Geodesics in Schwarzschild metric

[PeterDonis]

Do you mean the spherically symmetric property basically amounts to the existence of a spacetime foliation with a 'two-fold' family of 2-sphere ?

Yes, one way of stating "spherically symmetric" is "the spacetime can be foliated by 2-spheres". The fact that spacetime is 4-dimensional and the 2-spheres are 2-dimensional means that the foliation must have two parameters, which we can adopt as the other two coordinates as soon as we figure out what those parameters are for the particular spacetime we are considering.

but...standard $\theta$ and $\phi$ coordinates should not make sense just for 2-sphere in the 'space' slice (namely in the 3D spacelike hypersurfaces of constant coordinate time $t$) ?

Any 2-sphere can be coordinatized by $\theta$ and $\phi$. Go back and re-read @Orodruin's post #26. What he said there applies to an embedding in any higher dimensional manifold, not just Euclidean 3-space.

 

[cianfa72]

Any 2-sphere can be coordinatized by $\theta$ and $\phi$. Go back and re-read @Orodruin's post #26. What he said there applies to an embedding in any higher dimensional manifold, not just Euclidean 3-space.

ok, so each of 2-spheres coordinatized by $\theta$ and $\phi$ has the spacelike metric as above in post #32 for a given value of the parameter $r$. The embedding is actually in the 4D spacetime manifold (or in the 3D spacelike hypersurface slice we get for a given value of coordinate time $t$).

What we said is about the mathematical model of spacetime. From a physical point of view we interpret $\theta$ and $\phi$ as the standard spherical coordinates on immaginary spherical shells 'built' around the singular point having $r=0$.

 

[PeterDonis]

each of 2-spheres coordinatized by $\theta$ and $\phi$ has the spacelike metric as above in post #32 for a given value of the parameter $r$.

Yes.

The embedding is actually in the 4D spacetime manifold

Yes.

(or in the 3D slice we get for a given value of coordinate time $t$).

This only works for regions where $t$ is a valid coordinate. It isn't on the horizon, and inside the horizon, while there is a valid coordinate chart with a coordinate called $t$ and the metric you give, that chart is disconnected from the chart on the exterior region, and in the interior $t$ is not even timelike.

From a physical point of view we interpret $\theta$ and $\phi$ as the standard spherical coordinates on immaginary spherical shells

Yes.

'built' around the singular point having $r=0$.

This interpretation doesn't really work physically, because the locus $r = 0$ is a spacelike line, which physically is intepreted as an instant of time, not a point in space.

 

[cianfa72]

This interpretation doesn't really work physically, because the locus $r = 0$ is a spacelike line, which physically is intepreted as an instant of time, not a point in space.

So, to put it simple, immaginary spherical shells are 'built' around the center of the massive body source of the gravitational field.

 

[PeterDonis]

So, to put it simple, immaginary spherical shells are 'built' around the center of the massive body source of the gravitational field.

No. Go read what you quoted from my post again, carefully. What do you think "doesn't really work" means?

 

[cianfa72]

No. Go read what you quoted from my post again, carefully. What do you think "doesn't really work" means?

Sorry, maybe I didn't get the point. Why we cannot build such immaginary spherical shells (outside the horizon) ?

 

[Orodruin]

Sorry, maybe I didn't get the point. Why we cannot build such immaginary spherical shells (outside the horizon) ?

Outside the horizon, yes. But there is no real ”center” here if we are talking about the full Schwarzschid spacetime. The coordinate r in no way relates to a distance from such a center. It is just a coordinate that labels the concentric shells with their area.

 

[cianfa72]

Outside the horizon, yes. But there is no real ”center” here if we are talking about the full Schwarzschid spacetime. The coordinate r in no way relates to a distance from such a center.

Definitely, that is samehow related to the fact that 3D slices (i.e. the spacelike hypersurfaces of constant coordinate time $t$) are not Euclidean.