However, in the case of "energy meridians", of which I am also a skeptic, there does remain the fact that acupuncture is an AMA-approved treatment for several ailments now... even though it cannot be explained with our current understanding, even by the placebo effect. Several points:
1. Even if acupuncture worked, that doesn't constitute evidence for "energy meridians". You can describe a real effect with a completely bogus "explanation".
2. IMHO, AMA approval may not mean what it used to, given the infiltration of non-evidence based medicine into medical schools (e.g., here).
3. I don't know of any "ailments" for which acupuncture is AMA approved, if by that you mean actual diseases. It may be approved as some kind of alternative therapeutic pain relief program (see below), but I'm not sure about that either.
Regarding point 3, about a year ago I did a literature search on PubMED about a year ago, to find out for myself what was the evidential support in for or against acupuncture's efficacy. I found:
4. Acupuncture has not been reliably shown to have any efficacy, beyond a placebo effect, concerning the treatment of any disease.
5. Acupuncture does appear to have a weak but statistically significant ability to temporarily relieve certain kinds of chronic pain. (IIRC, they now even have experimenter-blind versions of the experiment, where they use Hollywood-like mock needles, so the experimenter doesn't know whether they're actually being inserted or not.)
6. However, this appears to have nothing to do with "energy meridians": "fake acupuncture" experiments in which the needles are placed at random or in other patterns, instead of at acupuncture points, also show an equal effect.
7. Nobody knows what is responsible for the acupuncture effect, if it exists (although it's not what acupuncturists say it is), but some have speculated that puncturing the skin releases chemicals which help with pain, or perhaps stimulates nerves in some way to mask the pain.
Antimatter has positive mass and energy. If it had negative mass-energy, then when it annihilates with matter, it would not produce anything; it would cancel out the positive mass-energy of the matter, leaving nothing. Instead, what happens is that a ton of energy is liberated, because the matter and antimatter both have positive mass-energy.
"I've never been 100% clear on this. Is the weak force really infinite but just drops off to effectively-zero faster than electricity and gravity to?"
Not really. Yes really. The Yukawa potential is technically infinite ranged, in that it never drops to zero at any finite distance. But it decreases much faster than a 1/r potential, so it "effectively" drops off to zero after a few e-foldings.
I know demographic factors are crucial and I agree with TFA that some such overlooked factor is likely responsible for the apparent discrepancies.
As for town size, I don't know why you say it's irrelevant. Urban vs. rural districts often have very different political preferences, and town size is a proxy for that distinction.
Example: What if the precincts with higher proportions of Obama supporters happen to be those with hand counted ballots? This is well within the realm of possibility, and from a statistical standpoint, just as likely a hypothesis as wrongdoing. TFA mentioned that some of these analyses have controlled for potential underlying causal variables, e.g., population density, and I think income and education.
The analysis you posted didn't even do that much.
So I think it's premature for you to say anything about whether this is "overblown" or not, until you've done an analysis at least as good as the (admittedly still crude) analyses that have been already done.
So, the gravitational curve that one would normally associate with orbital mechanics and acceleration is (roughly) as I calculated, but other relativistic effects mean that the acceleration needed to stay stationary diverges? I don't know if I would put it that way.
Fundamentally, what is going on is that spacetime inside a black hole becomes so curved that it is no longer stationary. Stationary observers cannot exist within a black hole. The horizon is the limiting case: light can remain stationary there, but no (timelike, massive) observer can. An observer near the horizon will have to boost harder and harder to remain stationary, because he's fighting the spacetime curvature. At the horizon, he loses, because only light can remain there, and no matter how hard he accelerates, he can't reach light speed.
So, although the spacetime curvature remains finite both near and at the horizon, I wouldn't say that the gravitational effects are similar to what you calculated. The acceleration required to stay stationary is much larger near a black hole than your Newtonian calculation suggests. Stable orbits don't even exist near a black hole; orbital mechanics close to a black hole are quite different from Newtonian theory. The spacetime curvature ("tidal forces", "gravitational gradient"), though finite, is very non-Newtonian near an event horizon.
So, what I hear you saying is that the limit of the value of the acceleration as you approach the horizon is finite, but the the value at the horizon is infinite (and the coordinate system fundamentally changes once you enter)(?) No.
Schwarzschild coordinates fundamentally change once you enter the horizon, but not all coordinate systems do (see, e.g., Kruskal-Szekeres coordinates), and the statement I made about the proper acceleration is independent of any coordinate system, by definition.
As you approach the horizon, the proper acceleration diverges — increases without bound — has no finite limit.
(This is true for static observers, i.e., ones who are hovering at a fixed location. Freely falling observers will experience zero proper acceleration, i.e., they are weightless. When I say "as you approach the horizon", I don't mean physically "fall into it", but rather a sequence of static observers at locations successively closer to the horizon. It is that limit that diverges: a static observer at r=2.5 M will feel an acceleration greater than another static observer further out at r=3M; a r=2.3 M observer will feel a still greater acceleration, and so on: an observer at the horizon, r=2M, will experience an infinite proper acceleration. Just graph the function I listed earlier and see its behavior as r->2M.)
However, the gravitational field is finite at the horizon. By "gravitational field" I do not mean a Newtonian force field: I mean the spacetime curvature. (Curvature is coordinate dependent, but you can construct curvature invariants which are not, and they are finite.) The curvature may be interpreted physically as tidal forces, or loosely as a "gravitational gradient".
Dark Energy is really just a term to describe the fact that we can't come up with an expanding universe even if we add up all the known effects that could cause an expansion. You mean an accelerating expansion, not an expanding universe. An expanding universe (with decelerating expansion) is easy to explain with known physics.
I've been looking for a calculation online which goes through in detail the calculation I sketched, so I don't have to rederive it myself. I found this page. The author derives it just with the line elements of the metric, without using the Christoffel symbols as I did.
The formula I gave appears below the text which reads, "Substituting this expression for dr into the above formula gives the proper local acceleration of a stationary observer", except he's using geometric units in which G=c=1. The preceding text is the derivation of that formula.
He also discusses at some length how the gravitational field can be finite at the horizon, yet the proper acceleration of a stationary observer can be infinite, as I stated. A freely falling observer will experience zero proper force at the horizon (or anywhere else where spacetime is not singular). But it requires an infinite boost to go from a freely falling frame to a stationary frame, when you're at the horizon, because the horizon is lightlike, not timelike: only light can be stationary there.
If the surface gravity of a black hole is always infinite, then how does hawkings radiation word? It depends on the fact that the gravity gradient at the event horizon is much larger for small black holes than it is for large ones. The gravity gradient (tidal force) at the event horizon is finite, and decreases for large black holes. However, the surface gravity (proper acceleration of a static observer) is infinite.
I think that you're confusing the fact that equations diverges (which rather implies a problem with the chosen coordinate system) with the actual gravity that would be experienced. No. The expression I gave is the proper acceleration, which is a coordinate independent invariant.
Think about it: if you could hover at the horizon with a finite acceleration g, then with any acceleration a>g, you could escape the horizon. But you cannot escape an event horizon.
Here's a sketch of the calculation leading to the answer I gave:
The acceleration 4-vector 'a' is given by,
a = D_u u
where 'u' is the 4-velocity vector and D_u is the covariant derivative contracted in the 'u' direction. In a coordinate system, this expands to:
a^mu = d(x^mu)^2/dtau^2 + G^mu_nu_rho u^nu u^rho
where x^mu is spacetime position, tau is proper time, G^mu_nu_rho is the Christoffel symbol, and there is an implicit summation over repeated indices.
You can work out the position as a function of time easily enough (spatial position is constant for a static observer) as well as the 4-velocity (only a time component, normalized with the Schwarzschild metric). You need the Chrisftoffel symbols for the metric, which can be calculated in a straightforward if laborious way, or you can look them up.
This will give you the components of the 4-acceleration vector, and then you compute its norm with respect to the Schwarzschild metric to get the coordinate invariant proper acceleration, resulting in the expression I gave.
Way back when I calculate 3 light-months as being the size that would have a surface gravity of 1G. I just don't remember the mass needed... It now looks like something near 800billion solar masses would do the trick. As I just explained, the surface gravity of a black hole is always infinite. Unless now you're talking about non-black holes now.
Well, the Standard Model of particle physics is kind of messy, which indeed is one reason why many theorists would prefer a grand unified theory.
The electromagnetic and weak interactions are actually already unified, into the electroweak interaction. A theory which unifies electroweak with the strong interaction is called a "grand unified theory", or GUT. A theory which unifies a GUT with gravity is called a "theory of everything" (TOE).
There are a number of GUT candidates out there, basically all of them are ordinary Yang-Mills theories like electroweak and strong, but with larger symmetry groups. These more symmetric theories contain electroweak and strong within them, and the two interactions appear distinct from each other at lower energies via spontaneous symmetry breaking. This is analogous to how the electromagnetic and weak interactions look different at low energies but unify into one electroweak theory at higher energies.
As for TOEs, string theory is pretty much the only workable candidate right now.
(There were other, earlier attempts at TOEs using extra dimensions in Kaluza-Klein theory (KK theory), but they were as hard to quantize as general relativity is — which is to say, nobody knows how to do it despite tons of effort, and it may be impossible without modifying the theory as in string theory — and Witten proved that KK theory cannot contain a symmetry group large enough to embed the entire Standard Model.)
Second Creation by Crease and Mann is an excellent history of particle physics which, IIRC, gives some treatment of GUTs at the end. The New Physics, edited by Paul Davies, is a collection of scientific essays on these and many other topics in modern physics. (Well, "modern" as of two decades ago...) The essays are of various difficulty, some aimed at laymen and many aimed at other physicists who aren't in that specific subfield. You might skim it for more of a flavor for what these theories are really about. For string theory, you'll want to read Brian Greene's The Elegant Universe.
The simplest theory of dark energy is just Einstein's cosmological constant, which doesn't make it a new interaction, but rather a peculiar long distance aspect of gravitation. It has also been interpreted as the gravitational effect of the quantum zero point energy.
There are fancier dynamical theories of dark energy in which dark energy is due to a new kind of particle (e.g., quintessence), but there it is still ultimately due to the gravitational effects of this new quantum field.
No. It doesn't take infinite thrust to hover at the Horizon. Yes, it does. This is not a hard calculation, although it takes a while unless you already have the Christoffel symbols handy.
I could rederive it for you if you really wanted, but in general relativity the answer works out to be:
g = GM/r^2 / sqrt(1-2GM/(rc^2))
where g is the "surface gravity" experienced by a stationary observer, r is the Schwarzschild distance, G is Newton's constant, and c is the speed of light.
This expression diverges as r->2GM/c^2, which is the Schwarzschild radius, so the "surface gravity" is infinite there. (I put "surface gravity" in quotes since of course there is no physical surface.) For r much larger than the Schwarzschild radius, it reduces to the Newtonian expression g = GM/r^2.
The Universe may be one meta-supermasive black hole. No. The observable universe may be contained within a supermassive black hole, but the entire universe itself cannot be a black hole; you need to have an inside and an outside to define an event horizon, and there is no "outside the universe".
If your premise were accurate, we should all have been violently sucked into the center of The Universe a long time ago. No. I didn't say that the gravitational pull was infinite everywhere outside the black hole; I said that the proper acceleration of an observer hovering at the horizon itself is infinite.
For a blackhole this size, the event horizon is going to have something in the range of a one light-year radius. Try calculating the Gravity for 18billion solar masses at a distance of one Light year, and see what you get. Your calculation is purely Newtonian, and therefore inapplicable to real black holes.
No. If you try to create gravitational waves (or light waves) and sent them back out through the horizon, they instead fall into the singularity (albeit more slowly than you yourself do as you fall, so you still see them traveling away from you).
The surface gravity at the event horizon of a black hole is always infinite: it requires infinite thrust to hover at a horizon. (Technically: proper acceleration for a stationary observer diverges at the Schwarzschild radius.) I think you were doing a Newtonian calculation, not a relativistic calculation.
(The tidal force, however, is finite at the horizon, and of course you experience no weight if you're freely falling through the horizon.)
Well, some people are wrong. Tom van Flandern is a well known crackpot from Usenet, sci.physics.relativity. Numerous posters there explained his mistakes there, if you want to search Google Groups. It turns out that you can apply his argument for FTL gravity to similarly "prove" that light itself travels faster than light in Maxwellian electromagnetism. Both conclusions are based on the same error ignoring the source velocity depedence of the field.
That means that you concede because I persist in paying attention to what you're saying, No, jackass. Every time you've responded, you've completely ignored what I said.
Well, since then you've been talking to me, and if I wanted to say something to that "earlier poster" you haven't specified, I'd be replying to its lies instead of to yours. The earlier poster is the one I first responded to, moron. Right here.
To recap:
Original poster implied that dark matter can have no detectable gravitational effects. I replied that the whole point of dark matter is that if it exists, it has gravitational effects. You responded, irrelevantly, that dark matter may not exist. Which is not my point. My point is, for the fourth time, that IF DARK MATTER EXISTS, IT HAS GRAVITATIONAL EFFECTS. I don't care about whatever stupid point you'd rather make instead. It has nothing to do with the point I was trying to make. Rinse, repeat.
You keep responding to a claim that I never made. So thank you for threadjacking my response to someone else with random B.S. that has nothing to do with anything I was talking about.
It is to you, now, to support the assertions that your pet theory more robustly explains those phenomena than does Wiltshire's. Dark matter is not my "pet theory". And it's easy to demonstrate, since now that I recall what TFA was originally about, Wiltshire's theory isn't even an alternative to dark matter: it's an alternative to dark energy, therefore it accounts for NONE of the phenomena that dark matter does.
The curvature outside the event horizon is influenced by the curvature at the horizon itself; the horizon can be treated as the external boundary condition. Nothing that happens inside changes the horizon, but the "frozen" horizon influences the gravity outside.
Matter traveling near the event horizon's surface can distort the horizon's shape and therefore the external gravity, but you don't need any matter to "keep the horizon intact and spacetime curved". Once the spacetime at and outside the horizon is curved, it will stay curved.
Hey, I'm not saying the science is wrong, I'm just saying I don't understand it. Ok, but I don't know what's so different about gravitons that makes you not understand them, if you can understand photons.
Photons I can understand. They are created at light sources, can be diffracted through gases, bounce off of solids, you can direct beams of light with a mirror, focus them with a lens, split them out into individual colors with prisms, etc. You can do a lot of those things with gravitons, too. You can diffract and refract them, scatter them, etc. Things are somewhat different though because electromagnetic fields can both attract and repel while gravitational fields only attract, so gravity interacts with matter differently than does electromagnetism.
But now when you say that gravity is transmitted by a particle, just how does that happen? The same way that electromagnetism is transmitted by a particle: matter changes its energy state when it adsorbs or radiates a graviton (or photon).
An atom of hydrogen is radiating graviton particles? Sure, just like it radiates photons. A hydrogen atom has mass, which means that it has gravity, which means that it can radiate gravitons. If it's just sitting there, it doesn't radiate any detectable gravitons, just virtual gravitons. If you shake it, it radiates gravitational waves in the form of real gravitons (in the graviton picture, anyway). That's the same as a charged particle sitting there radiating virtual photons which create its electric field, vs. shaking a charged particle to radiate electromagnetic waves which are light.
Do gravitons have mass? No. (Nor do photons.)
How do they attract another atom of hydrogen to the first? The two atoms exchange gravitons which alter their momentum/energy.
But with the graviton being a particle, can you reflect it with a graviton mirror? Ordinary matter doesn't reflect gravity much, but you can do it with a black hole... most gets scattered, some gets absorbed, a little gets reflected.
Gravity is really weak so you need a strong gravitational field to affect gravitons much.
Can you focus it with a lens? An ordinary lens would do little, but you could again use a black hole. Black holes focus both photons and, presumably, gravitons.
Can you create gravitons artificially, independent of simply producing a sufficient mass of matter? I don't know what you mean by creating gravitons "artificially". Can you create photons artificially?
Any time you shake a charged particle, it creates photons. Any time you shake any kind of particle, it creates gravitons. (Well, technically, you need a time-varying dipole moment for the former, and quadrupole moment for the latter.)
It's the same confusion I have when they talk about photons behaving as waves and particles. I understand that's what the scientists are saying, I can put the answer down on the test, but the fact is fundamentally confusing. Feynman has a nice description of that in the book Six Easy Pieces which might help you understand it a little better.
If that's the way it works, that's the way it works, but it doesn't make any sense to me. Again, I'm not sure what's not making sense to you. If you replaced the Sun-Earth system with an atom and asked about how the orbit of an electron would change due to the nucleus disappearing, would you expect it to happen instantly or only after a lightspeed delay? (That would replace gravity with electric force and gravitons with photons.) i.e., is your conceptual problem with gravity vs. electromagnetism, or is it with relativity, or what?
Gravitational fields do have mass, but your overall point is right: the curvature of spacetime outside the event horizon doesn't depend on the curvature inside the horizon; there is nothing that is propagating or "escaping" from inside to outside.
Slashdot Mirror
1. Even if acupuncture worked, that doesn't constitute evidence for "energy meridians". You can describe a real effect with a completely bogus "explanation".
2. IMHO, AMA approval may not mean what it used to, given the infiltration of non-evidence based medicine into medical schools (e.g., here).
3. I don't know of any "ailments" for which acupuncture is AMA approved, if by that you mean actual diseases. It may be approved as some kind of alternative therapeutic pain relief program (see below), but I'm not sure about that either.
Regarding point 3, about a year ago I did a literature search on PubMED about a year ago, to find out for myself what was the evidential support in for or against acupuncture's efficacy. I found:
4. Acupuncture has not been reliably shown to have any efficacy, beyond a placebo effect, concerning the treatment of any disease.
5. Acupuncture does appear to have a weak but statistically significant ability to temporarily relieve certain kinds of chronic pain. (IIRC, they now even have experimenter-blind versions of the experiment, where they use Hollywood-like mock needles, so the experimenter doesn't know whether they're actually being inserted or not.)
6. However, this appears to have nothing to do with "energy meridians": "fake acupuncture" experiments in which the needles are placed at random or in other patterns, instead of at acupuncture points, also show an equal effect.
7. Nobody knows what is responsible for the acupuncture effect, if it exists (although it's not what acupuncturists say it is), but some have speculated that puncturing the skin releases chemicals which help with pain, or perhaps stimulates nerves in some way to mask the pain.
Antimatter has positive mass and energy. If it had negative mass-energy, then when it annihilates with matter, it would not produce anything; it would cancel out the positive mass-energy of the matter, leaving nothing. Instead, what happens is that a ton of energy is liberated, because the matter and antimatter both have positive mass-energy.
Not really. Yes really. The Yukawa potential is technically infinite ranged, in that it never drops to zero at any finite distance. But it decreases much faster than a 1/r potential, so it "effectively" drops off to zero after a few e-foldings.
See this reprint of an 2006 article published in Science, in particular this graph.
Basically, only Turkey does worse than the U.S. in terms of what fraction of the public accepts that humans evolved from earlier species of animals.
I know demographic factors are crucial and I agree with TFA that some such overlooked factor is likely responsible for the apparent discrepancies.
As for town size, I don't know why you say it's irrelevant. Urban vs. rural districts often have very different political preferences, and town size is a proxy for that distinction.
The analysis you posted didn't even do that much.
So I think it's premature for you to say anything about whether this is "overblown" or not, until you've done an analysis at least as good as the (admittedly still crude) analyses that have been already done.
"Face Facts"? Your post is curiously lacking in them. Perhaps you'd like to offer some "facts" for us to "face"?
Fundamentally, what is going on is that spacetime inside a black hole becomes so curved that it is no longer stationary. Stationary observers cannot exist within a black hole. The horizon is the limiting case: light can remain stationary there, but no (timelike, massive) observer can. An observer near the horizon will have to boost harder and harder to remain stationary, because he's fighting the spacetime curvature. At the horizon, he loses, because only light can remain there, and no matter how hard he accelerates, he can't reach light speed.
So, although the spacetime curvature remains finite both near and at the horizon, I wouldn't say that the gravitational effects are similar to what you calculated. The acceleration required to stay stationary is much larger near a black hole than your Newtonian calculation suggests. Stable orbits don't even exist near a black hole; orbital mechanics close to a black hole are quite different from Newtonian theory. The spacetime curvature ("tidal forces", "gravitational gradient"), though finite, is very non-Newtonian near an event horizon.
Schwarzschild coordinates fundamentally change once you enter the horizon, but not all coordinate systems do (see, e.g., Kruskal-Szekeres coordinates), and the statement I made about the proper acceleration is independent of any coordinate system, by definition.
As you approach the horizon, the proper acceleration diverges — increases without bound — has no finite limit.
(This is true for static observers, i.e., ones who are hovering at a fixed location. Freely falling observers will experience zero proper acceleration, i.e., they are weightless. When I say "as you approach the horizon", I don't mean physically "fall into it", but rather a sequence of static observers at locations successively closer to the horizon. It is that limit that diverges: a static observer at r=2.5 M will feel an acceleration greater than another static observer further out at r=3M; a r=2.3 M observer will feel a still greater acceleration, and so on: an observer at the horizon, r=2M, will experience an infinite proper acceleration. Just graph the function I listed earlier and see its behavior as r->2M.)
However, the gravitational field is finite at the horizon. By "gravitational field" I do not mean a Newtonian force field: I mean the spacetime curvature. (Curvature is coordinate dependent, but you can construct curvature invariants which are not, and they are finite.) The curvature may be interpreted physically as tidal forces, or loosely as a "gravitational gradient".
I've been looking for a calculation online which goes through in detail the calculation I sketched, so I don't have to rederive it myself. I found this page. The author derives it just with the line elements of the metric, without using the Christoffel symbols as I did.
The formula I gave appears below the text which reads, "Substituting this expression for dr into the above formula gives the proper local acceleration of a stationary observer", except he's using geometric units in which G=c=1. The preceding text is the derivation of that formula.
He also discusses at some length how the gravitational field can be finite at the horizon, yet the proper acceleration of a stationary observer can be infinite, as I stated. A freely falling observer will experience zero proper force at the horizon (or anywhere else where spacetime is not singular). But it requires an infinite boost to go from a freely falling frame to a stationary frame, when you're at the horizon, because the horizon is lightlike, not timelike: only light can be stationary there.
Think about it: if you could hover at the horizon with a finite acceleration g, then with any acceleration a>g, you could escape the horizon. But you cannot escape an event horizon.
Here's a sketch of the calculation leading to the answer I gave:
The acceleration 4-vector 'a' is given by,
a = D_u u
where 'u' is the 4-velocity vector and D_u is the covariant derivative contracted in the 'u' direction. In a coordinate system, this expands to:
a^mu = d(x^mu)^2/dtau^2 + G^mu_nu_rho u^nu u^rho
where x^mu is spacetime position, tau is proper time, G^mu_nu_rho is the Christoffel symbol, and there is an implicit summation over repeated indices.
You can work out the position as a function of time easily enough (spatial position is constant for a static observer) as well as the 4-velocity (only a time component, normalized with the Schwarzschild metric). You need the Chrisftoffel symbols for the metric, which can be calculated in a straightforward if laborious way, or you can look them up.
This will give you the components of the 4-acceleration vector, and then you compute its norm with respect to the Schwarzschild metric to get the coordinate invariant proper acceleration, resulting in the expression I gave.
Well, the Standard Model of particle physics is kind of messy, which indeed is one reason why many theorists would prefer a grand unified theory.
...) The essays are of various difficulty, some aimed at laymen and many aimed at other physicists who aren't in that specific subfield. You might skim it for more of a flavor for what these theories are really about. For string theory, you'll want to read Brian Greene's The Elegant Universe.
The electromagnetic and weak interactions are actually already unified, into the electroweak interaction. A theory which unifies electroweak with the strong interaction is called a "grand unified theory", or GUT. A theory which unifies a GUT with gravity is called a "theory of everything" (TOE).
There are a number of GUT candidates out there, basically all of them are ordinary Yang-Mills theories like electroweak and strong, but with larger symmetry groups. These more symmetric theories contain electroweak and strong within them, and the two interactions appear distinct from each other at lower energies via spontaneous symmetry breaking. This is analogous to how the electromagnetic and weak interactions look different at low energies but unify into one electroweak theory at higher energies.
As for TOEs, string theory is pretty much the only workable candidate right now.
(There were other, earlier attempts at TOEs using extra dimensions in Kaluza-Klein theory (KK theory), but they were as hard to quantize as general relativity is — which is to say, nobody knows how to do it despite tons of effort, and it may be impossible without modifying the theory as in string theory — and Witten proved that KK theory cannot contain a symmetry group large enough to embed the entire Standard Model.)
Second Creation by Crease and Mann is an excellent history of particle physics which, IIRC, gives some treatment of GUTs at the end. The New Physics, edited by Paul Davies, is a collection of scientific essays on these and many other topics in modern physics. (Well, "modern" as of two decades ago
The simplest theory of dark energy is just Einstein's cosmological constant, which doesn't make it a new interaction, but rather a peculiar long distance aspect of gravitation. It has also been interpreted as the gravitational effect of the quantum zero point energy.
There are fancier dynamical theories of dark energy in which dark energy is due to a new kind of particle (e.g., quintessence), but there it is still ultimately due to the gravitational effects of this new quantum field.
By the way, an 18 billion solar mass black hole has a Schwarzschild radius of 0.00561932487 lightyears, or 355.36426 AU (calculation).
I could rederive it for you if you really wanted, but in general relativity the answer works out to be:
g = GM/r^2 / sqrt(1-2GM/(rc^2))
where g is the "surface gravity" experienced by a stationary observer, r is the Schwarzschild distance, G is Newton's constant, and c is the speed of light.
This expression diverges as r->2GM/c^2, which is the Schwarzschild radius, so the "surface gravity" is infinite there. (I put "surface gravity" in quotes since of course there is no physical surface.) For r much larger than the Schwarzschild radius, it reduces to the Newtonian expression g = GM/r^2. The Universe may be one meta-supermasive black hole. No. The observable universe may be contained within a supermassive black hole, but the entire universe itself cannot be a black hole; you need to have an inside and an outside to define an event horizon, and there is no "outside the universe". If your premise were accurate, we should all have been violently sucked into the center of The Universe a long time ago. No. I didn't say that the gravitational pull was infinite everywhere outside the black hole; I said that the proper acceleration of an observer hovering at the horizon itself is infinite. For a blackhole this size, the event horizon is going to have something in the range of a one light-year radius. Try calculating the Gravity for 18billion solar masses at a distance of one Light year, and see what you get. Your calculation is purely Newtonian, and therefore inapplicable to real black holes.
No. If you try to create gravitational waves (or light waves) and sent them back out through the horizon, they instead fall into the singularity (albeit more slowly than you yourself do as you fall, so you still see them traveling away from you).
The surface gravity at the event horizon of a black hole is always infinite: it requires infinite thrust to hover at a horizon. (Technically: proper acceleration for a stationary observer diverges at the Schwarzschild radius.) I think you were doing a Newtonian calculation, not a relativistic calculation.
(The tidal force, however, is finite at the horizon, and of course you experience no weight if you're freely falling through the horizon.)
P.S. Here's a paper detailing van Flandern's error — one which, nearly a decade after it was disproven, he is still relying upon.
Well, some people are wrong. Tom van Flandern is a well known crackpot from Usenet, sci.physics.relativity. Numerous posters there explained his mistakes there, if you want to search Google Groups. It turns out that you can apply his argument for FTL gravity to similarly "prove" that light itself travels faster than light in Maxwellian electromagnetism. Both conclusions are based on the same error ignoring the source velocity depedence of the field.
To recap:
Original poster implied that dark matter can have no detectable gravitational effects.
I replied that the whole point of dark matter is that if it exists, it has gravitational effects.
You responded, irrelevantly, that dark matter may not exist. Which is not my point. My point is, for the fourth time, that IF DARK MATTER EXISTS, IT HAS GRAVITATIONAL EFFECTS. I don't care about whatever stupid point you'd rather make instead. It has nothing to do with the point I was trying to make.
Rinse, repeat.
You keep responding to a claim that I never made. So thank you for threadjacking my response to someone else with random B.S. that has nothing to do with anything I was talking about. It is to you, now, to support the assertions that your pet theory more robustly explains those phenomena than does Wiltshire's. Dark matter is not my "pet theory". And it's easy to demonstrate, since now that I recall what TFA was originally about, Wiltshire's theory isn't even an alternative to dark matter: it's an alternative to dark energy, therefore it accounts for NONE of the phenomena that dark matter does.
The curvature outside the event horizon is influenced by the curvature at the horizon itself; the horizon can be treated as the external boundary condition. Nothing that happens inside changes the horizon, but the "frozen" horizon influences the gravity outside.
Matter traveling near the event horizon's surface can distort the horizon's shape and therefore the external gravity, but you don't need any matter to "keep the horizon intact and spacetime curved". Once the spacetime at and outside the horizon is curved, it will stay curved.
Gravity is really weak so you need a strong gravitational field to affect gravitons much. Can you focus it with a lens? An ordinary lens would do little, but you could again use a black hole. Black holes focus both photons and, presumably, gravitons. Can you create gravitons artificially, independent of simply producing a sufficient mass of matter? I don't know what you mean by creating gravitons "artificially". Can you create photons artificially?
Any time you shake a charged particle, it creates photons. Any time you shake any kind of particle, it creates gravitons. (Well, technically, you need a time-varying dipole moment for the former, and quadrupole moment for the latter.) It's the same confusion I have when they talk about photons behaving as waves and particles. I understand that's what the scientists are saying, I can put the answer down on the test, but the fact is fundamentally confusing. Feynman has a nice description of that in the book Six Easy Pieces which might help you understand it a little better. If that's the way it works, that's the way it works, but it doesn't make any sense to me. Again, I'm not sure what's not making sense to you. If you replaced the Sun-Earth system with an atom and asked about how the orbit of an electron would change due to the nucleus disappearing, would you expect it to happen instantly or only after a lightspeed delay? (That would replace gravity with electric force and gravitons with photons.) i.e., is your conceptual problem with gravity vs. electromagnetism, or is it with relativity, or what?
Gravitational fields do have mass, but your overall point is right: the curvature of spacetime outside the event horizon doesn't depend on the curvature inside the horizon; there is nothing that is propagating or "escaping" from inside to outside.