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How to get Killing vectors in a form of generators of $SO(3)$ group symmetry?

By using Killing equations for metric $ds^{2} = d\theta^{2} + \sin^{2}(\theta^{2}) d\varphi^{2}$ I got $$ \varepsilon_{\theta } = C_{1}\cos (\varphi ) + C_{2}\sin(\varphi ), $$ $$ \varepsilon_{\varphi} = \frac{1}{2}\sin(2 \theta )(C_{1}\cos(\varphi ) - C_{2}\sin(\varphi )) + C_{3}\sin^{2}(\theta ). $$ So, if I set $C_{1,2}$ to zero, I will get $$ \varepsilon^{\mu}_{1} = C_{3}(0, sin^{2}(\theta )), $$ and for another combinations $$ \varepsilon^{\mu}_{2} = \left(C_{2}sin(\varphi ), \frac{C_{2}}{2}cos(\varphi)sin(2 \theta)\right), $$ $$ \varepsilon^{\mu}_{3} = \left(C_{1}cos(\varphi ), -\frac{C_{1}}{2}sin(\varphi)sin(2 \theta)\right). $$ But the correct result refers to $$ \varepsilon^{\mu}_{1} = C_{3}(0, 1), \quad \varepsilon^{\mu}_{2} = \left(C_{2}sin(\varphi ), C_{2}cos(\varphi)ctg(\theta )\right), $$ $$ \varepsilon^{\mu}_{3} = \left(C_{1}sin(\varphi ), -C_{1}sin(\varphi)ctg(\theta )\right). $$ Why do I must multilpy the results for $\varphi $component on $\frac{1}{\sin^{2}(\theta )}$?

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    $\begingroup$ Hint : You have a difference between $\varepsilon_{\varphi}$ and $\varepsilon^{\varphi}$ $\endgroup$ – Trimok Nov 8 '13 at 17:37
  • $\begingroup$ @Trimok . Thank you, I forgot about it. But why do I must use exactly $\varepsilon^{\varphi}$ if I want to show the equivalence of Killing vectors and operator's representation of SO(3) group generators? $\endgroup$ – John Taylor Nov 8 '13 at 17:40
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    $\begingroup$ Any vector (so Killing vectors too) may be written as a differential operator $\varepsilon = \varepsilon^\mu \partial_\mu$. These differential operators must satisfy $so(3)$ algebra. $\endgroup$ – Trimok Nov 8 '13 at 17:47
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    $\begingroup$ @Trimok: Why don't you make that an answer? If the question doesn't get any votes for a long time, and remains unanswered, it will get auto - deleted. $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 21 '13 at 6:32
  • $\begingroup$ @JohnTaylor: Related/duplicate(?) physics.stackexchange.com/questions/67556/… I think this should answer your question. $\endgroup$ – user7757 Dec 14 '13 at 13:29

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