Questions about acceleration in SR

[Hans de Vries]

So we can write

<br /> \Lambda_{\phi} = e^{Z\phi}<br />​

Indeed

Regards, Hans

 

[yuiop]

So we can write

<br /> \Lambda_{\phi} = e^{Z\phi}<br />​

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 &amp; 0 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> \end{array}\right)

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

and therefor:

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