Gravity Profile of a Black Hole

Status
Not open for further replies.
J

jgreimer

Guest
One normally thinks of gravity obeying the inverse square law or as the profile being equivalent to 1/r^2. However since space is noticeably Lorentz contracted in strong gravitational fields, one would expect that to a distant observer, the gravitational field would appear to diminish much more rapidly around a black hole. How would one describe the gravity profile mathematically close to the event horizon of a black hole?
 
P

pyoko

Guest
Not known. <div class="Discussion_UserSignature"> <p> </p><p> </p><p><span style="color:#ff9900" class="Apple-style-span">-pyoko</span> <span style="color:#333333" class="Apple-style-span">the</span> <span style="color:#339966" class="Apple-style-span">duck </span></p><p><span style="color:#339966" class="Apple-style-span"><span style="color:#808080;font-style:italic" class="Apple-style-span">It is by will alone I set my mind in motion.</span></span></p> </div>
 
D

docm

Guest
Agreed. The universality, or lack of same, of the inverse square law is actually one of the big questions in physics right now. In fact many of the theories that try to incorporate gravity into Quantum Mechanics depend on inverse square law violations. <div class="Discussion_UserSignature"> </div>
 
J

jgreimer

Guest
It would seem that at distances very close to the event horizon and space-time becomes extremely contracted, that nearly all the BH's gravitational field would be contained in a very thin region (as seen by a distant observer) around the event horizon. This would make it seem that the BH is essentially without gravity, yet I know this can't be true because we have photos of matter orbiting BHs. Can anyone resolve this apparent contradiction?<br /><br />Surely the gravity profile could be estimated by measuring the amount of red or blue shift of orbiting matter at different radii from a BH.
 
P

primordial

Guest
jgreimer! Part of your idea is correct, as you stated, near the Black Hole space-time becomes contracted, but contracted is what increases the gravational effect. However I would like to present another aspect to the problem of understanding Black Holes; that being this, if light travels at the velocity of C and Gravity travels at the velocity of C, both are Interactions, and the requirment for escaping a Black Hole is, an ESCAPE VELOCITY greater than C, how then can a Black Hole have a Gravitational field, especialy one that can become greater; unless black holes are either, matter that has the Unified Field, intact, or a Black Hole that is not completely Black; one that can radiate very long wave photons, and have changes in its gravational field? Just Think About It.
 
R

rpmath

Guest
For black holes the 1/r^2 is corrected as 1/(r-rs)^2<br />where rs is the Schwarzschild Radius of the black hole:<br />rs = 2*G*M/c^2<br /><br />The problem is that "r" is not exactly a distance some one can measure.
 
J

jgreimer

Guest
Primordial, thanks for your post. I did spend some time thinking about it. First, space-time only appears contracted to distant observers, not to those observers inside the contracted space, so gravity would have to propagate though what to it is a much greater distance than it seems to distant observers. In fact if we are to believe Lorentz, at the event horizon space is infinitely contracted and gravity would be trapped just light light.<br /><br />On the other hand we know that BHs have gravity because we can measure the radiation of the matter falling into it. <br /><br />I've seen diagrams of BHs represented by a hyperbolic depression in a membrane with the singularity at center but I don't think this is correct. Since infinite contraction occurs at the event horizon, everything at the event horizon must be infinitely depressed. Think of the hyperbolic depression converging on a infinitely deep cylinder having the radius of the event horizon instead of converging to a point. This means there is no inside to a black hole. Infalling matter just keeps on falling and falling without ever quite reaching the event horizon. Thus the BH still has gravity outside the EH without gravity ever having to escape from within.<br />
 
U

usn_skwerl

Guest
why is a BH represented as a disk, or cylinder, and not a sphere? its always shown as the typical black hurricane shape. granted i understand the concept of the fabric of spacetime being warped intot he cone or cylinder, but logically, if a star gets to 99.99% repeating, and for example a hapless comet floats just a hair too close and poof; critical mass = black hole (basically). <br /><br />this BH should maintain the sphere the star originally was, even if you DO take into account the stars rotation. something kind of like a mobius ball. it should be a semi-perpetually collapsing cavity, i would think. <div class="Discussion_UserSignature"> </div>
 
A

alokmohan

Guest
Gravity in Einstein theory of relativity is call clearly not known to some posters.It is slightly different from Newtons law of gravity.
 
D

dragon04

Guest
<font color="yellow">why is a BH represented as a disk, or cylinder, and not a sphere?</font><br /><br />For one thing, we've never been able to view any BH from all possible angles.<br /><br />Secondly, they're modeled on what we observe. The Sun rotates, and all the planets, asteroids and, Kuiper Bodies (with at least one notable exception) more or less orbit the Sun on the same plane of inclination.<br /><br />Certainly, if we view a BH as a "sphere", gravity in a BH that had no "spin" would exert itself equally and omnidirectionally.<br /><br />But considering that all matter we observe (repeatedly) revolves around the central massive BH's at the core of every galaxy, that implies that Gravity exerts more force along a plane.<br /><br />I also think that any BH implies at least a 4th physical dimension that we can't directly observe.<br /><br />BH's are always depicted in 2 or 3 dimensions as a "curving funnel", but in relation to a sphere, one would see that "curved funnel" at every possible point of view.<br /><br />Sorry to sound so obscure, but imagine for a moment one of those long funnels you use to put transmission fluid on a car. Now, take as many of them as you have to to make a sphere out of them that connects the skinny ends at a central point.<br /><br />That would be what I think to be a 3 dimensional representation of a BH. <div class="Discussion_UserSignature"> <em>"2012.. Year of the Dragon!! Get on the Dragon Wagon!".</em> </div>
 
P

primordial

Guest
jgreimer ! Yes you are correct the only gravity must come from the mass outside the event horizon sort of frozen in a three spacial + one time dimension that is approaching 2 spacial + zero time dimensions with the progression of time to all observers external to the black hole. To those observers inside the black hole space is infinite and light inside can only approach the event horizon from the inside as its time passes, much like our expanding universe. What do you think?<br />
 
J

jgreimer

Guest
I don't understand how anything can cross the event horizon. It seems to me that as space becomes more and more contracted, matter just continues to fall towards the event horizon but never reaching it.<br /><br />I've heard the argument that matter doesn't have to cross the event horizon, that as more and more matter falls towards the event horizon the event horizon grows outwards past infalling matter. This also doesn't make sense to me. To infalling matter, the gravity of more infalling matter behind it i.e. farther from the BH would only tend to counter the gravity of the BH thus slowing its fall and ironically shrinking the diameter of the EH to it. To infalling matter farther away, the EH has indeed expanded, so each infalling particle sees a unique EH which is expanded by infalling matter ahead of it but shrunk by infalling matter behind it. Nevertheless I see no method by which matter could actually cross the EH.<br /><br />What about the collapsed star itself? Isn't it inside the EH? The Lorentz contraction is a result of the intensity of the gravitational field and where is the gravitational field most intense - at the surface no? Something at the center of a star is weightless isn't it? So even though the matter of the star would still tend to gravitate to the center of the star, space would only be infinitely contracted at the surface of the star. Space inside the EH would actually be less contracted than at the EH and in that sense would have to be considered outside the EH. (Is this making any sense?) Light trying to escape the star would have to pass through the thin shell of EH around it so still could never be seen.
 
A

alokmohan

Guest
You are confused.Understand event horizon first.Then we can have fruitful discussion.Your thoughts are mixed up frankly.
 
P

primordial

Guest
alokmohan ! How does the event horizon form relative to an observer positioned in an inertial platform 20 light-years away from the event as it occurs?
 
P

primordial

Guest
jgreimer ! Much of what you say is the same as my conception of Black Holes. I have trouble with the interaction of gravity and changes in gravity of a Black propagating from inside the Black Hole. One of my basic problems was with the original change in density relative to an observer outside the collapsing colapsar that is necessary to form a Black Hole, because the (density) is the cause, of the effect( increase in time dilation) that in turn exponenentially slows the rate of the compression of the mass along with other problems such as degeneracy pressure. It appears to me that if stellar Black Holes do form they should all form at the same size because the consistancy of the matter is homogeneous, so the radius should be the same relative to an inertial platform positioned the same distance from the event horizon. Just think about it.<br />
 
J

jgreimer

Guest
I may very well be confused and mixed up, but good etiquette when offering criticism is to explain how the person is confused or mixed up and to enlighten them with the correct interpretation.<br /><br />Would you please explain the places where I am confused and which thoughts are mixed up and then help me to understand the parts I am lacking about the event horizon?
 
A

alokmohan

Guest
Event horizon is the place from where you can see no radiation.On earth we can never see beyond horizon.Analogically you call it event horizon.
 
D

docm

Guest
Unless the BH is rapidly rotating, which could 'wind up' the event horizon and expose the singularity (a "naked singularity"). If they exist they would give most physicists a migraine. <div class="Discussion_UserSignature"> </div>
 
Status
Not open for further replies.

Latest posts