elfmotat

Come on now, you're being silly.

I'm not really sure what your point is. MTW defines a new tensor which is equal to the Einstein tensor plus the CC term. So what? This doesn't change any physics like you claimed in your last post, it just changes the notation a bit. The form of the field equations remains the same. No, it certainly doesn't. This equation is only correct if you are using MTW's new "G" tensor in place of the usual Einstein tensor. It's quite useless (and needlessly non-general) to declare the metric to be something specific in a general tensor equation. It's also impossible for the metric to be globally flat (which you are declaring it to be) with nonzero CC.

I'm not really sure what you mean. I've only ever seen one definition of the Einstein tensor: [math]G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R[/math]. Those aren't the field equations, and that equation is wrong. What do you mean? Why are you assuming the metric is the Minkowski metric? The purpose of the field equations is that it's relation between the metric and the energy-momentum distribution of spacetime (the stress-energy tensor), analogous to how Poisson's equation gives a relation between the potential and the mass distribution in space. Also, why are you including the CC term in your equation if you're letting it equal zero anyway? That's wrong and doesn't make any sense. Again, I have no idea what you're talking about. What "redefinition?" I've never heard of anything like this in all of the textbooks and papers that I've read.

He means "contradict" in the sense that the theory's predictions are completely different than those of the established theory / experimental data. He doesn't mean "contradict" in the sense that it produces slightly more accurate predictions than the established model. "Contradict" here is a bad thing.

Of course! Nobody is claiming otherwise. It's called "a better model." General relativity is a better model than Newtonian gravity. What we're saying is that the new model is going to look somewhat like the old model (because we already know the old model agrees well with experiment).

That sounds like a good thing to me.

The form of the field equations with zero cosmological constant are: [math]G_{\mu \nu}=\kappa T_{\mu \nu}[/math] Compare this with the analogous Poisson equation for Newtonian gravity: [math]\nabla^2 \phi =4\pi G\rho [/math] The field equations with nonzero CC are: [math]G_{\mu \nu}=\kappa T_{\mu \nu}-g_{\mu \nu}\Lambda[/math] Because the metric is analogous to the Newtonian potential, this suggests a modified Poisson equation of the form: [math]\nabla^2 \phi =4\pi G\rho-\phi \Lambda[/math] However, if we assume the gravitational field is weak (which is the only time when Newtonian gravity is accurate anyway) and we linearize the field equations with [math]g_{\mu \nu }=\eta_{\mu \nu }+h_{\mu \nu }[/math] and [math]|h_{\mu \nu}|\ll 1[/math], then we arrive at the following field equations: [math]G_{\mu \nu}=\kappa T_{\mu \nu}-\eta_{\mu \nu}\Lambda[/math] where [math]G_{\mu \nu}=-\frac{1}{2} \partial^{\alpha} \partial_{\alpha}(h_{\mu \nu}-\frac{1}{2}\eta_{\mu \nu} h^{\sigma}_{~\sigma })[/math], though the form of the Einstein Tensor isn't really important for our purposes. These field equations imply (because the relevant components of the Minkowski metric are of magnitude 1) a Newtonian Poisson equation of the following form: [math]\nabla^2 \phi =4\pi G\rho-\Lambda[/math] This is clearly much simpler than the previous form we considered. We can now apply Gauss' Law to a spherically symmetric mass to obtain a modified version of Newton's Law: [math]\nabla^2 \phi=\nabla \cdot (\nabla\phi )=\nabla \cdot(-\mathbf{g})[/math] [math]\int \nabla\cdot \mathbf{g}dV=\int \mathbf{g}\cdot d\mathbf{A}=\int (-4\pi G \rho +\Lambda )dV[/math] [math]g\int dA=-4\pi G\int \rho dV + \int \Lambda dV[/math] [math]4\pi r^2g=-4\pi GM + \frac{4}{3}\pi r^3 \Lambda [/math] [math]g=-\frac{GM}{r^2} + \frac{\Lambda r}{3} [/math]

The metric is a perturbation of the Minkowski metric, yes. That has nothing to do with the cosmological constant. A nonzero CC means nonzero vacuum Ricci curvature. A quick contraction of the vacuum equations yields R=4Λ, so another form of the vacuum equations would be: [math]R_{\mu \nu}=g_{\mu \nu}\Lambda[/math] With such a large value for Λ, you're going to get some funky geometry. When you're deriving the low-energy metric, you assume Λ=0. What you do is define the metric as [math]g_{\mu \nu}=\eta_{\mu \nu} + h_{\mu \nu}[/math], where [math]|h_{\mu \nu}|\ll 1[/math] because we're assuming the field is weak. The geodesic equation, [math]\frac{d^2 x^\mu}{d\tau^2}=-\Gamma^{\mu}_{\lambda \sigma} \frac{dx^\lambda}{d\tau} \frac{dx^\sigma}{d\tau}[/math], can be simplified under the condition that we're considering velocities much less than that of light, i.e. [math]\frac{dx^i}{d\tau} \ll \frac{dt}{d\tau }[/math]. This reduces it to: [math]\frac{d^2 x^\mu }{d\tau^2}=-\Gamma^{\mu}_{00}\left ( \frac{dt}{d\tau } \right )^2[/math] We can expand the remaining Christoffel symbols as the following if we assume the metric is static: [math]\Gamma^{\mu}_{00}=-\frac{1}{2}g^{\mu \lambda}\partial_{\lambda}g_{00}=-\frac{1}{2}\eta^{\mu \lambda}\partial_{\lambda}h_{00}[/math] The geodesic equation then becomes: [math]\frac{d^2 x^\mu }{d\tau^2}= \frac{1}{2}\eta^{\mu \lambda}\partial_{\lambda}h_{00} \left ( \frac{dt}{d\tau } \right )^2[/math] The 0-component obviously gives us [math]d^2t/d\tau^2=0[/math]. Applying the chain rule for the remaining components gives us: [math]\frac{dx^i}{dt^2}=\frac{1}{2}\partial_ih_{00}[/math]. We see here that if we set [math]h_{00}=-2\Phi [/math] where [math]\Phi[/math] is the Newtonian potential then we have reproduced Newtonian gravity. Since the remain components of [math]h_{\mu \nu }[/math] don't affect our differential equation, it is sufficient to set them equal to zero. So the way we represent Newtonian gravity with curved spacetime is with the following metric: [math]ds^2=-(1+2\Phi )dt^2+dr^2+r^2d\Omega ^2[/math] where [math]d\Omega^2=d\theta^2+sin^2\theta d\phi ^2[/math].

No, we're arbitrarily setting it to zero. Setting it to 1 would result in some funky vacuum geometry.

I'm not really sure what your point is, rah. Do you really think a theory constructed by someone with little/no knowledge of accepted physics is likely to accurately describe nature? Sure it's possible, but I don't know of a single instance of some Joe Shmoe physics illiterate pulling a theory out of his rectal cavity and having it coincidentally describe some phenomenon with better accuracy than the model that was in use before. If you want your theory to describe nature well, a good place to start is obviously with theories that we already know describe nature well. You new theory has to look at least somewhat like the old one, because we already know the old one works. For example, our first description of gravity was Newton's Law of Gravitation. Now we have General Relativity which agrees with experiment better than Newtonian gravity; however, by taking the low-energy limit of GR we recover good old Newtonian gravitation.

"Useful theoretical implications" simply means it's a useful concept that could assist in the development of a theory. The equivalence principle certainly has useful theoretical implications, considering Einstein used it when developing General Relativity. A "meaningless philosophical question" is a question that cannot be answered by science, yet people attempt to answer anyway. It's a question where even if someone had the right answer, they could never know for sure. It's a question that's pointless to attempt to answer. An example would be, "what is the correct interpretation of quantum mechanics?" Interpretations are just that: interpretations. There's no test or experiment that could be performed that could answer this question.

Very true, [math]\sqrt{-g}=det(J)[/math] where [math]g=det(g_{\mu \nu })[/math]. This is useful, for example, in the Hilbert action for deriving General Relativity: [math]S=k\int R\sqrt{-g}~d^4x[/math]

Well, it's certainly useful when doing multiple integrals in funky coordinates, because dV = det(J) dx1dx2...dxn, where dV is the volume element. For example, let's say we want to calculate the area of a circle (area is really just 2-dimensional volume). Now we could do this in Cartesian coordinates, but it's going to be a lot messier than if we do it in polar coordinates. Now, of course in Cartesian coordinates the 2-volume (area) element is just given by dA=dxdy. What is dA in polar coordinates? First we compute the Jacobian: Computing the determinant, we find that det(J)=r. So now we know that dA=det(J)drdϕ=rdrdϕ, and we can now evaluate the following integral:

What is v?

That might be the most meaningless nonsense I've ever read in a single post.

My analogy explains this as well. Notice how, as the particles move forward in time, the distance between the particles decreases at an increasing rate (i.e. they are accelerated toward each other). This is purely a consequence of the geometry of the particular spacetime - no forces are required. The curvature certainly changes as you get closer/farther from a large mass. Why would this imply gravity is a force?

Um, no. "Logically" the correct value to extrapolate to is c, not 2c. Say a light source emits a photon. Boosting from the rest frame of the light source to a frame traveling at velocity v in the opposite direction of the photon yields a photon velocity in the new frame of: This makes sense, because one of the core assumptions of relativity is that light travels at c in all inertial frames. Now if you want to extrapolate this to a nonexistent frame traveling at c relative to the light source (and I don't see why you would to be honest; it doesn't have any physical meaning) then can just take the limit of u as v->c, which is still c.

Yes, you're right. The x-axis great circles aren't perpendicular to the geodesics after t=0.

No, it's physically meaningless. For example, consider the coordinate transformation [math]x \rightarrow x+k[/math]. Now the "poles" (place where the x-axes intersect) have been shifted a distance k. This is just a coordinate change; nothing physical occurred, yet the poles are now at different locations.

There isn't any. They're coordinate singularities that can be removed by diffeomorphisms. And, like I said, a sphere isn't a realistic spacetime in the first place.

You're focusing far too much on the words being used, and you're asking meaningless philosophical questions that have no bearing on the physics. The only reason I replied to you was because you implied the equivalence principle was meaningless. What you do or do not decide to classify as "feeling gravity" is not something I'm interested in discussing.

Yes. They intersect at the left and right "poles" of the sphere.

They're not the same because one of them has useful theoretical implications, whereas the other does not. That was your original question: "It may be true, but is it meaningful?"

The x-axis at any time t is a great circle perpendicular to both geodesics.

They're not the same thing at all. The point is that if you were inside a box with no windows floating in space, you wouldn't be able to tell if you were free-falling towards a massive object or just floating around in the absence of anything. It's called the equivalence principle, and it's what helped Einstein come up with general relativity. It's what allows gravity to be represented as a consequence of geometry. Einstein himself called it his "happiest thought."