Severian

I don't understant what you mean. What s the 'vacuum fluctuation energy'?

Do you know the definition of velocity and acceleration?

OK, I'll bite. From your abstract: So let's start with the first sentence. What makes you think Electromagnetic theory does not model the electron correctly? Do you have any experimental (or theoretical) objection to Quantum Electrodynamics?

Who thinks field theory is dead? It is very much alive! Indeed Quantum Field Theroy is the foundation of all of modern particle physics!

How many universes would you like?

OK, lets take a look. a) you have done (I presume correctly - I am not going to put in the numbers). b) The Heisenburg Uncertainty Principle is telling us that if we know the position of the particle very well then its energy is rather uncertain (and vice versa). Do we know the position of the particle in this example very well? Can you make any statement at all about where it is? And if you can, what does this say about its energy? c) Notice the statement about room temperature? What would the kinetic energy of the particle be if it was at room temperature? How does this compare with the energy difference between the two levels? And what does this mean?

I think herein lies the root of your problems. The vacuum, is (if the Higgs meachanism is correct) not nothing, as you say. Mathematically this is entirely reasonable. The self interaction of the Higgs field means that having them around costs less energy than removing them. So the vacuum naturally has these fields present. This is contrary to the usual perception of the vacuum as being 'nothing' so may feel 'wrong' to you but this gut-feeling is simply a consequence of using everyday experiences to intuitively think abot the quantum world. In fact, although the Higgs mechanism has not been proven (yet), there are other systems which behave this way, most noticably supersonductors. In a superconductor, the lowest energy state is 'not-nothing' too.

Yes, it is proven.

I don't understand what you mean by the 'not-vacuum'.

There are two important ingredients to the Higgs mechanism: 1. A lowest energy state (i.e. vacuum) where the field content is non-zero. Usually you think of 'lowest energy' as being when there is nothing there. This needn't be the case, and in Higgs models the lowest energy state has a Higgs field present. 2. The field which is present in the vacuum state couples to ordinary particles. This coupling is as if the vacuum sort of holds onto the particles. The Higgs field is always there, and grabs the particles, making them appear massive. In fact, the it is the strength of the coupling which determines the mass. The more strongly coupled to the Higgs field a particle is, the more massive it will be.

Allow me. The Maxwell Equations are: [math] \nabla \cdot \mathbf{E} = 4 \pi \rho \qquad \qquad \nabla \times \mathbf{E} = -\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t}[/math] [math] \nabla \cdot \mathbf{B} = 0 \qquad \qquad \nabla \times \mathbf{B} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}+ \frac{4\pi}{c} \mathbf{J}[/math] where [math]\mathbf{E}[/math] and [math]\mathbf{B}[/math] are the electric and magnetic fields, [math]\mathbf{J}[/math] is the current and [math]\rho[/math] is the charge density. For simplicity lets imagine that the Electric field is constant with time, so that [math] \nabla \times \mathbf{B} = \frac{4\pi}{c} \mathbf{J}[/math] Now, remember that the cross-product of two vectors is at right angles to both of the vectors, so [math] \mathbf{B}[/math] must be at right angles to [math] \nabla \times \mathbf{B}[/math] and therfore must also be at right angles to [math]\mathbf{J}[/math]. In other words, the magnetic field is at right angles to the current. A more interesting question is where do these laws come from....?

[math]\Omega_k[/math] is (I think) simply how much [math]\Omega_{\rm tot}[/math] is deviating from 1 (I think it is [math]\Omega_{tot}=1-\Omega_k[/math] but I would have to check...). So, it is showing more of a deviation than before, but it is not statistically significant.

I could have parked the second one just as fast as she did, but in the traditional manner. One thing I have learned from living in a car crowded city is that I can get my car into pretty much any space in which it will physically fit.

Funnily enough I just posted about the Cosmological constant in Jim's thread. To explain further, you can just add a gravitational constant in my hand, just as Einstein did. Lets call that [math]\Lambda_0[/math]. But, since electroweak symmetry is broken the vaccum-energy associated with the Higgs field should also provide an effective contribution to the cosmological constant. when you work it out this contribution is [math] \Lambda_H = (10^{19} {\rm ~GeV})^4 = 10^{76} {\rm ~GeV}^4[/math] (1019 GeV is the Planck scale) so the total cosmological constant is [math] \Lambda_{\rm tot} = \Lambda_0 + \Lambda_H[/math] But we know that [math]\Lambda_{\rm tot} \sim 10^{-47} {\rm GeV}^4[/math] so we have to put in by hand a [math]\Lambda_0[/math] which has the opposite sign of [math]\Lambda_H[/math] but has the same magnitude up to the 123rd significant figure, where it must start to deviate. This is incredibly hard to justify. Having the same magnitude exactly would be fine, (you just need a symmetry) so this was not a problem when the cosmological constant was zero, but getting it not quite zero is a big problem.

Hopefully it will teach them that drinking alcohol (in moderation) isn't something to be ashamed of.

I do. In fact, out of all the states which have nuclear capability, I think Israel is the one most likely to use them. Given that they are rather geographically close to home, I am not comfortable at all. (Although, to be fair, they would probably be fired away from Europe... I hope.)

That would be consistant with a flat universe, so I am not sure what you mean? If it is not exactly flat then you will have another fine tuning problem - why would it be so close to 1 if it is not equal to 1?

Pretty much. They are slightly different in that their weighting when you measure a distance is with the opposite sign for time as compared to time, but that is a detail.

No, no no! These were nominations, not votes. So it doesn't make any difference how many times someone is nominated - it is only important that they are. So the vote for most helpful for example should now allow voting between 5614, Bascule, Cap'n Refsmmat, Dak, ecoli and IMM. At least that was how I understood 'nominations' to function!

Yes, I think it is. As I said before it is not terribly clear what the question is after. Since [math]\Psi = A[/math] is a solution of the Schroedinger equation, technically the correct answer is 'yes, they are solutions, with values of k and w given by ....'. But I suspect the questioner wants you to say 'no' because [math]\Psi=A[/math] is not physical. The 'eigenstate stuff' is just terminology. By Eigenstate, I mean a solution of the equation with a definite value of w and k. So your [math]\Psi=A[/math] is an eigenstate with w=k=0.

Black holes suck.

Neither is [math] \psi = A e^{i(kx-\omega t)}[/math] but it is a solution.