Questions about acceleration in SR

[Hans de Vries]

So we can write

$$\Lambda_{\phi} = e^{Z\phi}$$​

Indeed

Regards, Hans

 

[yuiop]

So we can write

$$\Lambda_{\phi} = e^{Z\phi}$$​

That looks a LOT like the polar equation for a logarithimic spiral http://mathworld.wolfram.com/LogarithmicSpiral.html . Do you think there is any connection?

 

[Hans de Vries]

That looks a LOT like the polar equation for a logarithimic spiral http://mathworld.wolfram.com/LogarithmicSpiral.html . Do you think there is any connection?

Note that Z is a matrix here...$$e^{Z\phi} ~=~ I~+~Z\phi ~+~ \frac{1}{2!}Z^2\phi^2~+~ \frac{1}{3!}Z^3\phi^3~+~ \frac{1}{4!}Z^4\phi^4~+~ \frac{1}{5!}Z^5\phi^5 ...$$

But higher powers of Z simply become either Z or Z^2

$$Z~ = Z^3 = Z^5 = ... = \left(\begin{array}{cccc}<br /> 0 & 0 & 0 & 1 \\<br /> 0 & 0 & 0 & 0 \\<br /> 0 & 0 & 0 & 0 \\<br /> 1 & 0 & 0 & 0 \\<br /> \end{array}\right)$$

$$Z^2 = Z^4 = Z^6 = ... = \left(\begin{array}{cccc}<br /> 1 & 0 & 0 & 0 \\<br /> 0 & 0 & 0 & 0 \\<br /> 0 & 0 & 0 & 0 \\<br /> 0 & 0 & 0 & 1 \\<br /> \end{array}\right)$$

and therefor:

$$e^{Z\phi} ~=~ I ~+~ \sinh(\phi)\left(\begin{array}{cccc}<br /> 0 & 0 & 0 & 1 \\<br /> 0 & 0 & 0 & 0 \\<br /> 0 & 0 & 0 & 0 \\<br /> 1 & 0 & 0 & 0 \\<br /> \end{array}\right) ~+~ (\cosh(\phi)-1)\left(\begin{array}{cccc}<br /> 1 & 0 & 0 & 0 \\<br /> 0 & 0 & 0 & 0 \\<br /> 0 & 0 & 0 & 0 \\<br /> 0 & 0 & 0 & 1 \\<br /> \end{array}\right) ~=~ \left(\begin{array}{cccc}<br /> \cosh\phi & 0 & 0 & \sinh\phi \\<br /> 0 & 1 & 0 & 0 \\<br /> 0 & 0 & 1 & 0 \\<br /> \sinh\phi & 0 & 0 & \cosh\phi \\<br /> \end{array}\right)$$
Regards, Hans