Now, I should be clear about this before I start: I don't know much about relativity, either special or general.

However, when I read your first document, the basic logic seems correct:

1) The gravitational potential becomes singular near a black hole, as the distance from the singularity can approach zero.

2) There is, therefore, a particular radius at which the energy of creation of particle-antiparticle pairs, 2mc^2, exactly equals the energy decrease in gravitational energy due to the particle of mass m.

3) At this point, one particle falls into the black hole, while its antiparticle escapes

It seems to me that what you are describing is normally called Hawking radiation - the energy loss in the black hole due to particle/antiparticle pair creation at the event horizon.  You forgot to take into account the 2 - since you must create particle/antiparticle pairs, the energy of creation is 2mc^2, rather than mc^2. 

So I think you're describing a valid phenomenon, but with the incorrect conclusion that the mass of the black hole increases, rather than the conclusion ala Hawking that the black hole diminishes with time because the antiparticles that escape carry energy with them.

What do you think of this?

Your derivation finds the radius at which the mass-energy of a particle exactly equals its gravitational energy.  This alone is not sufficient for spontaneous creation of particles; it merely says that, if a particle is created, it will fall into the black hole rather than, say, recombine. 

Particle/antiparticle creation happens everywhere in vacuum over short times.  If this spontaneous creation occurs within the critical radius (when you include the factor of 2 for pair production mass energy, your critical radius should become the Schwarzchild radius), then both the particle and its antiparticle should fall into the black hole (because the gravitational energy exceeds the mass energy).  However, this process does not result in an increase in the black hole's mass, because the particle/antiparticle pair has net mass zero (i.e., the antiparticle has "negative mass").  The only place where you can avoid this, so that one particle falls in but the antiparticle escapes, is at the Schwarzchild radius itself. 

Now, it seems to me that it should be equally likely for a particle or its antiparticle to fall into the black hole at the Schwarzchild radius.  This means that, on average, the net gain in mass of the black hole is zero (because there are equal amounts of particles and antiparticles flowing into it), so the mass should not increase.  However, the particles that escape carry kinetic energy with them, and thus, by conservation of energy, must result in a decrease in the energy of the black hole, which is exactly Hawking radiation.

So, I think your reasoning is right, except that the energetics are not sufficient for particle creation.

I also reiterate: even though it may be energetically favorable to create a particle that would then fall into the black hole, this is not sufficient for the particle to actually be created.  Particle creation always happens in particle/antiparticle pairs, or else all kinds of conservation laws (charge, baryon/lepton number, etc.) would be violated.  In fact, in Dirac's view of relativistic quantum mechanics, particle/antiparticle production does not actually "create" anything - it is the promotion of a negative-energy quantum state to a positive-energy quantum state.  The result is that a positive-energy particle is released, and a hole in the negative band results, which behaves like an antiparticle.  Thus there is no creation of particles alone.

A miscellaneous point: you referred to photosynthesis as an example of transforming energy into mass.  I cannot see how this is correct.  Photosynthesis is a process by which one form of energy (light) is converted to another (chemical/electrostatic energy).  Mass/energy conversions generally only occur in large-energy regimes.

Phase Theory wrote:

Is there no massive particle with zero charge and zero lepton number, etc. - zero all of those 'conserved' properties, but nonzero mass?
Alternatively, a particle pair could be created with equal and opposite charge, lepton, baryon etc. but not opposite mass...?  After all, surely all combinations of mass/charge/lepton number etc. should be theoretically possible.

Actually, no, this is not the case.  The standard model of particle physics lists some 21 fundamental particles (by my count).  These fall into the classes:

Quarks - all quarks are charged, massive, and have baryon number 1/3

Leptons - half of the leptons are charged, and half uncharged; all leptons are massive (although the neutrinos have extremely small masses), and all leptons have lepton number 1

Bosons (force carriers) - The W-bosons are massive and charged, the Z-boson is massive but uncharged, and the photon and the gluon are both uncharged (electrically) and massless. 


All fermions (quarks and leptons) must be produced in particle/antiparticle pairs.  The same is even true of the bosons, except that a boson is its own antiparticle.  For example, in electron/positron annihilation, two photons are produced, with the same frequency, but moving in opposite directions.  This is because the antiparticle to a photon is another photon with the same frequency, moving in the opposite direction.  Hence even photons (and phonons, etc.) are created in particle/antiparticle pairs.

Phase Theory wrote:

E.g. - There is a particle with zero charge: the neutron.  There is a particle with zero (some property that neutrons have nonzero of).  That being the case, there should be a particle with zero in both of those properties.
If I'm wrong (I probably am), please explain why.

Your logic here is quite reasonable, and it touches on one of the deepest mysteries in modern physics: why the particles we see should have only the properties that they do.  There is no derivation (that has been found) that can explain why the electron has the mass it does, for instance.  And it would seem that physics should work just fine with particles that have any combination of these properties - like the kind of particle you were trying to construct.  We have no good explanation why these are the only particles found in nature.

I suppose I should defer the photosynthesis question to a biologist (or a biophysicist, preferably).  I had not heard of any differences in mass of a plant, but I suppose that it's possible, although the mass difference such a small amount of energy would produce would be nigh-unmeasurable.

As for your question on special relativity, I'm not sure I fully understand what you're trying to convey.  Let me paraphrase and ask you to clarify what I've missed:

1.)  Consider two objects moving away from Earth in opposite directions with speed in Earth's rest frame.

2.)  Viewed from the Earth's rest frame, the speed of either object is , and the separation speed, as you have called it, is .

Now here I don't know what you mean by, "If body A was absolutely at rest, since body B could not be travelling faster than c, ss could not exceed c."  I read this sentence as, "in the rest frame of A, the speed of B cannot exceed c," or, equivalently, "the separation speed, viewed from the rest frame of A, cannot exceed c."  You then argue that the separation speed is greater than c, but this is only true in an intermediate frame (like the Earth's in this example.)  By exactly the reasoning you just gave, the separation speed will be less than c if viewed from the rest frame of either A or B.

You then consider the case where .  You then say, "since A is receding at c, the observer cannot be (absolutely) travelling in the direction of A, else A would be exceeding c."  This is not true.  If A is receding with speed c in any frame, it will be observed from all frames to have speed c, irrespective of the motion of the observer.  In other words, the form of the velocity transformation (which you have written earlier in the document) dictates that, if A is viewed from Earth to be receding with speed c, you cannot infer anything about the (absolute?) speed of Earth.

Your mirror example is also quite confusing to me.  You make the claim that if, say, , that light emitted from B can never reach A (i.e., they are outside of one another's light cone.)  You reach this conclusion by considering the motion from the Earth's rest frame.  But the physical conclusions must be the same in all frames, so let's consider the physics in the rest frame of B.  In this frame, the speed of A is less than c, while light emitted from B, of course, travels at c.  Therefore light emitted from B will catch up to A, and reflect off your mirror (and will, actually, eventually return to B at some later time).  In fact, I think the procedure this describes is called the Einstein clock synchronization procedure, and is true for all particles moving with speed .

Phase Theory wrote:

Thankyou, M@Man, you appear to have answered the first conundrum - clearly I know very little about particle physics; I wouldn't have suspected that particle physics would solve that relativity problem.

Sure, no problem, Phase Theory.  Your reasoning seems to be quite good; there just happen to be facts you don't know.

Phase Theory wrote:

E's frame says 'the light doesn't get there'.  Surely this contradicts the Principle of Relativity!

This is the part that confuses me; it seems to me that, in the Earth's rest frame, the light does get there.  If, in Earth's rest frame, B emits light (traveling at c), and if A is traveling in this frame with speed , then the light from B will reach it.  The fact that the separation speed, as you have defined it (which is something of an awkward quantity - it is the velocity of B relative to A viewed from the rest frame of Earth) is , and can exceed c doesn't mean that light can't catch up to A.  When B emits the light, A has a head start on the light, but it's moving slower.  Therefore the light will catch up to A, even in the Earth's rest frame.  This is equivalent to saying that light doesn't have to "overcome" the velocity of B, which is moving in the opposite direction.  This would be true in Galilean relativity (classical physics), but not here.  Is this clear?

Phase Theory wrote:

Lastly, with regard to the case v->c, because of limits etc., v =/= c.  So surely its observed speed need not be c.
I thought that the 'always observed at c' conclusion only applied to massless particles at c, not to massive particles at lim (v->c).  I'm probably wrong in this - please explain it to me.

You are correct in saying that only massless particles can travel at the speed of light.  However, in these discussions about velocity transformations and twin paradoxes, etc., we usually don't need to specify whether we're talking about a massive or massless particle.  Generally, we're just talking about how certain quantities (like the velocity of A relative to B - your separation speed) appear in different reference frames.  So, really, "A," "B," and "E," need not refer to actual particles, but merely 3 different reference frames that are moving relative to one another. 

As to the validity of taking the limit as for a massive particle, this works just fine.  This is referred to (at least in the electromagnetism course I'm taking) as the ultra-relativistic limit, and you find that the behavior of massive particles in the ultra-relativistic limit is almost identical to the behavior of photons.

Phase Theory wrote:

Incidentally, thankyou for having the patience to go through all this with me.  I hope it doesn't look like I'm trying to argue with you - I believe Relativity is valid (within our current set of observations etc), I just want to understand how.

Not at all; I love hashing this kind of stuff out with other people.  And I can tell that you're really interested in the answers for their own sake; you're not just trying to show off.  In fact, just the last line in your PDF files told me that much:

"Although it is to be expected that this problem has been considered, and furthermore solved, already – and may even be viewed as trivial – the author would, nonetheless, wish to hear of the nature of such a solution."

So, please, bring on your questions.  I won't promise I can answer them, but I'd love to discuss them with you.

-Matt

Friction has nothing to do with this.

Nicholas wrote:

If vacuum particles are forming everywhere and at all times in space then they would create friction to anything moving in it. You would absorb the space energy into your moving mass.

I had to ponder this for a little while, but I think I see what you're saying now.

Actually, a similar phenomenon happens in condensed matter physics.  In a crystal, for instance, you can get conduction of the electrons in the atoms.  But these conduction electrons are not observed with the same mass as free (vacuum) electrons.  Rather, they behave as if they had an "effective mass" rather than .  The reason for this is that the electrons are interacting with all the atoms in the crystal, and these interactions knock the electrons around.  The result is that, if you incorporate the interactions in a kind of average way (mean field theory), the electron "drags a cloud of interactions" with it, which effectively increases its mass.  In your terms, Nicholas, this behaves as a kind of "friction."

With this in mind, your statement about friction against virtual particles in vacuum makes sense.  If there is a menagerie of virtual particles being generated everywhere in free space, then any particle will interact with the virtual particles, dragging a "cloud of interactions" with it, resulting in an effective mass.  The difference between the interactions with the crystal and interactions with the virtual particles is that it is impossible to isolate a particle from the virtual particles in vacuum.  In other words, we know that the mass of the electron, as observed in the crystal, is really an effective mass due to the interactions.  On the other hand, if we measure the mass of a particle in vacuum - supposedly in isolation from the rest of the universe - it really has all kinds of interactions with the virtual particles.  So when we say, "the mass of the electron is ," this is really a statement about the "vacuum effective mass" of the electron interacting with the virtual particles. 

In other words, mass as we normally define it can be equivalently defined as the effective mass due to interactions with the vacuum.  Your "friction" is simply ordinary inertia as observed by Newton.

This, of course, raises the question: if you could isolate a particle from interactions with virtual particles, would it have any mass at all?  I don't know.  Perhaps the question is meaningless, since isolation from virtual particles is impossible.