elfmotat

[math]\eta_{ij}T^{i0}T^{j0} = \delta_{ij}T^{i0}T^{j0} = \sum_i(T^{i0})^2 \neq (T^{i0})^2[/math] There is no implied summation in Ti0 by itself - "i" is still a free index. You either need an explicit summation, or you can do what I did with the metric. The energy density of a magnetic field is [math]\frac{1}{2\mu_0}|\bold{B}|^2[/math]. Dividing by c2 yields what I have in my first reply. As for units, energy density (above) is in J/m3 while "mass" density (first post) is in kg/m3.

I think the expression you're looking for is: [math]\rho = \sqrt{(T^{00})^2-\eta_{ij}T^{i0}T^{j0}}[/math]

That expression doesn't make any sense. It's dimensionally inconsistent (due to the square root), and your indices don't match up.

It's tricky to know what exactly you're asking for. Of course, you could define a "mass" density with units of kg/m3 by dividing its energy density by c2. In that case it would just be: [math]\rho =\frac{1}{2}\epsilon_0|\bold{B}|^2[/math]

Sure, you could synchronize all clocks to agree by correcting for velocity and gravitational time dilation, but this doesn't have any real physical meaning. It doesn't stop initially synchronized clocks from ticking at different rates according to altitude or velocity.

http://soliton.ae.gatech.edu/labs/windtunl/classes/hispd/hispd06/ns_eqns.html They list the same equation under "Summary."

I'm not a quantum gravity expert by any stretch of the imagination, but I believe your impression that the singularity should obey uncertainty principle is valid. General relativity is a classical theory, and as such does not account for such effects. A good quantum gravity theory would need be invented to make reasonable predictions about quantum effects near the singularity.

The radial acceleration due to rotation is ω2R. At the equator, this is (7.2921150 * 10-5 rad/s)2(6,378137.0 m) = 0.033916 m/s2. Compared to the gravitational acceleration of 9.79 m/s2 at the equator, it is almost entirely negligible. The difference is about what you would get from moving from one of the poles (g=9.83 m/s2) to the equator (g=9.79 m/s2).

No objections here. It's certainly possibly to measure the components of the Riemann curvature tensor. Macroscopically, I don't think we'll get anything better than metric theories of gravity (though this is just my opinion of course). At large scales, we'll probably always work with a theory that relates curvature to stress-energy in some form. Quantum gravity will almost certainly describe gravity in terms of spin-2 gauge bosons (gravitons). You describe its motion as a result of the local geometry of spacetime. There are, though, well observed instances of non-locality in QM. Speculations abound. Some comments on your ill-defined theory: -Gravitons are massless -Weakly interacting means that the particles only interact through the weak force + gravity. It doesn't mean they're hard to measure. -Massive does not mean subject to gravity. Massless particles (e.g. photons) also interact gravitationally.

Are you willfully ignoring what everyone else is saying, or are you just too impatient to read the replies to your thread? Your posts make no sense.

No, time "speeds up" in space. The closer you get to a gravitational source, the "slower" time passes.

Well yes, "stationary" is technically the accurate word. "Least action" and "maximal proper time" are the usual names used, and are almost always accurate names. That's pretty cool that you got to work with Taylor and Wheeler.

Action principles have always fascinated me. Least action, Fermat's principle, maximal proper time, etc.

True. My initial Lagrangian would only be valid if the other particles in the system were held static. Here's what it would be for the system: [math]L=\frac{1}{2} \sum_k m_k|\dot{\bold{x}}_k|^2+ \frac{1}{2} G \sum _{i,j} \frac{\mu_i M_j}{|\bold{x}_i-\bold{x}_j|} [/math] The reason this works is because [math]\mu_1 M_2= \mu_2 M_1[/math], or else Newton's third law fails. I think this is reason enough to conclude that passive and active gravitational mass are equivalent, or at least the constant of proportionality between them is a universal constant which is absorbed into G. In that sense I don't see how you could perform any experiment to determine whether or not they differed, so I don't know why it would be useful to think of them as not being equivalent. The first term is actually an external potential, so it's zero in the case we're considering.

Nope, I wasn't summing the first term. I was actually considering the Lagrangian of a single particle acted on by others.

Classical mechanics traditionally refers to continuous (rather than quantum) physics, which includes relativity. Sometimes though, I use it to mean non-relativistic mechanics.

I wasn't sure whether or not you were confused about swansont's post, but clearly you weren't. Relativistically, no, I'd rather not even try. Doing it with good old classical mechanics isn't too hard though. The Lagrangrian of the ith particle with inertial mass mi, passive g-mass μi and active g-mass Mi would be: [math]L = \frac{1}{2} m_i |\bold{\dot{x}}_i|^2 + G \mu_i \sum_{j} \frac{M_j}{|\bold{x}_i-\bold{x}_j|}[/math] The sum is obviously over j where j≠i.

Conservation of momentum is a result of translational symmetry in the Lagrangian. It's easy to see: consider a Lagrangian which satisfies the following property: [math]L(x^i,\dot{x}^i,t)=L(x^i+\delta x^i,\dot{x}^i,t)[/math] This means that [math]\partial L / \partial x^i=0[/math], so the Euler-Lagrange equation reduces to: [math]\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{x}^i} \right )=0[/math] Now, [math]\partial L / \partial \dot{x}^i =p^i[/math] by definition, therefore momentum doesn't change over time.

I'm not really sure what your question is...

Just some quick comments on your page: -Poisson's equation is that the Laplacian of the potential is proportional to the mass density, not the gradient (i.e. ∇2ϕ=4πGρ). The latter doesn't make any sense because a gradient is a vector whereas ρ is a scalar. -The proportionality constant in the EFE's you give us are incorrect: Gμν= (8πG/c4) Tμν As for what you write above, you haven't given a definition of active/passive gravitational mass. It seems like, from what you say about four-momentum, you consider PG-mass to be equal to rest/invariant mass. But then you go on to say that invariant mass doesn't equal the magnitude of the four-momentum, so I'm not sure what to make of it.

Did you read any of the previous posts?

Has anyone done this? If it differs significantly from GR then it has already been falsified. The predictions of GR have already been tested and verified to high accuracy. This theory would have to give as good, or better, predictions than GR in the areas where it's already been tested.

I don't speak French so I can't read it, but: Does this make predictions that differ from GR? If so, in what way? It's likely that the theory has already been falsified.

That's where time curvature comes into play. (Actually, the time component of the curvature almost always plays the most important role. Until you start getting up toward the speed of light the space curvature affects you very little. For photons, time and space curvature affect them equally.) There's a pretty simple analogy. Take a spacetime diagram of two objects at rest with respect to each other in flat spacetime. The objects trace out two parallel lines that extend vertically to infinity. Now try drawing a spacetime diagram of the same situation, but on a curved surface (such as a ball). This time, the objects start out parallel (at rest), but as you extend their worldlines they eventually intersect. This is because in curved space there are no permanently parallel lines.