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Relative energy of a black hole. 
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#1
Feb1112, 09:04 PM

P: 2,468

Does kinetic energy of a body contribute to the stressenergy tensor of a black hole.
If I move with respect to a black hole I would perceive it to have more energy. So would it have a stronger G field in my frame. Would it have a larger Schwarzschild radius. My assumptions are probably wrong. I dont know that much about GR but would I use the Schwarzschild metric to try to solve this. How would I factor in the relative velocity. Dont make your response to complicated because I don't know that much about GR. Any help will be much appreciated. 


#2
Feb1212, 09:53 AM

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P: 4,160

cragar, It's true that if you are in a moving frame you will perceive the black hole to have greater energy. The key word here is "perceive". It's easy enough to write the Schwarzschild solution in a moving frame. However the attributes of the hole including its Schwarzschild radius are intrinsic and won't be affected. You will of course perceive them differently.



#3
Feb1212, 11:03 AM

P: 5,632

Say you have an atom, [or a black hole] and you HEAT that atom, now the constituent particles [or degrees of freedom]gain energy, move faster, and so even in the rest frame of the atom there is additional kinetic energy: that kind of kinetic energy, in the rest frame of the center of mass, DOES contribute to additional gravitational curvature. Another example, would be the energy required to compress a spring: such energy exists in the rest frame of the spring and so it's equivalent 'mass' increases. Another, equivalent way to picture this: Assume you have a flat sheet of graph paper to represent two dimensional space without gravity. [You can think of time as a third dimension if you like.] When we introduce gravitation, the paper itself becomes curved. (Curvature that cannot be "flattened" without distortion. Gravitational "spacetime curvature" refers to this curvature of the graph paper, regardless of observer, whereas visible/perceived curvature in space is related to distorted, nonsquare grid lines drawn on the graph paper, and depends on the frame choice of the observer...." Found another related discussion: http://www.physicsforums.com/showthr...=1#post3661242 


#4
Feb1212, 11:31 AM

Mentor
P: 17,543

Relative energy of a black hole.



#5
Feb1212, 11:58 AM

P: 152

When you are speeding up, accelerated by the black hole, your clock would start ticking slower for that reason contributing to your possible feeling that the black hole has gained mass. At the same time light travels slower close to massive bodies (Shapiro effect) so if you are measuring your velocity somehow in relation to the speed of light, you may come to other conclusions. A black hole is per definition black, but if there is some heated gas close to the black hole I think you will be perceiving that gas to be hotter the closer you get to the black hole, there will be lesser gravitational redshift. Of course if you travel faster and faster towards the black hole there will be more and more doppler shift contributing that you will be perceiving the gas as being more energetic. Those are the physical effects I can think of... Regarding G and the Schwarzshild radius I do not know, maybe it depends upon how you are attempting to determine those... 


#6
Feb1212, 08:52 PM

P: 2,468

Naty 1 said if I heat the atom it will contribute to the gravitational curvature.
So If I have a rotating black hole it will contribute to the curvature. What If I was going around in a circle outside the black hole using rocket power. But I guess I would know I was accelerating and I wouldn't think the black hole was rotating. And also if I had something orbiting a black hole that would contribute to its gravitational curvature. 


#7
Feb1912, 02:50 AM

P: 611

[tex]M^2=\frac{J^2}{4M_{ir}^{2}}+\left(\frac{Q^2}{4M_{ir}}+M_{ir}\right)^2[/tex] [tex]M_{ir}=\frac{1}{2}\sqrt{\left(M+\sqrt{M^2Q^2a^2}\right)^2+a^2}[/tex] where [itex]M=Gm/c^2,\,Q=C\sqrt(Gk_e)/c^2[/itex] and [itex]J=aM[/itex] where [itex]a=j/mc[/itex] where [itex]j[/itex] is angular momentum in SI units. [itex]M_{ir}[/itex] is the mass you would have left if all the charge and spin were extracted (i.e. a Schwarzschild BH). Source http://www.ece.uic.edu/~tsarkar/Good...ack%20Hole.pdf (page 12) 


#8
Feb1912, 04:44 PM

P: 2,468

interesting thanks for that reply. So if a Black hole is charged It will have more gravitational curvature because of the Energy in the electric field. Does a charged rotating BH create a B field. But does the B field only exist inside the event horizon or can it exist outside.



#9
Feb2112, 02:31 AM

P: 611




#10
Feb2112, 08:53 AM

P: 1,115

Further, just how can there be a necessarily *continuous* virtual particle exchange process giving rise to an exterior electric field? Seems especially problematic to me given that all temporal processes for any and all objects at or interior to the BH EH have come to a screeching halt wrt the BH exterior, where the E field supposedly effortlessly extends. It surely takes two to tango if an exchange process is to occur. You can have a genuine exchange if at one end of the telegrapher's line the telegrapher is in deep suspended animation?! In other words, if the exchange *rate* is necessarily zero, how is any exchange at all taking place between an external 'hovering' entity and charged matter at or interior to the EH? What deep principle am I missing here? 


#11
Feb2212, 07:30 AM

P: 95

http://arxiv.org/abs/0908.1780 


#12
Feb2212, 07:54 AM

P: 1,115

And btw folks, it would be nice if some GR buffs here actually responded to my #10! Or is it all too boring? 


#13
Feb2212, 08:53 AM

P: 95




#14
Feb2212, 10:53 AM

P: 1,115




#15
Feb2212, 11:07 AM

P: 95




#16
Feb2212, 01:00 PM

P: 1,115

But why am I contemplating the above, given general misgivings expressed in #10? Just for fun! 


#17
Feb2212, 01:11 PM

Emeritus
Sci Advisor
P: 7,663

It does say that a pair of particles colliding can become a black hole (which is true)  but it does not say that a single particle moving at high speed will become a black hole  because this statement, as previously mentioned, is false. Relative to some observers, you are right now moving at 99.999999% of the speed of light. But you are not a black hole. Not to yourself, and not to the observer at which you are moving at such a high velocity  because being a black hole is frame indepenent. The correct description of the gravitational field as seen by a rapidly moving observer is given by the AichelbergSexl ultraboost. See for instance http://arxiv.org/abs/grqc/0110032 It may take a little bit of advanced knowledge to interpret the comonents of the field tensor, given by the Riemann, in semiNewtonian terms. Basically, certain components of the Riemann tensor correspond to the gradient of the Newtonian gravitatioanl field, i.e. the Newtonian Tidal tensor. Other components of the Riemann include gravitomagnetic effects, and effects that have no direct Newtonian counterparts (the topogravitic part of the tensor). The analogous case of the field of a moving charge is simpler to understand, and very helpful. Basically, the field of the moving charge becomes in the relativistic limit an impulsive plane wave, and something very similar happens to a moving mass. 


#18
Feb2312, 03:21 AM

P: 1,115

Too late to edit my #16, but a few extra considerations to round that off:
1: From considerations there it is now seems obvious there is no real possibility of glancing motion ever generating conditions favouring 'BH' formation in adjacent matter, even if the ultrarelativistic mass was a 'BH' in it's own frame. Certainly out of the question for a mere elementary particle such as an electron. Accretion onto an existent BH is the best that could be hoped for. 2: Even assuming it were possible for an ultrarelativistic elementary particle to collapse nearby matter into a BH state, by reciprocity of reference frame, it would of necessity be a Kamikaze affair ending in mutual 'BH' creation  not just one acting unilaterally on the other. And this is the last plug here re entry #10. Why no takers? It only fuels suspicions there are no decent and believable answers. So come on, someone give it a shot please. 


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