Is 'charged black hole' an oxymoron?

[PeterDonis]

That makes sense if M does not just represent the neutral mass but is in some way inclusive of the total mass/energy contribution from Q.

I think that's right; I may not have been fully clear about it in earlier posts.

So do we interpret R-N expression as that M = M_n+M_Qtot (M_n the strictly neutral mass, M_Qtot the net contribution from Q)?

I'm not sure you can separate out M_n, because the "contribution from Q" varies with radius. At any finite radius, there will be *some* amount by which the Q term in the metric reduces the effect of the M term. And the Q term continues to change all the way down to r = 0; so there's no point at which you could say that *all* of the Q contribution has been "subtracted out" and what's left is M_n.

Well, ok, if 'fossil' EM field is exempt from definition of residue!

If a charged object were falling in, yes, the EM field "left behind" in the exterior vacuum region would not be "residue" in the sense I am using the term here. It would be analogous to the Weyl curvature left behind by the object's energy. Basically, the stuff that isn't "residue" is the stuff that is allowed to be "hair" on a BH, which means conserved quantities: mass-energy, charge, and angular momentum.

 

[stevendaryl]

Yes - I'm sceptical based on the implications as I see it for e.g. scenarios in #1 or say that charged-balloon-and-mat example in last para of #20. No higher maths, just following a comparative trend to it's logical conclusions. Others can come in and either translate or refute that using higher maths - and Internal cosistency is king.

I don't believe that you have pointed out any internal inconsistency in General Relativity or the theory of electromagnetism, or the two together. Charge and mass are actually different, because charge is a scalar, while mass is not. The total is just the sum of the charges of the pieces, but that's not true of mass, because interactions change the mass.

As said last time, afaik there is no empirical evidence either way as to charge invariance holding when gravity is involved.

That's a weird way of putting things. I would say exactly the opposite---we have plenty of evidence of charge invariance in gravitational fields, and no evidence to the contrary.

Agreed about charge counting. As to correctness of various integrals, it must imo be guided by or at least concur with logical conclusions from the sort of scenarios I have given.

I really don't think that the scenarios you have brought up have any bearing on
the invariance of charge. The current theory, which does have invariance of charge, works perfectly fine in all cases that we are capable of analyzing--there is no reason to believe that there is any problem. Of course, that doesn't mean that our current theories are correct, but there is no indication of any inconsistency in them (which is what I thought you were claiming).

As per my closing comments in #21, on a straight basis (gravitational acceleration at constant r = 'gravitational flux density') the divergence law does fail for 'gravitational charge'.

There is no such thing as "gravitational charge", according to General Relativity. The source of gravity is not mass, it is energy-momentum. That makes a technical difference. In the nonrelativistic case, gravity seems a lot like the electric field, but they are very different in the relativistic case.

Right - and we are stuck for the forseeable future with use of internal consistency arguments as said before.

Well, there is no consistency problem with invariance of charge.

Which most here consider to be superfluous. You have made no comment on examples I gave you last time,

I can't really make sense of them, but they seem to be motivated by a misunderstanding
of relativity theory. So it seems to me that it would be more productive for you to understand relativity better, rather than for me to understand your misunderstandings.

nor attempted to detail where any previous scenarios go wrong. Which is hardly unique to yourself here but is a source of some personal chagrin. If my conclusions are wrong, it should not be too hard to show precisely how in each case. Still waiting.

I wish that you could put your argument in a form that is a logical deduction of whatever contradiction you think you've discovered. I didn't see a contradiction in anything you said. For a black hole with charge, the charge Q is equal to the sum of all the charges that have fallen into the black hole. The mass M is not equal to the sum of all the masses that have fallen in, because mass is not conserved, while charge is. But what do you think is contradictory about that?

 

[stevendaryl]

BUT the pulling force that the positive charge exerts on a negative charge at higher altitude is not reduced. Conservation of charge requires this.

BUT at lower altitude it is observed that charges that are being lifted seem to gain more charge, and charges that are being lowered kind of lose charge. Seems the charge of outside universe may change as seen from a gravity well. Interesting. If this needs some justification, then I'll think up some justification.

That's really not correct. Charge is not affected by being dropped into a black hole.

 

[stevendaryl]

I apologize if my response was a little grumpy. I guess I'm being a little impatient. I guess I should read what his been said more carefully before replying.

 

[stevendaryl]

1: Matter of proper mass m gently lowered towards the EH reduces in coordinate measure as m' = fm, with f the usual redshift expression f = √(1-2GM/(rc²)). In keeping with that conservation of energy applies and work is being extracted in the lowering process. Now suppose that m also carries a charge q. it makes no difference to the net reduction in m as all forms of energy reduce the same. Locally there is no variation in the proper charge-to-mass ratio q/m. How can that local invariance of q/m (no free-fall case) not be also reflected as remotely observed - i.e. q' = fq? Certainly the locally invariant q/m will show remotely as a proportionately equally redshifted reduction in the Newtonian gravitational and Coulombic forces of attraction/repulsion between two adjacent such charged masses. The implication is obvious - as vanishes coordinate mass, so vanishes coordinate charge.

I guess the reason that I didn't respond specifically to this scenario is because it seems to start with a number of assumptions that I don't think are true at all. First, what does it mean to say that when you lower a mass, it "reduces in coordinate measure as m' = fm, with f the usual redshift expression". Where did you get that idea, that mass reduces by the redshift formula? I think that you are starting with an inconsistent notion of what GR says about mass.

 

[stevendaryl]

A balloon is tethered on a string to the centre of a large mat. Both are electrically charged with the same sign of charge - just sufficient to have the balloon float against Earth's gravity. For a distant observer, the mass of Earth and balloon are redshifted wrt to that locally observed

Where are you getting this idea that mass is redshifted? That doesn't make any sense to me.

 

[PeterDonis]

First, what does it mean to say that when you lower a mass, it "reduces in coordinate measure as m' = fm, with f the usual redshift expression". Where did you get that idea, that mass reduces by the redshift formula? I think that you are starting with an inconsistent notion of what GR says about mass.

If "mass" is interpreted to mean "energy at infinity", then it does "reduce" as described if you lower a mass slowly, by means of a rope, let's say, as opposed to letting it free-fall. I agree that "mass" is not a good term for this.

 

[jartsa]

That's really not correct. Charge is not affected by being dropped into a black hole.

Let me explain:

1: A hydrogen atom is lowered into a deep gravity well. Then a photon of visible light is dropped onto the atom, which becomes ionized, although visible light does not normally ionize hydrogen. That happened because the field that keeps the atom together weakened as the atom was lowered.

2: A hydrogen atom and a photon of visible light are both gently lowered into a gravity well. Now the photon does not ionize the atom. The reason for this is that the electric field of the photon weakened as the photon was being lowered, as did the electric field in the atom.

So a photon in a gravity well, observing an atom being lowered into the gravity well, will say that the force that keeps the atom together is weakenig, kind of like the charges in the atom were becoming smaller.

(if photon does not have an electric field, then replace photon with a EM-wave pulse) (photon can be lowered using a mirror lined elevator)

 

[PeterDonis]

1: A hydrogen atom is lowered into a deep gravity well. Then a photon of visible light is dropped onto the atom, which becomes ionized, although visible light does not normally ionize hydrogen. That happened because the field that keeps the atom together weakened as the atom was lowered.

No, it happened because the photon was blueshifted as it dropped into the gravity well. A visible light photon emitted locally, at the same altitude as the atom, won't ionize it, so the field of the atom is not "weakened" at all according to local measurements. The difference that lowering the atom gently makes is that it is at rest deeper inside the well, so it "sees" the blueshift of the photon. To see why that's important, consider an alternate experiment where you let both a hydrogen atom and a visible light photon free-fall into the gravity well, in such a way that they meet up somewhere much deeper into the well than where you released them (you time the release of the atom and the photon from your much higher altitude to ensure this). Will the photon ionize the atom? No, because the atom is not at rest in the field; it is falling inward at a high speed, so there is a large Doppler redshift when it absorbs the photon that cancels the gravitational blueshift.

2: A hydrogen atom and a photon of visible light are both gently lowered into a gravity well. Now the photon does not ionize the atom. The reason for this is that the electric field of the photon weakened as the photon was being lowered, as did the electric field in the atom.

No, the reason is that, by hypothesis, the photon of visible light did *not* blueshift, because it was "gently lowered" instead of being allowed to free fall. (Actually, I'm not sure how you would "gently lower" a photon. What do you do, attach a rope to it? But for purposes of a thought experiment I'm fine with assuming you can somehow do it.)

 

[jartsa]

No, it happened because the photon was blueshifted as it dropped into the gravity well. A visible light photon emitted locally, at the same altitude as the atom, won't ionize it, so the field of the atom is not "weakened" at all according to local measurements. The difference that lowering the atom gently makes is that it is at rest deeper inside the well, so it "sees" the blueshift of the photon. To see why that's important, consider an alternate experiment where you let both a hydrogen atom and a visible light photon free-fall into the gravity well, in such a way that they meet up somewhere much deeper into the well than where you released them (you time the release of the atom and the photon from your much higher altitude to ensure this). Will the photon ionize the atom? No, because the atom is not at rest in the field; it is falling inward at a high speed, so there is a large Doppler redshift when it absorbs the photon that cancels the gravitational blueshift.
No, the reason is that, by hypothesis, the photon of visible light did *not* blueshift, because it was "gently lowered" instead of being allowed to free fall. (Actually, I'm not sure how you would "gently lower" a photon. What do you do, attach a rope to it? But for purposes of a thought experiment I'm fine with assuming you can somehow do it.)

Well this is what I think:

A person that has spend some time in a deep gravity well will be observed having aged abnormally slowly, after he has been winched up from the well. He has produced some EM-waves by waving his slightly charged hands. Not very many EM-wave crests have been observed outside the gravity well, and the person does not report that he waved his hands very many times.

So the EM-waves that came up from the gravity well had just a few wave crests, because the person waved just a few times. And this is the reason the EM-waves that came up were slowly waving waves: Few crests during a long time, because of the slowness of the wave source.

The waves become longer at higher altitude, because light moves faster at higher altitude.

The case of free falling atom and photon I have to think about a bit.

 

[stevendaryl]

If "mass" is interpreted to mean "energy at infinity", then it does "reduce" as described if you lower a mass slowly, by means of a rope, let's say, as opposed to letting it free-fall. I agree that "mass" is not a good term for this.

I would not describe that situation with the terminology "Mass is reduced according to the redshift formula". I would say, rather, that if you extract energy from a system, then the total energy of that system becomes smaller. There's nothing surprising about that. In a situation in which energy is being extracted, but charge is not, I don't understand why anyone would believe that the ratio Total Energy/Total Charge would remain constant.

 

[stevendaryl]

Let me explain:

1: A hydrogen atom is lowered into a deep gravity well. Then a photon of visible light is dropped onto the atom, which becomes ionized, although visible light does not normally ionize hydrogen. That happened because the field that keeps the atom together weakened as the atom was lowered.

I don't think that's a very good explanation at all. I would not say that "the field that keeps the atom together weakened."

Look, the same thing happens in Newtonian gravity. If I throw a small stone at a car, it won't do much damage. But if I drop that stone from a great height, it can do a lot of damage. I would not describe this as "the forces holding the car together get weaker when the car is deep in a gravitational well".

The only meaningful notion of the "strength of forces holding an object together" is what you measure at the object, in the (local) inertial frame in which the object is at rest. That is not changed by dropping the object into a deep gravitational well.

What changes between high in the well and lower in the well is the translation of vectors. A velocity vector that is a small velocity high up in the well translates to a much larger velocity deeper down.

Of course, General Relativity understands gravity in a different way than Newtonian physics, but in neither case is it appropriate to say that "forces holding an object together get weaker when the object is lowered in a gravitational field".

 

[stevendaryl]

Well this is what I think:

A person that has spend some time in a deep gravity well will be observed having aged abnormally slowly, after he has been winched up from the well. He has produced some EM-waves by waving his slightly charged hands. Not very many EM-wave crests have been observed outside the gravity well, and the person does not report that he waved his hands very many times.

General Relativity is about curved spacetime. There are two different ways to think about curved spaces, and each is useful for different purposes. The first way is to use curvilinear coordinates. For example, on the surface of the Earth, you can use longitude and latitude as coordinates. This is convenient, except that there are some weird effects: For example, in terms of longitude, objects at the North Pole seem "stretched out" compared with the same object near the equator, because the same object can cover more lines of longitude. Another effect is that if a plane attempts to fly as straight as possible, his path in terms of latitude will look "curved"; the shortest path from Montreal to Seattle is not to fly straight west, but to go northwest half the trip and southwest the second half.

An alternative way of thinking about curved spaces is in terms of gluing together lots of approximately flat spaces. Imagine taking the surface of the Earth and breaking it up into lots of triangles of size 10 miles on a side. Now, within each little triangle, Euclidean geometry works fine. Straight lines look like straight lines on a map for that triangle. The "curvature" part is captured by how the various triangles are glued together. If a traveler leaves one triangle, you need to figure out which triangle he enters next, and what angle his trajectory makes in the second triangle.

So there is a "translation" process to translate vectors from one region of a curved space to another. In GR, a photon's momentum is a certain vector, that has to be translated when that photon moves from one region to another. That's what the redshift/blueshift formula is doing.

 

[jartsa]

I don't think that's a very good explanation at all. I would not say that "the field that keeps the atom together weakened."

Look, the same thing happens in Newtonian gravity. If I throw a small stone at a car, it won't do much damage. But if I drop that stone from a great height, it can do a lot of damage. I would not describe this as "the forces holding the car together get weaker when the car is deep in a gravitational well".

The only meaningful notion of the "strength of forces holding an object together" is what you measure at the object, in the (local) inertial frame in which the object is at rest. That is not changed by dropping the object into a deep gravitational well.

What changes between high in the well and lower in the well is the translation of vectors. A velocity vector that is a small velocity high up in the well translates to a much larger velocity deeper down.

Of course, General Relativity understands gravity in a different way than Newtonian physics, but in neither case is it appropriate to say that "forces holding an object together get weaker when the object is lowered in a gravitational field".

Nobody seems to know how the total energy of a falling rock changes. Obviously that energy that is responsible of car crushing does increase.

Light is more simple: When light enters a gravity field of a planet, the light slows down and the planet starts to move in the same direction as the light. So light loses energy when falling. Or as an extremely good aproximation the energy stays the same.

There was a long and technical discussion in this forum about the energy change of falling light, and those were the conclusion. The light losing energy is my own idea though.

So it must be the objects becoming weaker at lower altitude, which causes them to break more easily, when light from above hits them.

Hey I have one more scenario again: A charge in a gravity well is accelerated from 0 m/s to 100 m/s. Radiation energy is proportional to velocity change. As seen from higher altitude the velocity change was smaller than 100 m/s, and there is the reason why the radiation energy coming from the gravity well is smaller too.

 

[Dale]

So do we interpret R-N expression as that M = M_n+M_Qtot (M_n the strictly neutral mass, M_Qtot the net contribution from Q)?

Hi Q-reeus, I would not over-interpret the M and the Q as representing some particular mass or charge. I would think of them simply as parameters of the metric. The M term can include rest mass, energy (including energy in EM fields), pressure, stress, etc. And Q could be an E-field boundary condition at the edge of the manifold rather than some charged particles actually located in the manifold.

 

[Q-reeus]

I guess the reason that I didn't respond specifically to this scenario is because it seems to start with a number of assumptions that I don't think are true at all. First, what does it mean to say that when you lower a mass, it "reduces in coordinate measure as m' = fm, with f the usual redshift expression". Where did you get that idea, that mass reduces by the redshift formula? I think that you are starting with an inconsistent notion of what GR says about mass.

I'm ignoring your earlier posting #37 and later one in #41 (but accept the nice sentiments in #39), as it all hinges on getting right what you say here. Maybe you have already changed again - I say 'again' because in #22 there is "I was mistaken in my first response--allowing a dropping mass to do work does change the total mass/energy." - which was a seeming backflip from your earlier position, and amazingly you seem to have done another 180 and it's back to the original stance. [Stop press: just now read your #46. Evidently then you believe the energy gain/loss is to be considered a wholly delocalized affair - shared amongst the entire system more or less equally?] Then consider the following:

Suppose that mass/energy in a slowly lowered mass m is in the form of an unstable matter/antimatter doublet that self-annihilates and escapes entirely 'to infinity' as gamma rays. We surely agree that in escaping the gravitational potential well of central mass M, those rays - which carry all the energy tied up originally in m(r), are redshifted in coordinate measure. Annihilate the same matter/antimatter doublet out there in distant space, and obviously the gamma rays are not redshifted at all. This little experiment of the mind imo nicely indicates it is proper to consider the energy loss/gain in lowering/raising matter/energy of mass m(r) in a grav well as essentially confined to just that mass m(r) - provided m(r) << M. Unless that is one wishes to argue transporting matter/energy from a resting position at one potential to a resting position at another potential can be anything but a path independent process - assuming of course central mass M is taken as static. Again if one argues those gamma rays are not 'resting' just remember we are free to have them absorbed by nuclei in a massive block of matter etc. which then constitutes resting position.

You might also care to look again at the parallel-plate capacitor situation of last para in #24. We agree that locally, it only makes sense to ascribe energy loss (transfer of energy to higher up via electrical line) upon capacitor discharge to be confined essentially to the capacitor field-charge system, not say the planet of mass M as a whole. Again, haul the charged capacitor up to higher potential, then discharge. More power available - gained via the hauling process - and to be located in the capacitor field. It's only when masses are roughly comparable that delocalization/sharing becomes significant.

There has been some nitpickery over fine distinctions between mass and energy but let's recall the scenario 1: in #1 to which you have at least specifically addressed in #40 as quoted above - it's one of slow lowering. For which it's then perfectly appropriate to treat the mass/charge as momentarily resting at a given height. What then is the distinction between rest mass and total stress-energy? Sure there must be stress in the matter involved (it feels 'g' forces) and yes technically that comes under a different part of the stress-energy tensor contribution to total gravitating mass m(r). It will though be typically extremely small compared to the rest energy part, and anyway acts here simply as a scalar additive term. So can we please just take it that m(r) is inclusive of rest energy and stress without further fuss and ado?

Re your's and to some extent PeterDonis's criticism of jartsa over his description of atoms being 'weaker' further down. The local perspective is not the only legitimate one and from a coordinate viewpoint I would agree with his thrust. There is something similar in SR. A slab of dielectric lies immersed in an E field of a parallel plate capacitor. Both dielectric and capacitor are stationary in lab frame S', and E' there is below dielectric breakdown value. Now propel that slab to a relativistic velocity in S', normal to direction of E' such that in proper frame S of slab, E exceeds breakdown value and there is catastrophic failure and discharge. It is not legitimate to say that in S' the slab can be viewed as having 'weakened', since E has not changed in S'? Notice too that in jartsa's scenario if an atom spontaneously decays radioactively releasing a gamma ray, it is redshifted as seen from outside of potential well. That can legitimately qualify it as a 'weaker' atom with weaker internal EM and nuclear fields in my book - it's all a matter of pov.

There is a potential can of worms I'm uncovering or rather rediscovering in working through all the ways of looking at how mass, energy, momentum, field etc. should be evaluated in coordinate measure - more later. Maybe much later.

 

[Q-reeus]

Hi Q-reeus, I would not over-interpret the M and the Q as representing some particular mass or charge. I would think of them simply as parameters of the metric. The M term can include rest mass, energy (including energy in EM fields), pressure, stress, etc. And Q could be an E-field boundary condition at the edge of the manifold rather than some charged particles actually located in the manifold.

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]

The local perspective is not the only legitimate one

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]

Interesting perspective there DaleSpam. A bit out of my league and maybe Peter is best one to comment further there.

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]

Nobody seems to know how the total energy of a falling rock changes. Obviously that energy that is responsible of car crushing does increase.

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.

Light is more simple: When light enters a gravity field of a planet, the light slows down and the planet starts to move in the same direction as the light. So light loses energy when falling. Or as an extremely good aproximation the energy stays the same.

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.

There was a long and technical discussion in this forum about the energy change of falling light, and those were the conclusion.

That's a very misleading conclusion.

So it must be the objects becoming weaker at lower altitude, which causes them to break more easily, when light from above hits them.

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.

Hey I have one more scenario again: A charge in a gravity well is accelerated from 0 m/s to 100 m/s. Radiation energy is proportional to velocity change. As seen from higher altitude the velocity change was smaller than 100 m/s, and there is the reason why the radiation energy coming from the gravity well is smaller too.

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]

A slab of dielectric lies immersed in an E field of a parallel plate capacitor. Both dielectric and capacitor are stationary in lab frame S', and E' there is below dielectric breakdown value. Now propel that slab to a relativistic velocity in S', normal to direction of E' such that in proper frame S of slab, E exceeds breakdown value and there is catastrophic failure and discharge. It is not legitimate to say that in S' the slab can be viewed as having 'weakened', since E has not changed in S'?

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]

Nobody seems to know how the total energy of a falling rock changes.

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.)

Obviously that energy that is responsible of car crushing does increase.

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

Light is more simple: When light enters a gravity field of a planet, the light slows down and the planet starts to move in the same direction as the light.

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.

So light loses energy when falling. Or as an extremely good aproximation the energy stays the same.

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.

There was a long and technical discussion in this forum about the energy change of falling light, and those were the conclusion.

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

 

[jartsa]

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

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

 

[stevendaryl]

I'm ignoring your earlier posting #37 and later one in #41 (but accept the nice sentiments in #39), as it all hinges on getting right what you say here. Maybe you have already changed again - I say 'again' because in #22 there is "I was mistaken in my first response--allowing a dropping mass to do work does change the total mass/energy."

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.

- which was a seeming backflip from your earlier position, and amazingly you seem to have done another 180 and it's back to the original stance.

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.

[Stop press: just now read your #46. Evidently then you believe the energy gain/loss is to be considered a wholly delocalized affair - shared amongst the entire system more or less equally?]

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).

Then consider the following:

Suppose that mass/energy in a slowly lowered mass m is in the form of an unstable matter/antimatter doublet that self-annihilates and escapes entirely 'to infinity' as gamma rays. We surely agree that in escaping the gravitational potential well of central mass M, those rays - which carry all the energy tied up originally in m(r), are redshifted in coordinate measure.

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).

Annihilate the same matter/antimatter doublet out there in distant space, and obviously the gamma rays are not redshifted at all.

Right.

This little experiment of the mind imo nicely indicates it is proper to consider the energy loss/gain in lowering/raising matter/energy of mass m(r) in a grav well as essentially confined to just that mass m(r) - provided m(r) << M.

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.

Unless that is one wishes to argue transporting matter/energy from a resting position at one potential to a resting position at another potential can be anything but a path independent process - assuming of course central mass M is taken as static.

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.

Re your's and to some extent PeterDonis's criticism of jartsa over his description of atoms being 'weaker' further down. The local perspective is not the only legitimate one and from a coordinate viewpoint I would agree with his thrust.

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.

There is something similar in SR. A slab of dielectric lies immersed in an E field of a parallel plate capacitor. Both dielectric and capacitor are stationary in lab frame S', and E' there is below dielectric breakdown value. Now propel that slab to a relativistic velocity in S', normal to direction of E' such that in proper frame S of slab, E exceeds breakdown value and there is catastrophic failure and discharge. It is not legitimate to say that in S' the slab can be viewed as having 'weakened', since E has not changed in S'?

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

Notice too that in jartsa's scenario if an atom spontaneously decays radioactively releasing a gamma ray, it is redshifted as seen from outside of potential well. That can legitimately qualify it as a 'weaker' atom with weaker internal EM and nuclear fields in my book - it's all a matter of pov.

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]

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.

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]

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.

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]

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?

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]

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.

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

 

[PeterDonis]

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

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

When light enters a gravity field of a planet, the light slows down and the planet starts to move in the same direction as the light. So light loses energy when falling. Or as an extremely good aproximation the energy stays the same.

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 $k^{a}$ 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 $u^{b}$:

$$E = g_{ab} k^{a} u^{b}$$

So even if $k^{a}$ is unchanged as the photon travels, its observed energy can still change if either the metric $g_{ab}$ or the observers' 4-velocity $u^{b}$ 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 $u^{b}$ changes relative to that of the emitter, which determines the photon's 4-momentum $k^{a}$.

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 $g_{ab}$ 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]

We both agreed earlier that there is legitimately a reduction in mass/energy when matter is lowered into a potential well.

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.

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?

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

 

[Q-reeus]

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?

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]

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.

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'!

 

[PeterDonis]

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.

If you are right that there is higher E in frame S, so that the force on q is higher in frame S, then it must be higher in frame S' as well. The force on q, meaning the proper acceleration it experiences times its rest mass, must be invariant. Instead of waving your hands, you should actually do the calculation.

 

[PeterDonis]

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

See my new post; somehow "submit" got clicked while I was still thinking it over.