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I've spent the last couple of hours trying to derive the Lorentz Transformation from Maxwell's Equations. What I ended up with is $$L_{\nu}^{-1}=\left(\begin{array}{ll} \frac{1}{\sqrt{1-v^{2}}} & \frac{-v}{\sqrt{1-v^{2}}} \\ \frac{-v}{\sqrt{1-v^{2}}} & \frac{1}{\sqrt{1-v^{2}}} \end{array}\right)$$

Which matches exactly with the Transformation as described in my textbook. And yet, when I search up Lorentz Transformations online, I find no matrix of the above form. What have I actually derived? Am I anywhere close to the Lorentz Transformations? Please advise.

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    $\begingroup$ Looks like the boost matrix for 1-D motion and $c=1$ to me. $\endgroup$
    – DanDan0101
    Sep 3, 2020 at 1:16
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    $\begingroup$ What did you assume happens to $\mathbf{E}$ and $\mathbf{B}$, and where did the other two dimensions go? $\endgroup$
    – G. Smith
    Sep 3, 2020 at 1:48

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As @DanDan0101 points out in their comment to your question, if you define

$$\gamma = \frac{1}{\sqrt{1-v^2}},$$

where $v$ is the velocity measured in units of $c$ (otherwise this wouldn't make any sense dimensionally), then your matrix is just

$$L = \begin{pmatrix}\gamma & -\gamma v \\ -\gamma v & \gamma\end{pmatrix},$$

the "forward" Lorentz Transformation matrix, see this answer to What is a Lorentz boost and how to calculate it?

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This is the Lorentz transformation for an object in a universe with one spatial dimension and a time dimension. Notice that in most texts

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

and it seems like in your particulat textbook the authors have opted to go with units in which the speed of light, $c=1$. Your answer is therefore correct, it's just that some authors prefer using these units to simplify calculations.

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As @Philip points out in their answer to your question, if you define, $v=\tanh\theta$ (so that $ \gamma=\cosh\theta$), then $$ L = \begin{pmatrix}\cosh\theta & -\sinh\theta \\ -\sinh\theta& \cosh\theta\end{pmatrix} $$ which is the rapidity form of a Lorentz boost

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