- #1

stevebd1

Gold Member

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Curved space induces velocity, according to-

[tex]\gamma\equiv\frac{1}{\sqrt{1-(v/c)^2}}=\frac{1}{\sqrt{1-(r_s/r)}}[/tex]

which results in

[tex]v_{rel}=-\sqrt{\frac{r_s}{r}}\ c[/tex]

The velocity induced by the curvature at the surface of the Earth is 11.185 km/s, for the Sun 617.746 km/s which, in each case, matches the escape velocity for each object. These velocity's aren't relativistic and you would, with enough power, be able to hover above Earth without feeling any adverse effects. Around a black hole, the velocity induced by curvature increases relativistically, reaching the speed of light at the horizon. Curvature also increases distance within a confined radial distance (inverse of the time dilation), increasing to infinity at the event horizon. If you travel at the velocity induced by the curvature then dr will always equal dr (i.e. not increasing as suggested by Schwarzschild metric). My question is, if you were able to stop yourself deep within the gravitational field of a black hole (outside the EH), would you 'feel' the ultra-relativistic streaming of space around you, trying to make you pick up speed? I'm not talking about tidal forces but the actual curving of space itself, the gradient of which increases the closer you got to the EH. On one or two occasions, I've seen curvature described as a form of pressure which might be a fair comparison.

For example, the stress-energy tensor at the event horizon of a 10^12 sol mass black hole, if such a black hole could exist, would be negligible but the metric tensor (curvature) would induce a free-fall velocity that approached c at the EH. If you were to stop just outside the EH, what would be the resistance you would feel, it wouldn't be from the stress-energy tensor (I understand that gravity is a result of both stress-energy and metric tensors but in most cases, the stress-energy tensor is more significant, is there a difference to what would be experienced when the metric tensor is more significant?)

A comparison might be that in order to remain static within the gravity field, you would have to travel equal to the escape velocity for that specific radius, which might be compared to traveling at ultra-relativistic speed through flat space (hence the inclusion of the Lorentz factor above).

On a slightly different note regarding rotating black holes and frame dragging, the Killing surface gravity equation suggests that gravity reduces with higher spin-

[tex]\kappa_\pm=c^2\frac{r_+-r_-}{2(r_\pm^2+a^2)}[/tex]

is this in some way related to centripetal acceleration induced by frame-dragging cancelling out the gravitational pull? Also, if you were an observer with zero angular momentum that fell from infinity radially towards a rotating black hole, due to frame dragging, you would orbit the BH before crossing the event horizon. Would you 'feel' the rotation or would the fact that it is the frame that was rotating and that technically, you are relatively static within the frame mean you would not feel any centripetal/centrifugal force? I imagine that some centripetal acceleration is induced by frame dragging but the question is how much.

[tex]\gamma\equiv\frac{1}{\sqrt{1-(v/c)^2}}=\frac{1}{\sqrt{1-(r_s/r)}}[/tex]

which results in

[tex]v_{rel}=-\sqrt{\frac{r_s}{r}}\ c[/tex]

The velocity induced by the curvature at the surface of the Earth is 11.185 km/s, for the Sun 617.746 km/s which, in each case, matches the escape velocity for each object. These velocity's aren't relativistic and you would, with enough power, be able to hover above Earth without feeling any adverse effects. Around a black hole, the velocity induced by curvature increases relativistically, reaching the speed of light at the horizon. Curvature also increases distance within a confined radial distance (inverse of the time dilation), increasing to infinity at the event horizon. If you travel at the velocity induced by the curvature then dr will always equal dr (i.e. not increasing as suggested by Schwarzschild metric). My question is, if you were able to stop yourself deep within the gravitational field of a black hole (outside the EH), would you 'feel' the ultra-relativistic streaming of space around you, trying to make you pick up speed? I'm not talking about tidal forces but the actual curving of space itself, the gradient of which increases the closer you got to the EH. On one or two occasions, I've seen curvature described as a form of pressure which might be a fair comparison.

For example, the stress-energy tensor at the event horizon of a 10^12 sol mass black hole, if such a black hole could exist, would be negligible but the metric tensor (curvature) would induce a free-fall velocity that approached c at the EH. If you were to stop just outside the EH, what would be the resistance you would feel, it wouldn't be from the stress-energy tensor (I understand that gravity is a result of both stress-energy and metric tensors but in most cases, the stress-energy tensor is more significant, is there a difference to what would be experienced when the metric tensor is more significant?)

A comparison might be that in order to remain static within the gravity field, you would have to travel equal to the escape velocity for that specific radius, which might be compared to traveling at ultra-relativistic speed through flat space (hence the inclusion of the Lorentz factor above).

On a slightly different note regarding rotating black holes and frame dragging, the Killing surface gravity equation suggests that gravity reduces with higher spin-

[tex]\kappa_\pm=c^2\frac{r_+-r_-}{2(r_\pm^2+a^2)}[/tex]

is this in some way related to centripetal acceleration induced by frame-dragging cancelling out the gravitational pull? Also, if you were an observer with zero angular momentum that fell from infinity radially towards a rotating black hole, due to frame dragging, you would orbit the BH before crossing the event horizon. Would you 'feel' the rotation or would the fact that it is the frame that was rotating and that technically, you are relatively static within the frame mean you would not feel any centripetal/centrifugal force? I imagine that some centripetal acceleration is induced by frame dragging but the question is how much.

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