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My textbook says:

We take the most general transformation relating the coordinates of a given event in the two systems to be of the form:

$$x' = Ax +Bt$$ $$y' = y$$ $$z' = z $$ $$ t' = Cx + Dt $$

I understand why the $y'$ and $z'$ have to be the same as $y$ and $z$ and that the equations have to be linear for having same acceleration in all frames of reference, but what is the reason for the other to equations of this form to the most general form for the transformation? Can there be any other forms?

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  • $\begingroup$ the textbook is Kleppner and Kolenkov. $\endgroup$ – user175667 Jun 3 at 15:26
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Just wanted to provide a geometrical interpretation of what's going on here, perhaps without resorting to an exact form of a Lorentz transformation! In special relativity, we are ultimately analysing the geometry of $\mathbb{R}^{1, 3}$ under the Minkowski metric. That is to say, we are looking at time dimension and space dimensions and putting them on the same footing as each other but with a measure of distance defined as $$ ds^ 2 = - d t ^ 2 + d x ^ 2 + d y ^ 2 + d z ^ 2 $$ This means that given two 4-vectors in this space, say $ x ^ \mu $ and $ y ^ \mu $, we treat their contraction much like an inner product in Euclidean space, but instead we are computing $ x \cdot y = \eta _{\mu \nu } x ^ \mu y ^ \nu $ where $$ \eta_{\mu \nu } = diag(- 1 , 1 , 1, 1 ) $$ So, we have a meaningful notion of the norm of $ x $, where $$ x \cdot x = x ^ \mu x ^\nu \eta_{\mu \nu } $$ Now to answer the question. A Lorentz transform is defined as a transformation which preserves the norm of any four vector. So, if we do the Lorentz transform $ x ^\mu \to \Lambda ^ \mu _{ \ \ \nu} x ^ \nu$, where $\Lambda $ is some matrix, it must preserve our norm for arbitrary $ x ^ \mu $, so $$ \Lambda ^ \mu _{ \, \, \rho } \eta ^{ \rho \sigma } \Lambda^ \nu _{\, \, \sigma } = \eta ^{\mu \nu}. $$ In other words, it's any member of the group $ O(3, 1 ) $, which is the space of transforms which preserve norms under the Minkowski metric.

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You are correct. The most general form of the Lorentz transformation is the matrix $$\begin{bmatrix} \gamma&-\gamma\beta_x&-\gamma\beta_y&-\gamma\beta_z\\ -\gamma\beta_x&1+(\gamma-1)\frac{\beta_x^2}{\beta^2}&(\gamma-1)\frac{\beta_x \beta_y}{\beta^2}&(\gamma-1)\frac{\beta_x \beta_z}{\beta^2}\\ -\gamma\beta_y&(\gamma-1)\frac{\beta_x \beta_y}{\beta^2}&1+(\gamma-1)\frac{\beta_y^2}{\beta^2}&(\gamma-1)\frac{\beta_y \beta_z}{\beta^2}\\ -\gamma\beta_z&(\gamma-1)\frac{\beta_x \beta_z}{\beta^2}&(\gamma-1)\frac{\beta_y \beta_z}{\beta^2}&1+(\gamma-1)\frac{\beta_z^2}{\beta^2}\\ \end{bmatrix}$$ where $\begin{bmatrix} \beta_x \\ \beta_y \\ \beta_z \end{bmatrix}$ is the relative velocity (divided by $c$) between the 2 frames. As JoshuaTS has rightfully pointed out, this general form is rarely used because it is convenient to choose coordinates such that relative velocity lies along the $x$-axis. In this case, the matrix reduces to the simple textbook version $$\begin{bmatrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

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Your textbook is most likely saying that this is the general transformation because you can freely choose the $x$-axis to be in any direction. If the velocity of one reference frame relative to the other is $\mathbf{v}$, then you can choose the $x$-axis to lie along $\mathbf{v}$. There definitely can be more complicated transformations. Any rotation of the spacial coordinates gives a valid inertial reference frame. You can also boost along any arbitrary direction.

However, any arbitrary (continuous) Lorentz transformation can be performed by: 1) rotating the axes so that the $x$-axis lies along the direction of the boost, 2) performing the transformation from your textbook, and 3) rotating the axes again to wherever you want them to go (and translating the origin if you want). There are also discontinuous Lorentz transformations where you flip the direction of one of the spacial axes or the time axis. These transformations also give valid inertial reference frames. The most general discontinuous Lorentz transformation could be performed using a combination of the transformation from your textbook and rotations of the spacial axes.

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  • $\begingroup$ Ok, I understand your answer but one little thing, why the particular form x' = Ax + Bt and t' = Cx + Dt ? $\endgroup$ – user175667 Jun 3 at 17:02
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    $\begingroup$ I'm not sure I understand your question. $x'=Ax+Bt$ and $t'=Cx+Dt$ is the most general linear transformation (as long as we exclude rotations). It has to be linear if the reference frames are inertial. $\endgroup$ – JoshuaTS Jun 3 at 17:08

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