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I have two systems of coordinates: $$\{x^{\mu}\} = \{t, x, y, z\}$$

$$\{x^{\alpha'}\} = \{t', x', y', z'\} = \{t, x+y, x-y, z\}$$

And I have to find the Lorentz $\Lambda$ matrix of the transformation.

What I know is

$$x^{\mu} = \Lambda^{\mu}_{\alpha'} x^{\alpha'}$$

and hence

$$\Lambda^{\mu}_{\alpha'} = \frac{\partial x^{\mu}}{\partial x^{\alpha'}}$$

So it should be easy but then there is something I cannot understand: how to compute, for example ?

$$\frac{\partial x^1}{\partial x^{1'}} = \frac{\partial x}{\partial (x+y)}$$

The result should be anyway

$$ \Lambda = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & -1/2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

And I don't understand why, since I did the calculation and instead of all the $1/2$ terms, I get $1$.

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  • $\begingroup$ Why not just solve the matrix equation? $\endgroup$ – Kyle Kanos Nov 4 '17 at 17:16
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    $\begingroup$ It may be easier to first compute $\Lambda^{-1}$, and then invert the matrix. Or use the chain rule. $\endgroup$ – AccidentalFourierTransform Nov 4 '17 at 17:16
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As Henry Turing said, your answer is correct. However, your $\Lambda$ should not be called a Lorentz matrix. For a matrix to belong to the Lorentz group it's determinant must be 1, and it must leave the metric diag(-1,1,1,1) invariant. The matrix you called $\Lambda$ does neither.
$$ det\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & -1/2 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}=-1/2 $$ and $$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & -1/2 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & -1/2 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}= \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 0 \\ 0 & 0 & 1/2 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} $$ Your matrix $\Lambda$ is part of the larger group GL(4).

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  • $\begingroup$ "As Henry Turing said..." you do realize that Henry Turing also asked the question, no? $\endgroup$ – Kyle Kanos Nov 4 '17 at 19:02
  • $\begingroup$ Nope, I didn't notice that Henry was playing both roles! But that's good he figured out the answer to his own question and told us about it. $\endgroup$ – Gary Godfrey Nov 4 '17 at 19:10
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To compute $\frac{\partial x}{\partial x'}$ one needs $$ x=\frac{1}{2}(x'+y') $$ so that $\frac{\partial x}{\partial x'}=\frac{1}{2}$. Likewise one need $y$ as a function of $x'$ and $y'$ to compute $\partial y/\partial y'$.

Note that the transformation $x\to x'=x+y, y\to y'=x-y$ is not a Lorentz transformation so cannot be expected to yield a metric-preserving transformation. The simplest way to see this is to note that the determinant of the suggested transformation is given by $$ \hbox{Det}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right)=-2 $$ so this transformation does not preserve the space-only length of a vector.

The correct transformation to the new coordinate system should be $$ x\to x'=\frac{1}{\sqrt{2}}(x+y)\, ,\qquad y\to y'=\frac{1}{\sqrt{2}}(x-y) $$ and corresponds to a rotation in the $xy$ plane by $\pi/4$ and a reflection.

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  • $\begingroup$ While true, this doesn't answer the question. $\endgroup$ – Kyle Kanos Nov 4 '17 at 18:50
  • $\begingroup$ @KyleKanos I fixed it. $\endgroup$ – ZeroTheHero Nov 4 '17 at 19:00
  • $\begingroup$ Eh, not sure I'd agree that it should be migrated to MSE, but you've got enough rep to make the vote if you want to start the petition to do so. $\endgroup$ – Kyle Kanos Nov 4 '17 at 19:01
  • $\begingroup$ Right... In the end I don't think is so wise either so I deleted the comment. $\endgroup$ – ZeroTheHero Nov 4 '17 at 19:02
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Your answer is correct.

Te matrix you found is actually the inverse matrix of the transformation, hence when you find your $\Lambda$ you have to invert it, et voilà: the result appears.

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