Consider a frame $S$ which is at rest. The frams $S'$ at $t=0$ coincides with $S$ and then start moving with velocity $v$ in the $+x$ direction.
In Galilean transformation we can easily see that $y'=y$
But while deriving the transformation formulas for Lorentz transformation, we take $y'=y$
If we consider $y'=y'(x,y,z,t)$ then definitely it is linear with respect to all its arguments so that Newton's law holds.
So, we can write $y'=a_1x+a_2y+a_3z+a_4t\tag{1}$
I am trying to prove $y'=y$ from $y'=y(x,y,z,t)$

If the particle is on $x-axis$ in $S$ then it remains on $x-axis$ on $S'$
So, $y'=y=0$ and $z'=z=0$
From (1),
As $x$ and $t$ are independent of each other.
We can conclude that $a_1=a_4=0$
Thus (1) becomes $y'=a_2y+a_3z \tag{2}$
If the particle is in $x-z$ plane in $S$ at $y=0$ the particle remains in $x-z$ plane at $y'=0$ otherwise we get a contradiction that $S'$ moves in $y$ direction also.
So, (2) becomes, $0=a_3z,\;\;\forall z$
$\implies a_3=0$

Thus (2) becomes, $y'=a_2y\tag{3}$

But I have trouble in proving $a_2=1$.
Some authors use the concept of symmetry, but I am not able to understand their argument completely.
Please help in understanding how $a_2=1$?
I am very confused.


2 Answers 2


enter image description here

I think we must use the symmetry concept.

So, for notation convenience consider that your frames $\,\rm S,S'\,$ are $\,\rm S_1,S_2\,$ respectively as in above Figure-01(1). Your equation (3) is then(2) \begin{equation} y_2\boldsymbol{=}a_2\,y_1 \tag{A-01}\label{A-01} \end{equation} Below this configuration consider two frames $\,\rm S_3,S_4\,$ as follows : The system $\,\rm S_3\,$ is at rest with respect to $\,\rm S_2\,$ with reverse the $\,x-,y-\,$ axes so

\begin{align} x_3\boldsymbol{=-}x_2 \tag{A-02a}\label{A-02a}\\ y_3\boldsymbol{=+}y_2 \tag{A-02b}\label{A-02b}\\ z_3\boldsymbol{=-}z_2 \tag{A-02c}\label{A-02c} \end{align} The system $\,\rm S_4\,$ is at rest with respect to $\,\rm S_1\,$ with reverse the $\,x-,y-\,$ axes so

\begin{align} x_4\boldsymbol{=-}x_1 \tag{A-03a}\label{A-03a}\\ y_4\boldsymbol{=+}y_1 \tag{A-03b}\label{A-03b}\\ z_4\boldsymbol{=-}z_1 \tag{A-03c}\label{A-03c} \end{align}

The configuration of frames $\,\rm S_3,S_4\,$ is exactly that of $\,\rm S_1,S_2$ : that is the second frame of a pair is moving along the common $\,x-$axis with velocity $\,\boldsymbol{\upsilon}\,$ with respect to the first frame of the pair. So, corresponding to equation \eqref{A-01} we have \begin{equation} y_4\boldsymbol{=}a_2\,y_3 \tag{A-04}\label{A-04} \end{equation} which by equations \eqref{A-03b}, \eqref{A-02b} yields \begin{equation} y_1\boldsymbol{=}a_2\,y_2 \tag{A-05}\label{A-05} \end{equation} Combining equations \eqref{A-01}, \eqref{A-05} we have \begin{equation} y_1\boldsymbol{=}a^2_2\,y_1 \tag{A-06}\label{A-06} \end{equation} so \begin{equation} a^2_2\boldsymbol{=}1 \quad \boldsymbol{\Longrightarrow} \quad a_2\boldsymbol{=\pm}1 \tag{A-07}\label{A-07} \end{equation} We choose $a_2\boldsymbol{=+}1$ in order to exclude space inversion.

The Figure-02 below is a back view of Figure-01.

enter image description here


(1) This Figure is extracted from my answers here Schutz's geometrical proof that spacetime interval is invariant and here Special Relativity - Reference Frames S and S′ relative velocity.

(2) To be precise the coefficient $\,a_2\,$ must be considered as dependent on the velocity $\,\boldsymbol{\upsilon}\,$ and we must write $\,a_2(\boldsymbol{\upsilon})$. This fact doesn't modify the final conclusion.

  • 1
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    – SG8
    May 11, 2021 at 5:02
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    $\begingroup$ @SG8 : With access to moderation tools I'll do the best to help moderators and users to improve the site where and when necessary. $\endgroup$
    – Frobenius
    May 11, 2021 at 5:09
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    $\begingroup$ Great :) I'm sure about this. Warm regards $\endgroup$
    – SG8
    May 11, 2021 at 5:15
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    $\begingroup$ @Sebastiano : \boldsymbol is not an alternative to \mathbf. Both tools could be used to convert Latin characters to boldface for vectors etc but look the difference \mathbf{xyz}=$\mathbf{xyz}$, \boldsymbol{xyz}=$\boldsymbol{xyz}$. Greek characters are converted to boldface only with \boldsymbol look at this \boldsymbol{\upsilon\rho\sigma}=$\boldsymbol{\upsilon\rho\sigma}$. But \mathbf{\upsilon\rho\sigma}=$\mathbf{\upsilon\rho\sigma}$ $\endgroup$
    – Frobenius
    May 13, 2021 at 6:14
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    $\begingroup$ Numerals 12345, relation symbols (=,<,>,$\ne$,$\equiv$) and many symbols are converted to boldface only with \boldsymbol, look at this \boldsymbol{123=><\ne\equiv\Rightarrow}=$\boldsymbol{123=><\ne\equiv\Rightarrow}$. $\endgroup$
    – Frobenius
    May 13, 2021 at 6:20

the lines element are

$$s^2=-c^2 t^2+x^2+y^2+z^2\\ {s'}^2=-{c'}^2 {t'}^2 + {x'}^2 + {y'}^2 + {z'}^2\tag 1$$

with this Ansatz:

$$c=c'\\ x'=A\,(x-v\,t)\\ t'=B\,x+C\,t\\ y'=a_2\,y\\ z'=z$$

and ${s'}^2-s^2=0$ you obtain:

$$\underbrace{\left(1-A^2 + c^2 B^2\right)}_{eq_1=0} x^2 + \underbrace{\left(2\,A^2 v + 2\,c^2 CB\right)}_{eq_2=0} t\,x + \\ \underbrace{\left(-c^2 - A^2 v^2 + c^2 C^2 \right)}_{eq_3=0} t^2 +\underbrace{(1-a_2^2)}_{eq_4=0}\,y^2= 0.$$

You have now four equations for the four unknowns $A~,B~,C~,a_2$.

The solution

$$A = \gamma\\B = -\frac{\beta}{c}\,\gamma\\ C = \gamma\\a_2=1$$


$$\gamma=\frac{1}{\sqrt{1-\beta^2}}\quad \beta = \frac{v}{c}$$ $\Rightarrow$

\begin{align*} \begin{bmatrix} t' \\ x' \\ y' \\ z' \\ \end{bmatrix} = &\left[ \begin {array}{cccc} {\gamma}&-{\frac {{\gamma}\,v}{{c}^{2}}}&0 &0\\ -{\gamma}\,v&{\gamma}&0&0\\ 0 &0&1&0\\ 0&0&0&1\end {array} \right]\, \begin{bmatrix} t \\ x \\ y \\ z \\ \end{bmatrix} \end{align*}

  • $\begingroup$ How you havewritten eq (1). I think $s^2=x^2+y^2+z^2$. Also how can we conclude that $s^2=s'^2$? $\endgroup$
    – Iti
    May 10, 2021 at 16:37
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    $\begingroup$ @Iti we are in Minkowski space and $~s^2=s'^2$ is requirement of the Lorentz transformation $\endgroup$
    – Eli
    May 10, 2021 at 16:47
  • $\begingroup$ thanks for the reply. But I have not studied about Minkowski space. I am an undergraduate student and have studied Newtonian mechanics. Can you please explain accordingly? Some authors lso used the concept of symmetry. But I have trouble understandng their arguments. $\endgroup$
    – Iti
    May 10, 2021 at 16:54
  • $\begingroup$ how come your question is about Lorentz transformation? $\endgroup$
    – Eli
    May 10, 2021 at 17:16
  • $\begingroup$ In special relativity books for begineers, they use the transformation which takes into account the postulates of special theory of relativity (Laws of physics remains invariant in all inertial referance frames and the speed of light is constant) as Lorentz transformation. $\endgroup$
    – Iti
    May 10, 2021 at 17:21

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