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Sorry Halc but you are most definitely wrong in your thinking. I have already given you the reference three times in this note. Look at it. It is a very well respected java tool. http://xaonon.dyndns.org/hawking/
In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which must be treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface. This is because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity. Because of this, a renormalized value is used that corresponds to the Newtonian value in the non-relativistic limit. The value used is generally the local proper acceleration (which diverges at the event horizon) multiplied by the gravitational time dilation factor (which goes to zero at the event horizon). For the Schwarzschild case, this value is mathematically well behaved for all non-zero values of r and M.When one talks about the surface gravity of a black hole, one is defining a notion that behaves analogously to the Newtonian surface gravity, but is not the same thing. In fact, the surface gravity of a general black hole is not well defined. However, one can define the surface gravity for a black hole whose event horizon is a Killing horizon.
Question:"... about the actual force (in Newtons) that an object with mass m would receive from a Black Hole with Schwarzschild radius if the object was on the Event Horizon"Answer:"Hi Peterix, your question has been answered here. – John Rennie Apr 20 '14 at 8:05Where, I can't find it? Anyway is F=mc4/4GM right? – Peterix Apr 20 '14 at 9:17 No, it's F=ma=GMmr/r² 1/(√(1-2GM/c²r)) – John Rennie Apr 20 '14 at 9:25"
The event horizon is at the point where the escape velocity is equal to the velocity of light.
Gravitational acceleration is not directly related to escape velocity
Coming to your spaceship analogyThe most efficient propulsion system possible is that created by converting mass directly into energy of acceleration
Halc You are still thinking incorrectly. The reference you give from Wikipedia refers to what you would "see"and measure from a distant viewpoint as things vanished into the event horizon.
am not talking about that. What you see from a distance is not what is observed by the particles falling through a gravitational Field.
The viewpoint I am taking is what you would see and measure if you were travelling along with the particles and observing what is happening to other particles nearby
let us now try to come to some points of agreement.I presume that you agree that the radial position of the black hole event horizon in a spherical gravitational field created by an isolated (point) mass is defined by the point where the escape velocity equals the velocity of light according to the inverse square law
Yes, that gets you the Schwarzschild radius according to the escape velocity formula,
since I don't understand the meaning of a frame of reference of something at a singularity
When I say event horizon I mean the Schwarzschild radius for a simple isolated non rotating black hole. That is the point at which no further communication to a distant observer is possible. This is more complex in the Kerr or rotating case
Quote from: Halcsince I don't understand the meaning of a frame of reference of something at a singularityIt appears you are talking about what happens near a theoretical "singularity" my arguments are essentially that this never happens. What prevents it from happening. In the first simple arguments it is just that a stasis can be reached via the energy outflow caused by the increasing gravity gradient at the "surface" (hawking radiation) overcomes the energy created by any further contraction towards the singularity. I tend to equate this with what other writers have called the quantum firewall.
It is also known that the laws of physics are very finely balanced to allow stars to make complex atoms that then explode to distribute these to catalyse the formation of stars and stellar mass black holes.
Long lived Complex nuclei at least up to the complexity of iron (the last exothermal synthesis nucleus) are essential for the development of stars as we know them today and the nucleosynthesis processes that lead to core collapse in supernovae that create black holes.
If there were no stable nuclei beyond say lithium. Stars would be very different. Smaller ones would fade out as the proton proton reaction ran out of fuel. Large stars would evaporate to become smaller as the radiation pressure dispersed them by their stellar winds.
The only black holes would be created by extreme mass stars (100s of solar masses) which can (in theory) collapse directly into black holes without first dispersing themselves by the pressure of the high radiation outflows.
You are incorrect Halc and what you say is easily disproved.
But consider a spaceship trying to accelerate away from there even if it could manage the thrust needed ?
This is done with a simple total energy calculation that is independent of the method used to generate thrust.Any particle falling freely from a great distance directly towards the black hole would have gained one rest mass of energy as it passed the horizon at something less than the speed of light. ie Mgh +Mv² ie Mc² where h is very bigThat comes from integrating the inverse square law to this point .
The real point I am trying to make at this stage of the argument is that there is absolutely nothing special about the Scwartschild radius it is just somewhere in space near a big mass that particles can only travel through in one direction.