Is 'charged black hole' an oxymoron?

PeterDonis

I didn't miss it, as my last post should make clear. But I can't even try to answer it until you define more precisely what you mean by "what happens to the field lines". The gravitational lensing has an unambiguous observable: the direction from which the light comes to my detector changes when the lensing object is in the path. What is the unambiguous observable that tells me whether or not something has happened to the field lines?

DaleSpam

This is not true in *all* cases. It is only true in cases where a quantity of energy exactly equal to (m-m')c^2 is extracted from the system. There are many ways to add an additional mass that do not involve radiating this quantity of energy away.

In those situations where the energy was radiated away it seems strange to call that process "redshift", but regardless of what you call it, it seems more like a statement about energy flux across a surface than about any gravitational effect.

How do you conclude this? The change in the global mass, whether Schwarzschild M, Bondi, ADM, or Komar, tells you nothing about a local interaction like this. I actually think that you can make a case for this concept using parallel transport, but I would encourage you to think about this rigorously.

Not true.

I agree. Again, this concept is related to parallel transport.

I agree that the conclusion is ridiculous.

DaleSpam

Which makes your redshift concept a statement about net energy flux rather than gravitation.

Austin0

Originally Posted by Austin0

Is that actually the case?

The book creates deformation stresses on the table. The table necessarily counters this force through electrostatic and nuclear forces (Van der Walls etc). This is a continuing condition so implies a continuing flow of energy.
How to analyze this in terms of energy conservation is beyond me as the electrostatic and nuclear forces seem to be effectively inexhaustible but it seems that there has to be energy in play. Just as simply standing in gravity requires additional energy.
We consider that the table surface is accelerating upward even though there is n o coordinate displacement. Doesn't acceleration imply force/energy?
SO comparably shouldn't we view the book as having inertial momentum downward , exerting force on the table , even though in this case also there is no coordinate motion??

DaleSpam

This is an incorrect understanding of energy. Once the small deformation is done there is no motion, so no work is being done. Energy does not need to continually flow in order to provide a static force.

Austin0

Oh i certainly understand the normal view of energy you are presenting here , having thought exactly the same until encountering this situation .
So if there can be acceleration without motion perhaps it is possible to have energy expenditure without action. At this point I am just looking at the situation from all sides without any conclusions. In the end I may end up right where I started, agreeing with your view.

So are you saying that the book ceases to exert downward force on the table once the initial adjustment is made?
If so i would say my rear contact with my chair seat disagrees with you.

If the book is hovering under thrust at an equivalent height this would necessitate continuing energy of acceleration.This also implies that the book is effecting an equivalent continuing counter force ( momentum?). yes?
The EP would seem to suggest, as the book's force is the same in both cases, the upward acceleration/force would be equivalent .Isn't this the basis of the EP ? An accelerometer doesn't measure the downward force but rather the upward force acting against the inertia of some internal mass of the instrument?
Maybe it is my understanding of the EP that is lacking? ;-(

TrickyDicky

This would be correct only if the table is considered an inertial object (as it is done in most practical physics exercises, there is a lab frame considered to be at rest,it is an idealization that works great for most practical problems), but we know that is not the case in reality, the table is non-inertial and in continuous motion so there is work done. An accelerometer in the surface of the earth measures proper acceleration.

TrickyDicky

Your rear is entitled to disagree.

DaleSpam

No, I am most definitely not saying that. I explicitly said "Energy does not need to continually flow in order to provide a static force." I.e. there is a continued force, it does not require energy.

In the thrust example the KE of the exhaust is being increased (a lot), therefore energy is being used. None of that energy is going into the book whose KE and PE are remaining constant.

Probably it is more a misunderstanding of physics in non-inertial frames. Please see my response to Tricky Dicky below.

DaleSpam

No, I am not considering the table an inertial object, if the table were inertial then the force on the book would be 0.

You misunderstand how energy works in a non-inertial frame. In a non-inertial frame, like the usual frame attached to the surface of the earth, there is a fictitious force. The fictitious force, in this case, is equal to mg and directed downwards. We know that there is a fictitious force on the book precisely because an accelerometer measures a proper acceleration of g directed upwards and yet the book is not accelerating relative to our frame.

Now, work is f.d, so as an object is moved upwards against this fictitious force it requires an amount of work equal to f.d=mgh. Conversely, as an object is lowered the fictitious force does an amount of work equal to mgh. So, the ficitious force has an associated potential energy. (remember, energy is frame variant)

So, the book, sitting on the table at rest, has no change in KE. It also has no change in PE. No work is being done on it. There is no change to its internal structure or temperature nor anything else where energy could go. There is no change in any energy associated with the book. Despite the fact that there is a force on the book (two forces actually) and the book measures a proper acceleration and is therefore non-inertial.

TrickyDicky

Not exactly, KE is frame dependent, energy is a conserved quantity in newtonian mechanics (I understand you are restricting this to the newtonian mechanics POV since you were talking about fictitious forces).
You are forgetting here that work, like KE is a frame dependent quantity. I know the example of the book and the table is considered to have 0 net moment, with net moment defined as change in KE, that is not what I referred to when I said work was being done.
The conventional Newtonian treatment with fictitious forces is fine, I have nothing against it as long as one understands its range of validity.
I was pointing to a more realistic treatment, more like GR's Schwarzschild solution (so energy is not frame dependent) for instance. In which the table would be preventing the book from following its geodesic path. And in order to do that Work in the form of a quantity proportional to the EM force in the table material times distance from the theorical geodesic path of the book, must be done.
I know is not the conventional way work is referred to in Newtonian physics(wich is the way you were defending it in your post, and as I said I perfectly understand the use of fictitious forces and work in newtonian physics). But I think by introducing some GR ingredients in the situation it gets closer to reality.

DaleSpam

All energy is frame dependent, not just KE. And yes, energy is conserved (in SR generally, in GR in static spacetimes). I think you may be mixing up conservation with frame invariance, they are two separate concepts. Energy is conserved, but frame variant.

No, all of my comments above were wrt GR, not Newtonian physics. What makes you think that GR doesn't have ficitious forces? In fact, GR provides a very easy and consistent mechanism for determining fictitious forces through the Christoffel symbols.

http://en.wikipedia.org/wiki/Curvili...ar_coordinates
http://www.mathpages.com/home/kmath641/kmath641.htm
http://theoretical-physics.net/dev/s...nertial-frames

This is an incorrect definition of work. If you believe otherwise then please provide a mainstream scientific reference that defines work as the force times the "distance from the theorical geodesic path".

No, all of my comments above are GR comments. In GR there are fictitious forces, just like in Newtonian mechanics. See the references above.

The difference is that in Newtonian mechanics gravity is considered a real force and in GR it is considered fictitious. I refered to mg as a fictitious force, so I was definitely discussing from the GR perspective. You are clearly misunderstanding how non-inertial frames and fictitious forces are treated in GR.

TrickyDicky

Great then, if you are considering gravity as a fictitious force then you indeed are using a GR perspective, sorry I didn't notice it. Yor conclusions made me think you were only considering the Newtonian view. But then you are reinforcing my point, thanks.
From wikipedia page on Fictitious forces:
"Fictitious forces can be considered to do work, provided that they move an object on a trajectory that changes its energy from potential to kinetic."
If the EM forces of the table (and the earth) didn't hold the book, it would spontaneously fly towards its geodesic path.

DaleSpam

Far from reinforcing your point, this completely contradicts your point. See the last two paragraphs of 316.

TrickyDicky

Those two paragraphs look wrong to me. You have the fictitious force of gravity on the book acting in an upward and downward direction at the same time on the same object. That is not possible.

DaleSpam

Nonsense. Where did I say that?

The fictitious force of gravity always acts downwards. In fact, it is what defines the direction "down". I never said anything to the contrary.

TrickyDicky

Here maybe it is your wording but you seem to have the book going up and down in a strange way, and you even transform the fictitious force into ¡two forces¡. Make up your mind, is the fictitious force of gravity one or two forces? is the book going up or down? I'm afraid it can't go up and down at the same time.