# Lorentz $\Lambda$ matrix problem

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 $1$.

• Why not just solve the matrix equation? – Kyle Kanos Nov 4 '17 at 17:16
• It may be easier to first compute $\Lambda^{-1}$, and then invert the matrix. Or use the chain rule. – AccidentalFourierTransform Nov 4 '17 at 17:16

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).

• "As Henry Turing said..." you do realize that Henry Turing also asked the question, no? – Kyle Kanos Nov 4 '17 at 19:02
• 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. – Gary Godfrey Nov 4 '17 at 19:10

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.

• While true, this doesn't answer the question. – Kyle Kanos Nov 4 '17 at 18:50
• @KyleKanos I fixed it. – ZeroTheHero Nov 4 '17 at 19:00
• 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. – Kyle Kanos Nov 4 '17 at 19:01
• Right... In the end I don't think is so wise either so I deleted the comment. – ZeroTheHero Nov 4 '17 at 19:02

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.