elfmotat

Bell's Theorem applies to local hidden variable theories. The article discusses possible ways around this, with non-local variables etc.

Then you're arguing with a point that I wasn't making.

They're not just interpretations. Bell rules out all local hidden variable theories, i.e. theories in which the state of particles is determined by some unobservable deterministic variables. I believe Feynman said something along the lines of "Nature itself doesn't know what the answer is going to be."

I've been putting this off because I'm not sure how useful you'll find it. The purpose of showing you Newton's law with nonzero CC was to intuitively demonstrate how gravity is attractive when things are close together, but repulsive when the are very far away. In General Relativity, gravity is the geometry of spacetime. The solution to Einstein's equations with nonzero CC which describes a spherical mass [math]M[/math] is called the de Sitter-Schwarzschild metric: [math]ds^2=-\left (1-\frac{2GM}{c^2r}+\frac{\Lambda r^2}{3} \right )dt^2+\frac{dr^2}{1-\frac{2GM}{c^2r}+\frac{\Lambda r^2}{3}}+r^2d\theta^2+r^2sin^2\theta d\phi^2[/math] This solution isn't particularly useful because the effect that a nonzero CC has is very very small. Since the field equations are highly nonlinear the superpostion principle does not apply either, so you can't simply "add" the metric of another spherical mass to get a solution describing the spacetime around two masses. Nonetheless, this solution might be the key to putting your mind at ease: The metric tells you how things move in spacetime. If you restrict movement to be along the radial coordinate and calculate the equations of motion (called the geodesic equations) for particles moving near the mass [math]M[/math], you'll get some complicated-looking equations. But if you take the Newtonian limit where the field is weak and everything moves slow compared to light, you get the following equation: [math]\frac{d^2r}{dt^2}=-\frac{GM}{r^2}+\frac{c^2\Lambda r}{3}[/math] So with our approximation we can see that the CC term still acts as a "repulsive" term while the mass acts as an attractive term. One is a linear approximation of the other. So yes, there is a "dichotomy." They don't have equivalent predictions. Co-moving with what? I'm not sure whether or not you can perform a coordinate transformation on the FLRW metric so that the spacial components are independent of the scale factor. Regardless of if you can, I don't think it would be particularly useful or meaningful. I thought I pretty clearly demonstrated that adding a constant term to the Einstein Field Equations is just as valid as adding a constant term to the Newtonian Poisson Equation. So "galaxies receding due to gravity" is indeed predicted by Newtonian gravity. Maybe. But then that wasn't really my point, was it? My point was that GR provides the correct description of the expansion of the universe, while Newtonian gravity only supplies us with an approximation.

You might find my post here to be informative. Essentially the answer to your question is that the "correct" way of thinking about it is with the metric expansion of space. Metric expansion is a result derived from General Relativity, whose predictions about the way gravity behaves are much more accurate than the predictions of Newtonian gravity (especially with strong gravitational fields and when things are moving close to the speed of light). When you make some very simplifying approximations (everything is moving slow, all gravitational fields are weak) you can recover Newtonian gravity from Einstein's equations. In Newtonian gravity, the Cosmological Constant isn't a term which contributes to metric expansion (indeed the metric is not dynamical), but it turns up as a repulsive term which is linear with distance that is added to Newton's Law of Gravitation. This viewpoint is approximately correct ( it is correct up to first order). But still, what's "really" happening is that space is expanding. The two viewpoints aren't completely consistent because one of them is just a simplifying approximation of the other.

The Newtonian approximation of gravity with nonzero Cosmological Constant is well-known, and falls directly out of linearized General Relativity: [math]S=\int \left [\frac{1}{8\pi G}|\nabla \phi|^2+ \left (\rho-\frac{c^2\Lambda}{4\pi G} \right ) \phi \right ]d^4x[/math] This action yields the field equations for Newtonian gravity: [math]\nabla^2 \phi +c^2\Lambda =4\pi G\rho[/math] If you assume that all of the mass in the universe is just a single point particle at the origin of our coordinate system, i.e. [math]\rho =M\delta (\mathbf{r})[/math], then by applying Gauss' Law to the field equations with the point particle density (and assuming the equations of motion for a test particle are [math]\frac{d^2 \mathbf{r}}{dt^2}=-\nabla \phi[/math]) you arrive at the following force law between the gravitating mass [math]M[/math] and a test particle of mass [math]m[/math]: [math]\frac{\mathbf{F}}{m}=-\frac{GM}{|\mathbf{r}|^3}\mathbf{r}+\frac{c^2\Lambda}{3}\mathbf{r}[/math] Since our field equations are linear, the superposition principle applies and the force contribution from multiple point particles may be simply added with vector addition. From this, the shell theorem follows, etc., and we see that in a universe filled with complicated arrangements of matter ([math]\rho(\mathbf{r})[/math] can be any complicated function) all matter acts on all other all other matter according to the above force law. So clumps of matter which are close enough together will be attracted to each other, while clumps spread out far enough will actually repel each other. This means that in the context of Newtonian gravitation, the presence of a positive nonzero Cosmological Constant is interpreted as a repulsive term in the force law. This means that, as far as Newton is concerned, gravity is indeed repulsive once you're far enough away from sources of gravitation (masses). So in this context "Dark Energy" is simply the result of a modification of Newtonian gravity. The observation that far-away galaxies are accelerating away from us is all explained by our force law. In full-fledged General Relativity, the action is: [math]S=\int \left [ \frac{c^4}{16\pi G}(R-2\Lambda )+\mathcal{L}_{matter} \right ]\sqrt{-g}d^4x[/math] This yields the following (nonlinear) field equations: [math]G_{\mu \nu}+\Lambda g_{\mu \nu }=\frac{8\pi G}{c^4}T_{\mu \nu}[/math] From these field equations you can get exact solutions for the universe (using assumptions like isotropy and homogeneity) such as the FLRW metric. In this solution the spacial part of the metric (i.e. how spacial distances between points are determined) is dependent on a function of time [math]a(t)[/math] called the "scale factor." Plugging this solution into the field equations yields a few useful equations (called the Fiedmann equations), one of which is: [math]\frac{\ddot a}{a} = - \frac{4\pi G}{3}\left(\rho + \frac{3p}{c^{2}}\right) + \frac{\Lambda c^{2}}{3}[/math] where [math]\ddot{a}=\frac{d^2}{dt^2}[a(t)][/math]. Here, we can see that ordinary energy density and pressure are acting to "contract" the distances between points (i.e. trying to make [math]\ddot{a}[/math] negative). On the other hand, the term involving the Cosmological constant is working to increase the distances between points. We can also see from the Friedmann equations that the Cosmological Constant term acts as some sort of energy with negative pressure (which is dubbed "Dark Energy"). So in the context of GR, the Cosmological Constant is something which works to expand the universe itself. The point of this post was to show you that your idea that Dark Energy is actually a form of "antigravity" actually does work, to a certain approximation. But what's "really" going on is that the distances between points in the universe is increasing. GR is a much more accurate model of gravity than Newtonian gravitation, especially when things are moving fast compared to light and when gravitational fields are very strong. But the effects of a nonlinear gravity theory are also apparent within our own solar system, for example with the precession of Mercury. Applying Newtonian gravity on the cosmological scale is bound to give lots of inaccurate predictions. Your idea works to develop an intuition about the expansion of space, but it's bound to fail if you try to use it to make predictions about the behavior of the universe.

It means its frequency is Doppler-shifted.

"Now" is a local concept. It has no meaning for separated observers near different gravitating bodies.

As you say, introducing the Cosmological Constant is a natural generalization of the Field Equations. Essentially the simplest, most general metric theory of gravity has the Lagrangian [math]\mathcal{L}=k(R+C)\sqrt{-g}[/math]. This yields the Field Equations with [math]\Lambda =C/2[/math]. Arbitrarily setting [math]C=0[/math] results in a less general theory, because it is a tunable parameter which we should fit with observation. From the Field Equations with non-zero Cosmological Constant, plus some simplifying approximations (everything is moving slowly compared to light, the gravitational field is weak, etc.) we can derive a modified form of the Newtonian Poisson equation for gravity: [math]\nabla^2 \phi =4\pi G\rho -c^2\Lambda[/math] Apply Gauss' Law to that and you obtain the modified force law in my previous post. Also, interestingly, Newton himself realized that the shell theorem only applied to gravity forces which varied either linearly or inversely as the square of distance, so a logical next-step would be a superposition of the two: [math]F=ar+\frac{b}{r^2}[/math] The following article is rather interesting: http://www.homepages.ucl.ac.uk/~ucapola/CLrev.pdf

Thought I might add this: You can modify Newton's Law with a Cosmological Constant term to approximate weak-field gravity: [math]F=m\left (-\frac{GM}{r^2} + \frac{c^2 \Lambda}{3}r \right )[/math] Since the CC term is proportional to distance while the mass term is inversely proportional to the square of distance, you can see that eventually the second term will overwhelm the first. This means that at some distance from a mass gravity actually becomes repulsive. As a quick example, let's say that the Milky Way is the only galaxy in the universe. The MW has a mass on the order of 6×1042 kg. We know the cosmological constant is of the order of 10-52 m-2. Using these values, we can determine that at a distance of about 5.4 million light-years = 1.7 Mpc away from the MW gravity becomes repulsive.

If you measure the state of particle A then you immediately know the state of B. Since the two measurements are non-local, you can only confirm that B is in the state you expect when the two observers come together to compare results. It doesn't matter "when" each observer did the measurements because the results always agree. Indeed you can find valid frames in which the order of measurements is reversed and the results still agree. "When" the states of each particle were measured has no bearing on the measurements.

You'd need at minimum some experience with Vector Calculus, Linear Algebra, and Special Relativity. After that Bernard Schutz's intro GR textbook is probably your best bet.

Who said you have the "same time" in different frames? What do you think is going on that would require several frames to share ideas of simultaneity?

What exactly do you think is happening "at the same time?"

What, exactly, do you think is "simultaneous?" Perhaps this analysis will put you at ease: A neutral pion at rest decays into a positron and electron. The pion had spin-0, so due to conservation of angular momentum we know that one of the remaining particles must be spin-up and the other spin-down. We have two detectors set up in opposite directions which will tell us the spin state of each particle. Say detector A (which will measure the electron) is a bit closer to the original location of the pion, so in the center of mass reference reference frame the electron's spin state is measured first. Let's say the electron turns out to be spin-up. Now we automatically know that the positron is spin-down, which can be confirmed after it is measured and the two observers compare data. Even though the particles themselves didn't know which state they were going to be in once measured, as soon as one of them is measured you immediately also know the state of the other particle. But relativity tells us that with an instantaneous "interaction" (I use the word interaction loosely) we will be able to find a valid reference frame in which the positron was actually measured first. So in this frame the positron's state is measured, and we find that it's spin-down. In this frame we automatically know that the electron's state is spin-up. But this scenario is entirely identical to the original, in which the electron was measured first! Nothing changed from changing the order in which states were measured. Since the scenario is symmetrical it does not violate relativity in any way.

Well, as I said, I'm not really sure what your question is. I thought the analogy might be relevant. Yes, I believe I mentioned that.

No. Gravity is far too weak to explain such a phenomenon.

I'm not sure what your question is. I posted this in another thread: The analogy isn't perfectly accurate (Bell's inequalities, etc.), but gets the idea across.

The internal pressure required to "counteract" gravity approaches infinity as the radius of a spherical object approaches R=9GM/4c2.

I was actually just thinking about posting your video in this thread. Funny coincidence .

You actually only need to compress it to 9/8 its Schwarzschild radius to ensure collapse.

At the center of the Earth all of the mass is around you, so the net gravitational field is actually zero. What's unique about a black hole is that all of its mass is concentrated in a very small volume.

E2-p2corresponds to the mass of a particle in its rest frame, sure. But that doesn't mean that the definition of E2-p2 is "the mass of a particle in its rest frame." It is what it is: E2-p2 is just a quantity. For photons this quantity is zero, and there's nothing more to it than that. Whether or not photons have a rest frame does not determine whether or not we can measure E2-p2. As for experimental tests of the relation E2-p2=0 for photons, it is trivial to check that the energy and momentum relations required by experiments such as the Compton Effect yield a rest mass of zero: [math]\lambda f=c[/math] [math]E=hf[/math] [math]p=h\lambda^{-1}[/math] [math]E^2-c^2p^2=h^2 ( f^2-c^2\lambda^{-2})=h^2 ( f^2-f^2)=0[/math] There are numerous other reasons why the photon's mass must be zero, as outlined here.

You're using a different definition of mass than everyone else. Your definition of mass (relativistic mass) is proportional to energy: E=mrel. If you want to use relativistic mass, fine, but you need to clarify that that's what you're doing because it is not standard. The mass everyone else is using is defined by E2-p2=m2. We know that this quantity is zero for photons (or at least very very very close to zero) because their energy is directly proportional to their momentum. The fact that the quantity is called "rest mass" doesn't mean that particles need to be at rest for us to be able to measure it - it's just a name.