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For static dS metric we have $$x_{0}=\sqrt{H^{-2}-r^{2}}\sinh(Ht)$$$$x_{1}=\sqrt{H^{-2}-r^{2}}\cosh(Ht)$$ and the metric can be written as $$ds^{2}=-dx_{0}^{2}+dx^{2}_{1}+d\bar x$$ where the barred $x$ will be just the spherical part of the final metric lets focus on the first two term of above and calculate them first this can be done as follow (taking $g(r)=\sqrt{H^{-2}-r^{2}}$) we have $$dx_{0}=\frac{\partial x_{0} }{\partial t}dt+\frac{\partial x_{0} }{\partial r}dr=H\sqrt{H^{-2}-r^{2}}\cosh(Ht)dt-\frac{r\sinh(Ht)}{\sqrt{H^{-2}-r^{2}}}dr$$ and $$dx_{1}=\frac{\partial x_{1} }{\partial t}dt+\frac{\partial x_{1} }{\partial r}dr=H\sqrt{H^{-2}-r^{2}}\sinh(Ht)dt-\frac{r\cosh(Ht)}{\sqrt{H^{-2}-r^{2}}}dr$$ and hence $$-dx_{0}^{2}+dx^{2}_{1}=-H^{2}g^{2}dt^{2}+(g')^{2}dr^{2}=-(1-H^{2}r^{2})dt^{2}+\frac{H^{2}r^{2}}{1-H^{2}r^{2}}dr^{2}$$ Now in order to have the right form of static metric one shall do the following, $$\int(\frac{\partial}{\partial r }\frac{H^{2}r^{2}}{1-H^{2}r^{2}})dr=\frac{1}{1-H^{2}r^{2}}+C$$ and the final metric has a form of $$ds^{2}=-(1-H^{2}r^{2})dt^{2}+(1-H^{2}r^{2})^{-1}dr^{2}+r^{2}d\Omega^{2}$$ but as you can see when you integrate you will produce constant of integration which is C so this will not admit the final metric quite correctly, how shall we handle this? does one simply ignore the C? is it allowed or is there any alternative derivation of final metric?

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  • $\begingroup$ Can you explain what and why you are doing with the integral?! $\endgroup$ – MBN Jun 3 '15 at 14:06
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The short answer is your integration is both wrong and futile and your mistake was ignoring the other three parameters.

For a static dS metric we have $$x_{0}=\sqrt{H^{-2}-r^{2}}\sinh(Ht)$$$$x_{1}=\sqrt{H^{-2}-r^{2}}\cosh(Ht)$$ $$x_{2}=r\cos\theta$$ $$x_{3}=r\sin\theta\cos\phi$$ $$x_{4}=r\sin\theta\sin\phi.$$ and the metric can be written as $$ds^{2}=-dx_{0}^{2}+dx^{2}_{1}+dx^{2}_{2}+dx^{2}_{3}+dx^{2}_{4}.$$

You computed the first two terms correctly

$$-dx_{0}^{2}+dx^{2}_{1}=-(1-H^{2}r^{2})dt^{2}+\frac{H^{2}r^{2}}{1-H^{2}r^{2}}dr^{2}.$$ But in order to have the correct metric you need to compute the other three terms which will give you the surface area term but also another $dr^{2}$ term. A term that is exactly what you need to get the metric in the form $$ds^{2}=-(1-H^{2}r^{2})dt^{2}+(1-H^{2}r^{2})^{-1}dr^{2}+r^{2}d\Omega^{2}.$$

And as a final hint (in case you need another) the metric for $dx^{2}_{2}+dx^{2}_{3}+dx^{2}_{4}$ looks like the complete metric for euclidean 3 space (not just the surface area part).

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