elfmotat

As the others have mentioned, plugging relativistic mass into the Newtonian gravity equation is meaningless.

It depends on what you mean by "electricity." If you mean the power (energy per unit time) produced by an electric current (i.e. a current through a wire), then it's simply: [math]P=IV[/math] Where [math]I[/math] is the current in the wire and [math]V[/math] is the induced voltage in the wire. If you mean the energy-density (energy per unit volume) of the electric field, it's just: [math]u=\frac{1}{2}\epsilon_0 |\mathbf{E}|^2[/math] where [math]\epsilon_0[/math] is the vacuum permittivity and [math]|\mathbf{E}|[/math] is the magnitude of the electric field.

First of all, that site is a crackpot site, and I advise you not to take anything said there seriously. (As an aside, this is the second time I've seen you link to a blatant crackpot's page. Is there some reason you're reading nonsense like this?) To answer your question, I can see how one might naively think that the two could be related. After all, they both follow an inverse-square law. Since the coupling constant in Coulomb's Law is so much larger than the constant in Newton's Law of Gravitation, you might be lead to believe that gravity is caused by some small residual charge that adds up due to positives and negatives not completely canceling. But this immediately runs into problems. Since Coulomb's Law is repulsive for like-charges the obvious conclusion is that, if gravity were caused by EM forces, then we'd expect to see everything repelled by everything else. After all, why would charges add up differently in the Earth and in the Moon? They're made of roughly the same stuff. A better answer would be because the EM-field couples to a vector (rank-1 tensor) current, while gravity couples to a rank-2 tensor current. Newton's Law and Coulomb's Law are just static approximations of GR and EM. You can't get all of the effects in GR from a vector theory.

Ah, but that's the beauty of it: the length of the curve is invariant with respect to any choice of affine parameter. This should make sense - why would proper time (a measurable quantity) depend on how a physicist chooses to parametrize the clock's curve? So you can choose your parameters based on how well they simplify your calculations. So, for your example of proper time, yes it is completely valid to use proper time as the parameter: [math]\tau=\int \sqrt{-g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}~d\tau[/math] But if you can write [math]\tau[/math] so that it is a function of some other parameter (for example you can use coordinate time [math]t[/math], so [math]\tau = \tau (t)[/math]), then via chain rule we have: [math]\tau=\int \sqrt{-g_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}}~d\tau=\int \sqrt{-g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dt}{d\tau} \frac{dx^\nu}{dt} \frac{dt}{d\tau}} ~d\tau=\int \sqrt{-g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt}} ~\frac{dt}{d\tau} d\tau=\int \sqrt{-g_{\mu \nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt}} ~dt[/math] If you have a diagonalized metric, then the above form (parametrized W.R.T. coordinate time) can greatly simplify some problems. So, to answer you question more explicitly, yes you can use proper distance [math]ds[/math]. But you can also use any affine variable which might make your calculations simpler. I.e. if I was working in Schwarzschild coordinates, I would likely use the radial coordinate [math]r[/math] to parametrize radial distances.

I don't think anyone is going to write a textbook-length explanation of General Relativity. That's what textbooks are for. Informally, the basic idea is that the geometry of spacetime is connected to the energy-momentum distribution in spacetime by the following equation: [math]\begin{pmatrix} curvature\\ of\\ spacetime \end{pmatrix}=\begin{pmatrix} energy ~\\ momentum\\ density \end{pmatrix}[/math] The actual equation is called the Einstein Field Equation: [math]G_{\mu \nu}=\frac{8 \pi G}{c^4}~T_{\mu \nu}[/math] where [math]G_{\mu \nu}[/math] is the Einstein curvature tensor, [math]T_{\mu \nu}[/math] is the stress-energy tensor, [math]G[/math] is the gravitational constant, and [math]c[/math] is the speed of light. Once you know the geometry of spacetime, you can determine how things will move. If you remember Newton's first law: "If an object experiences no net force, then it travels along a straight line with constant speed." This law still applies, but now the idea of a "straight" line in curved spacetime becomes rather vague. The generalization of a straight line into curved space is called a "geodesic." It's essentially the "straightest" possible line you can draw. The modified version Newton's First Law in curved spacetime becomes: "If an object experiences no net force, then it travels along a geodesic in spacetime." This creates the illusion that objects are pulled toward massive objects due to some gravitational force. What's really happening is that everything is simply trying to follow a straight line in curved spacetime. That's pretty much the basic idea.

Locally, yes, it's always the same. However if you make your measurements over a large region of spacetime that is significantly curved, you may find that the measured speed of light will differ from the constant c.

This looks like a job for matlab.

I, for one, am particularly stunned that this discussion has lasted for well over 100 replies. It's obvious to anyone with a pulse that Universal Theory's posts are nothing but meaningless poppycock. He has long bouts of rambling word-salad mixed in with nonsensical "equations," and consistently fails to answer even basic questions in a coherent manner. Are you guys still responding because you're having fun reading his gibberish? Because there's just no way anyone could possibly expect anything cogent at this point.

I've heard of the names of derivatives up to snap, crackle, and pop before. The others (including "absement") I've never come across, though pop, lock, and drop made me smile.

In quantum mechanics observables like energy and momentum are replaced by operators. In particular, momentum is replaced by the operator: [math]p=-i\hbar \frac{\partial }{\partial x}[/math] This means that the kinetic energy operator is: [math]T=\frac{p^2}{2m}=-\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial x^2}[/math] The wavefunction [math]\Psi[/math] associated with a particle must satisfy the Schrodinger Equation (a partial differential equation), which simply tells us how the time derivative of the wavefunction is related to the Hamiltonian (total energy) of the particle: [math]i\hbar \frac{\partial \Psi}{\partial t}=H\Psi =(T+V)\Psi=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}+V\Psi[/math] What quantum mechanics also tells us is that, if we solve Schrodinger's Equation for the wavefunction, then the probability of finding the particle between the points x=a and x=b is given by: [math]P(a\leq x\leq b)=\int_a^b \Psi^* \Psi~dx[/math] where [math]\Psi^*[/math] is the complex conjugate of the wavefunction. Using this knowledge, we can determine the expectation value of kinetic energy. If you were to do a bunch of repeated identical experiments to determine the kinetic energy of the particle, you would find that the measurements "cluster" around the expectation value. It's a sort of average value. In line with that intuition that it's an average value, we determine the expectation value of the kinetic energy to be: [math]\langle T \rangle =-\frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \Psi^* \frac{\partial^2 \Psi}{\partial x^2}~dx[/math] If you're familiar with any linear algebra, QM also tells us that the value of a measured observable must always be one of the operator's eigenvectors (which are usually called "eigenstates" by most authors). So, for your example of kinetic energy, the values that it's allowed to take on are given by: [math]-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi }{\partial x^2}=\tau \Psi [/math] So the only values that the kinetic energy is allowed to have are the values [math]\tau[/math] which satisfy the above equation.

Something with units of [length]*[time] with little to no physical significance.

Consider the straight line (x,y,z)=(t,t,0) where t is some parameter. (In two dimensions this is just the line y=x.) Now let's change our coordinate system by rotating it +45 degrees around the z-axis. In this coordinate system the line is given by (x,y,z)=(t,0,0). So it's just a line which is restricted to the x-axis. In general, for a line you can always rotate or translate your coordinates in such a way that the line coincides with only one axis.

As I said, gravitons and curved spacetime make (AFAIK) equivalent predictions. So whether gravity is "really" caused by gravitons in flat spacetime or whether it's "really" caused energy-momentum changing the geometry of spacetime is not a question physics can answer.

Good question. What we find is that the "propagator" (which gives the probability [amplitude] of a particle to travel from one place to another) is never spread outside of the light-cone. In other words, if a photon is emitted at time t=0 at x=0, there is zero probability that the particle will be found at a location where x>ct.

The product of position and (canonical) momentum uncertainty must always be greater than or equal to [math]\hbar /2[/math]. So if you make (for example) the uncertainty in position close to zero then the uncertainty in momentum has to be very large for their product to be above the required number. If you take the limit where uncertainty in position goes to zero then the uncertainty in momentum grows to infinity.

Please keep crackpottery like this out of the discussion. Evidently that guy doesn't know the difference between proper time and coordinate time.

http://en.wikipedia.org/wiki/Pp-wave_spacetime

By "it's not" I meant "it's not 'real' mass." Total energy is certainly frame dependent.

Yes, with Quantum Electrodynamics (QED). You will probably find the following video series informative:

Taylor's book is very good. It's probably my favorite classical mechanics text. Halliday and Resnik is good as well.

Well, sort of. A curved spacetime is physically equivalent to a spin-2 massless field on a flat background spacetime.(For reference see http://arxiv.org/abs/astro-ph/0006423 ) They both predict the same things. The quantization of a spin-2 field would be gravitons, though there's no currently accepted model of quantum gravity. In string theory, a graviton (which mediates the gravitational force) can be removed by replacing it with a correction of the background spacetime. So the two are equivalent in this context as well. There's also the possibility that gravity isn't a fundamental force at all, and that its' actually an entropic phenomenon. A good place to start on this would be http://en.wikipedia.org/wiki/Entropic_gravity . Ultimately though, physics doesn't tell us what's "really" happening. It gives us a mathematical model which quantitatively describes phenomenon. So whether or not spacetime is "actually" curved cannot be answered with physics. The best we can do is say, "the data we observe agrees with what we should find if spacetime is curved."

Sea slugs enjoy carbon-based adult toys.

It's not. It's just total energy (which is frame-dependent) divided by c2. It's just measuring total energy in units of mass.

*sigh* you're still arguing a strawman. The hint that that wasn't meant to be taken literally was the quotes around the word "really." My only point was that GR is more accurate than Newton, which OP didn't seem to be aware of.

This is getting annoying now. YOU were the one who brought up how Bell rules out some hidden variable theories. The only hidden variables Bell applies to are local. Your original comment on the article had nothing to do with the 2007 experiment, and my response was completely valid. It seems like you're just trying to find nonsense to argue with.