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stevebd1

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(This was originally posted in https://www.physicsforums.com/showthread.php?t=248625&page=2" but I started a new thread as it was going slightly off topic)

Einsteins equation for gravity is g=

The equation for Newtonian gravity being (incorporating m=V

[tex]g=G\frac{m}{r^2}=G\frac{4\pi r_0^3}{3r^2} \rho[/tex]

and the equation for GR gravity being-

[tex]g=G\frac{4\pi r_0^3}{3r^2} \left(\rho+\frac{3P}{c^2}\right)[/tex]

The escape velocity of an object relates directly to g-

[tex]v_e=\sqrt{\frac{2Gm}{r}}=\sqrt{2gr}[/tex]

Which raises the issue of the effects of pressure on the collapse of a neutron star into a black hole as the collapse of the star is relative to the escape velocity exceeding c-

http://math.ucr.edu/home/baez/einstein/einstein.pdf" [Broken] by J C Baez and E F Bunn, page 7

GR escape velocity might result in a temporarily different Schwarzschild radius as the pressure would increase the gravity, therefore increasing the escape velocity and increasing the radius which v

Considering the escape velocity of a static 3 sol mass neutron star on the brink of collapse with a radius of 11 km and an average equation of state of ~1/7 -

Newtonian gravity and escape velocity-

g

GR gravity and escape velocity-

g

So if pressure is included the star would collapse sooner into a black hole. But would the 'new' Schwarzschild radius be temporary as the black hole settled down to 2Gm/c^2?

Einsteins equation for gravity is g=

*ρ*c^{2}+3P (basically energy density plus 3 times the pressure)-The equation for Newtonian gravity being (incorporating m=V

*ρ*)-[tex]g=G\frac{m}{r^2}=G\frac{4\pi r_0^3}{3r^2} \rho[/tex]

*where G is the gravitational constant, r*_{0}is the radius of the object of mass, r is the distance between the center of the object and the point at which gravity is being calculated and ρ is the density of the object of mass in kg/m^3 (Note when calculating gravity at the surface, r_{0}^3/r^2 can be reduced to simply r_{0})and the equation for GR gravity being-

[tex]g=G\frac{4\pi r_0^3}{3r^2} \left(\rho+\frac{3P}{c^2}\right)[/tex]

*as above but where P is pressure in N/m^3 and c is the speed of light*The escape velocity of an object relates directly to g-

[tex]v_e=\sqrt{\frac{2Gm}{r}}=\sqrt{2gr}[/tex]

Which raises the issue of the effects of pressure on the collapse of a neutron star into a black hole as the collapse of the star is relative to the escape velocity exceeding c-

*'There are a number of important situations in which ρ does not dominate P. In a neutron star, for example, which is held up by degeneracy pressure of the neutronium it consists of, pressure and energy density contribute comparably to the right-hand side of Einstein's equation. Moreover, above a mass of about 2 solar masses a nonrotating neutron star will inevitably collapse to form a black hole, thanks in part to the gravitational attraction caused by pressure.'*http://math.ucr.edu/home/baez/einstein/einstein.pdf" [Broken] by J C Baez and E F Bunn, page 7

GR escape velocity might result in a temporarily different Schwarzschild radius as the pressure would increase the gravity, therefore increasing the escape velocity and increasing the radius which v

_{e}=c.Considering the escape velocity of a static 3 sol mass neutron star on the brink of collapse with a radius of 11 km and an average equation of state of ~1/7 -

Newtonian gravity and escape velocity-

g

_{N}= 3.291x10^12 m/s^2, v_{e}= 2.691x10^8 m/s (0.898c)GR gravity and escape velocity-

g

_{GR}= 4.625x10^12 m/s^2, v_{e}= 3.190x10^8 m/s (>c)So if pressure is included the star would collapse sooner into a black hole. But would the 'new' Schwarzschild radius be temporary as the black hole settled down to 2Gm/c^2?

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