Topic by Tub

Posted: 6/1/2012 4:21 PM

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Tub

Joined: 6/25/2005
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Warp wrote:

My original question was about the expanding universe and overlapping observable universes.

I haven't been able to find the original question in the last few pages, but there's nothing wrong with expansion within relativity, and the quote you provided is a good explanation of light cones and causality. Now you're asking what happens if they move faster than c. The answer is: they don't. Not within relativity. If they do, you're using a different set of definitions and formulas, and everything you know about relativity is wrong; the diagram doesn't apply any more. So you're either looking for an explanation within relativity, in which there is no receding faster than c. Or you're looking for an explanation in which they recede faster than c, in which case you won't find the answer in relativity, but in the formulas outlined behind the link I gave you. Let's try within relativity. You could sit on the star in the middle and notice that A is receding with 0.9c and B is receding with 0.9c in the opposite direction. That doesn't mean they're receding with 1.8c from each other - if an observer on A could see B, it'd still measure a speed less than c. So since we can see A and B, we're obviously in the uAB-part of the diagram you linked, and A and B are in our past. Let's say that the "probe" is actually a radio signal and travels with c - it would reach us along the light. (If it's a probe below, but sufficiently close to c, it'll reach us a bit later.) We can retransmit the signal, and eventually it would reach a B' in the future. Unless B accelerates away from us. B will always be slower than the radio signal, yet the signal will never reach B. How is that possible? Let's do an example in simple newtonian physics. You have a person running away at 900 km/h and accelerating to a top speed of 1000 km/h, halving the distance to his top speed each hour. So he'd start at 900 km/h, reach 950 km/h after one hour, 975 after 2, 987.5 after 3 etc. We're firing a bullet at 1000 km/h. It's unstoppable, will never slow down, and is heat seeking. Screw the guy. During the first hour, the bullet is at most 100 km/h faster and can catch up at most 100km. During the second hour, the bullet is at most 50 km/h faster, and can catch up at most 50km. During the third hour, 25km. 12.5km. 6.25km, ... This is an infinite sum, but with a finite value. Adding these up, you'll find a limit of 200km. So if the guy had a 200 km head start, he'd live. Now replace "1000km" with "c" in your head, and you have an idea why the observable universe can shrink. It's not because objects move away from us faster than c (they don't), but it's because they're accelerating away from us. Now queue someone to bash me on the head and substitute the real formulas.

Warp wrote:

Can't someone just answer the questions and help me understand the answers?

Sure, provided there is a short and easy answer..

m00

Posted: 6/1/2012 6:37 PM

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Warp

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Tub wrote:

Now you're asking what happens if they move faster than c.

They do not move faster than c. The distance between them grows faster than c. There's a crucial difference.

The answer is: they don't. Not within relativity.

What do you mean by "relativity"? General Relativity allows (and predicts) the distance between two objects increasing faster than c, for example because of the metric expansion of the universe, and this is accepted to be happening to our universe right now (it's estimated that the majority of the universe is so far from us that it's receding from us faster than c). This was a question about General Relativity.

Posted: 6/1/2012 8:06 PM

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p4wn3r

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I also didn't see your original question, Warp, so I thought you were asking about the spacetime diagram. Anyway, that diagram you quoted will only work if (1) you're in an inertial frame; and (2) gravitational effects aren't significant enough to distort spacetime. One could develop some expansion model for SR, but it would contradict observational data. However, this doesn't involve things moving faster than c. SR holds for global inertial frames, GR was developed because those frames were shown not to exist. In fact, for some galaxies, SR can be a good local approximation. The result that c is the maximum speed in SR occurs only if you're at an inertial frame. If you assume that everything is expanding with some acceleration, our frame here at Earth is being accelerated and, since the postulates of SR hold for inertial ones, observing something receding faster than c does not contradict SR. Period. Does this mean that we can observe something outrun a light beam? No, if we see something moving away at -v, then light sent towards us at c - v. Now, you could argue that if the receding speed v is larger than c, light is moving away from us, so we don't receive it. However, this doesn't necessarily happen. Look at this:

You only need to look at the red line, the light cone (or observable universe, it's corrected to take into account GR and our acceleration) and the line called hubble distance (it should have been drawn symmetrically on the other side, but the author omitted it). Anyway, all events below the Hubble line are receding from us faster than c. Notice though, that some of these are inside the light cone and, thus, can send light beams at us. I haven't read your apparent paradox thoroughly, but it seems that it arises because you assume that the light cone must be contained above the Hubble line. A more intuitive description is that the Hubble sphere describes what's receding from us exactly at c. As time passes, the sphere will recede and so will the photons that were sent from a galaxy receding faster than c. However, the sphere can recede faster than these photons, when the sphere passes them, they'll be inside a zone that's moving from us at v < c, so they acquire positive speed and can get to us. Notice how this agrees with the diagram, when the hubble line cuts the light cone (the worldline of a photon), the world line that was moving away from the origin starts to move towards it. Reference: http://arxiv.org/pdf/astro-ph/0310808v2.pdf

Posted: 6/2/2012 9:03 AM

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Warp

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In another forum I got the answer that if the premise is that the universe expands at such a rate that the distance between planet A and planet B increases at a constant rate (that's slightly faster than c), and if A sends a probe towards B, the probe will eventually reach B, regardless of how unintuitive that might sound. The most intuitive explanation for this was the "snail on a stretching rubberband" analogy. However, there's a limit to the rate of the metric expansion before sending the probe becomes impossible (in the exact same way as with the snail-on-a-rubber-band analogy). From some rate of expansion up the distance between A and B accelerates so much that the probe will never reach the latter. (I'm not sure what this limit is, but I think that a constant metric expansion of the universe is enough for this.) I'm assuming that in the former case (where the probe does reach B), from A's perspective it will look as if the probe accelerates until it surpasses c and hence exits A's observable universe. (B will see the opposite, ie. the probe appearing inside its observable universe and decelerating.)

Posted: 6/4/2012 3:40 PM

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Warp

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HHS wrote:

Likewise, the earth does not actaully "move" through time. It's a 4-cylinder that has a beginning and an end.

If we don't move along the time axis, why do inertial objects change position due to a curved spacetime? If they change position, that implies that they are moving along a geodesic.

Posted: 6/25/2012 6:49 AM

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Warp

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A very interesting presentation on quantum mechanics. (Don't get put off by the title. The title is tongue-in-cheek.) http://www.youtube.com/watch?v=dEaecUuEqfc

Posted: 6/29/2012 8:48 AM

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Warp

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Suddenly I got thinking: GR predicts rotating black holes... but what exactly is rotating in such a black hole? The singularity has zero size, so by preservation of angular momentum its rotation speed would be infinite. There's "nothing else" in the black hole to rotate. (The event horizon is not something physical. It's just a curvature of spacetime.) But then I remembered reading that the equations predict that in a rotating black hole the singularity is actually not a point, but a ring (with a finite major radius and zero minor radius). Then I realized the answer: It's this ring singularity that rotates (at a finite speed). Is my understanding correct? Followup question: If the singularity is a ring, not a point, then what's the shape of the event horizon?

Posted: 6/29/2012 9:27 AM

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marzojr

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It is correct: it is a rotating ring singularity. The event horizon and static limit are oblate spheroids in the coordinate system where the singularity disk is a "point", so they remain more-or-less that shape.

Marzo Junior

Posted: 6/29/2012 12:26 PM

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p4wn3r

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Warp wrote:

A very interesting presentation on quantum mechanics. (Don't get put off by the title. The title is tongue-in-cheek.) http://www.youtube.com/watch?v=dEaecUuEqfc

That was a really great presentation, his result that entanglement leaves classically correlated particles is amazing, and a good argument to believe that entanglement and measurement are equivalent. Some of his points didn't convince me, though: 1) He considers the Copenhagen Interpretation scientifically untenable because you can undo a measurement. It just happens that in his example, putting a polarizing filter at 45º should in fact collapse the wave function again, and could be considered as some kind of measurement. A common quantum description of the phenomenon is that a vertically polarized photon, when encountering the filter will have 50% chance of collapsing to a state that passes through and gets polarized at 45º and another 50% chance of collapsing to a state that's reflected back. That justifies the half intensity he later finds through mathematical formulas. His example does, in fact, shows something bad about Copenhagen, the fact that a "measurement" is loosely defined, the fact that you can collapse the function with two measurements though is perfectly consistent with it. 2) The EPR paradox was coined up by Einstein as a critique of QM because his view of QM was discarded by most physicists of the time, who preferred Bohr's interpretation. His thought experiment shows that either QM violates locality (predicts faster-than-light action) or is an incomplete theory, because it fails to predict information about the particles at the moment of their separation. The most accepted solution to the paradox is Bell's theorem. Bell showed that if we go by Einstein's assumptions of how QM should be (local with hidden variables) then entangled particles should be much less correlated than we observe them to be. Thus, Einstein's assumptions are inconsistent with experimental data and his paradox is not a valid critique of quantum mechanics. When he said "something must be wrong with this story", I thought he'd give an interpretation that made Bell's result more intuitive, but his "solution" seems to come from the fact that since measurement and entanglement are the same, it's virtually impossible to disentangle particles that are far apart. I wish he had explained this better, because it seems to hope for a theory with only local effects, which was already shown to conflict with experimental data. And, regarding his final philosophical remarks, I find it hard to discuss them in a complete scientific way, so I'll just ignore them.

Posted: 6/30/2012 7:24 AM

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Warp

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If we have a typical-sized star (that's large enough to collapse into a black hole) rotating at a typical speed of revolution, how fast would the resulting ring singularity rotate (in revolutions per second)? How fast would the rotation speed be for a humongous star such as VY Canis Majoris when it finally collapses? Also: What would the diameter of the ring be?

Posted: 6/30/2012 3:53 PM

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marzojr

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I will start with the easier question: if the black hole is a rotating uncharged black hole with mass M and angular momentum per unit mass A then the radius of the ring singularity is A/c. From that, you can infer the diameter of the ring. Although to be honest, things aren't really as simple as I made them seem to be: this is coordinate radius; how it relates to what is usually meant by "radius" is hard to say, as the geometry within the disk is not exactly euclidean. As for the rotational speed of the singularity: there is no good way to answer this question: saying "pie" is probably as good an answer as anything else. The issue here is that we are dealing with a singularity -- the coordinates fail at these points and, being an intrinsic singularity, means that all coordinate systems will exhibit this failure. The coordinates are not the ones that fail -- the equations of GR also fail there (they are, in fact, what causes the failure of the coordinate systems). The singularity is off-limits for GR, plain and simple.

Marzo Junior

Posted: 6/30/2012 4:17 PM

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Warp

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marzojr wrote:

I will start with the easier question: if the black hole is a rotating uncharged black hole with mass M and angular momentum per unit mass A then the radius of the ring singularity is A/c.

The units seem to match. (The unit of angular momentum seems to be kg·m²s⁻¹, so if we divide that by the mass and the unit of c, ie. m/s, we end up with just meters.) However, I'm having trouble understanding how the angular momentum of a rotating body is calculated. Wikipedia cites the formula for the angular momentum of a single particle to be the cross-product of its distance from the center and its linear momentum. While this gives the angular momentum of one single point-like particle, how do you calculate the angular momentum of a(n almost) spherical object? (Thinking about it, since angular momentum deals with vectors, and the total angular momentum of a spherical object would be sum of the angular momentums of all of its points... wouldn't they sum up to be zero?)

The issue here is that we are dealing with a singularity -- the coordinates fail at these points and, being an intrinsic singularity, means that all coordinate systems will exhibit this failure. The coordinates are not the ones that fail -- the equations of GR also fail there (they are, in fact, what causes the failure of the coordinate systems). The singularity is off-limits for GR, plain and simple.

This doesn't sound right to me. If dealing with the rotation speed of a black hole would be impossible, then wouldn't it likewise be impossible to calculate the Kerr metric in the first place, since it deals with a rotating black hole? I don't quite understand.

Posted: 6/30/2012 6:28 PM

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marzojr

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Warp wrote:

The units seem to match. (The unit of angular momentum seems to be kg·m²s⁻¹, so if we divide that by the mass and the unit of c, ie. m/s, we end up with just meters.) However, I'm having trouble understanding how the angular momentum of a rotating body is calculated. [snip] This doesn't sound right to me. If dealing with the rotation speed of a black hole would be impossible, then wouldn't it likewise be impossible to calculate the Kerr metric in the first place, since it deals with a rotating black hole? I don't quite understand.

For a black hole, it is one of the parameters of the solution: a variable you plug in, very much like the mass of all black holes or the charge in a Reissner–Nordström or Kerr-Newmann black hole. For other bodies, preferrably non-black holes, it is more complicated:

Warp wrote:

Wikipedia cites the formula for the angular momentum of a single particle to be the cross-product of its distance from the center and its linear momentum. While this gives the angular momentum of one single point-like particle, how do you calculate the angular momentum of a(n almost) spherical object?

In the classical (nonrelativistic) world, the angular momentum is a vector that can be decomposed into two components: the "intrinsic" or "spin" angular momentum (which is the part of angular momentum strictly about the object's center of mass) and the object's orbital angular momentum (which the part of the angular momentum strictly about an external reference point). The orbital angular momentum is usually computed by calculating the total and spin angular momentum and subtracting the latter from the former. Both are computed by integrals: given a mass density rho(x,t), a position x, a velocity density v(x,t), using xcm(t) to denote the center of mass and O(t) to denote the point about which the object is orbiting, then the total angular momentum is: L(t) = integral over all space of (rho(x,t) * cross(x - O(t), v(x,t)) while the spin angular momentum is S(t) = integral over all space of (rho(x,t) * cross(x - xcm(t), v(x,t)) In relativity, you can't represent the angular momentum by a vector: the cross product is representable as a vector only in 1, 3 and 7 dimensions (this being related to the fact that the only normed division algebras with imaginary units are the complex, quaternion and octonion algebras), and you need to use a tensor to represent it in spaces with dimensions other than those. In relativity, the mass density is not a good candidate for use in the definition of angular momentum due to lack of invariance when changing coordinates; the stress-energy-momentum tensor is used instead. This gives a 3-tensor density which, when integrated, gives an asymmetric rank 2 tensor for angular momentum. The tensor integrals involved aren't over all space-time, but over a space-like slice of space-time -- basically, you select a observer and integrate over the set of all points he considers simultaneous. Spin, orbital and total angular momentum are all defined in an analogous way to the non-relativistic case, with spin being now related to the center of energy-momentum instead.

Warp wrote:

(Thinking about it, since angular momentum deals with vectors, and the total angular momentum of a spherical object would be sum of the angular momentums of all of its points... wouldn't they sum up to be zero?)

Nope: the cross product makes sure of that: all points rotating about a given point will have an angular momentum pointing in the same direction if they are rotating in the same direction.

Marzo Junior

Posted: 7/2/2012 6:09 AM

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Warp

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So what would the answer to my original question be? (In other words, if a typical star with a typical rotation speed were to collapse into a black hole, what would be the diameter of the ring singularity?) Our sun is not massive enough for this, but just to get some idea, let's assume that it were to somehow collapse into a black hole. Its mass is approximately 2*10³⁰ kg and its rotation velocity at the equator is approximately 7000 km/h. What would the diameter of the ring singularity be?

Posted: 7/2/2012 2:50 PM

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marzojr

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The problem you ask is harder than you think -- it depends on how much angular momentum is lost during collapse. If you assume no angular momentum is lost, you can get an upper bound. The issue then becomes that stars are messy -- they are not solid rotating spheres with uniform densities, so computing angular momentum accurately is difficult. Assuming the Sun is a solid sphere with uniform density, we can estimate its angular momentum; it could be wrong by more than one order of magnitude. The estimate uses the following formulas: I = (2/5)*m*r^2 L = Iω ω = v/r I is the moment of inertia of a solid rotating sphere of uniform density, m is the mass, r is the radius and ω is the angular velocity. Since we are interested on the angular momentum per unit mass (to compute the radius of the ring singularity), we want this: A = L/m = (2/5)*r^2*v/r = (2/5)*r*v so we can compute the singularity radius A/c. I will leave the number plugging to you.

Marzo Junior

Posted: 7/2/2012 10:39 PM

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amaurea

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http://arxiv.org/abs/1112.4168 wrote:

The angular momentum of a star is an important astrophysical quantity related to its internal structure, formation and evolution. On average, helioseismology yields S = 1.92 10^41 kg m^2 s^-1 for the angular momentum of the Sun.

Posted: 7/2/2012 11:26 PM

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marzojr

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So the solid sphere is (if I did the math right) wrong by a factor of about 5; this is much better than I would have thought it would be, actually.

Marzo Junior

Posted: 7/4/2012 6:43 AM

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Warp

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If the angular momentum of the Sun is S = 1.92*10⁴¹ kg m² s^-1 and the mass M of the sun is approximately 2*10³⁰ kg, then: radius = A/c = S/M/c = 320 m Would that be about correct? Question: A consequence of general relativity is that rotating objects "bend" their surrounding spacetime in such a peculiar way that they cause so-called frame dragging. (The spacetime is bent in such way that objects in the vicinity of the rotating body will tend to co-rotate with it.) The spacetime in the vicinity of the event horizon of a rotating black hole is bent in the same way as if the event horizon where an actual solid object (rather than just empty space with a certain geometry). We can measure the "rotation" of the event horizon by measuring the geometry of spacetime in its vicinity. In other words, we can say that the event horizon is effectively rotating as if it were an actual solid object. So the question is, does the event horizon "rotate" slower than the ring singularity, and by how much?

Posted: 8/25/2012 8:44 AM

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Warp

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This is not a question, but a commentary on a video, namely this: Link to video For the life of me I can't figure out if this is a spoof or an actual seriously-made "documentary". If it's a spoof, it's quite good at simulating those 70's-80's documentaries. Anyways, I couldn't help but be amused by the claim that ships look like they go over the horizon, and the Earth looking round from Space, as being caused by the light bending predicted by general relativity. Never mind that for the GR equations to predict such a strong curvature would require the Earth to have the density approaching a neutron star, that's not the most egregious error in that claim. The most egregious error is that the light is bending in the wrong direction for that to happen. If light were bending down due to gravity, it would actually make the sea and the ship with it look like it's curving up, when looked at from the shore. It would look like you were on the bottom of a depression. Likewise from space it would not make a flat earth look like a sphere; instead, it would make it look like you were looking into a bowl. (This is, in fact, what would happen if you were to fall into a neutron star of black hole: As you approach it, the surface will start looking bigger and bigger until at some altitude it will look like a plane that encompasses half of the universe. As you keep falling, it will start looking like the surface curves up.)

Posted: 10/9/2012 1:11 PM

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ElectroSpecter

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These questions might be elementary and ridiculous in many ways, but I figured this was an okay place for them to go: 1) If I had a knife and somehow sharpened it to an atomic edge, would I be able to actually see the edge of the knife or would it become transparent? 2) If I somehow got to this point where I had atomically sharpened a knife, would the edge instantly break from the slightest bit of movement or air resistance? 3) If the knife was unbreakable and I dropped it blade down (and there was no handle, to make my question less complicated), would it just keep cutting through the earth until it melted?

My Backloggery

Posted: 10/9/2012 6:12 PM

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Warp

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ElectroSpecter wrote:

1) If I had a knife and somehow sharpened it to an atomic edge, would I be able to actually see the edge of the knife or would it become transparent?

I have no idea, but I know that it's not possible to "see" individual atoms, and this has something to do with the fact that atoms are smaller than the wavelength of visible light. I don't know what happens if you have an entire layer of individual atoms. Would light pass through? (Btw, such a layer is called a monolayer. The article might have more info on that subject.)

3) If the knife was unbreakable and I dropped it blade down (and there was no handle, to make my question less complicated), would it just keep cutting through the earth until it melted?

A physically impossible object meeting a physical object... I don't know if there is an answer.

Posted: 10/9/2012 8:09 PM

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Tub

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ElectroSpecter wrote:

3) If the knife was unbreakable and I dropped it blade down (and there was no handle, to make my question less complicated), would it just keep cutting through the earth until it melted?

No. There's still only a limited force behind the drop, and sooner or later that force is expended. No matter how sharp or indestructible the knife is, it will not go through a rock. You might get it lodged into a rock (if dropped from a sufficient height), but then it'll get stuck by the horizontal pressure and friction. Try adding "frictionless", but that'll open another can of definitions about friction and whether or not a frictionless knife can even cut things. Shouldn't it glide off?

m00

Posted: 10/9/2012 8:30 PM

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Derakon

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A knife is more than just the edge of the blade. The knife widens out from the edge into something thicker -- otherwise you have a monolayer (2D) or monofilament (1D) construction. Larry Niven had monofilament melee weapons -- they were wires only one atom thick, held together by a stasis field, and so sharp they could cut through anything. Though one wonders why you couldn't just cut with the statis field directly... Anyway, dropping the blade into the ground would be like trying to drive a wedge through some wood. Even if there's no piercing resistance, you still have to push things sideways to make way for the thicker portion of the blade, and that costs energy. Sufficiently small objects can penetrate through matter more or less freely, though. Neutrinos pass through the entire Earth all the time because they interact so weakly with normal matter.

Pyrel - an open-source rewrite of the Angband roguelike game in Python.

Posted: 10/10/2012 7:28 AM

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Warp

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Derakon wrote:

Sufficiently small objects can penetrate through matter more or less freely, though. Neutrinos pass through the entire Earth all the time because they interact so weakly with normal matter.

Why do neutrinos pass easily through any amount of matter (with extremely little chance of being stopped), but neutrons can only pass through a rather moderate amount of matter before being stopped? Neutrons are electromagnetically neutral, so electromagnetic forces don't stop them. And why would they ever collide with an atomic nucleus, considering how vanishingly small they are compared to the space between nucleii? (I don't remember now the exact figures, but I have a faint memory that if atomic nucleii were the size of a golf ball, they would be separated in a typical molecule by at least the length of a football field, or even more.)

Posted: 10/10/2012 3:54 PM

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Derakon

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I'm not a particle physicist, so take this with an appropriately large grain of salt. A little googling says that neutrinos are only affected by the weak nuclear force and by gravity. Neutrons in contrast are also affected by the strong nuclear force. I would have to assume that this is the basis for why neutrons have so much less "penetrating power" than neutrinos; the only forces that interact with neutrinos are incredibly weak. Also, with respect to your scale analogy, let's plug some numbers in assuming that the golfball : football field analogy is vaguely accurate. A golfball has a radius of about 2cm (thus, projected, i.e. flat, area of about 12 square centimeters -- this is back-of-the-envelope stuff here); a non-American football field (since you're from Finland, I assume you didn't mean American football) has an area of about 7140 square meters (thus 71400000 square cm). A particle hitting the field at a random location would thus have about a 1 in 6 million (10^6) chance of "hitting" a nucleus. Hell, we can make this more generous -- say you were off by three orders of magnitude, and it's actually a 1 in 6 billion (10^9) chance. Now, remember how many molecules we're dealing with here. For example, a mol of iron typically weighs about 56g, and that's 6.02 * 10^23 atoms. A 1-mol sphere of iron would have a radius of a bit over 1cm (thus diameter about 2cm, or roughly the size of a golfball, for a neat bit of symmetry). Passing through a 2cm-block of iron requires passing a stupendous number of 1-in-6-billion chances. Eventually the odds will get you.

Pyrel - an open-source rewrite of the Angband roguelike game in Python.