elfmotat

It might be "strange" or "alien" or "unintuitive," but QM is certainly not "illogical."

Did you even read what he quoted, or were you just outright put-off by the fact that he bothered to disagree with your nonsense? The speed of light certainly isn't observer-dependent, as you claimed. This is the core assumption of relativity.

You need to be careful when you're talking about what's "observed" in relativity. "Observe" is not at all synonymous will "seen." What you can actually see with your eyes can drastically differ from what we know is "actually" happening. We can calculate that in-falling matter will cross the event horizon in finite proper time. What we see is light reflected from said matter until it is red-shifted enough to be undetectable. But if one believes GR is an accurate description of gravity at macroscopic scales, then one necessarily believes that the matter indeed crossed the event horizon. Science itself undoubtedly gets a bit fuzzy here, since we are extrapolating our model past the point where it makes predictions we can test - namely that objects cross the event horizon in finite time. We can't ever know for sure that this is true, because an object that has already crossed the horizon is forever causally disconnected with the rest of the universe. The problem with your question is that it begs another question: "flow rate of time" according to who? An observer "hovering" a short distance outside the event horizon? An observer free-falling toward the BH? An observer an infinite distance away from the BH? The second two are easy: The free-falling observer measures no change in the "rate of time" as he falls toward (and eventually through) the event horizon. The observer at infinity measures in-falling matter to experience asymptotically zero time as it approaches the horizon. The first is harder. I'll do the calculation later when I have time. Undoubtedly changes in the metric would propagate at c. I'd imagine the event horizon would move outward at c, perhaps even "before" the in-falling matter has crossed it (but don't quote me on that). Okay, according to an outside observer. I'm not saying "it doesn't matter that it should theoretically come to a stop," I'm saying that it doesn't theoretically come to a stop. The main point of my post, however, was that the Schwarzschild solution is no longer valid when in-falling matter is significant enough to perturb the metric. As for what the correct solution would look like for a massive body falling towards a black hole, I don't really have any idea. I imagine somebody's done research into the topic before, though I don't know of the papers myself.

What's your background in math? My recommendation for someone who knows vector calculus is going to be very different from my recommendation for someone who doesn't know what a derivative is.

I think you may be a little confused. I certainly don't know of any texts that take a difference in velocity to be the definition of "relative velocity." Einstein's certainly didn't, despite that quote you misunderstood. The Wikipedia page on relative velocity uses the same definition as me: http://en.wikipedia.org/wiki/Relative_velocity .

When people say "time stops according to an outside observer" what they are referring to is the fact the 00 component of the metric in Schwarzschild coordinates is zero at the event horizon. This is merely an artifact of the particular coordinate system, and is called a "coordinate singularity." The fact that Schwarzschild coordinates can't describe matter crossing the event horizon is physically meaningless. If you do the calculation in a coordinate system which covers the entire spacetime, such as Kruskal coordinates, you find that in-falling matter crosses the event horizon in finite proper time for an outside observer. Now, all of this was assuming the in-falling matter was small enough so that it didn't significantly affect the space-time in the first place. If the matter falling into the BH is significant enough to perturb the metric, what you really need is a new solution to the Einstein Field Equations which describes the space-time in the scenario. I.e. the typical vacuum BH solutions (Schwarzschild, Kerr, etc.) are no longer valid anyway, so now it really doesn't make sense to apply Schwarschild coordinate time to the scenario.

As far as electrons go, they have nonzero mass so they cannot possibly travel at c. For massless particles like photons which do travel at c, their momentum is independent of their velocity. Their momentum is related to their energy/frequency.

Separation velocity is the rate at which the distance between two objects increases in an arbitrary frame. Separation velocities adds linearly, like velocities did in Galilean relativity. For example, if a red ball is moving in the +x direction with speed vred, and a blue ball is moving in the -x direction with speed vblue, then the separation velocity between the red and blue balls is just vred+vblue. Since nothing can travel faster than c, separation velocity cannot possibly be equivalent to relative velocity because all we would need to do to have objects moving at >c is let the velocities of the balls be some significant fraction of c. The relative velocity between the balls (i.e. the speed of one ball as measured by the other) is found via Einstein velocity addition, and never exceeds c.

This simply isn't true. A cone is still "flat" because you can unfold it and it will lie flat on a table with no bumps. If you draw a triangle on a cone, its angles will add up to 180°. A sphere, on the other hand, has "intrinsic curvature." No matter how hard you try, you won't be able to make an unfolded sphere lie flat on a table. The curvature of General Relativity is intrinsic curvature. Spacetime doesn't need to be embedded in an extradimensional space for there to be curvature.

In cosmological models of the universe in General Relativity, for example the FLRW metric, positive "overall" curvature refers to spherical geometry and negative curvature refers to hyperbolic geometry. In a spherical universe, the angles in all triangles add up to greater than 180°. In a hyperbolic universe, the angles in a triangle add up to less than 180°. A flat universe's triangles have angles that add up to exactly 180°. So a way to figure out the "shape" of the universe would be to measure the angles in triangles. We've conducted such experiments, and so far we find that the universe is approximately flat. If the universe does have some curvature to it, it's too small to detect with current methods.

http://www.desy.de/user/projects/Physics/Relativity/BlackHoles/black_fast.html

This is one of the strangest things I've ever read...

You can't just plug relativistic mass into Newtonian formulas. That's not how gravity works at all relativistically. Objects do not collapse into black holes due to kinetic energy alone, or the paradox I mentioned above would ensue.

1. Pick an object traveling at .9999999999999c. 2. Boost to object's rest frame. 3. Object is at rest in this frame, and therefore has zero kinetic energy. So according to you, it collapses into a black hole in one frame but doesn't in another. Obviously this makes no sense.

Not really, no. Spacetime is certainly dynamical, but I don't think classifying it as a "substance" is either meaningful nor accurate.

While what you say is certainly true, the point of the exercise was to figure out how both perspectives can be correct according to SR, yet seemingly be contradictory. Since you are apparently unable to answer, I'll do it for you: in the meter stick's rest frame, the plate is tilted at an angle [math]\phi'=tan^{-1}(v_y v_{rel}\gamma_{rel}/c^2)[/math] with respect to the x-axis. This is because of the relativity of simultaneity - if all points on the plate cross the x-axis at t=0 in one frame, they cannot possibly do so in any other frame. In this frame the plate passes over the meter stick at a tilted angle, so it doesn't matter what is length contracted and what isn't. As someone who apparently "understands" relativity without understanding the math, why is it that you couldn't predict this phenomenon? I'm sure you already know that the answer to this is "at the event horizon." The proper acceleration required to "hover" outside of a black hole is: [math]a=\frac{GM}{r^2\sqrt{1-r_s/r}}[/math] You can look at this as the equivalent of the Newtonian "g" value (which at Earth's surface, for example, is ~9.8 m/s2). As you can see, as you approach the Schwarzschild radius (the event horizon), the value of [math]a[/math] asymptotically approaches infinity. There isn't some "mysterious" (as you like to call it) jump discontinuity where, all of a sudden, [math]a[/math] shoots to infinity - it's an asymptote. The free-falling object and the object accelerating away from the BH act completely differently from the very beginning. The free-falling object experiences zero proper acceleration, and is moving toward a region of spacetime with more and more curvature. The other object experiences nonzero proper acceleration and is approaching an asymptotically flatter region of spacetime. They're in no way equivalent. Time dilation and length contraction by themselves aren't well-defined in GR. You may get analogous effects in particular solutions to the EFE's, but there's no direct generalization of TD and LC to curved spacetime. They don't contradict each other. Proper time corresponds to what clocks actually measure in GR. It is the only physically meaningful type of time. All objects cross the even horizon in finite proper time according to all observers a finite distance from the BH. Schwarzschild "coordinate time" is not physically meaningful - it's just a coordinate. It actually corresponds to the proper time of an observer at an infinite distance away from the BH. A consequence of using this particular type of coordinate system is that there's a coordinate singularity at the event horizon. There's no such thing as "correct" coordinates, because coordinates are not physically meaningful. It obviously doesn't matter what I decide to call a particular set of points. Why would it? Treating coordinates as physical "things" is absurd! They're just labels! Your question is equivalent to asking how Cartesian and polar coordinates can both be "correct." It doesn't make any sense. I'm not really sure what you're asking here. Could you try rephrasing the question? Because what you just said is patently false. I really don't understand your logic here. More distant observers can't have objects crossing the horizon because then the objects would have to escape the horizon? What? How did you come up with that? The object, and the part of the rope that crossed the event horizon will fall toward the singularity. You'd be pretty hard-pressed to find a rope that can stay together under that much tension in the first place. As I already explained, all observers have the object crossing the even horizon in finite proper time. You're applying Newtonian gravity to a clearly relativistic scenario, so you shouldn't be surprised that you're getting nonsensical answers. I already answered this question in detail in #1. I have no idea what you're trying to say here. A black hole is a 3D sphere, not a cone (where'd you get that from?) or a 4D sphere. An angular time coordinate (which you are implying with your "4D sphere" nonsense) implies closed timelike curves. What? Why would it be gone? This is simply not true, as I've probably explained to you half a dozen times already. As I said, LC and TD are not well-defined in GR. An object's proper length and proper time are unaffected by your choice of coordinates. Since there's a coordinate singularity at the event horizon in Schwarzschild coordinates, trying to use them to describe an object crossing the horizon is nonsensical (which is what you're apparently trying to do). I've never heard of a "speed of light horizon" before, but I think I know what you're trying to say. There is no Rindler horizon for a free-falling object, because a free-falling object isn't really accelerating. A free-falling object is essentially inertial, and light also travels along geodesics. 1. Because it's not equivalent. Free-falling objects experience zero proper acceleration, whereas accelerating objects in flat spacetime obviously experience nonzero proper acceleration. 2. Because the Rindler horizon is an effect in flat spacetime. There's no reason to think that it would put any kind of restrictions on the behavior of objects in curved spacetime. Because they are physically different. Here are two experiments for you to try so you can prove it to yourself: 1. Get in car. 2. Accelerate. 3. Feel yourself being pushed back in your seat (i.e. nonzero proper acceleration). 1. Find cliff. 2. Jump off cliff. 3. Feel weightless (i.e. zero proper acceleration). The first experiment is a curved path in spacetime. The second scenario is geodesic motion in spacetime. In one of them you feel and acceleration, and in the other you don't. If free-fall was equivalent to proper acceleration then, for example, astronauts in orbit wouldn't feel weightless. Because it is self consistent? I have a better question: do you really think you're smarter than nearly 100 years of scientists researching and testing GR? Seriously, do you honestly believe that you, someone who barely even knows what a tensor is, just shattered a century's worth of scientific consensus? Or do you think it's maybe just a little more likely that you don't understand GR, that you don't know what you're talking about, and that you're not as smart as you think you are?

Not by itself, no.

Frame dragging is separate from gravitational radiation. A spherically symmetric rotating body will produce frame dragging effects, but will not radiate.

Here's a URL to the pic: http://i1.minus.com/jZct9VvGyOv6n.png Did you even read the question? In the rest frame of the meter stick, the plate's velocity has nonzero x-component and is therefore length-contracted in the x-direction. So the hole's diameter is shorter than one meter in the meter stick's rest frame. The point of the exercise is to reconcile the seemingly paradoxical disagreement between the two frames. . I probably should have mentioned that this exercise was taken from Taylor & Wheeler's Spacetime Physics.

That sounds about right to me.

Using the linearized field equations in a vacuum, we can get: [math]\left ( \nabla^2-\frac{\partial^2}{\partial t^2} \right )\bar{h}^{\mu \nu}=0[/math] where [math]\bar{h}^{\mu \nu}=h^{\mu \nu}-\frac{1}{2} \eta^{\mu \nu } h^\sigma_{~\sigma}[/math], and [math]g_{\mu \nu }=\eta_{\mu \nu}+h_{\mu \nu}[/math]. This is just the wave equation for [math]\bar{h}^{\mu \nu}[/math]. So we see that at first order, gravitational waves are just sinusoidal perturbations of the metric which propagate at c. Also note that gravitational waves are generated by quadrupoles. A rotating symmetric sphere will not generate gravitational waves, for example.

Tachyons (nonexistent particles that travel faster than c) have imaginary mass. But they don't actually exist (or there's absolutely zero evidence for them at least), so it's a moot point.

If you understand relativity then you should have no issues at all resolving the following "paradox": A meter stick lies along the x-axis of the laboratory frame and approaches the origin with velocity vrel. A very thin plate parallel to the xz laboratory plane moves upward in the y-direction with speed vy as shown in the figure. The plate has a circular hole with a diameter of one meter centered on the y-axis. The center of the meter stick arrives at the laboratory origin at the same time in the laboratory frame as the rising plate arrives at the plane y=0. Since the meter stick is Lorentz-contracted in the laboratory frame it will easily pass through the hole in the rising plate. Therefore there will be no collision between meter stick and plate as each continues its motion. However, someone who objects to this conclusion can make the following argument: "In the frame in which the meter stick is at rest the meter stick is not contracted, while in this frame the hole in the plate is Lorentz-contracted. Hence the full-length meter stick cannot possibly pass through the contracted hole in the plate. Therefore there must be a collision between the meter stick and the plate." Answer unequivocally the question, will there be a collision between the meter stick and the plate?

That's interesting. Sort of a "multiversal" evolution, where the black hole trait is selected for.

As far as we know, it doesn't. There's as yet no successful theory of quantum gravity, so this question can't really be answered in full. Your question can quite easily be answered without any QM, however. Classically, the strength gravitational field scales linearly with mass: [math]g=-GM/r^2[/math]. A small body 1 mile away from a large mass M will experience half of the force due to gravity that a small body 1 mile away from a mass 2M experiences. Also, while asking "why" questions, it's always useful to have watched this: .