Is 'charged black hole' an oxymoron?

Q-reeus

Interesting perspective there DaleSpam. A bit out of my league and maybe Peter is best one to comment further there. As you will be aware, my real focus is on logical reason for any external E for a BH. Cheers.

PeterDonis

We agree on the actual observable results, so I can't say that this view is "wrong". But it seems to me that taking this view often leads you into confusion, because it keeps you from applying common sense reasoning to disentangle causes. You have objects that "look different" whey they are far away than they do when they are close up, but instead of taking the obvious route of attributing the differences to the effect of the spacetime in between, you insist on saying that some "intrinsic" property of the objects has changed. And this prevents you from adopting a simple method of distinguishing the two possibilities: look at what local measurements say about the objects in their new location.

A simple analogy: two people, A and B, are standing next to a cube, and both of them agree that it looks white. Now the cube is moved to the other side of the room, and both of them agree that it now looks red. A says that the cube must have "changed color"; B says no, something about the space between must be altering the light reflected from the cube, changing it from white to red. They both agree on how the cube looks, but they disagree on why.

In one sense, the difference between A and B is just a "difference in pov"; after all, they both agree on all the experimental results. But suppose they now ask their friend C, who is standing on the other side of the room next to the cube, what color the cube looks to him. C answers that it looks white. B says, "You see? The cube is still white, but something about the space between us is making its apparent color change." How can A respond? If he tries to claim that the cube somehow "really has" changed color, even though it looks white to C, the one standing right next to it, won't he seem foolish? Wouldn't it be more reasonable for A (and B) to look for something in the middle of the room that could be changing the color of the light from the cube--a large red filter screen, perhaps? They could even shine some white light from their end of the room and ask C how it looks to him, and find that it looks red. In short, they could apply standard scientific techniques to figure out the causes of what they observe.

You can see the analogy, I hope. Consider the hydrogen atom that's slowly lowered into the gravity well. An observer, C, right next to it will not be able to ionize it with visible light; he will find its ionization energy to be exactly what it was when it was far away from all gravitating bodies and that energy was measured locally. Now observers A and B, at a much higher altitude, emit visible light and find that it ionizes the atom. A claims that the atom has somehow "weakened"; but B says no, it must be something about the spacetime between that is altering the light. After all, C finds the ionization energy to be the same as always. Furthermore, if they ask C how the "visible" light they are shining down looks to him, he will say it looks like gamma radiation; so obviously something about the spacetime between is changing the light. This is all standard scientific reasoning, but of course if you refuse to avail yourself of it, you will continue to be confused by these types of scenarios.

PeterDonis

I would agree with what DaleSpam said. It is true that you can measure M and Q by looking at the behavior of test objects at very large radii, as I described in an earlier post; but doing that does not commit you to any particular belief about "where" the "mass" or "charge" being measured is located. It just means you've measured global properties of the spacetime.

stevendaryl

What do you mean by that? Explanations come in layers--you explain phenomena at one level by showing how it works in terms of a more fundamental description. Sometimes that more fundamental description itself can be explained in terms of even more fundamental laws, but sometimes it can't.

We know how the energy of a rock changes as it falls, in the sense that it is perfectly described by General Relativity. We don't know, in any deep sense, why GR is true.

No, that's a very poor explanation, in my opinion. The General Relativity explanation, as I suggested, is in terms of parallel transport of vectors, and that applies both to a rock falling and a photon falling.

That's a very misleading conclusion.

That's not what General Relativity says. What you're doing is taking the predictions of GR, and reinterpreting them according to a different model. There certainly can be two physical models that produce the same predictions (an example is Special Relativity and certain variants of aether theory), but talking in terms of forces getting weaker in a gravitational field is completely contrary to the General Relativity way of understanding gravity.

Curved spacetime can be thought of as taking a bunch of little regions of flat spacetime and "gluing" them together on the edges. Inside each little region, the laws of physics work almost exactly the same as they would in gravity-free space. The "curvature" comes in when you try to glue neighboring regions together. The vector corresponding to a slow-velocity rock or a low-energy photon in one region becomes a high-velocity rock or high-energy photon in another region.

It's exactly like trying to describe the surface of the Earth using flat maps. If each map only covers a small region of the Earth, say 100 x 100 kilometers, then you don't notice the curvature. But when you try to glue one map together with a second map, you will find that vectors don't precisely match up. A vector that is vertical on one map corresponds to a vector that is slightly tilted from parallel on the second map.

I don't think that's a very good description at all. To go from "the light from the event is red-shifted" to "the velocity change must have been slow" is very dubious.

Imagine light from some event (say, a charged particle accelerating) comes to you from two different directions; for example, suppose there is a very massive star, or black hole between the event and you, and the light can go around in one direction, or the other. When the light gets to you, the two images can have different amounts of redshift. You can't explain that in terms of weaker or stronger electrical forces down in gravitational well. The way to explain it is to realize that light has to travel from the event to your eyes. Depending on the path it takes, the light is changed by its journey.

PeterDonis

This is a different type of scenario: the large relative motion between capacitor and dielectric is a local effect, not an "at a distance" effect. There is no "spacetime in between"; everything happens locally. Even if you make the capacitor and the dielectric each a light-year long, the local field at any point in the dielectric, in the dielectric's rest frame, can still be attributed to the "local" part of the capacitor; it doesn't have to be transmitted over a distance.

PeterDonis

Maybe you don't know, but that doesn't mean nobody knows. If the rock is freely falling, then its total energy is constant; as it falls, it gains kinetic energy and loses potential energy, but the sum of the two remains constant. (Strictly speaking, this is only true in certain spacetimes, such as Schwarzschild spacetime, but that's general enough to cover what we've been discussing.)

Yes, that's the kinetic energy. The potential energy decreases by the same amount.

If this happens with light, wouldn't it also happen with the falling rock? Why would the two be different?

One could adopt a viewpoint in which both the rock/light and the planet moved; this is just the center of mass frame. But the planet is so much more massive than either the rock or the light that the center of mass frame is not measurably different from the frame in which the planet is at rest. So that's the conventional approximation. The description I gave of the rock's motion above was in that approximation; and in that approximation, the light also "falls" towards the planet in such a way that its total energy, kinetic plus potential, remains constant. A light ray's kinetic energy is proportional to its frequency, so as the light falls, it blueshifts; or if it's rising, climbing out of a gravity well, it redshifts.

The total energy, kinetic plus potential, stays the same. You have the kinetic energy backwards: it increases when falling and decreases when rising, just as a rock's does.

Do you have a reference? I think you must have misinterpreted something.

jartsa

Here it is:
http://physicsforums.com/showthread.php?t=588937

stevendaryl

I agree that if you get work out of lowering a mass toward a black, then the total energy of the system (the black hole + the mass you are lowering) will be less than if you don't extract work from it. It's not correct to describe this as "mass is reduced by the redshift formula"--it's that expression that seems completely wrong.

At an informal level, the total energy of a black hole is equal to the energy you dropped into it, minus the energy you pulled out of it. The total charge of a black hole is equal to the charge you dropped into it.

I would say that it's a matter of trying to understand what you're saying. I would not say that dropping an electron into a black hole changes the mass of the electron. That's just a weird thing to say. Dropping an electron into a black hole changes the mass of the entire system, black hole + electron. Exactly how much the mass of the system is changed depends on how you lower the electron into the black hole. But I would not say that the mass of the electron changed. The definition of the mass of an electron is the total energy of the electron as measured in a local inertial frame in which the electron is at rest. That is not changed by lowering it into a black hole.

Yes, in general, energy is a function of an entire system. Mass is the energy as a measured in a frame in which the system is at rest (has zero total momentum).

Yes. The way I would say it is that the particle/antiparticle pair annihilate to produce a pair of photons with a characteristic frequency, as measured in the local rest frame of the pair. These photons then travel up out of their gravitational well and escape to infinite. Their frequencies are changed by their journey (according to the redshift formula).

Right.

No, I don't agree. When two particles annihilate, they produce two photons that go off in opposite directions. These two photons could have their paths warped by the gravity of a massive star or black hole, and then come back together. When they come back together, the two frequencies need not be the same. It doesn't make sense to attribute the difference in frequency to differences in the masses of the particles that produced the photons. Instead, the differences should be understood as a change in the momenta of the photons as they travel from where they are produced to where they are measured.

Parallel transport of vectors is certainly path-dependent. In general, you can't compute redshift by noting where the photons started from, you have to take into account the path taken by the photons.

You can certainly describe things using whatever coordinates you like, but you have to be careful not to attribute physical effects to artifacts of your choice of coordinates.

The relevant value of E is E as measured in the frame in which the dielectric is at rest.

That's a very bad way of looking at it, in my opinion. The insight that Einstein formulated as the equivalence principle is that in any small region of spacetime, most phenomena--the ticking of clocks, the decay of particles, etc.--work exactly the same as they would in the absence of gravity. The complexity comes in when you try to relate phenomena in one region of spacetime to phenomena in another region. That's where the technical tool of "parallel transport" comes into play.

Q-reeus

Are you picking on me now Peter? We both agreed earlier that there is legitimately a reduction in mass/energy when matter is lowered into a potential well. And it shows remotely - the gravitational contribution felt 'out there' is not that owing to plugging in the locally observed proper mass but the redshifted coordinate value. Agreed?

Q-reeus

Actually there are other factors to consider in that case and I jumped in too soon (dielectric polarization field transformation, plus effect of motion through the induced magnetic field in S'. There is cancellation but it gets messy). A cleaner example would be: same capacitor and field E' in S', but now there is a charge q lying at the end of a cantilever arm - oriented mutually orthogonal to both E' and later applied relative velocity v. We suppose initially the arm, at rest in S' barely resists the force on q from E'. Now set it in relative motion such that arm breaks under higher E in proper frame S of arm+charge. In S' we attribute no higher force on q owing to v. How to explain breakage? To say the material has relativistically weakened seems about ok to me. There is imo this much commonality with gravitational case - situation is viewed from differing proper frames in both cases.

stevendaryl

I don't think it's appropriate to talk about mass being redshifted. You can talk about the frequency of a photon being redshifted, and maybe that's what you mean?

stevendaryl

That seems like a very bad way of looking at it, in my opinion.

PeterDonis

I've read that thread, and I don't see how you got the following out of it:

There was no talk at all in that thread about the planet moving in the same direction as the light. You said the light losing energy when falling was your own idea, and indeed that's not in the thread either. Nor is the energy staying the same "to an extremely good approximation"; Jonathan Scott said that the frequency of the light stays the same, but he meant something different by "frequency", and he specified that that was relative to a particular coordinate system, the global Schwarzschild coordinates.

Let me try to summarize what the technical posts in that thread were actually saying. A photon emitted at a particular event in any spacetime will have a 4-momentum vector [itex]k^{a}[/itex] associated with it. Since photons travel on null geodesics, that 4-momentum vector will be parallel transported along the photon's worldline; this is the sense in which the photon "does not change" as it travels.

However, the observables associated with the photon are determined, not just by the photon's 4-momentum, but by geometric objects, vectors and tensors, associated with the observer. For example, the energy the photon is measured to have by that observer is the contraction of the photon's 4-momentum with the observer's 4-velocity [itex]u^{b}[/itex]:

[tex]E = g_{ab} k^{a} u^{b}[/tex]

So even if [itex]k^{a}[/itex] is unchanged as the photon travels, its observed energy can still change if either the metric [itex]g_{ab}[/itex] or the observers' 4-velocity [itex]u^{b}[/itex] changes. (We actually measure photon frequency, not energy, but the latter is just Planck's constant times the former.)

In the case of the standard Doppler shift, the measured energy (frequency) changes because the 4-velocity of the observer [itex]u^{b}[/itex] changes relative to that of the emitter, which determines the photon's 4-momentum [itex]k^{a}[/itex].

In the case of a photon falling into or climbing out of a gravity well, the energy (frequency) measured by static observers--observers who are "hovering" at a constant radius r--changes with r because the metric [itex]g_{ab}[/itex] changes. (The 4-velocity of "hovering" observers is the same at all r--all their 4-velocity vectors point "in the same direction".)

The "frequency staying the same" that Jonathan Scott was talking about was a different sense of "frequency": if I have a blinker, say, emitting flashes of light deep in a gravity well, such that it emits N flashes between Schwarzschild coordinate times t = 0 and t = 1, then an observer much higher up in the gravity well will also count N flashes between coordinate times t = 0 and t = 1. The two observers will differ in how much *proper* time they experience between those two coordinate times, so they will assign a different proper frequency (flashes per second of proper time) to the blinker; but the frequency relative to *coordinate* time is the same. This is all consistent with what I said above.

PeterDonis

I agreed that there was a reduction in *energy at infinity*, because you are extracting work during the lowering process. But if you measure the rest mass of the object locally, you will get the same answer after it has been lowered as before.

Yes, but that doesn't mean what you think it means.

Q-reeus

Taking on board your comments in #59 and what I just read in #63, it seems we just have fundamentally different outlooks in matters discussed and best to just agree to disagree methinks. Have a nice day.

stevendaryl

Okay, but I think that you are deeply confused about these topics, and you are interpreting your confusion as a contradiction in General Relativity. The contradiction is in your own head.

Q-reeus

Peter - where did your last post go - I so looked forward to saying 'gotcha'!