Flux of Lorentz trasformation in spacetime, infinitesimal generator

Question

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

Answer 1

In general, for a one-parameter group of operators $A(\rho)$, it's infinitesimal generator is just $A'(0)$. So, if I understood the question correctly, the infinitesimal generator is

$$\begin{pmatrix} -{1\over 3} && -{2 \over 3} && 0 && 0 \\ {2\over 3} && -{1 \over 3} && 0 && 0 \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 1 \\ \end{pmatrix}$$

Answer 2

See my comment, in that case: Do this kind of problem by writing this matrix as a sum of two matrices as follows:$$ W(\rho) = \left(\begin{array}{cccc} \frac{4- \cos(\rho)}{3} & \frac{2- 2\cos(\rho)}{3} & 0 & -\frac{\sin(\rho)}{\sqrt{3}} \\ \frac{2\cos(\rho) - 2}{3} & \frac{4\cos(\rho)-1}{3} & 0 & \frac{2\sin(\rho)}{\sqrt{3}}\\ 0 & 0 & 1 & 0 \\ -\frac{\sin(\rho)}{\sqrt{3}} & -\frac{2\sin(\rho)}{\sqrt{3}} & 0 & \cos(\rho) \\ \end{array}\right) = \\ \begin{pmatrix} \frac{4-\cos(\rho)}{3} & \frac{2-2\cos(\rho)}{3} & 0 & 0 \\ \frac{2\cos(\rho) - 2}{3} & \frac{4\cos(\rho)-1}{3} & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & \cos(\rho) \\ \end{pmatrix} - \sin(\rho)\begin{pmatrix} 0 & 0 & 0 & \frac{1}{\sqrt{3}} \\ 0 & 0 & 0 & -\frac{2}{\sqrt{3}}\\ 0 & 0 & 0 & 0 \\ \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\ \end{pmatrix} $$ Where I have explicitly pulled out the $\sin$ dependence. Now look at the matrix $$ u =\begin{pmatrix} 0 & 0 & 0 & \frac{1}{\sqrt{3}} \\ 0 & 0 & 0 & -\frac{2}{\sqrt{3}}\\ 0 & 0 & 0 & 0 \\ \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\ \end{pmatrix}$$ Notice that $u^3 = -u$ so that in particular, this matrix is easy to exponentiate. In fact Exp$(u) = W(1)$ is your original matrix evaluated at $\rho = 1$ so then you can prove $$\textrm{Exp}( \rho u) = W(\rho)$$ If memory serves me, in physics land they call $u$ the infinitesimal generator of $W$. But check with your instructor.