elfmotat

All you said was . I said it is by definition. If your problem is not with my definition, then what exactly is it? I know you didn't, but I was making explicit the fact that what I did was completely valid. Even if it weren't true in general (which it is), what I did would still be valid because of the commutativity of partial derivatives. I don't see how you could possibly think that's true. Classically it certainly wouldn't be true in general, and only would be when F=0. Regardless, the partial of the momentum operator is simply not zero in general as I showed above. I can't figure out why you think it should be.

I agree. Strange how no one has done that yet.

I fail to see the problem. In what way do you think it violates energy conservation?

In relativity an observer is just a coordinate system. It's sort of a lattice of rulers and clocks. In quantum mechanics, it's something that collapses the wave function. It could be, for example, a detector that uses EM radiation.

What are you talking about? I defined the operator F to be [math]\hat{F}\Psi \equiv -\nabla \hat{H}\Psi[/math]. As far as I can see that's a perfectly valid definition. Yes, partial derivatives commute: [math]\frac{\partial^2 f}{\partial x \partial t}=\frac{\partial^2 f}{\partial t \partial x}[/math] How so? Operators are associative, i.e. [math]\hat{A} (\hat{B}\Psi) = (\hat{A} \hat{B})\Psi[/math]. I see nothing invalid about what I did. How do you figure? [math]\frac{\partial \hat{p}}{\partial t}=-i\hbar(\frac{\partial^2 }{\partial t\partial x},\frac{\partial^2 }{\partial t\partial y},\frac{\partial^2 }{\partial t\partial z})[/math]

I added to my post to get that equation into a form you might find more familiar. And yes, H is the Hamiltonian.

I don't get it, why don't you just use [math]F=-\frac{\partial H}{\partial x}[/math]? You could take the gradient of the Schrodinger equation to get: [math]-i\hbar \frac{\partial }{\partial t}\nabla\Psi=\hat{F}\Psi [/math] , where F-hat is defined as a force operator -∇H. Then, using the fact that the momentum operator is [math]\hat{p}=-i\hbar \nabla[/math], everything simplifies to: [math]\frac{\partial \hat{p}}{\partial t}\Psi=\hat{F}\Psi[/math] This is all very messy and unnecessary by the way.

SR and GR are mathematically consistent classical theories, so the only real question is whether or not they agree with experiment. So far they do. Beware anyone who says things like "thousand of people built their career on [insert theory here] so they will refuse to accept my brilliance..." That's completely nonsensical. If someone were to falsify SR, they'd likely win the Nobel prize. In physics you receive accolades for successfully challenging common knowledge.

Sorry, I mean't to say it *wouldn't* be too difficult.

I suppose it wouldn't be too difficult to learn Newton's law of gravitation, centripetal forces, orbital and escape velocities, etc. You'll need to be familiar with some physics basics first, such as Newton's three laws, energy conservation, and how to solve problems using those concepts. The math background needed for modern astrophysics (namely General Relativity) is well beyond what the average 13 year-old has under his belt. You'll need to be fluent with algebra, trigonometry, basic calculus, vector calculus, classical mechanics (Lagrangian formulation), special relativity, etc., before you'll get anything out of reading a modern textbook. You've got a long road ahead, but don't give up .

Well, if there' no charge other than the ions then they're all going to cancel when you're calculating flux through any symmetrical Gaussian surface (I would use a cylinder).

The stress-energy tensor contains all of the information about the energy and momentum distribution in spacetime. The SET is denoted with the symbol [math]\bold{T}[/math] (bold symbols mean it's a tensor). The Einstein tensor, [math]\bold{G}[/math], contains the information about the curvature of spacetime. The Einstein field equation tell us that the SET is related to the Einstein tensor by the following: [math]\bold{G}=\frac{8\pi G}{c^4} \bold{T}[/math] where G is Newton's gravitational constant (even though it uses the same symbol as the Einstein tensor, they aren't related) and c is the speed of light. So if you know the energy and momentum distribution in spacetime, the Einstein field equation will tell you what the curvature of spacetime is.

I've gotten into this before and don't really want to again, nor do I want to derail the thread; however, I feel obligated to point out that what you're saying here is patently false. The equivalence principle is certainly present in GR, and is in fact one of its core principles. It is equivalent to the statement that spacetime is locally flat and the "comma-goes-to-semicolon rule" which is mentioned in many GR texts. The EP itself does not yield bad predictions - it is the EP + Newtonian gravity which does so.

How do you intend to transfer information by this effect? I.e. how would one group of people know when the other had opened their boxes?

Imagine you have two balls: one red and one blue. The balls are, without you watching, put in separate boxes so that you can't tell which ball is in which box. Now, if you were to open one of the boxes and discover that the ball inside was red, you would immediately know what the color of the ball in the other box is. It doesn't matter how far apart you separate the boxes before you open one of them. That's sort of what entanglement is like. No information is transferred in this process.

Negative pressure isn't negative energy. E.g. would you consider tension on a rope to be negative energy?

The Alcubierre metric would require negative energy - something which (outside of stuff like the Casimir effect) nobody is sure even exists.

You mean like this?: http://en.wikipedia.org/wiki/Vaidya_metric

It's not a problem so much as a simplifying approximation. Physics itself really only deals with the study of simple models. Taking the radiation of a star into account when making gravitational calculations is like taking into account dust particles when measuring the diameter of a basketball (actually, the dust would almost definitely be more significant in relation to the basketball than the radiation from the Sun would be gravitationally).

Yes, ∂tgμv=0, i.e. it is constant in time. This is the last time I'm going to explain it because I'm sick of repeating myself: First of all, GR is a classical theory which doesn't take into account QM effects. There is currently no good theory of quantum gravity, so talking about how quantum effects will change the metric is impossible at this point. The best you can currently do is specify a background metric and describe quantum effects within this background. If these effects are large enough to affect the background, then we have no idea how to deal with them. The Schwarzschild metric is the metric which describes a non-rotating, uncharged, spherically symmetric mass that doesn't change in time. That is all it can and does describe. If, for example, you want to describe a spherically symmetric body with angular momentum then you need an entirely different metric: the Kerr metric. If you want to describe a cloud of dust, you need something like the Lemaître–Tolman metric. If you want to describe really specific situations, you might not be able to solve the EFE's perfectly and will have to resort to numerical methods. The SC metric is not necessarily an accurate representation of reality if you're taking into account small details like radiation. It is useful, for example, to model the Sun with the SC metric. It's obviously not a 100% accurate model since the Sun is not perfectly symmetrical, it's giving off radiation, it's rotating a bit, etc. Nonetheless, the SC metric is a good approximation. The SC metric itself cannot and does not change over time (like I said above: take the time derivatives of its components if you don't believe me). It is derived by specifying the fact that it is t-symmetric. If you want to see a specific derivation, I like Carroll's: http://ned.ipac.calt...3/Carroll7.html. He talks about how the metric is static after equation (7.20).

This is becoming increasingly frustrating; it's starting to feel like I'm talking to a wall. I've already tried to explain to you (multiple times) why what you suggest is invalid. You need to specify how the mass is changing FIRST, and then you can determine the metric it produces. Talking about a changing SC metric is an oxymoron. I don't know how else I can explain it to you. Also, GR is a classical theory, and I don't see at all how this ties in with QM.

I still don't see your point. There are still quite a few differences. Too bad, math doesn't take into account your feelings. Take the time derivative of the SC metric components and you'll see that ∂tgμv=0. Whether or not it is reflective of reality is irrelevant. The point is that the SC metric doesn't change over time. This is a fact. Yes, you're right. I'm not used to working in conventional units, and I forgot the extra c2 factor on the tt component. Yep.

They are equivalent and equivalently useless. You're the one who decided to write the SC metric in terms of energy, not me. As I've said before, I don't see the point. I've already tried to explain this to you: the metric is related to the energy-momentum distribution in spacetime by the Einstein field equations. The SC metric is solved from the EFE's using the assumption that it does not change over time. If the central mass is changing, then the metric produced by that mass is not the SC metric. This doesn't mean that the central mass isn't allowed to change, it just means that the metric it produces will be different from the SC metric. You can't just say "so now we vary the energy..." It doesn't work that way. Agreed. They still look significantly different to me. Mine: Yours: See if you can spot the differences (there are quite a few).

Okay... so what? All you had to do was substitute M=E/c2 to get it in terms of rest energy. I don't see how this is useful. I never mentioned rest mass. My point is that the energy distribution in the SC metric can't change, or else it ceases to be the SC metric. He describes the energy associated with test particles in SC spacetime, not some sort of "metric self-energy" like you seem to be implying. Of course, there is automatically an energy-momentum distribution associated with any metric given by the field equations. The SC metric is a vacuum solution, so all of the components of the stress-energy tensor (including energy density) are zero. I also fail to see how this relates to your original post, or any of mine. Written in the form you have it, it should look like this: [math]c^2 d\tau^2 = (1 - 2\frac{G}{c^2} \frac{M}{r} )c^2 dt^{2} - \frac{dr^2}{(1-2\frac{G}{c^2} \frac{M}{r})} - r^2 d \phi^2-r^2sin^2\phi d\theta^2[/math]

In general, yes, the metric can be time dependent. It doesn't make any sense to talk about changes in the central mass in the Schwarzschild metric because it is derived using the explicit assumption that it does not change over time. If the metric is t-dependent then it is not the Schwarzschild metric. That's not the Schwarzschild metric. I've never seen anything that looks like that before. The Schwarzschild metric is: [math]ds^2=\left (1-\frac{r_s}{r} \right )dt^2-\left (1-\frac{r_s}{r} \right )^{-1}dr^2-r^2d\phi ^2-r^2sin^2\phi d\theta ^2[/math] where [math]r_s=\frac{2GM}{c^2}[/math] Yes, you can get the SC metric in terms of rest energy if you wanted to. No, you're not allowed to do this. If the energy distribution is changing then you need to go back to the Einstein field equations, insert this information into the stress-energy tensor, then re-solve the equations to obtain a completely new metric. What you're doing doesn't make sense. That pdf never once mentions any change in central mass in the SC metric. It is certainly possible to talk about the energy of particles in SC spacetime (so long as they're not energetic enough to disturb the metric), but this says nothing about what you're proposing.