Relative energy of a black hole.

cragar

Does kinetic energy of a body contribute to the stress-energy 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.

Bill_K

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.

Naty1

Here is one 'trick' to help resolve such questions:

And a further perspective:
This means that no matter how fast you see a particle whizzing by, it will never become a black hole. It turns out, however, that even though such speed does not add to gravitational curvature, you generally PERCEIVE the spacetime as being curved .....but it is not gravitational curvature.

Only if the KE exists in the rest frame of the body:

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:

And docAl provided me this explanation several years ago:[paraphrased]

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, non-square 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

DaleSpam

Yes. However, kinetic energy does not only contribute to the time-time component of the stress-energy tensor, but it also contributes to the spatial components. So the overall effect is not straightforward.

Agerhell

When you approach a black hole your own clock would start to tick slower due to what is known as gravitational time dilation. This could make you think that the black hole has gained energy(mass) and is thus able to accelerate you faster. (When your clock starts ticking slower you will think that you are speeding up)

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...

cragar

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.

stevebd1

You might be interested to know that a black holes gravitational curvature is a combination of its irreducible mass, spin and charge where (in geometric units)-

[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^2-Q^2-a^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)

cragar

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.

stevebd1

The charged field (and any consequential magnetic field) would reside outside the event horizon. Extract from 'How does the gravity get out of the black hole?'-

Source- http://math.ucr.edu/home/baez/physic...k_gravity.html

Q-reeus

And that same source goes on to say:
Nice try, but presenting the problems and pretending to answer them doesn't cut it imo. So the 'fossil field', be it gravitational or electric, is a source unto itself? Perhaps someone can enlighten me here. It is well known that in GR the stress-energy tensor T as source of gravity contains zero contribution from the field itself. Yet when we get to a notional BH, seems some kind of magic takes over and the 'fossil field' of necessity becomes it's own source. Pray tell how is this not an act of sweeping under the rug an embarrassing contradiction? Carefully avoiding the words 'field as source term' and vaguely substituting 'fossil field' is not simply calling a rose by another name?

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?

Sam Gralla

Nope, at least not for a boson star model of a particle. Choptuik and Pretorious showed a few years back that "center of mass frame energy" (including "kinetic energy") really does contribute to the total energy relevant for gravitational collapse.

http://arxiv.org/abs/0908.1780

Q-reeus

I will dare to speak on Naty1's behalf here, and suggest he was referring to a single massive object 'in flight', which is totaly different to the situation of head-on collision of two such massive objects. It is obviously true that a moving observer cannot influence the physics in some other reference frame merely by having relative motion to it. And that surely was the gist of Naty1's point.
And btw folks, it would be nice if some GR buffs here actually responded to my #10! Or is it all too boring?

Sam Gralla

Sure, if by "observer" you mean a limit where the mass of a real observer goes to zero. All I'm saying is that if you have any mass at all, then there will be a speed for which, when the electron comes whizzing by, you better get ready to become a black hole =).

Q-reeus

Agreed; and to be picky one should throw in 'at an arbitrarily large minimum separation distance'.
That's an interesting thought. So for instance a sufficiently ultra-ultra relativistic particle, boring straight through the earth say, would leave behind a black hole wake?!! Yikes.

Sam Gralla

Now that I think about it this may not be true. Intuitively, the energy has to stick around long enough in order for a black hole to actually form, and now that I think about it, I think this is what Choptuik and Prestorius found (and what Penrose originally argued, although I've never read that work). So the correct statement is probably that for a fixed size/mass of the earth and fixed size/mass of the particle coming whizzing by, there is a (possibly vanishing) finite range of speeds for which a black hole forms.

Q-reeus

By speed I guess you mean energy, since for ultra-relativistic v = c-(an absolute whisker). In the case of glancing motion rather than direct head-on collision, wouldn't it tend to be a case of a minimum product of (tidal 'g')*(impulse duration) for crushing nearby matter sufficiently? By my crude reasoning, at ultra-relativistic energies the particle and field becomes a 2D pancake whose thickness is inversely proportional to particle KE E, but whose gravitational curvature (read tidal acceleration) at a given transverse radial distance grows quadratically with E. Again crudely, as we therefore have an impulse duration dt ~ E^-1, tidal accelerations a ~ E², then using good old S = 1/2a*dt² (S being some 'critical crushing displacement' for a given blob of stationary matter), it appears to be a stalemate with no advantage in going beyond a presumed minimum energy. This does not take into account gravito-magnetic interaction but I would think it relatively insignificant. The real complications might be the time dilational factor as crushed matter approaches it's Schwarzschild radius, and have little clue how that should be factored in.
But why am I contemplating the above, given general misgivings expressed in #10? Just for fun!

pervect

There is nothing in this paper that says that a single particle moving at a high velocity will become a black hole.

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 Aichelberg-Sexl ultraboost. See for instance http://arxiv.org/abs/gr-qc/0110032

It may take a little bit of advanced knowledge to interpret the comonents of the field tensor, given by the Riemann, in semi-Newtonian 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.