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Author Topic: Where does an event horizon first form in a Black Hole?  (Read 769 times)

AndroidNeox

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Where does an event horizon first form in a Black Hole?
« on: 01/04/2016 01:09:52 »
A spherical, homogeneous planetary mass would have a characteristic escape velocity. If there were a vertical shaft reaching down to the center of the planet, there would be an escape velocity associated with this center of gravity. It would be the escape velocity from the surface plus the velocity needed to get from the center to the surface (equal speed & opposite direction to the velocity an object would achieve falling from the surface, down the shaft, to the planet's center.

An event horizon is thought to form when the escape velocity for some point(s) reaches the speed of light, c. So, if a homogeneous mass were to be uniformly compressed until an event horizon formed, the EH would first form at the center point and not at the surface.

Correct?
« Last Edit: 07/07/2016 07:57:11 by chris »

evan_au

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Re: Escape Velocity
« Reply #1 on: 03/04/2016 22:41:20 »
That is an interesting way of looking at the formation of a black hole.

Rather than defining the event horizon by the amount of mass inside a given radius, you look at it as the escape velocity "to infinity".

Of course, this raises the question of an observer who is not "at infinity", but is only partway out of the gravitational well. Does this observer the event horizon as being the point where escape velocity to the observer's location would equal the speed of light? If so, would this event horizon would have a smaller radius than one seen by an observer "at infinity"?

Perhaps this could be partially resolved by looking at oscillations at the inner edge of the accretion disk, as this plasma is effectively an observer very close to the event horizon? These oscillations should be visible as short-term variation in the X-Ray output of a black hole's accretion disk.

AndroidNeox

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Re: Escape Velocity
« Reply #2 on: 07/07/2016 05:44:18 »
That is an interesting way of looking at the formation of a black hole.

Rather than defining the event horizon by the amount of mass inside a given radius, you look at it as the escape velocity "to infinity".

Thank you. Since Schwarzschild's Radius is defined from the perspective of an observer at infinity, this seemed the only proper place to start. I don't think it's correct to take Schwarzschild's solution and apply it to the vicinity of a black hole without accounting for shifting the POV from an infinite distance to a finite distance.

The Shapiro delay shows that space is measurably stretched, dilated, in a gravity well. The degree of dilation between two points in space is proportional to the difference in gravitational potential energy between the two points, measurable in terms of change in wavelength of a light beam passing through a gravity well. From Schwarzschild's perspective, this radius will be unchanged and the time delay will appear to be light slowing down in the gravity well. Locally, the speed of light will remain unchanged. Time will pass more slowly for an observer in free fall deeper in a gravity well (lower gravitational potential energy) than for Schwarzschild's observer in free fall at an infinite distance.

The greater the change in gravitational potential energy, the greater the spacetime dilation. For all observers, for some massive object, the minimum gravitational potential energy will be at the center of gravity. The point of lowest gravitational potential energy is the point that's downhill from everywhere else. The total gravitational vector will be zero but the total field will be maximal.

Correct?

Note: I presume the Shapiro effect is not constrained to low field situations. There is nothing in Relativity to suggest it should be. A proper solution in General Relativity will keep the speed of light constant for all observers.

Of course, this raises the question of an observer who is not "at infinity", but is only partway out of the gravitational well. Does this observer the event horizon as being the point where escape velocity to the observer's location would equal the speed of light? If so, would this event horizon would have a smaller radius than one seen by an observer "at infinity"?

Since we're starting with Schwarzschild, we ought to determine how things change as one's POV changes from infinity, inward. For example, the Schwarzschild Radius for an object of one solar mass is about 3 km. Were the Sun compressed to nearly that size, neglecting spin, it would gravitationally collapse. As the total field at the center of gravity increased, spacetime would stretch without bound. I don't find any references to 'spacetime dilation', though that's exactly what a gravity well is... demonstrably (i.e. Shapiro)

The difference in gravitational potential energy between every point in space and any event horizon is infinite, equivalent to the redshift of a light beam shining upward from the EH, equivalent to the blueshift of a beam shown downward toward the event horizon.

For an object/observer in free fall into a black hole, the external universe will appear to speed up and contract. From the perspective of an observer farther out of the gravity well & stationary WRT the black hole, the infalling object will appear to contract radially and expand horizontally (perpendicular to the radius) within a contracting shell about the center of gravity. These are perfectly symmetrical & consistent observations. Also, from the perspective of the distant observer, a clock carried by the infalling observer will slow to a virtual halt. From the perspective of the infalling observer, the external universe (far enough out of the gravity well that Relativistic effects are negligible) will appear to speed up. The mutual observations will work out so that, as the infalling observer is seen to reach the event horizon, the infalling observer will see the universe end as it ages to infinity.

Because the relationship between time for the infalling observer and a distant inertial observer is exponential, though the infalling observer experiences finite time ("proper" time), *before* they can reach the event horizon, infinite time will have passed for external inertial observers. This would explain Einstein's insistence that event horizons cannot form. They require infinite time to pass *before* they can form... there aren't any yet and in 100 trillion years there still won't be any. In 100 trillion times that period there still won't be any.

Because the space dilation outside of the event horizon is infinite, I don't expect there to be any quantum effects. To my knowledge, as distance approaches infinite, quantum interactions approach zero. So, no Hawking radiation.

I set up a thought experiment, appended below, showing how I would measure the distance to an event horizon from some point in space.

Perhaps this could be partially resolved by looking at oscillations at the inner edge of the accretion disk, as this plasma is effectively an observer very close to the event horizon? These oscillations should be visible as short-term variation in the X-Ray output of a black hole's accretion disk.

Dunno. Not sure what would come out. Of course, every infalling trajectory would eject the infalling matter if not for collisions. So, all the momentum dropped in will, eventually, come back out. The kinetic energy will either come back out or not, in either case it can be converted to mass. Also, opposed momenta can cancel and become kinetic energy or mass.

But, collisions and radiance will remove energy so some fraction of the infalling matter will be destined to fall forever. Spacetime will stretch without bound as the infalling matter forever gets closer to the absolute maximum mass within the surface area Schwarzschild defined (maximum field across any surface).

Note: Though space stretches radially, it does not stretch tangentially. I.e. Were the Sun to be collapsed to a black hole, the Earth's orbital circumference would remain unchanged but the orbital radius would approach infinite distance. Likewise, the time required for a light beam to travel from any point in space to an event horizon will approach infinite.

I think the 'rubber sheet' model of gravity in spacetime is excellent. Ideally, the vertical distance will be proportional to the difference in gravitational potential energy between two points. Schwarzschild's observer would be at the edge, where the sheet was attached. An event horizon would be a vertical surface. The continuous curve of spacetime would, after an infinite distance, reach the event horizon where an infinitesimal change in altitude yields an infinite change in gravitational potential energy.

Were an event horizon to form, it wouldn't first form at the surface of a mass. It would first form at the point of highest total field. Center of gravity.

Correct?

As a non-spinning, chargeless collapsar formed, it's center would fall away at an exponentially increasing rate. I agree with Einstein that event horizons cannot form (in finite time).

In the thought experiment, below, the mirror is lowered at a constant rate toward the event horizon. By the time the mirror crosses the event horizon, the downward (and reflected) light beams will contain infinitely many light wave cycles. The light beam is generated at a constant frequency. Therefore, independent of location or velocity (up to and including c), and object moving toward an event horizon cannot reach the event horizon until an infinite time has passed for all observers outside of the event horizon.

Note: all components are idealized. The rope is infinitely strong and of negligible weight. The mirror is infinitely thin and strong and rigid. The laser generates a beam forever. The platform is perpetually stationary (constant gravitational potential energy and zero rotation) WRT the black hole.
« Last Edit: 07/07/2016 07:01:08 by AndroidNeox »

The Naked Scientists Forum

Re: Escape Velocity
« Reply #2 on: 07/07/2016 05:44:18 »