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I'm considering the following matrixs which I know that they form a flux of Lorentz trasformation in spacetime.

I want to know how to calculate the infinitesimal generator of this flux. Unfortunately I have no particular knowledge of Lie algebra for this reason I need an explanation that does not assume the whole knowledge of it.

$$\begin{pmatrix} \frac{4- \cos(\rho)}{3} & \frac{2- 2\cos(\rho)}{3} & 0 & -\frac{\sin(\rho)}{\sqrt{3}} \\ \frac{2\cos(\rho) - 2}{3} & \frac{4- \cos(\rho)}{3} & 0 & \frac{2\sin(\rho)}{\sqrt{3}}\\ 0 & 0 & 1 & 0 \\ -\frac{\sin(\rho)}{\sqrt{3}} & -\frac{2\sin(\rho)}{\sqrt{3}} & 0 & \cos(\rho) \\ \end{pmatrix}$$

Thank you so much for your help

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    $\begingroup$ Calculating the infinitesimal form typically requires you perform a Taylor expansion, introducing an infinitesimal parameter. Do you see how to proceed here knowing the expansions of sine and cosine? $\endgroup$ – JamalS Oct 31 '17 at 8:46
  • $\begingroup$ To be honest no. $\endgroup$ – Stefano Barone Oct 31 '17 at 9:03
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    $\begingroup$ The infinitesimal form of $e^{i\alpha}$is $1 + i\alpha$ for infinitesimal $\alpha$. Here you face the same problem except you have a matrix of functions rather than just a scalar but you still work entry by entry. $\endgroup$ – JamalS Oct 31 '17 at 10:02
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I'm a bit suspicious of the 22 entry of the matrix you write down , $$M=\begin{pmatrix} \frac{4- \cos(\rho)}{3} & \frac{2- 2\cos(\rho)}{3} & 0 & -\frac{\sin(\rho)}{\sqrt{3}} \\ \frac{2\cos(\rho) - 2}{3} & \frac{4- \cos(\rho)}{3} & 0 & \frac{2\sin(\rho)}{\sqrt{3}}\\ 0 & 0 & 1 & 0 \\ -\frac{\sin(\rho)}{\sqrt{3}} & -\frac{2\sin(\rho)}{\sqrt{3}} & 0 & \cos(\rho) \\ \end{pmatrix}$$ whose logarithm you are invited to take. I suspect that entry to be something like $(4\cos \rho -1)/3$ instead---see below. Your time appears to be in the 4th component, unlike the first one in the conventional notation.

In any case, observe $M=\mathbb{1}$ as $\rho\to 0$, so to find its logarithm, we expand in the first two powers of ρ, $$ M=\mathbb{1} -\frac{\rho}{\sqrt{3}}\begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & -2\\ 0 & 0 & 0 & 0 \\ 1 & 2& 0 & 0 \\ \end{pmatrix}+ \frac{\rho^2}{6} \begin{pmatrix} 1 & 2 & 0 & 0 \\ - 2 & 1& 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 &0 & 0 & -3 \\ \end{pmatrix} +Ο(\rho^3). $$

Let us call the first big matrix A and the second one B. Note $$A^2=\begin{pmatrix} 1 & 2 & 0 & 0 \\ - 2 & -4& 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 &0 & 0 & -3 \\ \end{pmatrix} . $$

Now, if the 22 entry of B were -4 instead of 1, we'd have $A^2=B$, and thus $A^3=-3A$, $A^4=-3A^2$, etc... so (glory!) you can confirm $$ M=\mathbb{1} -\frac{\rho}{\sqrt{3}}A+ \frac{1}{2} \left(-\frac{\rho A}{\sqrt{3}}\right)^2 +...=e^{-\frac{\rho }{\sqrt{3}}A}, $$ since the expansion of the exponential reduces to $$ =\mathbb{1} -\sin\rho ~ \frac{A}{\sqrt{3}} +\frac{1-\cos\rho}{3} A^2 , $$ by the above recursive rules.

You would then, indeed, call this logarithm $A/\sqrt{3}$ of the exponential, up to the parameter -ρ, the generator of the group element M . In your specific case, you see it is a linear combination of a spacetime rotation (antisymmetric elements) and a boost-like strain (symmetric elements).

However, as it stands, your B is problematic, which is why I am convinced it is wrong, and should be my proposed expression, instead.

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  • $\begingroup$ In fact A it is a linear combination of a "space-time rotation" (antisymmetric elements) and a "space-time strain" which is a boost boost (symmetric elements). Space-time rotations are part of GL(4) but not the Lorentz subgroup. $\endgroup$ – Gary Godfrey Mar 2 '18 at 19:49

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