)=(A^0)^2-(A^1)^2-(A^2)^2-(A^3)^2. Let’s calculate the value I(p^{\mu}), we will use the usual notation p^2=\vec{p}\cdot\vec{p} for the square of the magnitude of a three-dimensional vector.

(34)   \begin{equation*}\begin{split} I(p^{\mu})&=(E/c)^2-(p)^2\\ &=\gamma^2m_0^2c^2-\gamma^2m_0^2c^2\beta^2\\ &=\gamma^2m_0^2c^2\left(1-\beta^2\right)\\ &=m_0^2c^2 \end{split} \end{equation*}

In the last line in Eq. (34), we made use of the identity \gamma^2(1-\beta^2)=1, which is just another form of rewriting the definition of \gamma=1/\sqrt{1-\beta^2}.

Owing to the invariance of I, we will obtain the same value in any other inertial system I(p^{\prime\mu})=I(p^{\mu})=m_0c^2. In particular, in the rest frame of the particle, its spatial momentum \vec{p}=\vec{0} is zero, and therefore, for a particle at rest, we have E=m_0c^2. In general, from the first and the last line in Eq.(34) we obtain the identity (E/c)^2-p^2=m_0^2c^2, which can be rewritten in the form

(35)   \begin{equation*} E^2=m_0^2c^4+p^2c^2 \end{equation*}

This is the famous relativistic energy-momentum relation. In this relation, E is the total energy of a free particle.

Relativistic transformation of the acceleration

We can use the same procedure used to find the transformation of the velocity. First of all, the acceleration is the time derivative of the velocity, i.e., \vec{a}=d\vec{u}/dt. The same definition of acceleration is valid in the reference system S^{\prime}, i.e., \vec{a}^{\prime}=d\vec{u}^{\prime}/dt^{\prime}. Therefore, all what we have to do is to calculate the derivative of the expressions for each component of the velocity. For example

(36)   \begin{equation*} \begin{split} a_x^{\prime}&=\frac{du_x^{\prime}}{dt^{\prime}}\\ &=\frac{du_x^{\prime}}{dt}\frac{dt}{dt^{\prime}}\\ &=\frac{du_x^{\prime}}{dt}\left(\frac{dt^{\prime}}{dt}\right)^{-1} \end{split} \end{equation*}

After calculating the corresponding derivatives, and doing some algebraic manipulations, we arrive to the following expressions:

(37)   \begin{equation*} \begin{split} a_x^{\prime}&=\frac{a_x}{\gamma_0^2\left(1-\beta_0\beta_x\right)^3}\\ a_y^{\prime}&=\frac{a_y}{\gamma_0^2\left(1-\beta_0\beta_x\right)^2}\\ &\ \ +\frac{\beta_0\beta_ya_x}{\gamma_0^2\left(1-\beta_0\beta_x\right)^3}\\ a_z^{\prime}&=\frac{a_z}{\gamma_0^2\left(1-\beta_0\beta_x\right)^2}\\ &\ \ +\frac{\beta_0\beta_za_x}{\gamma_0^2\left(1-\beta_0\beta_x\right)^3} \end{split} \end{equation*}

The relativistic transformation of the acceleration is complicated and is not used so frequently as the  energy-momentum and space-time coordinates four-vectors. From the transformation law of the acceleration it is difficult to guess what kind of four-vector can be constructed. It will be easier to turn to the energy-momentum transformation laws, and construct a four-vector related to the force rather than to the acceleration. This analysis will be carried out in the next section.

Relativistic transformation of the force

Let’s consider first the energy-momentum relation E^2=m_0^2c^4+p^2c^2 given in Eq.(35). Taking the derivative of this relation with respect to time , we obtain

(38)   \begin{equation*} 2E\frac{dE}{dt}=2c^2\vec{p}\cdot\frac{d\vec{p}}{dt} \end{equation*}

Now, remember the definition given in Eq.(28) of relativistic total energy E=\gamma m_0c^2, and momentum \vec{p}=\gamma m_0\vec{u}, for a particle whose rest mass is m_0. Substitution of these relations in Eq.(38), and using the definition of force as the time derivative of the momentum \vec{F}=d\vec{p}/dt, leads to

(39)   \begin{equation*} \frac{1}{c}\frac{dE}{dt}=\vec{\beta}\cdot\vec{F}. \end{equation*}

In what follows it is convenient to write the relation between the time derivative operators d/dt^{\prime} in S^{\prime}, and d/dt in S:

(40)   \begin{equation*} \begin{split} \frac{d}{dt^{\prime}}&=\frac{dt}{dt^{\prime}}\frac{d}{dt}\\ &=\frac{1}{\gamma_0\left(1-\beta_0\beta_x\right)}\frac{d}{dt} \end{split} \end{equation*}

With help of the relation \gamma^{\prime}=\gamma_0\gamma(1-\beta_0\beta_x), the latter equation can be rewritten in the form

(41)   \begin{equation*} \gamma^{\prime}\frac{d}{dt^{\prime}}=\gamma\frac{d}{dt} \end{equation*}

If we apply Eq.(40) to Eq.(31) and Eq.(32), we obtain the following set of transformations for the force components and for the quantity (1/c)dE/dt=\vec{\beta}\cdot\vec{F}:

(42)   \begin{equation*} \begin{split} \vec{\beta}^{\prime}\cdot\vec{F}^{\prime}&=\frac{\vec{\beta}\cdot\vec{F}-\beta_0F_x}{1-\beta_0\beta_x}\\ F_x^{\prime}&=\frac{F_x-\beta_0\vec{\beta}\cdot\vec{F}}{1-\beta_0\beta_x}\\ F_y^{\prime}&=\frac{F_y}{\gamma_0\left(1-\beta_0\beta_x\right)}\\ F_z^{\prime}&=\frac{F_z}{\gamma_0\left(1-\beta_0\beta_x\right)} \end{split} \end{equation*}

The four quantities given in Eq.(42), do not form a four-vector because the relation between the primed and unprimed quantities, is not through a Lorentz transformation as in Eq.(4). We can construct a four-vector related to the force, by multiplying both sides of Eq.(42) by \gamma^{\prime}, and using the relation \gamma^{\prime}=\gamma_0\gamma(1-\beta_0\beta_x) which was given in Eq.(22). Equivalently, we can apply Eq.(41) to both sides of Eq.(31) and Eq.(32). In either of these forms we obtain the four-vector that can be called power-force:

(43)   \begin{equation*}\begin{pmatrix}\gamma^{\prime}\vec{\beta}^{\prime}\cdot\vec{F}^{\prime}\\\gamma^{\prime}F_x^{\prime}\\\gamma^{\prime}F_y^{\prime}\\\gamma^{\prime}F_z^{\prime}\end{pmatrix}=\Lambda_x\begin{pmatrix}\gamma\vec{\beta}\cdot\vec{F}\\\gamma F_x\\\gamma F_y\\\gamma F_z\end{pmatrix} \end{equation*}

where \Lambda_x is the matrix appearing in Eq(4) and that corresponds to a boost in the x direction.

The meaning of the invariant I(F^{\mu}), as well as the relation between force and acceleration,  will be analyzed in another article.

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