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Comment author: 09 June 2012 08:17:48AM 1 point [-]

Not sure you're the right person to ask that to, but there have been two questions which bothered me for a while and I never found any satisfying answer (but I've to admit I didn't take too much time digging on them either) :

1. In high school I was taught about "potential energy" for gravity. When objects gain speed (so, kinetic energy) because they are attracted by another mass, they lose an equivalent amount of potential energy, to keep the conservation of energy. But what happens when the mass of an object changes due to nuclear reaction ? The mass of sun is decreasing every second, due to nuclear fusion inside the sun (I'm not speaking of particles escaping the sun gravity, but of the conversion of mass to energy during nuclear fusion). So the potential energy of the Earth and all other planets regarding to gravity is decreasing. How is this compatible with conversation of energy ? It can't be the energy released by the nuclear reaction, the fusion of hydrogen doesn't release more energy just because Earth and Jupiter are around.

2. Similarly for conservation issue, I always have been bothered with permanent magnet. They can move things, so they can generate kinetic energy (in metal, other magnets, ...). But where does this energy comes from ? It's stored when the magnet is created and depleted slowly as the magnet does it's work ? Or something else ?

Sorry if those are silly questions for a PhD physicist as you are, but I'm a computer scientist, not a physicist and they do bother me !

Comment author: 09 June 2012 10:43:39AM *  3 points [-]

The mass of sun is decreasing every second, due to nuclear fusion inside the sun (I'm not speaking of particles escaping the sun gravity, but of the conversion of mass to energy during nuclear fusion).

IMO “conversion of mass to energy” is a very misleading way to put it. Mass can have two meanings in relativity: the relativistic mass of an object is just its energy over the speed of light squared (and it depends on the frame of reference you measure it in), whereas its invariant mass is the square root of the energy squared minus the momentum squared (modulo factors of c), and it's the same in all frames of references, and coincides with the relativistic mass in the centre-of-mass frame (the one in which the momentum is zero). The former usage has fallen out of favour in the last few decades (since it is just the energy measured with different units -- and most theorists use units where c = 1 anyway), so in recent ‘serious’ text mass means “invariant mass”, and so it will in the rest of this post.

Note that the mass of a system isn't the sum of the masses of its parts, unless its parts are stationary with respect to each other and don't interact. It also includes contributions from the kinetic and potential energies of its parts.

The reason why the Sun loses mass is that particles escape it; if they didn't, the loss in potential energy would be compensated by the increase in total energy. The mass of an isolated system cannot change (since neither its energy nor its momentum can). If you enclosed the Sun in a perfect spherical mirror (well, one which would reflect neutrinos as well), from outside the mirror, in a first approximation, you couldn't tell what's going on inside. The total energy of everything would stay the same.

Now, if the Sun gets lighter, the planets do drift away so they have more (i.e. less negative) potential energy, but this is compensated by the kinetic energy of particles escaping the Sun... or something. I'm not an expert in general relativity, and I hear that it's non-trivial to define the total energy of a system when gravity is non-negligible, but the local conservation of energy and momentum does still apply. (Is there any theoretical physicist specializing in gravitation around?)

As for 2., that's the energy of the electromagnetic field. (The electromagnetic field can also store angular momentum, which can leading to even more confusing situations if you don't realize that, e.g. the puzzle in The Feynman Lectures on Physics 2, 17-4.)

Comment author: 09 June 2012 11:44:41PM 3 points [-]

I'm not an expert in general relativity, and I hear that it's non-trivial to define the total energy of a system when gravity is non-negligible, but the local conservation of energy and momentum does still apply. (Is there any theoretical physicist specializing in gravitation around?)

Sean Carroll has a good blog post about energy conservation in general relativity.

Comment author: 09 June 2012 10:26:31AM *  2 points [-]

I'm not Rolf (nor am I strictly speaking a physicist), but:

1. There isn't really a distinction between mass and energy. They are interconvertible (e.g., in nuclear fusion), and the gravitational effect of a given quantity of energy is the same as that of the equivalent mass.

2. There is potential energy in the magnetic field. That energy changes as magnets, lumps of iron, etc., move around. If you have a magnet and a lump of iron, and you move the iron away from the magnet, you're increasing the energy stored in the magnetic field (which is why you need to exert some force to pull them apart). If the magnet later pulls the lump of iron back towards it, the kinetic energy for that matches the reduction in potential energy stored in the magnetic field. And yes, making a magnet takes energy.

[EDITED to add: And, by the way, no they aren't silly questions.]

Comment author: 09 June 2012 10:59:41AM 1 point [-]

Hum, that's a reply to both you and army1987; I know mass and energy aren't really different and you can convert one to the other; but AFAIK (and maybe it's where I'm mistaken), if massless energy (like photons) are affected by gravity, they don't themselves create gravity. When the full reaction goes on in the Sun, fusing two hydrogen into an helium, releasing gamma ray and neutrinos in the process, the gamma ray doesn't generate gravity, and the resulting (helium + neutrino) doesn't have as much gravitational mass as the initial hydrogen did.

The same happen when an electron and a positron collide, they electron/positron did generate a gravitation force on nearby matter, leading to potential energy, and when they collide and generate gamma ray photons instead, there is no longer gravitation force generated.

Or do the gamma rays produce gravitation too ? I've pretty sure they don't... but I am mistaken on that ?

Comment author: 09 June 2012 02:48:26PM 5 points [-]

Or do the gamma rays produce gravitation too ? I've pretty sure they don't... but I am mistaken on that ?

They do. In Einstein's General Relativity, the source of the gravitational field is not just "mass" as in Newton's theory, but a mathematical object called the "energy-momentum tensor", which as it name would indicate encompasses all forms of mass, energy and momentum present in all particles (e.g. electrons) and fields (e.g. electromagnetic), with the sole exception of gravity itself.

Comment author: 03 July 2012 08:18:55PM *  1 point [-]

with the sole exception of gravity itself.

I’ve seen this said a couple of times already in the last few days, and I’ve seen this used as a justification for why a black hole can attract you even though light cannot escape them. But black holes are supposed to also have charge besides mass and spin. So how could you tell that without electromagnetic interactions happening through the event horizon?

Comment author: 03 July 2012 09:10:48PM 1 point [-]

That is a good question. There is more than one way to formulate the answer in nonmathematical terms, but I'm not sure which would be the most illuminating.

One is that the electromagnetic force (as opposed to electromagnetic radiation) is transmitted by virtual photons, not real photons. No real, detectable photons escape a charged black hole, but the exchange of virtual photons between a charge inside and one outside results in an electric force. Virtual particles are not restricted by the rules of real particles and can go "faster than light". (Same for virtual gravitons, which transmit the gravitational force.) The whole talk of virtual particles is rather heuristic and can be misleading, but if you are familiar with Feynman diagrams you might buy this explanation.

A different explanation that does not involve quantum theory: Charge and mass (in the senses relevant here) are similar in that they are defined through measurements done in the asymptotic boundary of a region. You draw a large sphere at large distance from your black hole or other object, define a particular integral of (respectively) the gravitational or the electromagnetic field there, and its result is defined as the total mass/charge enclosed. So saying a black hole has charge is just equivalent to saying that it is a particular solution of the coupled Einstein-Maxwell equations in which the electromagnetic field at large distances takes such-and-such form.

Notice that whichever explanation you pick, the same explanation works for charge and mass, so the peculiarity of gravity not being part of the energy-momentum tensor that I mentioned above is not really relevant for why the black hole attracts you. Where have you read this?

Comment author: 05 August 2012 11:42:46AM *  0 points [-]

Hi Alejandro, I just remembered I hadn’t thanked you for the answer. So, thanks! :-)

I don’t remember where I’ve seen the explanation (that gravity works through event horizons because gravitons themselves are not affected), it seemed wrong so I didn’t actually give a lot of attention to it. I’m pretty sure it wasn’t a book or anything official, probably just answers on “physics forums” or the like.

For some reason, I’m not quite satisfied with the two views you propose. (I mean in the “I really get it now” way, intellectually I’m quite satisfied that the equations do give those results.)

For the former, I never really grokked virtual particles, so it’s kind of a non-explanatory explanation. (I.e., I understand that virtual particles can break many rules, but I don’t understand them enough to figure out more-or-less intuitively their behavior, e.g. I can’t predict whether a rule would be broken or not in a particular situation. It would basically be a curiosity stopper, except that I’m still curious.)

For the latter, it’s simply that retreating to the definition that quickly seems unsatisfying. (Definitions are of course useful, but less so for “why?” questions.)

The only explanation I could think of that does make (some) intuitive sense and is somewhat satisfactory to me is that we can never actually observe particles crossing the event horizon, they just get “smeared”* around its circumference while approaching it asymptotically. So we’re not interacting with mass inside the horizon, but simply with all the particles that fell (and are still falling) towards it.

( * : Since we can observe with basically unlimited precision that their height above the EH and vertical speed is very close to zero, I can sort of get that uncertainty in where they are around the hole becomes arbitrarily high, i.e. pretty much every particle becomes a shell, kind of like a huge but very tight electronic orbital. IMO this also “explains” the no-hair theorem more satisfyingly than the EH blocking interactions. Although it does get very weird if I think about why should they seem to rise as the black hole grows, which I just dismiss with “the EH doesn’t rise, the space above it shrinks because there are more particles pulling on it”, which is probably not much more wrong than any other “layman” explanation.)

Of course, all this opens a different** can of worms, because it’s very unintuitive that particles should be eternally suspended above an immaterial border that is pretty much defined as no-matter-how-hard-you-try-you'll-still-fall-through-it. But you can’t win them all, and anyway it’s already weird that falling particles see something completely different, and for some reason relativity always seemed to me more intuitive than quantum physics, no matter how hairy it gets.

(**: Though a more accurate metaphor would probably be that it opens the same can of worms, just on a different side of the can...)

Comment author: 08 August 2012 03:11:24AM *  0 points [-]

OK, here is another attempt at explanation; it is a variation of the second one I proposed above, but in a way that does not rely on arguing by definition.

Imagine the (charged, if you want) star before collapsing into a black hole. If you have taken some basic physics courses, you must know that the total mass and charge can be determined by measurements at infinity: the integral of the normal component of the electric field over a sphere enclosing the star gives you the charge, up to a proportionality constant (Gauss's Law), and the same thing happens for the gravitational field and mass in Newton's theory, with a mathematically more complicated but conceptually equivalent statement holding in Einstein's.

Now, as the star begins to collapse, the mass and charge results that you get applying Gauss's Law at infinity cannot change (because they are conserved quantities). So the gravitational and electromagnetic fields that you measure at infinity do not change either. All this keeps applying when the black hole forms, so you keep feeling the same gravitational and electric forces as you did before.

Comment author: 10 August 2012 07:23:55PM *  0 points [-]

Yeah, you’re right, putting it this way at least seems more satisfactory, it certainly doesn’t trigger the by-definition alarm bells. (The bit about mass and charge being conserved quantities almost says the same thing, but I think the fact that conservation laws stem from observation rather than just labeling things makes the difference.)

However, by switching the point of view to sphere integrals at infinity it sort of side-steps addressing the original question, i.e. exactly what happens at the event horizon such that masses (or charges) inside it can still maintain the field outside it in such a state that the integral at infinity doesn’t change. Basically, after switching the point of view the question should be how come those integrals are conserved, after the source of the field is hidden behind an event horizon?

(After all, it takes arbitrarily longer to pass a photon between you and something approaching an EH the closer it gets, which is sort of similar to it being thrown away to infinity the way distant objects “fall away” from the observable universe in a Big Rip, it doesn’t seem like there is a mechanism for mass and charge to be conserved in those cases.)

Comment author: 10 August 2012 10:36:43PM 0 points [-]

how come those integrals are conserved, after the source of the field is hidden behind an event horizon?

First, note that there are no sources of gravity or of electromagnetism inside a black hole. Contrary to popular belief, black holes, like wormholes, have no center. In fact, there is no way to tell them apart from outside.

Second, electric field lines are lines in space, not spacetime, so they are not sensitive to horizons or other causal structures.

it takes arbitrarily longer to pass a photon between you and something approaching an EH

This is wrong as stated, it only works in the opposite direction. It takes progressively longer to receive a photon emitted at regular intervals from someone approaching a black hole. Again, this has nothing to do with an already present static electric field.

Comment author: 06 August 2012 03:26:28AM 0 points [-]

I'm sorry that my explanations didn't work for you; I'll try to think of something better :).

Meanwhile, I don't think it is good to think in terms of matter "suspended" above the event horizon without crossing it. It is mathematically true that the null geodesics (lightray trajectories) coming from an infalling trajectory, leaving from it over the finite proper time period that it takes for it to get to the event horizon, will reach you (as a far-away observer) over an infinite range of your proper time. But I don't think much of physical significance follows from this. There is a good discussion of the issue in Misner, Thorne and Wheeler's textbook: IIRC, a calculation is outlined showing that, if we treat the light coming from the falling chunk of matter classically, its intensity is exponentially suppressed for the far-away observer over a relatively short period of time, and if we treat it in a quantum way, there are only a finite expected amount of photons received, again over a relatively short time. So the "hovering matter" picture is a kind of mathematical illusion: if you are far away looking at falling matter, you actually do see it disappear when it reaches the event horizon.

Comment author: 03 July 2012 08:52:37PM 0 points [-]

Interesting question, I never though about if there is any way to test a black holes charge. My guess is that we only can assume if it is there based on the theory right now

Comment author: 03 July 2012 09:05:44PM 0 points [-]

found a relevant answer at http://www.astro.umd.edu/~miller/teaching/questions/blackholes.html "black holes can have a charge if they eat up too many protons and not enough electrons (or vice versa). But in practice this is very unusual, since these charges tend to be so evenly balanced in the universe. And then even if the black hole somehow picked up a charge, it would soon be neutralized by producing a strong electric field in the surrounding space and sucking up any nearby charges to compensate. These charged black holes are called "Reissner-Nordstrom black holes" or "Kerr-Newman black holes" if they also happen to be spinning." -Jeremy Schnittman

Comment author: 03 July 2012 09:20:25PM 0 points [-]

Interesting question, I never though about if there is any way to test a black holes charge.

Calculate the black hole's mass. Put a charged particle somewhere in the vicinity of the black hole. Measure acceleration. Do math.

Comment author: 03 July 2012 10:15:37PM 0 points [-]

That much is obvious given an assumption that charged fields work proberly through a blackhole, which was not obvious particularily given aljandro's statement. After confirming that the charge of a blackhole can interact with being impeded by the singularity, there are a lot of obvious ways to check the charge

Comment author: 03 July 2012 09:30:06PM -1 points [-]

Will that work? Or to put it particle-ish-ly, how is the information about a charge inside an event horizon able to escape?

Comment author: 09 June 2012 07:33:00PM *  3 points [-]

Or do the gamma rays produce gravitation too ? I've pretty sure they don't... but I am mistaken on that ?

There is a lot of potential (no pun intended) for confusion here, because the subject matter is so far from our intuitive experience. There is also the caveat "as far as we know", because there have not been measurements of gravity on the scale below tenths of a millimeter or so.

First, in GR gravity is defined as spacetime (not just space) curvature, and energy-momentum (they are linked together in relativity) is also spacetime curvature. This is the content of the Einstein equation (energy-momentum tensor = Ricci curvature tensor, in the units where 8piG/c^2=1).

In this sense, all matter creates spacetime curvature, and hence gravity. However, this gravity does not have to behave in the way we are used to. For example, it would be misleading to say that, for example, a laser beam attracts objects around it, even though it has energy. Let me outline a couple of reasons, why. In the following, I intentionally stay away from talking about single photons, because those are quantum objects, and QM and GR don't play along well.

• Before a gravitational disturbance is felt, it has to propagate toward the detector that "feels" it. For example, suppose you measure the (classical) gravitational field from an isolated super-powerful laser before it fires. Next, you let it fire a short burst of light. What does the detector feel and when? If it is extremely sensitive, it might detect some gravitational radiation, mostly due to the laser recoiling. Eventually, the gravitational field it measures will settle down to the new value, corresponding to the new, lower, mass of the laser (it is now lighter because some of its energy has been emitted as light). The detector will not feel much, if any, "pull" toward the beam of light traveling away from it. The exact (numerical) calculation is extremely complicated and requires extreme amounts of computing power, and has not been done, as far as I know.

• What would a detector measure when the beam of light described above travels past it? This is best visualized by considering a "regular" massive object traveling past, then taking a limit in which its speed goes to the speed of light, but its total energy remains constant (and equal to the amount of energy of the said laser beam). This means that its rest mass is reduced as its speed increases. I have not done the calculation, but my intuition tells me that the effects are reduced as speed increases, because both the rest mass and the amount time the object remains near the detector go down dramatically. (Note that the "relativistic mass" stays the same, however.)

There is much more to say about this, but I've gone on for too long as it is.

EDIT: It looks like there is an exact solution for a beam of light, called Bonnor beam. This is somewhat different from what I described (a short pulse), but the interesting feature is that two such beams do not attract. This is not very surprising, given that the regular cosmic strings do not attract, either.

Comment author: 09 June 2012 08:11:52PM *  1 point [-]

8piG

How comes no-one has come up with a symbol (say G-bar) for that, as they did with ħ for h/2pi when they realized ħ was a more ‘natural’ constant than h? (or has anybody come up with a single symbol for 8piG?)

Comment author: 09 June 2012 10:05:23PM *  1 point [-]

There aren't many people who do this stuff for a living (as is reflected in exactly zero Nobel prizes for theoretical work in relativity so far), and different groups/schools use different units (most popular is G=1, c=1), so there is not nearly as much pressure to streamline the equations.

Comment author: 10 June 2012 12:44:44AM *  1 point [-]

The notation kappa = 8 pi G is sometimes used, e.g. in this Wiki article. However, it is much less universal than ħ.

Comment author: 09 June 2012 10:41:06PM 1 point [-]

They are not silly questions, I asked them myself (at least the one about the Sun) when I was a student. However, it seems army1987 got there before I did. So, yep, when converting from mass-energy to kinetic energy, the total bending of spacetime doesn't change. Then the photon heads out of the solar system, ever-so-slightly changing the orbits of the planets.

As for magnets, the energy is stored either in their internal structure, ie the domains in a classic iron magnet; or in the magnetic field density. I think these are equivalent formulations. An interesting experiment would be to make a magnet move a lot of stuff and see if it got weaker over time, as this theory predicts.

Comment author: 10 June 2012 07:26:08PM 2 points [-]

An interesting experiment would be to make a magnet move a lot of stuff and see if it got weaker over time, as this theory predicts.

If you're not thinking of moving a lot of stuff at once, every time you pull a piece of the stuff back off the magnet where it was before you're returning energy back to the system, so the energy needn't eventually be exhausted. (Though I guess it still eventually be if the system is at a non-zero temperature, because in each cycle some of the energy could be wasted as heat.)