# Transformation of $4-$velocity

Notation: a greek index indicates four labels; spacetime coordinates $$\mu = (0,1,2,3)$$. A latin index indicates three labels; spatial coordinates $$i = (1,2,3)$$.

$$* * *$$

A quantity, to be identified as a (contravariant) vectorial quantity, must satisfy the following rule (using Einstein's summation convention):

$$A'^{\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}} A^{\nu} \tag{1}$$

So, the jacobian matrix in $$(1)$$ express the concept of invariance under any transformation rule. Now, consider, respectively, the 3-velocity and 4-velocity vector components:

$$v^{i} \equiv \frac{\mathrm{d}x^{i}(t)}{\mathrm{d}t}\tag{2}$$

$$u^{\mu} \equiv \frac{\mathrm{d}x^{\mu}(\tau)}{\mathrm{d}\tau}\tag{3}$$

where $$\tau$$ is the proper time.

Some books prefer to introduce the 4-velocity directly by differential geometry concepts saying that the 4-velocity is the vector tangent to the curve:

$$\alpha (\tau) = \alpha(t(\tau),x(\tau),y(\tau),z(\tau))$$

Others just say that the 4-velocity is a invariant quantity because we divide $$\mathrm{d}x^{\mu}$$ by $$\mathrm{d}\tau$$ and it works because $$d\tau = ds^{2}/c^{2}$$ is a invariant quantity.

Now, a Minkowski vector is the quantity where the jacobian matrices are given by:

$$\frac{\partial x'^{\mu}}{\partial x^{\nu}} \equiv \Lambda^{\mu'}_{\nu} = \left[ {\begin{array}{ccccc} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right]$$

(henceforth consider that the jacobian matrix are the lorentzian boost matrix).

In other words, a Minkowski vector is the one who transforms like:

$$A'^{\mu} = \Lambda^{\mu'}_{\nu} A^{\nu} \tag{4}$$

Then, after applying the tranformation $$(4)$$, in general a vectorial quantity in Minkowski spacetime (in cartesian coordinates) have the form:

$$\begin{cases} A'^{0} = \gamma\Big(A^{0} - \frac{v}{c}A^{1} \Big) \\ A'^{1} = \gamma\Big(A^{1} - \frac{v}{c}A^{0} \Big)\\ A'^{2} = A'^{2}\\ A'^{3} = A'^{3} \end{cases}$$ Well, here begin my confusion about 4-velocity. It's well known that if we simply try to generalize the 3-velocity to spacetime (to form 4-quantities) as:

$$v^{\mu} \equiv \frac{\mathrm{d}x^{\mu}(t)}{\mathrm{d}t}\tag{5} = \Big(\frac{\mathrm{d}x^{0}(t)}{\mathrm{d}t},\frac{\mathrm{d}x^{1}(t)}{\mathrm{d}t},\frac{\mathrm{d}x^{2}(t)}{\mathrm{d}t},\frac{\mathrm{d}x^{3}(t)}{\mathrm{d}t}\Big)$$

won't work properly. But, if we take the quantity:

$$u^{\mu} \equiv \frac{\mathrm{d}x^{\mu}(\tau)}{\mathrm{d}\tau} = \Big(\frac{\mathrm{d}x^{0}(\tau)}{\mathrm{d}\tau},\frac{\mathrm{d}x^{1}(\tau)}{\mathrm{d}\tau},\frac{\mathrm{d}x^{2}(\tau)}{\mathrm{d}\tau},\frac{\mathrm{d}x^{3}(\tau)}{\mathrm{d}\tau}\Big)\tag{6}$$

Then we can introduce a validy notion of velocity in spacetime physics. Also, there is another technicality that is about the relationship between 4-velocity and 3-velocity.

$$u^{\mu} \equiv \frac{\mathrm{d}x^{\mu}(\tau)}{\mathrm{d}\tau} = \Big(\frac{\mathrm{d}x^{0}(\tau)}{\mathrm{d}\tau},\frac{\mathrm{d}x^{1}(\tau)}{\mathrm{d}\tau},\frac{\mathrm{d}x^{2}(\tau)}{\mathrm{d}\tau},\frac{\mathrm{d}x^{3}(\tau)}{\mathrm{d}\tau}\Big) = \Big(\gamma\frac{\mathrm{d}x^{0}(t)}{\mathrm{d}t},\gamma\frac{\mathrm{d}x^{1}(t)}{\mathrm{d}t},\gamma\frac{\mathrm{d}x^{2}(t)}{\mathrm{d}t},\gamma\frac{\mathrm{d}x^{3}(t)}{\mathrm{d}t}\Big) \tag{7}$$

Question: So, my doubt is about the verificantion of Minkowskian vectorial character of $$(7)$$ and the non-vectorial character of $$(5)$$. Because,there's a difference between the construction of 4-velocity (motivated by any argument that you want) and the verification as a true Minkowski vector.

I simply didn't get the right results. Firstly,I did applying the tranformation rule $$(4)$$ (which is the most general picture to treat about vectors! -At least to do the calculations of basic special/general relativity-)

$$u'^{\mu} \equiv \frac{\mathrm{d}x'^{\mu}(\tau)}{\mathrm{d}\tau} = \frac{\partial x'^{\mu}}{\partial x^{\nu}} \frac{\mathrm{d}x^{\nu}(\tau)}{\mathrm{d}\tau}= \frac{\partial x'^{\mu}}{\partial x^{\nu}} \Big[\gamma\frac{\mathrm{d}x^{\nu}(t)}{\mathrm{d}t}\Big] \equiv \frac{\partial x'^{\mu}}{\partial x^{\nu}} u^{\nu}\tag{8}$$

And I got the results:

$$\frac{\mathrm{d}x'^{0}(\tau)}{\mathrm{d}\tau} = \gamma\Big(\Big[\gamma\frac{\mathrm{d}x^{0}(t)}{\mathrm{d}t}\Big] - \frac{v}{c}\Big[\gamma\frac{\mathrm{d}x^{1}(t)}{\mathrm{d}t}\Big]\Big) =\gamma \Big(\frac{\mathrm{d}x^{0}(\tau)}{\mathrm{d}\tau} - \frac{v}{c} \frac{\mathrm{d}x^{1}(\tau)}{\mathrm{d}\tau}\Big)$$

$$\frac{\mathrm{d}x'^{1}(\tau)}{\mathrm{d}\tau} = \gamma\Big(\Big[\gamma\frac{\mathrm{d}x^{1}(t)}{\mathrm{d}t}\Big] -v\Big[\gamma\frac{\mathrm{d}x^{0}(t)}{\mathrm{d}t}\Big]\Big) =\gamma \Big(\frac{\mathrm{d}x^{1}(\tau)}{\mathrm{d}\tau} -v\frac{\mathrm{d}x^{0}(\tau)}{\mathrm{d}\tau}\Big)$$

$$\frac{\mathrm{d}x'^{2}(\tau)}{\mathrm{d}\tau} = \frac{\mathrm{d}x'^{2}(\tau)}{\mathrm{d}\tau}$$

$$\frac{\mathrm{d}x'^{3}(\tau)}{\mathrm{d}\tau} = \frac{\mathrm{d}x'^{3}(\tau)}{\mathrm{d}\tau}$$

Which clear aren't in the form of a Minkowski vector.

On the other hand if we simply take the matrix product:

$$\left[ {\begin{array}{ccccc} \gamma \frac{\mathrm{d}x'^{0}}{\mathrm{d}t}\\ \gamma \frac{\mathrm{d}x'^{1}}{\mathrm{d}t}\\ \gamma \frac{\mathrm{d}x'^{2}}{\mathrm{d}t}\\ \gamma \frac{\mathrm{d}x'^{3}}{\mathrm{d}t}\\ \end{array} } \right] = \left[ {\begin{array}{ccccc} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right] \left[ {\begin{array}{ccccc} \gamma \frac{\mathrm{d}x^{0}}{\mathrm{d}t}\\ \gamma \frac{\mathrm{d}x^{1}}{\mathrm{d}t}\\ \gamma \frac{\mathrm{d}x^{2}}{\mathrm{d}t}\\ \gamma \frac{\mathrm{d}x^{3}}{\mathrm{d}t}\\ \end{array} } \right]$$

I got the right results:

$$\gamma \frac{\mathrm{d}x'^{0}}{\mathrm{d}t} = \gamma \Big[ \gamma \frac{\mathrm{d}x'^{0}}{\mathrm{d}t}\Big] - \gamma \frac{v}{c}\Big[ \gamma\frac{\mathrm{d}x'^{1}}{\mathrm{d}t}\Big]$$

$$\gamma \frac{\mathrm{d}x'^{1}}{\mathrm{d}t} = \gamma \Big[ \gamma \frac{\mathrm{d}x'^{1}}{\mathrm{d}t}\Big] - \gamma \frac{v}{c}\Big[ \frac{\gamma\mathrm{d}x'^{0}}{\mathrm{d}t}\Big]$$

$$\gamma \frac{\mathrm{d}x'^{2}}{\mathrm{d}t} = \gamma \frac{\mathrm{d}x^{1}}{\mathrm{d}t}$$

$$\gamma \frac{\mathrm{d}x'^{3}}{\mathrm{d}t} = \gamma \frac{\mathrm{d}x^{3}}{\mathrm{d}t}$$

Which is in the form of a Minkowskian vector.

Despite all of this, if I take the same matrix product using a the (wrong) attempt to 4-velocity given by $$(5)$$ I still get an Minkowskian vector:

$$\left[ {\begin{array}{ccccc} \frac{\mathrm{d}x'^{0}}{\mathrm{d}t}\\ \frac{\mathrm{d}x'^{1}}{\mathrm{d}t}\\ \frac{\mathrm{d}x'^{2}}{\mathrm{d}t}\\ \frac{\mathrm{d}x'^{3}}{\mathrm{d}t}\\ \end{array} } \right] = \left[ {\begin{array}{ccccc} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right] \left[ {\begin{array}{ccccc} \frac{\mathrm{d}x^{0}}{\mathrm{d}t}\\ \frac{\mathrm{d}x^{1}}{\mathrm{d}t}\\ \frac{\mathrm{d}x^{2}}{\mathrm{d}t}\\ \frac{\mathrm{d}x^{3}}{\mathrm{d}t}\\ \end{array} } \right]$$

I got a minkowskian vector form:

$$\frac{\mathrm{d}x'^{0}}{\mathrm{d}t} = \gamma \Big[\frac{\mathrm{d}x'^{0}}{\mathrm{d}t}\Big] - \gamma \frac{v}{c}\Big[\frac{\mathrm{d}x'^{1}}{\mathrm{d}t}\Big]$$

$$\frac{\mathrm{d}x'^{1}}{\mathrm{d}t} = \gamma \Big[\frac{\mathrm{d}x'^{1}}{\mathrm{d}t}\Big] - \gamma \frac{v}{c}\Big[ \frac{\mathrm{d}x'^{0}}{\mathrm{d}t}\Big]$$

$$\frac{\mathrm{d}x'^{2}}{\mathrm{d}t} = \frac{\mathrm{d}x^{1}}{\mathrm{d}t}$$

$$\frac{\mathrm{d}x'^{3}}{\mathrm{d}t} = \frac{\mathrm{d}x^{3}}{\mathrm{d}t}$$

But this doesn't make much sense because this quantity are not a 4-velocity.

So, how can I verify that the 4-velocity transforms properly under a lorentz transformation and 3-velocity does not? (Using the formula $$(4)$$ ).

• You already proofed $u^{\mu}$ being a Minkowski 4-vector in eq. (8) since $u^{\mu}$ transforms like $A^{\mu}$ in eq. (4) (taking into account the definition of $\Lambda^{\mu}_\nu$ one line above). However, $v^{\mu}$ does not transform like a Minkowski 4-vector because $v'^\mu= \frac{dx^{\mu}}{dt'}$ (note the little prime) and depending on the reference system $dt =d\tau/\gamma$ or NOT. However in your equation it is simply assumed that $dt =d\tau/\gamma$ always. That's the point. (On the other hand as you already said $d\tau$ is always the same independent on the reference system). Feb 15, 2019 at 14:39
• Equation (1) is missing a prime on the left hand side. Feb 15, 2019 at 17:56

SECTION A : The 3+1-Lorentz transformation in the Standard Configuration

I'll try to answer your questions and moreover in the more general 3+1 case.

So, let an inertial system $$\:\mathrm S'\:$$ translated with respect to the inertial system $$\:\mathrm S\:$$ with constant velocity $$$$\mathbf{u}\boldsymbol{=}\left(\mathrm u_1,\mathrm u_2,\mathrm u_3\right) \,, \qquad \Vert \mathbf{u}\Vert \boldsymbol{=} \mathrm u \in \left(0,c\right) \tag{01}\label{01}$$$$ as shown in Figure-01.

The 3+1-Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t \tag{02a}\label{02a}\\ c\,t^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,t\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{x}}{c}\right) \tag{02b}\label{02b}\\ \gamma_{\mathrm u} & \boldsymbol{=} \left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{02c}\label{02c} \end{align} in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathrm d\mathbf{x}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \left(\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,\mathrm dt \tag{03a}\label{03a}\\ c\, \mathrm dt^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(c\,\mathrm dt\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c}\right) \tag{03b}\label{03b} \end{align} and in matrix form $$$$\mathbf{X}^{\boldsymbol{\prime}} \boldsymbol{=} \begin{bmatrix} \mathbf{x}^{\boldsymbol{\prime}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ c t^{\boldsymbol{\prime}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \mathrm I\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \mathbf{u}\mathbf{u}^{\boldsymbol{\top}} & \boldsymbol{-}\dfrac{\gamma_{\mathrm u}}{c}\mathbf{u} \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \boldsymbol{-}\dfrac{\gamma_{\mathrm u}}{c}\mathbf{u}^{\boldsymbol{\top}} & \hphantom{-}\gamma_{\mathrm u}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \mathbf{x}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ c t\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \mathrm L\mathbf{X} \tag{04}\label{04}$$$$ where $$\:\mathrm L\:$$ the real symmetric $$\:4\times 4\:$$ matrix $$$$\mathrm L \boldsymbol{\equiv} \begin{bmatrix} \mathrm I\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \mathbf{u}\mathbf{u}^{\boldsymbol{\top}} & \boldsymbol{-}\dfrac{\gamma_{\mathrm u}}{c}\mathbf{u} \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \boldsymbol{-}\dfrac{\gamma_{\mathrm u}}{c}\mathbf{u}^{\boldsymbol{\top}} & \hphantom{-}\gamma_{\mathrm u}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \tag{05}\label{05}$$$$

$$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$$

SECTION B : Transformation of the velocities 3-vectors

Suppose now that a point particle $$\:\mathrm P\:$$ is moving at a given instant $$\:t\:$$ with velocity $$\:\mathbf w\:$$ with respect to system $$\:\mathrm S$$ $$$$\mathbf{w} \boldsymbol{=}\dfrac{\mathrm d\mathbf x}{\mathrm dt} \qquad (\text{velocity of point particle \:\mathrm P\: with respect to system \:\mathrm S\:}) \tag{06}\label{06}$$$$ For its velocity $$\:\mathbf w'\:$$ with respect to system $$\:\mathrm S'\:$$ $$$$\mathbf{w'} \boldsymbol{=}\dfrac{\mathrm d\mathbf x'}{\mathrm dt'} \qquad (\text{velocity of point particle \:\mathrm P\: with respect to system \:\mathrm S'\:}) \tag{07}\label{07}$$$$ we simply divide equations \eqref{03a} and \eqref{03b} side by side and so we find that $$$$\mathbf{w'} \boldsymbol{=}\dfrac{\mathbf{w}\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}\left(\mathbf{u}\boldsymbol{\cdot}\mathbf{w}\right)}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\mathbf{u}\boldsymbol{-}\gamma_{\mathrm u}\mathbf{u}}{\gamma_{\mathrm u}\left(1\boldsymbol{-}\dfrac{\mathbf{u}\boldsymbol{\cdot}\mathbf{w}}{c^2}\right)} \tag{08}\label{08}$$$$ Above equation, beyond to be the transformation law for 3-velocities, is the law of relativistic addition of 3-velocities, more exactly of $$\:\mathbf w\:$$ and $$\:\boldsymbol{-}\mathbf u \:$$. If these two velocities $$\:\mathbf w,\mathbf u\:$$ are collinear with magnitudes $$\:\mathrm w,\mathrm u\:$$ respectively then equation \eqref{08} gives the known one-dimensional result
$$$$\mathrm{w'} \boldsymbol{=}\dfrac{\mathrm{w}\boldsymbol{-}\mathrm{u}}{1\boldsymbol{-}\dfrac{\mathrm{u}\mathrm{w}}{c^2}} \tag{09}\label{09}$$$$

Now, we agree that a 4-dimensional real vector $$$$\mathbf{A} \boldsymbol{=}\left(\mathbf a, \alpha\right) \qquad \mathbf a \in \mathbb{R}^3, \alpha \in \mathbb{R} \tag{10}\label{10}$$$$ is a 4-vector (or Minkowski vector in terms of the question) if it is Lorentz transformed as the 4-dimensional position vector $$\:\mathbf{X} \boldsymbol{=}\left(\mathbf x, c\,t\right)\:$$ in Minkowski space, that is according to \eqref{04}
$$$$\mathbf{A}^{\boldsymbol{\prime}} \boldsymbol{=} \begin{bmatrix} \mathbf{a}^{\boldsymbol{\prime}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \alpha^{\boldsymbol{\prime}}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \mathrm I\boldsymbol{+}\dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)} \mathbf{u}\mathbf{u}^{\boldsymbol{\top}} & \boldsymbol{-}\dfrac{\gamma_{\mathrm u}}{c}\mathbf{u} \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \boldsymbol{-}\dfrac{\gamma_{\mathrm u}}{c}\mathbf{u}^{\boldsymbol{\top}} & \hphantom{-}\gamma_{\mathrm u}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \begin{bmatrix} \mathbf{a}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\\ \alpha\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \end{bmatrix} \boldsymbol{=} \mathrm L\mathbf{A} \tag{11}\label{11}$$$$ written explicitly as in equations \eqref{02a} and \eqref{02b}
\begin{align} \mathbf{a}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{a}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{a}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}\alpha \tag{12a}\label{12a}\\ \alpha^{\boldsymbol{\prime}} & \boldsymbol{=} \gamma_{\mathrm u}\left(\alpha\boldsymbol{-} \dfrac{\mathbf{u}\boldsymbol{\cdot} \mathbf{a}}{c}\right) \tag{12b}\label{12b} \end{align}

The construction of a velocity 4-vector $$$$\mathbf{W} \boldsymbol{=}\left[\boldsymbol \eta\left(\mathbf w\right), \xi\left(\mathbf w\right)\right] \qquad \boldsymbol \eta\left(\mathbf w\right) \in \mathbb{R}^3, \xi\left(\mathbf w\right) \in \mathbb{R} \tag{13}\label{13}$$$$ from the velocity 3-vector $$\:\mathbf w\:$$ according to the transformation \eqref{08} encounters two problems :

1. The velocity 3-vector $$\:\mathbf w\:$$ is not transformed as the position 3-vector $$\:\mathbf x\:$$ does, see equation \eqref{02a}, because of the denominator in the rhs of equation \eqref{08}. So in equation \eqref{13} $$\:\boldsymbol\eta\left(\mathbf w\right)\ne\mathbf w$$, that is the velocity 3-vector $$\:\mathbf w\:$$ could not be the $$^{\prime\prime}$$space$$^{\prime\prime}$$ part of the velocity 4-vector $$\:\mathbf W$$.

2. The unknown vector function $$\:\boldsymbol\eta\left(\mathbf w\right)\:$$ must be determined and paired together with the unknown scalar function $$\:\xi\left(\mathbf w\right)$$, the $$^{\prime\prime}$$time$$^{\prime\prime}$$ part of the velocity 4-vector $$\:\mathbf W$$.

These two problems are solved and the construction of an acceptable velocity 4-vector $$\:\mathbf W\:$$ is achieved through a relation between the $$\gamma-$$factors $$\gamma_{\mathrm u},\gamma_{\mathrm w},\gamma_{\mathrm w'}\:$$ as it will be discussed in the next SECTION C.

$$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$$

SECTION C : A relation between the $$\gamma-$$factors $$\gamma_{\mathrm u},\gamma_{\mathrm w},\gamma_{\mathrm w'}\:$$

Between the $$\gamma-$$factors $$\gamma_{\mathrm u},\gamma_{\mathrm w},\gamma_{\mathrm w'}\:$$ the following equation is valid

$$$$\boxed{\:\: \gamma_{\mathrm w'} \boldsymbol{=}\gamma_{\mathrm u}\gamma_{\mathrm w}\left(1\boldsymbol{-}\dfrac{\mathbf{u}\boldsymbol{\cdot}\mathbf{w}}{c^2}\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{14}\label{14}$$$$ This relation is proved as follows :

Let $$\:\mathrm S^{\mathrm P}\:$$ the rest system of the particle $$\:\mathrm P$$. In this system $$\:\mathrm S^{\mathrm P}\:$$ the time is the proper one $$\:\tau$$. The rest system $$\:\mathrm S^{\mathrm P}\:$$ is moving with velocity $$\:\mathbf w\:$$ with respect to system $$\:\mathrm S\:$$ so according to the Lorentz transformation between these systems we have $$$$\dfrac{\mathrm dt}{\mathrm d\tau}\boldsymbol{=}\gamma_{\mathrm w} \tag{15}\label{15}$$$$ On the same step, since the rest system $$\:\mathrm S^{\mathrm P}\:$$ is moving with velocity $$\:\mathbf w'\:$$ with respect to system $$\:\mathrm S'\:$$ we have $$$$\dfrac{\mathrm dt'}{\mathrm d\tau}\boldsymbol{=}\gamma_{\mathrm w'} \tag{16}\label{16}$$$$ On the other hand from the Lorentz transformation between the systems $$\:\mathrm S\:$$ and $$\:\mathrm S'\:$$ we have, see \eqref{03b} $$$$\dfrac{\mathrm dt'}{\mathrm dt}\boldsymbol{=}\gamma_{\mathrm u}\left(1\boldsymbol{-}\dfrac{\mathbf{u}\boldsymbol{\cdot}\mathbf{w}}{c^2}\right) \tag{17}\label{17}$$$$ From equations \eqref{15},\eqref{16} and \eqref{17} the relation \eqref{14} is proved, that is $$$$\gamma_{\mathrm w'}\boldsymbol{=}\dfrac{\mathrm dt'}{\mathrm d\tau}\boldsymbol{=}\dfrac{\mathrm dt'}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm d\tau}\boldsymbol{=}\gamma_{\mathrm u}\gamma_{\mathrm w}\left(1\boldsymbol{-}\dfrac{\mathbf{u}\boldsymbol{\cdot}\mathbf{w}}{c^2}\right) \tag{18}\label{18}$$$$

$$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$$

SECTION D : Construction of the velocity 4-vector $$\:\mathbf W$$

The denominator of the rhs of equation \eqref{08} is from the relation \eqref{14} $$$$\gamma_{\mathrm u}\left(1\boldsymbol{-}\dfrac{\mathbf{u}\boldsymbol{\cdot}\mathbf{w}}{c^2}\right)\boldsymbol{=}\dfrac{\gamma_{\mathrm w'}}{\gamma_{\mathrm w}} \tag{19}\label{19}$$$$ Replacing this in equation \eqref{08}, moving the primed variables to the lhs and the unprimed ones to the rhs we have the following equation \eqref{20a} while equation \eqref{14} multiplied by $$c$$ gives equation \eqref{20b} \begin{align} \left[\gamma_{\mathrm w'}\mathbf{w'}\right] & \boldsymbol{=} \left[\gamma_{\mathrm w}\mathbf{w}\right]\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \left[\gamma_{\mathrm w}\mathbf{w}\right]\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}\left[\gamma_{\mathrm w}c\right] \tag{20a}\label{20a}\\ \left[\gamma_{\mathrm w'}c\right] &\boldsymbol{=}\gamma_{\mathrm u}\left(\left[\gamma_{\mathrm w}c\right]\boldsymbol{-}\dfrac{\mathbf{u}\boldsymbol{\cdot}\left[\gamma_{\mathrm w}\mathbf{w}\right]}{c^2}\right) \tag{20b}\label{20b} \end{align}

Comparing the pair of equations \eqref{20a},\eqref{20b} with that of equations \eqref{02a},\eqref{02b} we note that the variables $$\:\gamma_{\mathrm w}\mathbf{w}\:$$ and $$\:\gamma_{\mathrm w}c\:$$ are transformed as the variables $$\:\mathbf{x}\:$$ and $$\:c\,t\:$$ of the position 4-vector $$\:\mathbf{X}$$. This means that the 4-dimensional vector $$$$\boxed{\:\: \mathbf{W} \boldsymbol{\equiv}\left(\gamma_{\mathrm w}\mathbf w, \gamma_{\mathrm w}c\right)\boldsymbol{=}\gamma_{\mathrm w}\left(\mathbf w, c\right)\vphantom{\dfrac{\frac{a}{b}}{\frac{a}{b}}}\:\:} \tag{21}\label{21}$$$$ is a 4-vector, the constructed velocity 4-vector in Minkowski space.

• how to get eq.(02a)? Mar 30 at 8:22