What distinguishes the interior and exterior Schwarzschild metrics in Maxima?

[BWV]

playing around with ctensor & the Schwarzschild metric in Maxima

what is the difference between interior and exterior Schwarzschild metrics?

exteriorschwarzschild 4 [t,r,theta,phi] Schwarzschild metric
interiorschwarzschild 4 [t,z,u,v] Interior Schwarzschild metric

also when with the exterior Schwarzschild metric the scalar curvature is zero - this cannot be right, can it?

Function: scurvature ()
Returns the scalar curvature (obtained by contracting the Ricci tensor) of the Riemannian manifold with the given metric.

 

[George Jones]

what is the difference between interior and exterior Schwarzschild metrics?

The exteior solution is the vacuum solution outside a spherically symmetric mass distribution. The interior solution represents the interior of a static constant density mass. See

https://www.physicsforums.com/showthread.php?t=323684.

also when with the exterior Schwarzschild metric the scalar curvature is zero - this cannot be right, can it?

Any vacuum solution has $R = 0$. This does not necessarily mean that the curvature tensor is zero. Exterior Schwarzschild has $R = 0$ with non-zero curvature tensor.

 

[BWV]

The exteior solution is the vacuum solution outside a spherically symmetric mass distribution. The interior solution represents the interior of a static constant density mass. See

https://www.physicsforums.com/showthread.php?t=323684.


Any vacuum solution has $R = 0$. This does not necessarily mean that the curvature tensor is zero. Exterior Schwarzschild has $R = 0$ with non-zero curvature tensor.

but the scalar curvature of zero means that the Einstein tensor reduces to the Ricci tensor in the field equation?

 

[George Jones]

but the scalar curvature of zero means that the Einstein tensor reduces to the Ricci tensor in the field equation?

Right (for zero cosmological constant), and vacuum means that the stress-energy tensor is zero, so the vacuum Einstein equation, of which exterior Schwarzschild is one solution, is

$$R_{\mu \nu} = 0.$$

 

[BWV]

Right (for zero cosmological constant), and vacuum means that the stress-energy tensor is zero, so the vacuum Einstein equation, of which exterior Schwarzschild is one solution, is

$$R_{\mu \nu} = 0.$$

I'm confused - isn't the stress-energy tensor what reduces to the mass in the Newtonian approximation? If there is no mass (and no curvature), how can the Schwarzschild equation have been useful in calculating Mercury's perihelion, which was its first important use, right?

Also, for the Ricci Tensor to be zero means that Schwarzschild can be reduced to a cartesian coordinate system, so I guess that when r_s is zero this reduces to the Minkowski metric?

 

[George Jones]

Consider the Sun-Mercury system, and ignore Mercury's contribution to spacetime curvature as negligible compared that of the Sun. Outside the body of the Sun, there is (for the purposes of this exercise) a vacuum, and the stress-energy and Ricci tensors are zero outside the Sun, even though the Riemann curvature is non-zero outside the Sun. Since spacetime curvature is non-zero outside the Sun, there is no extended inertial coordinate system outside the Sun. The amount of spacetime curvature outside the Sun depends on the mass of the Sun, and, consequently, the amount of change perihelion position depends on the mass of the Sun.

Inside the body of the Sun, the stress-energy and Ricci tensors are non-zero.