Is relativistic event horizon half of Newtonian event horizon?

Is Relativistic event horizon half of Newtonian event horizon?

relativistic escape velocity formula (from $m\phi=E-E_0$) is $v_e=\sqrt{2\phi-(\frac{\phi}{c})^2}$ and the Newtonian version of the formula (from $m\phi=\frac{1}{2}mv_e^2$) is just $v_e=\sqrt{2\phi}$, in both cases $v_e$ is the escape velocity, and $\phi$ is the gravitational potential. (for black hole $v_e=c$)

$R=\frac{GM}{c^2}$ Relativistic event horizon

where R is the radius of the black hole, or the distance from its center to its Relativistic event horizon. This formula gives exactly half the value as that of the standard Newtonian formula $R=\frac{1}{2}R_s$

$R_s=\frac{2GM}{c^2}$ Newtonian event horizon

or

Is the Schwarzschild radius twice the distance from the center of a black hole to its real event horizon?

or

are black holes only half as large as previously believed (according to relativistic effects)?

• Am I correct in assuming this question was spawned by the comparison of the deflection angle of light in Newtonian gravity versus GR? – zibadawa timmy Sep 21 '14 at 10:25
• possible duplicate of Can a black hole be explained by newtonian gravity? – zibadawa timmy Sep 21 '14 at 10:28
• See physics.stackexchange.com/q/19405/55483 and the answers in particular. – zibadawa timmy Sep 21 '14 at 10:28
• @Achmed - where did you get the relativistic equation for the radius of the black hole? It must be wrong. For non-rotating black holes, Einstein and Newton agree that the horizon radius is $2GM/c^2$. – Johannes Sep 21 '14 at 11:58
• I suspect the question is based on this article, or one like it. – John Rennie Sep 21 '14 at 15:10

1 Answer

No. You must have made an error in determining the GR horizon radius.

The radius of the event horizon determined from Newtonian theory (simply determining the distance from a point mass at which the escape velocity equals the speed of light) happens to be the same as the radius rigorously derived from the General Relativistic equations.

This, by the way, is a 'coincidence'. Would you calculate the radius of the photon sphere from Newtonian theory (determining the distance from a point mass at which the centripetal force equates the force of gravity for a test particle moving at the speed of light), you would find a distance that is three times too small compared to the correct General Relativistic expression.

• It would be better if you say exactly where I'm wrong – Achmed Sep 21 '14 at 21:51
• @Achmed - you do not specify how you arrive at your value for the radius of the event horizon. You can conclude that it is wrong also from the values for the escape velocity you presented: in leading order (lowest order in the expansion in powers of $\phi/c^2$) these are the same for the Newtonian and the relativistic (post-Newtonian) case. – Johannes Sep 27 '14 at 11:40
• how you arrive at your value for the radius of the event horizon? from $m\phi=E-E_0$ one can find $\phi=c^2$ – Achmed Sep 27 '14 at 13:07
• You apparently work in a post-Newtonian approximation to GR, and you have derived $v_e=\sqrt{2\phi (1-\frac12 (\frac{\phi}{c})^2 + ...)}$. Now set $v_e = c$ and find the value of the radial coordinate $r$ for which the potential $\phi$ satisfies the resulting equation. – Johannes Sep 28 '14 at 2:02
• Please ignore the typo in above (expansion is in powers of $(\phi/c^2)$, not in powers of $(\phi/c)^2$. – Johannes Sep 28 '14 at 2:12