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Kerr Interior Solution

Posted Apr27-09 at 11:19 AM by stevebd1
Updated May3-09 at 02:20 AM by stevebd1

Information compiled from this source

In all cases a=J/mc and M=Gm/c2

Schwarzschild interior solution-

$$d\tau^{2}=\left( \frac{3}{2}\sqrt{1-\frac{2M}{r_0}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{r_0^{3}}}\right) ^{2}dt^{2}-\left( 1-\frac{2Mr^{2}}{r_0^{3}}\right) ^{-1}dr^{2}-r^{2}\left( d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right)$$

where r0 is the closed 2-surface radius (i.e. star surface). r0>9M/4 (minimum 2.25M) 'in order to have positive tangential pressures.'

$$g_{00}=\left( \frac{3}{2}\sqrt{1-\frac{2M}{r_0}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{r_0^{3}}}\right)^{2}$$

solve f0 for g00=(1-2f0(r)/r^2)-

\begin{flalign} &f_0(r)= \frac{r^2}{4} \left[ \frac{Mr^2}{r_0^3} +\frac{9M}{r_0} +3 \sqrt{ \left( 1-\frac{2M}{r_0} \right) \left( 1-\frac{2Mr^2}{r_0^3} \right)} -3 \right]\12mm] &g_{11}=\left( 1-\frac{2Mr^{2}}{r_0^{3}}\right)^{-1} \end{flalign} solve f1 for g11=(1-2f1(r)/r^2)-1- $$f_1(r)=\frac{Mr^4}{r_0^3}$$ Interior solution at closed 2-surface- f0=f1=Mr0 ⇒ Mr Kerr interior solution (incorporating f0 and f1)- $$d\tau^2=\frac{\Delta_0-a^2sin^2\theta}{\rho^2}dt^2\,-\,\frac{\rho^2}{\Delta_1}dr^2\,-\,\rho^2d\theta^2\,-\,\frac{\Sigma^2}{\rho^2}sin^2\theta d\phi^2\,+\,\frac{4af_0(r)\,sin^2\theta}{\rho^2}dtd\phi$$ where $$ \Delta_i= r^{2}+a^{2}-2f_i(r)\ \ \ \ \ \ \ \ \ \ \ i=0,1,\\ \\ \rho^2=r^2+a^2 cos^2\theta\\ \\ \Sigma^2=(r^2+a^2)^2-a^2\Delta_0 sin^2\theta \[$$

where again, r0>9M/4

The interesting thing here is that while in the static interior solution, proper distance (dr') decreases from a peak at r0 to 1 at r=0, in the Kerr interior solution, when θ>0, dr'=dr=1 occurs at a specific radius within the rotating star, the radius dictated by the quantity of spin (if you were to take a cross-section through the star, this boundary would resemble an 8 on its side, the 3D shape resembling a closed (horn) torus). Within this boundary, the space term actually begins to dilate in unison with the time component, the dilation increasing significantly towards the centre, already giving some indication of the nature of the singularity that would occur if the star was to collapse into a black hole. The other interesting aspect is approaching from the pole, dt'=dt=1 occurs at r=0 (as apposed to the time dilation peaking at r=0 in the static solution) which if gravity is synonymous with time dilation would imply that matter within the star would move away from the centre towards the poles (but stopping short of the surface). For the Kerr solution, the time component (dt') is similar to the static solution in the equatorial plane and the space component (dr') is similar to the static solution in the polar plane. There does appear to be a coordinate singularity at exactly r=0 but 0 can be replaced with very small radii.

In the Kerr interior solution, r0 is the equatorial radius for the closed 2-surface only. Solution for all radii to the close 2-surface (from Hogan metric)-

$$r=\frac{r_0}{2}\left(1+\sqrt{1-\frac{4a^2cos^2\theta}{r_0^2}\right)$$

which produces an oblate spheroid for a>0 and a sphere for a=0.

reduced circumference $(\varpi)$ to closed 2-surface-

$$\varpi=\Sigma/\rho$$

redshift-

$$z=\frac{\Sigma}{\rho}(\Delta_0)^{-1/2}-1$$

The metric moves from exterior to interior and static (a=0, Schwarzschild) to rotating (a>0, Kerr) solutions smoothly

sources-

'Simple Tests for Proposed Interior Kerr Metrics' by P. COLLAS
http://www.csun.edu/%7Evcphy00d/PDFP...as%20tests.pdf

'Trapped Null Geodesics in a Rotating Interior Metric' by P. COLLAS and J. K. LAWRENCE
http://www.csun.edu/%7Evcphy00d/PDFP...s-Lawrence.pdf
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