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.