Black Hole thought experiment

[doubletruncation]

<blockquote><font class="small">In reply to:</font><hr /><p>I'm still curious as to how the coontracted space close to the event horizon affects the gravitaional gradient as seen by a distant observer. Surely the gradient would appear to be much steeper than 1/r^2 close to the EH and would affect the apparent (to a distant observer) orbits of objects around the BH.<p><hr /></p></p></blockquote><br /><br />Close to the black hole gravity certainly doesn't behave like a 1/r^2 force, but I'm not sure that thinking of it as an effective force that increases more rapidly than 1/r^2 is that helpful either except for in the intermediate limit (I could easily be wrong here though, so I don't want to make any sort of pronouncement <img src="/images/icons/smile.gif" /> - does someone else know?). The orbits do behave quite differently than how they do in the Newtonian limit. For one thing, there is actually a minimum distance from the black hole for which you could have a stable orbit, that is at 3 times the event horizon radius. What that means is that if you were on a circular orbit that's any closer to the black hole than 3 times the event horizon radius then if you are nudged the slightest bit you will fall into the black hole unless you can quickly counter with exactly the force needed to put you back on that orbit. For Newtonian gravity all circular orbits are stable in the sense that if you nudge a circular orbit just a little it'll become a slightly eccentric orbit, you won't just drop into the star (or whatever you're orbiting). If you do consider the case of an object orbiting in the last stable circular orbit, it turns out that an observer at infinity will see the object orbit with a period identical to the Newtonian value. However, the object will be suffering time dilation from the observer's point of view, so an observer orbiting with the object would note that they actually take less time to orbit the black hole than they might calculate assuming Newtonian gravity. If you h <div class="Discussion_UserSignature"> </div>

 

[barrykirk]

Wow!!! I was not aware of that. Photons of light can have a stable orbit around a Black Hole and that orbit is well outside the event horizon.<br /><br />That is just too cool..

 

[skeptic]

Thanks again for your explanation.<br /><br />Pehaps you can help me with another issue I've never understood, that of how an object can actually cross the EH. There seem to be two problems.<br /><br />The first is as we discussed before that of space becoming contracted to zero at the EH. To a distant observer, though the photons would be extremely red shifted, would it not appear that the object was slowing down as it approached the EH because of the contracted space? In the time frame of the distant observer, the object would never cross the EH, correct?<br /><br />The second problem is the infalling velocity. It would seem that the infalling velocity of an object would be very close to the escape velocity at every point along the trajectory. Thus at the EH the object would be traveling at c in its own frame of reference. I don't think this is as heretical as it sounds. Since a freely falling object is at rest, its mass would not increase as we are accustomed to thinking for normal acceleration. The object would still experience length contraction and time dilation. To distant observers the object would travel zero distance (due to length contraction) in an infinite amount of time (due to time dilation).<br /><br />To the object itself, everything seems normal. It reaches c at the EH and continues accelerating past c until it reaches the center.<br /><br />Note: I am aware of various paradoxes that result if an object travelling greater than c interacts with objects travelling less than c. Those paradoxes are avoided because once inside the BH the object can no longer interact with any object moving at less than c.<br /><br />Please doubletruncation, correct all the errors in my reasoning.

 

[doubletruncation]

<blockquote><font class="small">In reply to:</font><hr /><p>Photons of light can have a stable orbit around a Black Hole and that orbit is well outside the event horizon. <p><hr /></p></p></blockquote><br /><br />yeah, it is pretty neat! I think you can think of it as extreme gravitational lensing. <div class="Discussion_UserSignature"> </div>

 

[doubletruncation]

I think the problem again is one of whose point of view you're considering.<br /><br /><blockquote><font class="small">In reply to:</font><hr /><p>The first is as we discussed before that of space becoming contracted to zero at the EH. To a distant observer, though the photons would be extremely red shifted, would it not appear that the object was slowing down as it approached the EH because of the contracted space? In the time frame of the distant observer, the object would never cross the EH, correct?<p><hr /></p></p></blockquote><br /><br />Right, if the distant observer was watching someone fall into the black hole they would see the person falling in assymptotically slow down to no speed at all and also assymptotically approach infinte redshift. One way to think about this is that the light that the person falling in emits back to the distant observer takes longer and longer to reach the distant observer as the person gets closer to the event horizon, the light that the person emits just as they are crossing the event horizon takes infinitely long to go back to the distant observer (In some sense it's not moving in the distant observer's coordinates). In fact if a second person were to fall along the exact same path across the event horizon the second person would see the first person crossing the event horizon just as the second person is falling across the horizon herself. You might say that in the time frame of the distant observer the falling person never reaches the event horizon, but it's important to note that the distant observer's time frame is really only good for that observer at that particular location in space/time. He can project the coordinates to the rest of the universe to figure out what he will see, but his time coordinates/space coordinates are only really good for himself locally and they don't necessarily have any meaning for other observers. If somehow there was a wormhole inside the event horizon and the person falling in went through that wormhole <div class="Discussion_UserSignature"> </div>

 

[skeptic]

Thanks for a very good explanation. I think you anticipated the direction I was going. <br /><br />I have wondered if the concept of time going backwards at velocities greater than c was strictly crackpot or if there is some basis for it. In fact I think backwards flowing time was the basis for some of the paradoxes resulting when faster than c objects interact with slower than c ones. Instead of going backwards, it always seemd to me that time would must become imaginary, but I have very little understanding of what that would mean.<br /><br />Instead of being in a parallel universe, could the white hole be time shifted in our own universe i.e. the big bang. E.g. Particles falling into a BH, fall backwards in time to the center of the BH which corrosponds to the Big Bang?

 

[doubletruncation]

<blockquote><font class="small">In reply to:</font><hr /><p>I have wondered if the concept of time going backwards at velocities greater than c was strictly crackpot or if there is some basis for it. In fact I think backwards flowing time was the basis for some of the paradoxes resulting when faster than c objects interact with slower than c ones. Instead of going backwards, it always seemd to me that time would must become imaginary, but I have very little understanding of what that would mean. <p><hr /></p></p></blockquote><br /><br />If we consider a flat universe where special relativity holds globally you can see that an object going faster than light will to some observers also be moving backwards in time. To see this you need to use the equations that transform positions and times as measured from one observer to those measured by another observer. Suppose an object leaves from position 0 at time 0 (call this the Earth), and in some observer's reference frame is moving at velocity u which is faster than the speed of light. Then when that observer sees the faster than light object travel a distance l (say Alpha Centauri), she will say that it took the object a time l/u to travel that distance. Now suppose another observer moving in the same direction but at velocity v (less than the speed of light) also called the initial position 0 and time 0, they would conclude that the object travels a distance of l*(1-v/u)/sqrt(1-v^2/c^2) and arrives at a time l/c^2 * (c^2/u - v)/sqrt(1 - v^2/c^2). Since u is greater than c, it is possible to find a v less than c such that c/u < v/c, and to such an observer the object would appear to have travelled backwards in time and arrived at Alpha Centauri before it left the Earth. So if an object can travel faster than the speed of light, under the special relativity transformation laws some observer will conclude that the object is travelling backwards in time as well. This can give rise to a number of paradoxes: suppose when the objec <div class="Discussion_UserSignature"> </div>

 

[skeptic]

Thank you very much for your generous answers. Questions like mine usually just bring sneers in the astronomy and physics newsgroups.<br /><br />I would like to post a question from my now retired astronomy professor.<br /><br />In reponse to many assertions that gas falling towards a BH becomes very hot and radiates in the Xray or gamma ray spectrum he asks, by what means does the gas become so hot? He disputes that the gas experiences that much compression or friction.

 

[doubletruncation]

<blockquote><font class="small">In reply to:</font><hr /><p>In reponse to many assertions that gas falling towards a BH becomes very hot and radiates in the Xray or gamma ray spectrum he asks, by what means does the gas become so hot? He disputes that the gas experiences that much compression or friction.<p><hr /></p></p></blockquote><br /><br />I'm no expert in this, so please take what I say with a grain of salt. I may be mistaken, but I think the way people approach this is first of all empirically - that is, we see x-ray sources co-located with stars (e.g. Cygnus X-1) that have large radial velocity variations (that is, you look at the optical spectrum of the star, and you see the absorption lines in the spectrum doppler shift over time). The whole spectrum of the star-light would be shifted together, so whatever the star is orbiting does not provide optical light. From the doppler shifts you can get a mass for whatever the star is orbiting, and you find that it's something quite massive to not be contributing any optical light - so the interpretation is that it's a black hole, or neutron star or white dwarf (depending on the inferred mass). So where are all these x-rays coming from? The interpretation is that there is an accretion disk (the stellar companion provides a nice source for the accreting material). Now if you just put some material in orbit around a black hole it's not just going to fall into it (so long as it's beyond the last stable orbit), it would just orbit the black hole with whatever angular momentum/kinetic energy that it has initially. So how does the material end up falling onto the black hole? Well, it has to lose energy in some way to do that, and a natural way would be for the gas to radiate away some of the energy (radiation will cool the gas, the total energy drops and the material will fall to a slightly lower orbit). In order for the material to radiate it needs to be heated up in some way, so there would have to be some viscosity. Last I hea <div class="Discussion_UserSignature"> </div>

 

[agnau]

It was my thought that as a particle crosses the EH, due to mass/engery conservation (even in a less than balanced state leading to entropy) a virtual particle is released, this particle could interact with the accretion disk and therefore raise the temperature of the nearby material, which then passes the heat outward.

 

[doubletruncation]

Interesting idea. It should be possible to calculate such a process. It seems that the black hole should be emitting thermal radiation from this process with a temperature of something like 10^-7 * (Msun/M_bh) kelvin (the original reference for this is 1974, Nature, 248 30.). So it probably won't be nearly enough energy to heat up the surrounding gas, but nonetheless I think it's a good thought. <div class="Discussion_UserSignature"> </div>

 

[agnau]

The thought is that there would be more than one virtual particle release -- perhaps a number near to the surface area of a fully active black hole (under the idea of spherical absorption) or a eqatorial belt of such absorption release pairs. Add to this the idea that a black hole may be spinning and increasing the disk's agitation.