Why Can We Feel the Gravitational Field of a Black Hole?

[Dale]

The annihilation scenario was simply a round about way of showing depressed charge rest energy, when at rest in a grav potential.

But the charge dipole rest energy is not depressed:

If p is the four-momentum of the dipole then g_{\mu\nu}p^{\mu}p^{\nu} is the same at infinity and on the surface.

 

[Q-reeus]

But the charge dipole rest energy is not depressed:

A gravitating spherical shell composed of N particles undergoes partial collapse to some reduced radius. Those same N particles are now collectively in a reduced gravitational potential. Work was done during the collapse, released as heat, light, which escapes. The total final rest energy of the shell is undeniably less than before. There are still the same N particles present. By my reckoning that means each particle has a reduced rest energy, owing to the changed gravitational potential. Should I exempt dipoles from this basic logic for some strange reason? Are you sure the 4-momentum argument is sound? Are you sure it applies to the *final* rest state of the dipole after heat etc has dissipated to infinity, and not just up to the moment where in your scenario it's free-fall from infinity abruptly ends with a crash on to the NS surface? I would have preferred a gentler winching down procedure, but whatever.

 

[Q-reeus]

In #30 things weren't expressed all that well. By analogy with the notions of active and passive gravitational mass, I would now put it as the problem of a split between 'active charge' q_a, as source of E field, and 'passive charge' q_p, as measure of response to an applied E. It is then the product q_aq_p that drops by the factor (1+phi). According to Bowler, by this parlance q_p is the invariant wrt gravitational potential phi, and q_a is the one gravitationally depressed. That implies failure of charge invariance consistent with previous arguments re clock rate effect on virtual photon exchange. However it is not the consensus view as per Wikipedia entry given in #12. So again, any textbook or preferably online links showing from first principles a contradictory finding that static EM fields are gravitational invariants?

 

[Dale]

The total final rest energy of the shell is undeniably less than before.

Yes.

By my reckoning that means each particle has a reduced rest energy, owing to the changed gravitational potential.

No. Each particle has its original rest energy. Did you read the wikipedia link on binding energy that I had given earlier:
http://en.wikipedia.org/wiki/Binding_energy

"A bound system typically has a lower potential energy than its constituent parts". So the rest energy of a system of bound particles is less than the rest energy of the constituents.

Are you sure it applies to the *final* rest state of the dipole after heat etc has dissipated to infinity, and not just up to the moment where in your scenario it's free-fall from infinity abruptly ends with a crash on to the NS surface?

Yes, I am sure about it, but I am not the one making "contrarian" claims. The real question is if you are sure that it does not apply. When you are challenging mainstream science you had better be very sure.

 

[Q-reeus]

...No. Each particle has its original rest energy. Did you read the wikipedia link on binding energy that I had given earlier:
http://en.wikipedia.org/wiki/Binding_energy
"A bound system typically has a lower potential energy than its constituent parts". So the rest energy of a system of bound particles is less than the rest energy of the constituents...

Yes I read it. A matter of careful definition - no point comparing apples with oranges. By 'constituent parts' I take it to mean the original state of gravitationally unaffected N constituent parts - ie when each part is 'at infinite separation' and in zero gravitational potential. After collapse, the very same but now gravitationally bound N constituent parts have reduced net energy. One can argue the semantics but by that definition everything gone over in #25 is logically coherent.

 

[Dale]

By 'constituent parts' I take it to mean the original state of gravitationally unaffected N constituent parts - ie when each part is 'at infinite separation' and in zero gravitational potential.

That would be "unbound constituents".

One can argue the semantics but by that definition everything gone over in #25 is logically coherent.

I don't think so. Can you provide a rigorous derivation of your claims in #25? If you are going to go about claiming that GR is wrong for this reason then you should be able to prove it mathematically in very careful detail. Your handwaving arguments have been handwavingly rebutted, if you believe that your arguments still have merit then you should formulate them in a more concrete manner.

 

[Q-reeus]

That would be "unbound constituents".

Hey - we are in agreement!

I don't think so. Can you provide a rigorous derivation of your claims in #25? If you are going to go about claiming that GR is wrong for this reason then you should be able to prove it mathematically in very careful detail. Your handwaving arguments have been handwavingly rebutted, if you believe that your arguments still have merit then you should formulate them in a more concrete manner.

Expected it to come to this - the ultimatum strategy. No, a fraction of my admittedly handwavy arguments were handwavy criticized (mischaracterized by yourself in #20,#22, #24) but not really effectively rebutted - depending on how one defines effectively. No I'm like many here an interested layman who thinks about things, and if 'controversial' points are raised in a reasoned manner, one [STRIKE]expects[/STRIKE] hopes experts will take it up in some detail. And feel free btw to provide any constructive feedback on #30 and #33 - which might just lead to some useful conclusions. Also feel free to explain, as one with I assume formal training in GR, just how a free-falling charge at or inside an EH can logically be emitting vp's at a rate less than infinite and have a well removed external entity receive those same vp's at a rate greater than zero. And how this ties in with relevance of SC's. Having hammered that matter incessantly I'm still waiting anxiously for some GR expert to explain. Seems a reasonable request, wouldn't you say? Lot's of other truly bizarre stuff like wormholes and GR 'time-machines' are enthusiastically tackled here all the time, so this one should be a relative cake-walk.

 

[Dale]

No, a fraction of my admittedly handwavy arguments were handwavy criticized (mischaracterized by yourself in #20,#22, #24) but not really effectively rebutted - depending on how one defines effectively.

And that is the key problem with handwavy arguments. IMO your arguments have been completely refuted, in your opinion they have not. Therefore this cannot be resolved by further handwaving arguments and requires something more rigorous. Nothing fruitful can be gained by further handwaving.

if 'controversial' points are raised in a reasoned manner, one [STRIKE]expects[/STRIKE] hopes experts will take it up in some detail.

And I did so. In as much detail as you did (i.e. handwavy).

In any case, the burden of proof is always on the controversial position. It is up to you to convince me that your objection is valid, which you have not done, and I have explained why not. If you do not wish to pursue your position, that is fine, but as it is it certainly does not meet the burden of a serious theoretical challenge to GR.

as one with I assume formal training in GR

I am also an interested layman with no formal training in GR. However, when I am concerned about something the first thing that I do is learn the math and figure it out for myself.

 

[Q-reeus]

And that is the key problem with handwavy arguments. IMO your arguments have been completely refuted, in your opinion they have not. Therefore this cannot be resolved by further handwaving arguments and requires something more rigorous. Nothing fruitful can be gained by further handwaving.

Kindly link to where you or anyone else has iyo completely refuted me - which entry(s), what point(s) of argument exactly? I can't recall any.

And I did so. In as much detail as you did (i.e. handwavy).

Not imo. There was a brief entry in #4 - but the worldline thing was never fleshed out in any detail. No attempt to place the statement there into any useful context - who's worldline, who's proper time, how this sensibly makes a connection between inner and outer events, etc. Otherwise, basically all I recall was the flat claim I'm wrong, period, until the string of mistaken criticisms in #20, #22, #24. After that was cleaned up, we come to the fairly pointless definition arguments over whether rest mass is or isn't depressed, and mine was the last word there.

In any case, the burden of proof is always on the controversial position. It is up to you to convince me that your objection is valid, which you have not done, and I have explained why not. If you do not wish to pursue your position, that is fine, but as it is it certainly does not meet the burden of a serious theoretical challenge to GR.

Disagree with the philosophy here. If I was some specialist presenting a paper then yes the burden of proof is on me. But this is a *forum*, yes? Lots and lots of lay folks turn up with various questions or controversial claims - and expect to obtain good feedback from the pros. They don't expect to be told to go and become a specialist before further discussion is warranted. This place is supposed to be a brains trust - ask the oracle! Works out most of the time too I'd say. But clearly not for everyone.

I am also an interested layman with no formal training in GR. However, when I am concerned about something the first thing that I do is learn the math and figure it out for myself.

Not so much about the maths on this issue as conceptual basis. What disappoints but doesn't surprise is what you have not tackled in #37. But that's your choice. Anyway off to some sleep. :zzz:

 

[Dale]

Kindly link to where you or anyone else has iyo completely refuted me - which entry(s), what point(s) of argument exactly? I can't recall any.

You first posted your reduced charge argument clearly enough for me to understand in post 25 and I immediately refuted it in post 27. IMO the refutation is complete in post 27. As outlined in 27 there is simply no missing energy to account for and therefore no reason to assume a reduced charge.

What disappoints but doesn't surprise is what you have not tackled in #37.

As far as I can tell there is nothing new in 37 that I didn't already address in 4.

Disagree with the philosophy here. If I was some specialist presenting a paper then yes the burden of proof is on me. But this is a *forum*, yes? Lots and lots of lay folks turn up with various questions or controversial claims - and expect to obtain good feedback from the pros.

And you have received good feedback, exactly in line with the quality of the questions and claims you have presented. If you want further feedback then simply present your argument in a form where further feedback is possible.

On a practical note, I have been quite bad at interpreting your arguments (e.g. taking 20 posts to understand your point) so the most likely result of any derivation of mine at this point is that I would solve a problem that didn't accurately reflect your argument. Then we would be back at the same point we are now, but I would have wasted a lot of time. Nobody else but you can take this discussion further. Even if you think that you do not have the burden of proof, there is no practical way that I can do it for you.

 

[Q-reeus]

In a sense I should be grateful - you have at least made a habit of responding on this issue. Precious little input from the other 650+ and still growing lookers-on. No point in continuing a slanging match for high moral ground. I'd say this thread is done, unfortunately without anything resolved imo. Maybe the OP can chime in and say if his query is answered though - nothing from him after #1.

 

[Dale]

OK.

One more quick point that indicates that the charge dipole rest energy is not depressed. If we take the received gamma rays and work backwards, blueshifting them, we find that at the annihilation site the energy is the usual flat-spacetime energy (I.e. single Doppler shift). Alternatively, any local measurement of the energy measures the usual rest energy. This is an example why we attribute the reduced binding energy to the system as a whole, and not specifically to any of the constituent parts.

 

[Q-reeus]

sorry to say We just don't see eye to eye on that one. However have to admit I'm still in the process of consistently figuring this whole issue out. Just since my last entry have realized that simply trying to equate E fields in and out of gravitational potential is somewhat missing the mark. A simple thought experiment involving a variously oriented 'test' capacitor static in a grav field shows that by the SC's there is a tensor relationship - radial and azimuthal E, as determined 'from afar' are equal but charge surface density is different. Moreover, it's not just the coordinate determined E field(s) further in that counts - it's how the field 'travels' out over a radial distance different from the Newtonian value - the inner charges are 'further away' and that imo counts too. My overall conclusion remains as before, but inapropriate pursuing it all here. May or may not post a new thread on this later, after dealing with more pressing matters ouside of PF.

 

[Dale]

. May or may not post a new thread on this later, after dealing with more pressing matters ouside of PF.

Good luck, I hope all is well.

 

[Q-reeus]

Good luck, I hope all is well.

Thanks!

 

[Zac Einstein]

If the gravity travels at the speed of light, it can't get out of the black hole...but this can't be true it's just my opinion : )

 

[DrGreg]

If the gravity travels at the speed of light, it can't get out of the black hole...but this can't be true it's just my opinion : )

It is actually changes to gravity that travel at the speed of light. In the case of a black hole, the gravity doesn't need to "get out" because it's already there, and has been ever since a star started collapsing to form the black hole.

 

[Zac Einstein]

It is actually changes to gravity that travel at the speed of light. In the case of a black hole, the gravity doesn't need to "get out" because it's already there, and has been ever since a star started collapsing to form the black hole.

Excuse me sir, you wrote "changes" do you mean the objects that travel near a black hole?

 

[WannabeNewton]

Excuse me sir, you wrote "changes" do you mean the objects that travel near a black hole?

His second sentence basically answers your question. The gravitational field of the black hole is formed as the star collapses and is therefore formed when the black hole is fully formed so there is no need for it to escape the black hole. Also, I think what Dr Greg stated in his first sentence is that in the dynamics of black hole formation, the irregularities or non - spherically symmetric parts of the collapse get propagated out as gravitational waves.