The shape of the Einstein-Rosen bridge is often visualized/modelized with the Flamm's paraboloid, and many other references have also stated clearly that it's a "surface of revolution of a parabola".(adsbygoogle = window.adsbygoogle || []).push({});

But as far as I can see, when we rotate the parabola w^2 = 8M(r-2M) (in natural units c=G=1) around the w axis (i.e r=0), we get no paraboloid.

The rotation process replaces r coordinate by sqrt(x^2+y^2), so that the resulting surface equation becomes

(w^2/8M+2M)^2 = x^2+y^2

This is even not a quadric surface, let alone a paraboloid (which looks like a bowl).

Only the rotation around its symmetric axis (i.e r axis) gives a paraboloid of revolution!

Instead, the shape of the Einstein-Rosen bridge rather looks like a hyperboloid of 1 sheet (although not rigorously).

What's annoying me is that I cannot find anyone / any article on the Internet highlighting the fact that the Flamm's paraboloid is NOT a paraboloid, am I wrong??

Thanks for any helpful comments (for this is my first post).

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# I Is Flamm's paraboloid a paraboloid?

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