elfmotat

Newtonian gravity is not very accurate when you're close to a strong field source like a black hole. The rough equivalent of the Newtonian "g" value actually blows up to infinity at the event horizon.

Once you're familiar with Vector Calculus, Linear Algebra, and Differential Equations you're ready to start tackling the more complicated fields of physics. Most physics textbooks assume you're already competent in those three areas and will introduce any additional math as needed. And, of course, you need to be familiar with Classical Mechanics, including the Lagrangian and Hamiltonian formulations. I like John Taylor's book for Classical Mechanics. You'll need to study Special Relativity before anything else. It's definitely the easiest to learn, and really doesn't require too much math. Taylor and Wheeler's Spacetime Physics would be my recommendation for SR. From a math standpoint, QM is probably the next easiest. Griffiths' book would be my recommendation. You don't actually need to know any differential geometry before you start studying GR (though it would be invaluable) because most intro textbooks will introduce the math as you go along. Schutz's book would be my recommendation for a beginner. Once you're familiar with all of those + Classical Field Theory (the short intro in Carroll's GR textbook should be enough CFT) you'll be ready to start tackling QFT. Peskin and Schroeder's text is generally recommended here.

I should also point out that the GEM equations aren't Lorentz invariant because mass density and mass-current density (momentum density) don't form a Lorentz invariant vector because they are some of the components of a rank-2 tensor: the stress-energy tensor. This is where gravity and EM differ, because charge-density and current density form an invariant four-vector (rank-1 tensor): four current. Thus, Maxwell's equations are compatible with relativity while the GEM equations are not (though they do agree to a good approximation). The "correct" form of linear gravity is Linearized GR, in which the metric is a perturbation of the flat Minkowski metric: [math]g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}[/math] If we compute the Einstein Field Equations in terms of this metric and drop all higher order terms of huv (and impose a gauge condition), then we obtain the Linearized Field Equations: [math]\square \bar{h}_{\mu \nu}=-\frac{16\pi G}{c^4}T_{\mu \nu }[/math] where [math]\square=\left (\nabla^2-\frac{\partial^2 }{\partial t^2} \right )[/math] is the d'Alembert operator and [math]\bar{h}^{\mu \nu}=h^{\mu \nu}-\frac{1}{2}\eta^{\mu \nu} h^\sigma_{~\sigma }[/math]. These field equations are Lorentz invariant. We also see that if we set the stress-energy tensor to zero (a vacuum) then we obtain a wave equation: [math]\square \bar{h}_{\mu \nu}=0[/math] So a gravitational wave is really just oscillating perturbations in the metric. We can actually derive the GEM equations from the Linearized Field Equations by specifying a coordinate system and setting various components of "h" equal to scalar and vector potentials.

If the shape of the tissue paper is changing in time then the center of mass will also change in time. If you know the mass distribution as a function of time [math]\rho (\mathbf{r},t)[/math] then you can calculate the center of mass as: [math]\mathbf{R}(t)=\frac{\int_V \rho (\mathbf{r},t)~\mathbf{r}~\mathrm{dV}}{\int_V \rho (\mathbf{r},t)~\mathrm{dV}}[/math]

You're over-simplifying. We know how to do quantum physics in curved backgrounds, but (as ajb said) when it comes to actual quantum gravity we don't know how it works yet.

Feynman had a knack for being able to intuitively explain many complicated ideas, but not even he could do that with quantum mechanics. He approached QM from the "shut up and calculate" perspective. In The Feynman Lectures he explains multiple times why the reader should stop asking "but what does this mean(?)" or trying to understand it intuitively and simply learn the math. He was big on emphasizing that he could tell you how nature works, but he could not explain it in a way that you will understand because he didn't understand it either.

I would find it incredibly difficult to believe that gravitational waves don't exist, because they're not only required by General Relativity (which has passed every experimental/observational test to date) but they're also required by much-less-complicated linear gravity theories such as Linearized GR and Gravitomagnetism - both of which are valid in the weak-field limit. If you're familiar with electromagnetism and Maxwell's equations, the easiest way to see this is with Maxwell-esque GEM equations. If frame-dragging exists (which Gravity Probe B has confirmed to ~15% uncertainty) then these effects can be encapsulated in a "gravitomagnetic field" Bg. The gravitomagnetic field is produced by mass-current in the same way that in EM the magnetic field is produced by charge-current. The gravitomagnetic field along with the traditional Newtonian "gravitational field" (which we'll denote by Eg) have four Maxwell-esque equations which determine how the fields are produced: [math]\nabla \cdot \mathbf{E}_\text{g} = -4 \pi G \rho[/math] [math]\nabla \cdot \mathbf{B}_\text{g} = 0 [/math] [math]\nabla \times \mathbf{E}_\text{g} = -\frac{\partial \mathbf{B}_\text{g} } {\partial t}[/math] [math]\nabla \times \mathbf{B}_\text{g} = -\frac{4 \pi G}{c^2} \mathbf{J} + \frac{1}{c^2} \frac{\partial \mathbf{E}_\text{g}} {\partial t}[/math] In a vacuum these reduce to: [math]\nabla \cdot \mathbf{E}_\text{g} =0[/math] [math]\nabla \cdot \mathbf{B}_\text{g} = 0 [/math] [math]\nabla \times \mathbf{E}_\text{g} = -\frac{\partial \mathbf{B}_\text{g} } {\partial t}[/math] [math]\nabla \times \mathbf{B}_\text{g} = \frac{1}{c^2} \frac{\partial \mathbf{E}_\text{g}} {\partial t}[/math] Now you apply the same method for finding the differential equations for gravitational waves that you would to find the equations for EM waves (using the curl of the curl "trick"): [math]\nabla^2 \mathbf{E}_\text{g}= \frac{1}{c^2} \frac{\partial^2 \mathbf{E}_\text{g}} {\partial t^2}[/math] [math]\nabla^2 \mathbf{B}_\text{g}= \frac{1}{c^2} \frac{\partial^2 \mathbf{B}_\text{g}} {\partial t^2}[/math] The solution to these equations are plane waves.

He's known mainly for singularity theorems in General Relativity and his work in Cosmology. He's also well known for Hawking radiation, which is a consequence of Quantum Field Theory in curved spacetime. I wouldn't classify his area of expertise as "quantum mechanics."

I'm sorry, but I really have no idea what you're asking. Maybe you could try rephrasing the question?

Imatfaal, the second term in the last of the four equations should be positive. If it's not positive then you can't get the wave equation out of Maxwell's equations.

I believe M-Theory which unifies the five different string theories requires 10+1 dimensions.

I was able to resolve this with an another (anonymous) individual. (1) The factor of 1/2 in front of the second term is incorrect. (2) it turns out that the second term on the RHS of the equation isn't gauge-invariant because the first term isn't either! A gauge transformation is equivalent to a change in the 5th coordinate, so the first term on the RHS becomes: [math]k F^{\lambda}_{~\mu} \frac{dx^{\mu}}{ds} \frac{dx^4}{ds} \rightarrow kF^\lambda_{~\mu} \frac{dx^\mu}{ds} \left(\frac{dx^4}{ds}+\frac{d \zeta}{ds}\right)[/math] The second term becomes: [math]k A_{\nu}F^{\lambda}_{~\mu}\frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds} \rightarrow k (A_{\nu}-\partial_{\nu} \zeta)F^{\lambda}_{~\mu} \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds} = kA_{\nu}F^{\lambda}_{~\mu} \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds}-kF^{\lambda}_{~\mu} \frac{dx^{\mu}}{ds} \frac{d \zeta}{ds}[/math] So the entire RHS is actually gauge-invariant. (3) The Noether conserved momentum about the "curled up" 5th dimension turns out to be: [math]k\left (\frac{dx^4}{ds}+A_\mu \frac{dx^\mu}{ds} \right )[/math] So if we identify this conserved quantity with electric charge, we obtain the standard Lorentz force law with no additional term.

You need to learn the basics of vector calculus and linear algebra before you start tackling subjects like GR. SR can be done by (mostly) algebra alone, so if you want to learn about basic time dilation due to relative velocity then it shouldn't be too difficult.

The other person is right - I didn't give it enough thought. The change in flux in the first scenario comes from the fact that the field in the cross-section of the coil is gradually decreasing, where in the second scenario the total flux through the cross section shouldn't change. So you shouldn't get any EMF.

But that wasn't your original point. You were saying that you need physical experiments to obtain valid equations, and swansont pointed out that this is true only when they contain tunable parameters. Obviously your equations need to agree with experiment for them to be meaningful, but nobody was arguing about that.

To give specific numbers you'd have to know the field B(x,y,z) that the magnet produces. I don't think the outcome of the experiment would be too different - you should get an induced EMF in the coil. EDIT: This is wrong - ignore.

Interestingly a satellite orbiting very close to the Earth's surface has the same period . Also, that equation assumes the Earth is of uniform density. For a more accurate discussion about the falling through a hole in the Earth, see DH's post in this thread: http://www.scienceforums.net/topic/64539-earth-hole/

This came up in another forum, and as of yet we weren't able to come up with an answer. Someone posted the following pdf about KK theory: http://www.weylmann.com/kaluza.pdf . In it, the Lorentz force law is derived from the 5D geodesic equation. What was puzzling us was the appearance of an additional term: [math]\frac{d^2x^\lambda }{ds^2}+\Gamma^\lambda_{\mu \nu }\frac{dx^\mu}{ds} \frac{dx^\nu}{ds}=-kF^\lambda_{~\mu}\frac{dx^\mu}{ds} \frac{dx^4}{ds}-\frac{1}{2}kA_\nu F^\lambda_{~\mu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}[/math] This is troubling for two reasons: (1) there's no experimental evidence to suggest deviations from the traditional Lorentz force law, and (2) it is directly dependent on the four-potential which suggests it is not gauge-invariant and that the potential is directly observable. Any insight would be appreciated.

We know that the rate of time is affected by gravitation and motion from experiment. Indeed every time you use a GPS system you are directly confirming the predictions of relativity, because without taking into account the affects of time dilation our GPS systems would be drastically off. You should stop talking about what should happen, or what seems logical to you. Nature doesn't care about your naive biases.

First of all, that's a terribly arbitrary value for the photon's frequency. Second, what you've derived is the relativistic mass of a photon, which as far as most people are concerned is basically useless. Your answer is actually frame-dependent because the frequency changes when you transform to another reference frame. The rest mass of a photon (which is what matters) is zero.

You seem to be making a rather odd distinction between "dimensionally sound" and "dimensionally correct." I'm not sure what the difference is between the two, so perhaps you could elaborate.

I'm quite sure that Feynman himself would agree that he was never the expert in anything. Physics has many contributors each with a different range of influence on the field. There is not, nor has there ever been (not in the past few hundred years at least), a single "figurehead" in the physics community whose opinion counted above all others.

Let's say that Tim is using a rocket ship to hover at a constant distance from the black hole for simplicity (or else you have to take into account time dilation due to velocity). As long as the black hole is significantly far away from Earth, the approximate relation between the times experienced by each person's clock is: [math]t_{Jim}=\frac{t_{Kim}}{\sqrt{1-R_\oplus/r_\oplus}}=\frac{t_{Tim}}{\sqrt{1-R_{BH}/(R_{BH}+r)}}[/math] where [math]R_\oplus[/math] is the Schwarzschild radius of the Earth, [math]r_\oplus[/math] is Earth's radius, [math]R_{BH}[/math] is the Schwarzschild radius of the black hole, and [math]r[/math] is (approximately) the distance from Tim's ship to the black hole's event horizon.

I don't know what you mean by "there's no time dilation." I don't see anything to indicate that in the figure you mention.