elfmotat

The big bang is irrelevant to the conversation and your argument makes absolutely no sense.

dt doesn't approach zero, that doesn't make any sense. Δτ approaches zero as v approaches c. All observers measure nonzero finite Δτ for Earth. And, as swansont said, an observer traveling at c doesn't make sense, and is not a valid frame.

Nature is essentially "lazy." If potential energy gets converted into kinetic energy (i.e. the potential does work on the particle) then this is reflected by a change in the Lagrangian: L=T-V. Nature doesn't want to do work, so the Lagrangian should change as little as possible over the particle's trajectory. The action is the changes in the Lagrangian added up over the entire trajectory, i.e. essentially how much work was done over the particle's path. So the path that minimizes the action is the "laziest" possible path.

To be honest, I'm not really sure what the topic is. As far as I can tell, it seems to be a fallacious argument about how Hawking is wrong. I have yet to see a good point a the from the OP. I also see no "bitching," as you describe it. Most threads of this quality would have been moved to speculations at this point.

Balulu, welcome! Unfortunately, I have no idea what you're talking about. There's no "minimim quantum energy of the universe."

That's one of the most facepalm inducing things I've ever read.

That was the point of my post.

The problem with your question is that interpretations can't be verified by experiment. Science does not and cannot answer the question of what's "really" happening. Using a certain interpretation may be advantageous when developing a theory, but that doesn't necessarily mean it's correct. Two theories based on vastly different interpretations can describe the same phenomena.

Do you need a PhD to be able to make valid commentary on topics like these? Certainly not. Do you need to have some basic knowledge of what you're talking about? Obviously. This is not stuff that laymen can "debunk" with common-sense.

It sounds like you're wondering how gravitons can exist, or why they're even needed, if general relativity models gravity as a result of the geometry of spacetime. There are two main interpretations that people take when approaching quantum gravity. One of these is that gravity "really is" a result of the spacetime geometry, so quantizing gravity should just be quantizing spacetime in some manner. This is the approach of Loop Quantum Gravity. The other interpretation that gravity is caused by a spin-2 massless field on a flat spacetime background, similar to the way electromagnetism is caused by a massless spin-1 field. The properties of a spin-2 field give the "illusion" that the background spacetime is curved. The excitations of this field are gravitons, just like photons are excitations of the electromagnetic field. This is the approach taken by string theory.

BILL! BILL! BILL! BILL! BILL! Bill Nye the Science Guy! Anyway, it depends on what you mean really. The "inertial coefficient" called "mass" can cause clocks to tick at different rates while they are placed near massive objects.

Which curvature tensor (Riemann, Ricci, Weyl)? And in what situation? In a vacuum where Tab=0, then by the field equations you get Rab=0. So in a vacuum the Ricci tensor vanishes. On the other hand, the Riemann and Weyl tensors do not generally vanish.

Time is what we measure on clocks. Try to go any deeper and you're no longer doing science, you're doing unfalsifiable philosophy.

One of the worst things ever made. See: http://en.wikipedia....ademic_reaction This as well:

Every calculator has a point at which they round off. Google calculator seems to round off earlier (less accurate), calculate for free rounds of a bit later (a little more accurate), and wolframalpha rounds off even later.

The best free online calculator you'll find is WolframAlpha: http://www.wolframalpha.com/input/?i=0.104384134+*+9.58

Uh... yes you can.

Here's how I'd think about it: the units we use are rather arbitrary. Humans (originally) defined the kilogram, the second, and the meter in terms of familiar mass/time/length scales. Since Newtons (force) are derived from those units, the Newton is also a rather arbitrary unit. The Law of Gravitation therefore pretty much requires a constant of proportionality in order to scale length, time, and mass into our arbitrary units. Planck units are much more fundamental: working in terms of Planck units means setting the speed of light, Planck's constant, and the gravitational constant all to one. Length, time, and mass units are defined in terms of fundamental constants. So, working in terms of Planck units, you can indeed say that simply F=m1m2/r2.

Quick thing: use \langle and \rangle when using bra-ket notation. Makes things nicer: [math]P(\omega )=|\langle\omega |\psi \rangle|^2[/math]

Dark matter isn't predicted by the standard model, and it's more than reasonable to think that DM may consist of particles.

Source? EDIT: Nevermind:http://blogs.scientificamerican.com/guest-blog/2011/08/23/a-higgs-setback-did-stephen-hawking-just-win-the-most-outrageous-bet-in-physics-history/

"Wave-particle duality" is often a misleading name. Electrons and photons are quantum mechanical particles. They do not behave in the same way that classical particles do, and indeed can't really be explained in terms of anything familiar to you. The particles have an associated "wave function," which is essentially a wave of probability. The wave function gives the probability of finding the particle in a given region of space, and it exhibits normal wave characteristics (i.e. it can interfere, etc.). Asking whether or not it is a particle or a wave is nonsensical. Whether or not it behaves more like a classical particle or a wave depends on its energy. The more energetic the photon, the more it will exhibit "particle-like" behavior, and vice versa. You pretty much answered your own question, didn't you? But yes, a de Broglie wave has momentum [math]\vec{p}=\hbar \vec{k}[/math] where k is its wave vector ([math]|\vec{k}|=2 \pi /\lambda[/math]), and ħ is the reduced Planck constant, h/2π. It is a particle whose associated wave function is a standing wave.

Except gravitationally, of course .

I've never heard of anything like this, but it seems like nonsense to me.

"unless " But that simply isn't true. by definition. Unless you in some way object to this definition, your "condition" makes absolutely no sense. Right, but we aren't exactly talking about Newton's law here, are we? We're talking about the partial time derivative of an operator, not a variable or a function. Don't you understand the distinction? Even if that weren't the case, you can't tacitly claim that ∂f/∂t=0. I.e. consider the momentum function p(t)=p0+kt (representing constant force). Do you think ∂p/∂t=0? First of all, the Ehrenfest theorem has to do with expectation values which isn't what we're talking about. Second, do you or do you not recognize that ?