Writing source four velocity for Lorentz boosted frame

Question

I am trying to derive the source four velocity for Lorentz boosted frame. If the source four velocity for rest frame is denoted as $U^{\alpha} = (1, \bar 0)$, then how do I write this $U^{\alpha}$ for a Lorentz boosted frame? Also could you please provide a good explanation why we write the four velocity as $(1, \bar 0)$? A good derivation for the boosted frame will also be very helpful for me.

Answer 1

The velocity 4-vector is \begin{equation} \mathbf{U}=\left(\gamma\, c, \gamma\, \mathbf{u}\right) \quad\text{where}\quad \gamma=\left(1-\dfrac{u^2}{c^2}\right)^{\bf -\frac12} \tag{01}\label{eq01} \end{equation} and $\:\mathbf{u}\:$ the velocity 3-vector.

In the rest frame of the particle $\:\mathbf{u}=\boldsymbol{0}\:$ and $\:\gamma=1\:$ so (for $\:c=1\:$) \begin{equation} \mathbf{U}_0=\left(1, \boldsymbol{0}\right) \tag{02}\label{eq02} \end{equation} Obviously $\:\bar 0\:$ is the symbol for the null 3-vector $\:\boldsymbol{0}\:$ and I think that $\:\bar u\:$ would be for the 3-vector $\:\mathbf{u}\:$ in general.

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Now, in above Figure-02 an inertial system $\:\mathrm S'\:$ is translated with respect to the inertial system $\:\mathrm S\:$ with constant velocity
\begin{equation} \boldsymbol{\upsilon}=\left(\upsilon_{1},\upsilon_{2},\upsilon_{3}\right)=\left(\upsilon \mathrm n_{1},\upsilon \mathrm n_{2},\upsilon \mathrm n_{3}\right)=\upsilon \mathbf n\,, \qquad \upsilon \in \left(-c,c\right) \tag{03}\label{eq03} \end{equation} The Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & = \mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x})\mathbf{n}-\gamma \boldsymbol{\upsilon}t \tag{04a}\label{eq04a}\\ t^{\boldsymbol{\prime}} & = \gamma\left(t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right) \tag{04b}\label{eq04b} \end{align} in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & = \mathrm d\mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm d\mathbf{x})\mathbf{n}-\gamma\boldsymbol{\upsilon}\mathrm dt \tag{05a}\label{eq05a}\\ \mathrm d t^{\boldsymbol{\prime}} & = \gamma\left(\mathrm d t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c^{2}}\right) \tag{05b}\label{eq05b} \end{align} and in matrix form \begin{equation} \mathbf{X}^{\boldsymbol{\prime}}= \begin{bmatrix} \mathbf{x}^{\boldsymbol{\prime}}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ c t^{\boldsymbol{\prime}}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}} \end{bmatrix} = \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma \end{bmatrix} \begin{bmatrix} \mathbf{x}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ c t\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}} \end{bmatrix} =\mathrm L\mathbf{X} \tag{06}\label{eq06} \end{equation} where $\:\mathrm L\:$ the real symmetric $\:4\times 4\:$ matrix \begin{equation} \mathrm L \equiv \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma \end{bmatrix} \tag{07}\label{eq07} \end{equation} and \begin{equation} \mathbf{n}\mathbf{n}^{\boldsymbol{\top}} = \begin{bmatrix} \mathrm n_{1}\vphantom{\dfrac{}{}}\\ \mathrm n_{2}\vphantom{\dfrac{}{}}\\ \mathrm n_{3}\vphantom{\dfrac{}{}} \end{bmatrix} \begin{bmatrix} \mathrm n_{1} & \mathrm n_{2} & \mathrm n_{3}\vphantom{\frac12} \end{bmatrix} = \begin{bmatrix} \mathrm n_{1}^{2} & \mathrm n_{1}\mathrm n_{2} & \mathrm n_{1}\mathrm n_{3}\vphantom{\dfrac{}{}}\\ \mathrm n_{2}\mathrm n_{1} & \mathrm n_{2}^{2} & \mathrm n_{2}\mathrm n_{3}\vphantom{\dfrac{}{}}\\ \mathrm n_{3}\mathrm n_{1} & \mathrm n_{3}\mathrm n_{2} & \mathrm n_{3}^{2}\vphantom{\dfrac{}{}} \end{bmatrix} \tag{08}\label{eq08} \end{equation}
a matrix representing the vectorial projection on the direction $\:\mathbf{n}$.

The velocity 3-vector $\:\mathbf{u}\:$ of a particle is transformed as follows \begin{equation} \mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}-\gamma \boldsymbol{\upsilon}}{\gamma \left(1-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{u}\vphantom{\frac12}}{c^{2}}\right)} \tag{09}\label{eq09} \end{equation} equation proved by dividing equations \eqref{eq05a}, \eqref{eq05b} side by side and setting $\:\mathbf{u}\equiv \mathrm d\mathbf{x}/\mathrm d t\:$, $\:\mathbf{u'}\equiv \mathrm d\mathbf{x'}/\mathrm d t'$.

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Hint :

Using above equations and especially \eqref{eq09} try to define a 4-dimensional quantity $\:\mathbf{U}\:$ that would be a (Lorentz) 4-vector and would be used as the velocity 4-vector.

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^{Related : Lorentz transformation of velocity 4-vector.}