Swartzchild Ricci tensor?

[Endercreeper01]

What is the christoffel symbol of the second kind them for the swartzchild metric?

[ajb]

What is the christoffel symbol of the second kind them for the swartzchild metric?

I am sure you can find that easy enough via google.

[xyzt]

What is the christoffel symbol of the second kind them for the swartzchild metric?

See here

[Widdekind]

A vacumm solution is when the energy-momentum tensor is zero in the field equations. This means that there are no masses and no fields other than the gravitational field on the space-time.

 

Now things like the Schwarzschild metric and so on describe the vacumm around some massive object. You need to also consider "interior solutions" carefully matched up with the vacumm solution to describe the whole space-time.

 

 

 

 

http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution

 

 

i tried re-deriving the Schwarzschild solution from the above, recouping the original R_ii terms (in the derivation above, the R_ii have been multiplied by factors vaguely resembling (4A²B, 4B²A, and 2A²B²). Then i set the SET to T_tt = rho c2, all the rest zero, assuming that time is orthogonal to xyz <----> r theta phi, so that the change of spatial coordinates does not affect the T_tt term in the corner (according to common convention, not the wiki article).

 

Then i used your dual equation:

 

http://blogs.scienceforums.net/ajb/?p=934

 

and tried to calculate the interior solution, for a spherically symmetric, constant mass-energy density ball...

 

now the RHS <> 0, and no more is AB = const...

 

combining all the equations, i got

 

B = stuff involving A's in numerator and denominator

 

then i assumed, that A® = dr² coefficient = 1/sqrt(1 - kr²) = constant closed curvature = cosmology...

 

then everything simplified, and

 

B = 3/4 (approx.)

 

actually

 

B = pi rho R³ / M (approx.)

 

which i think varies between 1/4 to 3/4 depending on the compactness of the object (dense objects stretch space more, and can pack more mass, for the same density, within the same coordinate radius)

 

but anyway B ~= 1 (of order)

 

so i essentially recouped FRW cosmological metric, for closed curvature (B=1, A=1/sqrt(1-kr²)).

 

such seems surprising... immediately above a compact object's surface... space is stretched to near-tearing (and time "thins out" to preserve AB=1, i.e. space-time volume is preserved, space stretches, but time thins out)...

 

but then immediately under said surface, time is nearly normal, and so is space, the stretchings and contractings are only of order 1...

 

if you consider the Flamm's paraboloid, above the object, space is super-stretched...

 

but w/in the body, space is only "bowed down" slightly... a diameter through the object ought to have a length 2R, but (assuming constant curvature) now stretches to pi*R = 50% increase = of order one... the following figure is qualitatively accurate, in conveying the concept, of constant closed curvature w/in the body, but (potentially) arbitrary spatial stretch above-but-near its surface...

 

http://cr4.globalspec.com/PostImages/201202/general_relativity_curved_space_time_sta_7A21E635-A1DF-6F5B-F2C7D945567BFAE0.jpg

 

so, is that a potentially plausible solution for a Schwarzschild-interior ?? in low-orbit above the body, time-and-space dilations are enormous... but inches under the same's surface, things are nearly normal ??

 

The above solution matches the spatial curvature, at the object's radius, from the Flamm's paraboloid curve above the body, to a constant closed curvature w/in the body... but how would one handle the discontinuous jump, in time ?? Time ticks super-slow just above the body, but nearly normally w/inside the same ?? That would require lines-of-simultaneity to jump discontinuously,

[ajb]

Write out the metric explicitly and we can see it it is a known solution. That said, I am not an expert in exact solutions and so may not recognise it.

Edited September 6, 2013 by ajb

[Widdekind]

attempting to combine and comport the conventions, of page vs. page,

 

[math]B \longrightarrow -B[/math]

 

[math]U^{\alpha} = \left( \begin{array}{c c c c} \frac{c}{\sqrt{B}} & 0 & 0 & 0 \end{array} \right)[/math]

[math]U_{\alpha} = \left( \begin{array}{c c c c} \frac{-c}{\sqrt{B}} & 0 & 0 & 0 \end{array} \right)[/math]

[math]g_{\alpha \beta} = \begin{bmatrix} -B & 0 & 0 & 0 \\ 0 & A & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2 sin^2(\theta) \end{bmatrix}[/math]

 

[math]T_{\alpha \beta} = \left( \rho + \frac{P}{c^2} \right) U_{\alpha} U_{\beta} + g_{\alpha \beta} P[/math]

 

[math]T_{\alpha \beta} = \begin{bmatrix} \rho c^2 B & 0 & 0 & 0 \\ 0 & P A & 0 & 0 \\ 0 & 0 & P r^2 & 0 \\ 0 & 0 & 0 & P r^2 sin^2(\theta) \end{bmatrix}[/math]

 

[math]T = T^{\alpha}_{\alpha} = Trace( g^{-1} T ) = \left( - \rho c^2 + 3 P \right)[/math]

 

[math]RHS = \left( \frac{8 \pi G}{c^4} \right) \left( T_{\alpha \beta} - \frac{T}{2} g_{\alpha \beta} \right)[/math]

 

[math]RHS = \left( \frac{8 \pi G}{c^4} \right) \begin{bmatrix} \frac{\rho c^2 + 3 P}{2}B & 0 & 0 & 0 \\ 0 & \frac{\rho c^2 - P}{2}A & 0 & 0 \\ 0 & 0 & \frac{\rho c^2 - P}{2}r^2 & 0 \\ 0 & 0 & 0 & \frac{\rho c^2 - P}{2}r^2sin^2(\theta) \end{bmatrix}[/math]

 

[math]LHS = \begin{bmatrix} \frac{R_{tt} \times (-1)}{4 r A^2 B} & 0 & 0 & 0 \\ 0 & \frac{R_{rr}}{4 r A B^2} & 0 & 0 \\ 0 & 0 & \frac{R_{\theta \theta}}{2A^2B} & 0 \\ 0 & 0 & 0 & \frac{R_{\theta\theta} sin^2(\theta)}{2A^2B} \end{bmatrix}[/math]

 

[math] \left( R_{tt} \times (-1) \right) + R_{rr} = 4B \left( AB \right)' = \left( \frac{8 \pi G}{c^4} \right) \left( \rho c^2 + P \right) \times \left( 4 r A^2 B^2 \right) [/math]

[math] \left( AB \right)' = \left( \frac{8 \pi G}{c^4} \right) \left( \rho c^2 + P \right) \times \left( r A^2 B \right) > 0 [/math]

 

Within material objects, space-time is "contractile" (with the words written by John Archibald Wheeler), as you move outward from the center of the object, the four-volume of space-time "voxels" [math]dV_4 = \sqrt{A} \sqrt{B} \; cdt \; dr \; r^2 d\Omega[/math] increases.

 

[math] R_{\theta \theta} = r \left( AB \right)' + 2 A B \left( A - 1 \right) - 2 r B' A = \left( \frac{8 \pi G}{c^4} \right) \left( \rho c^2 - P \right) \left( r^2 A^2 B \right)[/math]

 

[math] 2 A B \left( A - 1 \right) - 2 r B' A = \left( \frac{8 \pi G}{c^4} \right) \left( - 2 P \right) \left( r^2 A^2 B \right)[/math]

 

[math] B \left( 1 - A \right) + r B' = \left( \frac{8 \pi G}{c^4} \right) \left( P \right) \left( r^2 A B \right)[/math]

 

[math] B + r B' = \left( \frac{8 \pi G P}{c^4} \right) \left( r^2 A B \right) + \left( A B \right)[/math]

 

[math] \frac{1 + \frac{r B'}{B}}{1 + \frac{8 \pi G P}{c^4} r^2} = A[/math]

 

HSE:

 

[math]\frac{dP}{ds} = - \rho g[/math]

 

[math]dP = - \rho \left( - \Gamma^{r}_{tt} \right) dr \frac{ds}{dr} = \rho \left( \frac{B'}{2 A} \right) dr \sqrt{B}[/math]

 

which would seemingly bring about an integro-differential equation, for a self-consistent solution, even for the simplified situation, of incompressible constant density. Schwarzschild seemingly assumed (see below) that the space-time within a massive object, of constant density, would manifest closed constant curvature:

 

[math]B \equiv \frac{1}{1 - \left( \frac{r}{R_C} \right)^2}[/math]

 

If so, then

 

[math]B' = B^2 \left( 2 \frac{r}{R_C^2} \right)[/math]

 

[math]x \equiv \frac{r}{R_C}[/math]

 

[math]A = \frac{1 + 2 B x^2}{1 + \frac{8 \pi G P}{c^4} r^2} = \left( \frac{1 + x^2}{1 - x^2} \right) \frac{1}{1 + \frac{8 \pi G P}{c^4}r^2} [/math]

 

Now

 

[math]R_C^2 = \frac{R^3}{R_S}[/math]

 

so at the surface of the massive object

 

[math]x^2 = \frac{R_S}{R}[/math]

 

and the pressure P = 0, so

 

[math]A® = \frac{1 + \frac{R_S}{R}}{1 - \frac{R_S}{R}} \ne 1 + \frac{R_S}{R}[/math]

 

of the exterior solution. So the assumption of Schwarzschild seems (slightly) in error. Still, his interior solution (see below) appears, plausibly, to be qualitatively correct. Note, that time totally stalls out, into static stasis, inside the center of some massive object, which becomes compresses & compacted down, to R = 9/8 R_S. So, according to the interior solution of Schwarzschild, once some star collapses & compresses, to 9/8 R_S, its center stalls out into a stasis of timelessness. Per that picture, space is strewn w/ stars, which are technically still in the process of continuing collapse... but which infall will require infinite time to actually occur (as seen from afar). If so, then there are not actually any "black holes" anywhere within the fabric of space-time... all "complete collapses" stall out into timeless stasis, before full infall actually occurs (as seen from afar = HST).

 

Note that the low-gravity limit, of the gravitational acceleration, according to Schwarzschild's interior solution, is g® ~ r, which is the field from an isothermal sphere, with density decreasing as r^-2. So i know not where Schwarzschild derived his interior solution... still his has the advantage, of being continuous through the surface of the object, and so seems qualitatively correct.

 

http://www.rafimoor.com/english/GRE3.htm

http://en.wikipedia.org/wiki/Geodesics_in_general_relativity

http://cr4.globalspec.com/blogentry/17100/Spacetime-Curvature

[Widdekind]

...

 

[math] \left( R_{tt} \times (-1) \right) + R_{rr} = 4B \left( AB \right)' = \left( \frac{8 \pi G}{c^4} \right) \left( \rho c^2 + P \right) \times \left( 4 r A^2 B^2 \right) [/math]

[math] \left( AB \right)' = \left( \frac{8 \pi G}{c^4} \right) \left( \rho c^2 + P \right) \times \left( r A^2 B \right) > 0 [/math]

 

Within material objects, space-time is "contractile" (with the words written by John Archibald Wheeler), as you move outward from the center of the object, the four-volume of space-time "voxels" [math]dV_4 = \sqrt{A} \sqrt{B} \; cdt \; dr \; r^2 d\Omega[/math] increases.

 

[math] R_{\theta \theta} = 2 r A' B + 2 A B \left( A - 1 \right) - r (AB)' = \left( \frac{8 \pi G}{c^4} \right) \left( \rho c^2 - P \right) \left( r^2 A^2 B \right)[/math]

 

[math] 2 r A' B + 2 A B \left( A - 1 \right) = \left( \frac{8 \pi G}{c^4} \right) \left( 2 \rho c^2 \right) \left( r^2 A^2 B \right)[/math]

 

[math] \frac{r A'}{A} + \left( A - 1 \right) = \left( \frac{8 \pi G \rho}{c^2} \right) \left( r^2 \right)[/math]

 

the solution can be computed exactly, for constant density

 

[math]A(x) = \frac{1}{1-\frac{1}{3}x^2 + \frac{k}{x}}[/math]

 

[math]x \equiv \frac{r}{R_C}[/math]

 

[math]R_C^{-2} \equiv \frac{8 \pi G \rho}{c^2}[/math]

 

said solution must be matched, at r=R, to the exterior solution, which defines the constant of integration k. Plotting said solution shows that, similar to the suggestion of Schwarzschild, the spatial stretch factor does decrease inwards

[Widdekind]

...

 

[math] R_{\theta \theta} = r \left( AB \right)' + 2 A B \left( A - 1 \right) - 2 r B' A = \left( \frac{8 \pi G}{c^4} \right) \left( \rho c^2 - P \right) \left( r^2 A^2 B \right)[/math]

 

[math] 2 A B \left( A - 1 \right) - 2 r B' A = \left( \frac{8 \pi G}{c^4} \right) \left( - 2 P \right) \left( r^2 A^2 B \right)[/math]

 

[math] B \left( 1 - A \right) + r B' = \left( \frac{8 \pi G}{c^4} \right) \left( P \right) \left( r^2 A B \right)[/math]

 

[math] B + r B' = \left( \frac{8 \pi G P}{c^4} \right) \left( r^2 A B \right) + \left( A B \right)[/math]

 

[math]\frac{r B'}{B} = A \left( \epsilon x^2 + 1 \right) - 1[/math]

 

[math]\approx \frac{ \epsilon x^2 + 1 }{1 - \frac{1}{3} x^2 } - 1[/math]

 

making the simplifying assumption of Schwarzschild (seemingly)

 

[math]= \frac{ \left(\epsilon + \frac{1}{3}\right) x^2 }{1 - \frac{1}{3}x^2}[/math]

 

[math]\longrightarrow[/math] (if P = constant)

 

[math]B(x) = \left( \frac{1-\frac{R_S}{R}}{1-\frac{R_S}{R}\left(\frac{r}{R}\right)^2} \right)^{\frac{1+3 \epsilon}{2}}[/math][math]\times \left( 1 - \frac{R_S}{R} \right)[/math]

 

 

 

 

 

HSE:

 

[math]\frac{dP}{ds} = - \rho g[/math]

 

[math]dP = - \rho \left( - \Gamma^{r}_{tt} \right) dr \frac{ds}{dr} = \rho \left( \frac{B'}{2 A} \right) dr \sqrt{A}[/math]

 

which would seemingly bring about an integro-differential equation, for a self-consistent solution, even for the simplified situation, of incompressible constant density. Schwarzschild seemingly assumed (see below) that the space-time within a massive object, of constant density, would manifest closed constant curvature...

 

[math] B + r B' = \left( \frac{8 \pi G P}{c^4} \right) \left( r^2 A B \right) + \left( A B \right)[/math]

 

[math]\frac{r B'}{A} = B \left( x^2 \frac{P}{\rho c^2} + 1 - \frac{1}{A} \right)[/math]

 

[math] = B \left( \epsilon x^2 + 1 - \left( 1 - \frac{1}{3} x^2 + \frac{k}{x} \right) \right)[/math]

 

[math] \approx B \left( \epsilon + \frac{1}{3} \right) x^2[/math]

 

making the simplifying assumption of Schwarzschild (seemingly)...

 

these equations quickly become quite complex and cumbersome

 

[math]dP = \frac{\rho}{2} \frac{dr}{r} \sqrt{A} B \left(\epsilon + \frac{1}{3}\right) x^2[/math]

 

[math] \approx \frac{\rho}{2} \left( \epsilon + \frac{1}{3} \right) \left( x dx \right) \sqrt{\frac{1}{1 - \frac{1}{3}x^2}} [/math][math]\left( \frac{1-\frac{R_S}{R}}{1-\frac{R_S}{R}\left(\frac{r}{R}\right)^2} \right)^{\frac{1+3 \epsilon}{2}}[/math][math]\times \left( 1 - \frac{R_S}{R} \right)[/math]

 

but

 

[math]R_C^{-2} \equiv \frac{8 \pi G \rho}{c^2} \approx \frac{3 R_S}{R^3}[/math]

 

so

 

[math]x^2 \equiv \left( \frac{r}{R_C} \right)^2 \approx \frac{3 R_S}{R} \left( \frac{r}{R} \right)^2[/math]

 

and

 

[math] dP \approx \frac{\rho c^2}{2} \left( \epsilon + \frac{1}{3} \right) \left( x dx \right) \times[/math][math] \frac{ \left( 1-\frac{R_S}{R} \right)^{\frac{3+3\epsilon}{2} } }{\left(1-\frac{R_S}{R}\left(\frac{r}{R}\right)^2\right)^{\frac{2+3\epsilon}{2}} }[/math]

 

[math]= \frac{\rho c^2}{2} \left( \epsilon + \frac{1}{3} \right) \left( x dx \right) \times \sqrt{1 - \frac{R_S}{R}} \times[/math][math]\left( \frac{1-\frac{R_S}{R}}{1-\frac{R_S}{R}\left(\frac{r}{R}\right)^2} \right)^{\frac{2+3 \epsilon}{2}}[/math]

 

[math]= \frac{\rho c^2}{2} \left( 1 + 3 \epsilon \right) \left( \frac{R_S}{R} \right) \left( \left( \frac{r}{R} \right) d\left(\frac{r}{R}\right) \right) \times \sqrt{1 - \frac{R_S}{R}} \times[/math][math]\left( \frac{1-\frac{R_S}{R}}{1-\frac{R_S}{R}\left(\frac{r}{R}\right)^2} \right)^{\frac{2+3 \epsilon}{2}}[/math]

 

 

the obvious limiting cases would be [math]\epsilon = 0,1[/math], for which the function can be plotted, as an approximation, for the full solution, for which [math]\epsilon(x)[/math] varies w/ radius

[math]dP = \frac{\rho c^2}{2} \left( 1 + 3 \epsilon \right) \left( \frac{R_S}{R} \right) \sqrt{1-\frac{R_S}{R}} \left( y dy \right) \times \left( \frac{1-\frac{R_S}{R}}{1-\frac{R_S}{R}y^2} \right)^{\frac{2+3 \epsilon}{2}}[/math]

 

to lowest order in R_S, and for [math]\epsilon = 0[/math], the equation defaults to a common classical HSE

 

[math]\longrightarrow \frac{dP}{dr} \approx \frac{G M \rho}{R^2} \frac{r}{R}[/math]

 

which is the isothermal sphere equation [math]\rho \propto r^{-2}[/math]...

 

otherwise the equation seems easily integrable, by substitution [math] u = 1 - k y^2[/math]