Problem with the proof that for every timelike vector there exists an inertial coordinate system in which its spatial coordinates are zero

Question

I am reading lecture notes on special relativity and I have a problem with the proof of the following proposition.
Proposition. If $X$ is timelike, then there exists an inertial coordinate system in which $X^1 = X^2 = X^3 = 0$.
The proof states that as $X$ is timelike, it has components of the form $(a, p,mathbf{e})$, where $mathbf{e}$ is a unit spatial vector and $lvert a rvert > lvert p rvert$. Then one considers the following four four-vectors:
begin{align*}
frac{1}{sqrt{a^2 - p^2}}(a, p,mathbf{e}) & & frac{1}{sqrt{a^2 - p^2}}(p, a,mathbf{e}) & & (0, mathbf{q}) & & (0, mathbf{r}),,
end{align*}
where $mathbf{q}$ and $mathbf{r}$ are chosen so that $(mathbf{e}, mathbf{q}, mathbf{r})$ form an orthonormal triad in Euclidean space. Then the proof concludes that these four-vectors define an explicit Lorentz transformation and stops there.
For me this explicit Lorentz transformation is represented by the following matrix.
begin{bmatrix}
frac{1}{sqrt{a^2 - p^2}} a & frac{p}{sqrt{a^2 - p^2}} & 0 & 0 
frac{p}{sqrt{a^2 - p^2}} e^1 & frac{a}{sqrt{a^2 - p^2}} e^1 & q^1 & r^1 
frac{p}{sqrt{a^2 - p^2}} e^2 & frac{a}{sqrt{a^2 - p^2}} e^2 & q^2 & r^2 
frac{p}{sqrt{a^2 - p^2}} e^3 & frac{a}{sqrt{a^2 - p^2}} e^3 & q^3 & r^3 
end{bmatrix}
However, multiplying the column vector $(X^0, X^1, X^2, X^3)$ by the matrix above does not seem to yield a column vector whose spatial components are zero.
What did I miss?

A. Bordg · Accepted Answer

Actually, your matrix can be greatly simplified as
$$
M =
begin{bmatrix}
frac{1}{sqrt{a^2 - p^2}} a & frac{p}{sqrt{a^2 - p^2}} & 0 & 0 
frac{p}{sqrt{a^2 - p^2}} & frac{a}{sqrt{a^2 - p^2}} & 0 & 0 
0 & 0 & 1 & 0 
0 & 0 & 0 & 1 
end{bmatrix}
$$
since $(mathbf{e}, mathbf{q}, mathbf{r})$ forms an orthonormal triad, hence $e_1 = q_2 = r_3 = 1$ and the other coefficients are zero.
Now, as pointed out by Valter Moretti in his comment, the matrix you're looking for is the inverse of this matrix. An easy calculation gives the inverse.
$$
M^{-1} =
frac{1}{(a^2 - p^2)^{frac{3}{2}}} 
begin{bmatrix}
a(a^2 - p^2) & -p(a^2 - p^2) & 0 & 0 
-p(a^2 - p^2) & a(a^2 - p^2) & 0 & 0 
0 & 0 & (a^2 - p^2)^{frac{3}{2}} & 0 
0 & 0 & 0 & (a^2 - p^2)^{frac{3}{2}}
end{bmatrix}
$$
Finally, one easily checks the result as follows.
$$
M^{-1} times 
begin{bmatrix}
a 
p 
0 
0
end{bmatrix}
= 
begin{bmatrix}
sqrt{a^2 - p^2} 
0 
0 
0 
end{bmatrix}
$$
Note that taking $mathbf{q}$ and $mathbf{r}$ such that $(mathbf{e}, mathbf{q}, mathbf{r})$ forms an orthonormal triad is equivalent to doing a spatial rotation such that only $X^1$ is nonvanishing. Thus, a more geometric, less algebraic, proof will start with a spatial rotation then will proceed with a boost along the $x$-axis.

Frobenius · Answer

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{+}1right)}left(mathbf{u}boldsymbol{cdot} mathbf{x}right)mathbf{u}boldsymbol{-}dfrac{gamma_{mathrm u}mathbf{u}}{c}c,t tag{01a}label{01a} c,t^{boldsymbol{prime}} & boldsymbol{=} gamma_{mathrm u}left(c,tboldsymbol{-} dfrac{mathbf{u}boldsymbol{cdot} mathbf{x}}{c}right) tag{01b}label{01b} gamma_{mathrm u} & boldsymbol{=} left(1boldsymbol{-}dfrac{mathrm u^2}{c^2}right)^{boldsymbol{-}frac12} tag{01c}label{01c} end{align} and in differential form begin{align} mathrm dmathbf{x}^{boldsymbol{prime}} & boldsymbol{=} mathrm dmathbf{x}boldsymbol{+}dfrac{gamma^2_{mathrm u}}{c^2 left(gamma_{mathrm u}boldsymbol{+}1right)} left(mathbf{u}boldsymbol{cdot} mathrm dmathbf{x}right)mathbf{u}boldsymbol{-}dfrac{gamma_{mathrm u}mathbf{u}}{c}c,mathrm dt tag{02a}label{02a} c, mathrm dt^{boldsymbol{prime}} & boldsymbol{=} gamma_{mathrm u}left(c,mathrm dtboldsymbol{-} dfrac{mathbf{u}boldsymbol{cdot} mathrm dmathbf{x}}{c}right) tag{02b}label{02b} end{align}

$=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=$

For time-like vectors

Could you check in eqref{02a} what would be the quantity $mathrm dmathbf{x}^{boldsymbol{prime}}$ if in this same equation you replace begin{equation} mathbf{u}boldsymbol{longrightarrow}dfrac{mathrm dmathbf{x}}{mathrm dt} tag{03}label{03} end{equation} under the assumption begin{equation} leftVertdfrac{mathrm dmathbf{x}}{mathrm dt}rightVert <c tag{04}label{04} end{equation}

For space-like vectors

Could you check in eqref{02b} what would be the quantity $mathrm d t^{boldsymbol{prime}}$ if in this same equation you replace begin{equation} mathbf{u}boldsymbol{longrightarrow}dfrac{c^2}{leftVertdfrac{mathrm dmathbf{x}}{mathrm dt}rightVert^2}dfrac{mathrm dmathbf{x}}{mathrm dt} tag{05}label{05} end{equation} under the assumption begin{equation} leftVertdfrac{mathrm dmathbf{x}}{mathrm dt}rightVert >c tag{06}label{06} end{equation}

$=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=$

ADDENDUM

From eqref{02a} and eqref{03} if begin{equation} mathbf{X}boldsymbol{=} begin{bmatrix} x_1 x_2 x_3 x_4 end{bmatrix} boldsymbol{=} begin{bmatrix} mathbf{x} x_4 end{bmatrix},quad texttt{with } leftVertmathbf{X}rightVert^2boldsymbol{=}x^2_4boldsymbol{-}x^2_1boldsymbol{-}x^2_2boldsymbol{-}x^2_3boldsymbol{>}0 tag{07}label{07} end{equation} is a time-like 4-vector in an inertial frame $color{blue}{mathbf{S}}$, then in any inertial frame $color{blue}{mathbf{S}'}$ moving with velocity begin{equation} boxed{::mathbf{u}boldsymbol{=}dfrac{mathbf{x}}{x_4}cboldsymbol{=}left(dfrac{x_1}{x_4},dfrac{x_2}{x_4},dfrac{x_3}{x_4} right)c vphantom{tfrac{tfrac{a}{b}}{tfrac{a}{b}}}::} tag{08}label{08} end{equation} its space component is zero begin{equation} mathbf{X}'boldsymbol{=} begin{bmatrix} x'_1 x'_2 x'_3 x'_4 end{bmatrix} boldsymbol{=} begin{bmatrix} mathbf{x}' x'_4 end{bmatrix} boldsymbol{=} begin{bmatrix} boldsymbol{0} x'_4 end{bmatrix} ,quad texttt{with } x'^{2}_4boldsymbol{=}x^2_4boldsymbol{-}x^2_1boldsymbol{-}x^2_2boldsymbol{-}x^2_3 tag{09}label{09} end{equation}
From eqref{02b} and eqref{05} if begin{equation} mathbf{X}boldsymbol{=} begin{bmatrix} x_1 x_2 x_3 x_4 end{bmatrix} boldsymbol{=} begin{bmatrix} mathbf{x} x_4 end{bmatrix},quad texttt{with } leftVertmathbf{X}rightVert^2boldsymbol{=}x^2_4boldsymbol{-}x^2_1boldsymbol{-}x^2_2boldsymbol{-}x^2_3boldsymbol{<}0 tag{10}label{10} end{equation} is a space-like 4-vector in an inertial frame $color{blue}{mathbf{S}}$, then in any inertial frame $color{blue}{mathbf{S}'}$ moving with velocity begin{equation} boxed{::mathbf{u}boldsymbol{=}leftVertdfrac{mathbf{x}}{x_4}rightVert^{boldsymbol{-}2}dfrac{mathbf{x}}{x_4}cboldsymbol{=}leftVertdfrac{mathbf{x}}{x_4}rightVert^{boldsymbol{-}2}left(dfrac{x_1}{x_4},dfrac{x_2}{x_4},dfrac{x_3}{x_4} right)c vphantom{tfrac{tfrac{a}{b}}{tfrac{a}{b}}}::} tag{11}label{11} end{equation} its time component is zero begin{equation} mathbf{X}'boldsymbol{=} begin{bmatrix} x'_1 x'_2 x'_3 x'_4 end{bmatrix} boldsymbol{=} begin{bmatrix} vphantom{x'_1} mathbf{x}' vphantom{x'_3} x'_4 end{bmatrix} boldsymbol{=} begin{bmatrix} vphantom{x'_1} mathbf{x}' vphantom{x'_3} 0vphantom{x'_4} end{bmatrix} ,quad texttt{with } Vertmathbf{x}'Vert^2boldsymbol{=}x^2_1boldsymbol{+}x^2_2boldsymbol{+}x^2_3boldsymbol{-}x^2_4 tag{12}label{12} end{equation}

Javier · Answer

The matrix you wrote will take the standard basis ${(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}$ into the basis constructed from $mathbf{X}$. Therefore, to take $mathbf{X}$ into something proportional to $(1,0,0,0)$, you need to use the inverse matrix.

Answered by Javier on March 9, 2021

Problem with the proof that for every timelike vector there exists an inertial coordinate system in which its spatial coordinates are zero

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