1. Dec 18, 2008

Fleet

Hi all,

I have found the "generic" form of the FLRW metric:
$$ds^2=(cdt)^2-dl^2$$

And I have found the three-dimension spatial metric for euclidian space (K=0, spherical space K=1 and hyperboloid space (K=-1):

$$dl^2=a^2(dr^2+r^2d\Omega^2)$$

$$dl^2=a^2(\frac{dr^2}{1-r^2})+r^2d\Omega^2)$$

$$dl^2=a^2(\frac{dr^2}{1+r^2})+r^2d\Omega^2)$$

BUT how do I find the "general" form of the FLRW metric, how can I include the curvature parameter K?

Best regards.

2. Dec 18, 2008

marcus

What happens if you just stick a K into the formula, with the understanding that it can take on just those 3 values: -1,0,and 1? Don't you get the three cases you want?

$$dl^2=a^2(\frac{dr^2}{1-Kr^2})+r^2d\Omega^2)$$

3. Dec 18, 2008

Fleet

Thank you very much for you answer, I really appreciate it!

Yes, you are right I get the cases I want. But are you questioning to be ironical or are you sure it the correct way? :)

Love this forum, I'm gonna contribute

Best regards

4. Dec 19, 2008

cristo

Staff Emeritus
It is the correct way. The general form for the line element is the one that marcus gives. Plugging in values for k gives you the three specific line elements.