2
$\begingroup$

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)?

Black hole regions
(source: cloudfront.net)

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Am I correct in assuming this question was spawned by the comparison of the deflection angle of light in Newtonian gravity versus GR? $\endgroup$
    – zibadawa timmy
    Sep 21, 2014 at 10:25
  • 1
    $\begingroup$ possible duplicate of Can a black hole be explained by newtonian gravity? $\endgroup$
    – zibadawa timmy
    Sep 21, 2014 at 10:28
  • 3
    $\begingroup$ See physics.stackexchange.com/q/19405/55483 and the answers in particular. $\endgroup$
    – zibadawa timmy
    Sep 21, 2014 at 10:28
  • 4
    $\begingroup$ @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 $2GM/c^2$. $\endgroup$
    – Johannes
    Sep 21, 2014 at 11:58
  • 3
    $\begingroup$ I suspect the question is based on this article, or one like it. $\endgroup$
    – John Rennie
    Sep 21, 2014 at 15:10

2 Answers 2

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
7
  • $\begingroup$ It would be better if you say exactly where I'm wrong $\endgroup$
    – Achmed
    Sep 21, 2014 at 21:51
  • $\begingroup$ @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. $\endgroup$
    – Johannes
    Sep 27, 2014 at 11:40
  • $\begingroup$ how you arrive at your value for the radius of the event horizon? from $m\phi=E-E_0$ one can find $\phi=c^2$ $\endgroup$
    – Achmed
    Sep 27, 2014 at 13:07
  • $\begingroup$ 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. $\endgroup$
    – Johannes
    Sep 28, 2014 at 2:02
  • 1
    $\begingroup$ @Achmed That's only true as long as you do not consider gravity. Once you include gravity, special relativity is insufficient. And as Johannes has explained, you're not using special relativity, you're using a leading order post-Newtonian approximation for the velocity. Your value for the event horizon radius of a schwarzschild black hole is incorrect en.wikipedia.org/wiki/Schwarzschild_radius Since you did not show us how you obtained it, then we cannot tell you why it is incorrect.... $\endgroup$
    – Daddy Kropotkin
    Dec 14, 2020 at 13:24
3
$\begingroup$

Your question is meaningless. In Newtonian physics, there is no limit to the possible velocities. Light has velocity $c$ but higher velocities can exist. Therefore, you can send something from any distance from any mass fast enough to have escape velocity.

Black holes are a purely GR concept.

And it is easy to prove.

Consider a Newtonian compact object, of mass M and radius not necessarily zero, but small enough to be what you'd call a "Newtonian black hole", say radius less than $\frac {MG}{10c^2}$

Hovering very far from this object, I send in its direction a projectile with a reasonable velocity, much larger than escape velocity from its attraction at this distance. I aim extremely carefully, to have a hyperbolic trajectory that passes, say at a distance $\frac {MG}{5c^2}$, so it does not crash into the object. Well within what you would like to call its "horizon". Of course my projectile has acquired in the process a speed much higher than $c$, but we are in Newtonian physics, aren't we ? So it is OK.

Since the trajectory is hyperbolic, it will get out of the gravitational potential with more than the escape velocity.

There can be no horizon in Newtonian physics

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.