General metric with zero riemann tensor

[archipatelin]

A metric consistent with interval:
\mathrm{d}s^2=-\mathrm{d}\tau^2+\frac{4\tau^2}{(1-\rho^2)^2}\left(\mathrm{d}\rho^2+\rho^2\mathrm{d}\theta^2+\rho^2\sin(\theta)^2\mathrm{d}\varphi^2\right)
get zero for riemann's tensor, therefor must be isomorphic with minkowski tensor.
But I don't find thus transformation of coordinates from \tau,\,\rho\rightarrow t,\,r
so that after transformation is interval in minkowski form:
\mathrm{d}s^2=-\mathrm{d}t^2+\fmathrm{d}r^2+r^2\mathrm{d}\theta^2+r^2\sin(\theta)^2\mathrm{d}\varphi^2.

What's transformation?

 

[bcrowell]

Haven't checked carefully, but it looks like a Robertson-Walker metric with a(t)=t, \rho=r/2, and k=-1.

 

[archipatelin]

Haven't checked carefully, but it looks like a Robertson-Walker metric with a(t)=t, \rho=r/2, and k=-1.

Yes, it is special case of FLWR metric. But, still don't know this transformation.

 

[George Jones]

Hint:
https://www.physicsforums.com/showthread.php?p=1757634#post1757634.

 

[archipatelin]

Hint:
https://www.physicsforums.com/showthread.php?p=1757634#post1757634.

Thanks,
I found this transformation in form:
\rho\equiv{}\rho(t,r)
\tau\equiv{}\tau(t,r)
Solutions with respect to minkowski metric are:
\rho=\frac{2t}{r}\left(1\pm\sqrt{1-\left(\frac{r}{2t}\right)^2}\right)
\tau=\frac{1}{2}r\frac{1-\rho^2}{\rho}
But it's regular only for \left\|\frac{r}{2t}\right\|<1.
However minkowski metric is regulare everywhere. It's OK?

 

[archipatelin]

I want to make sure that all solutions of the equation

R_{\mu\nu\varkappa\lambda}=0​

for any dimension D are isomorphic with tensor in form

g_{\mu\nu}=\mbox{diag}(\pm{}1,\pm{}1,\dots,\pm{}1)​

Or are there other solutions?