Aeschylus

No Lorentz-Fitzgerald contraction is real by any objective standard, there is no reason to think of the proper diemnsions of the disc as it's actual dimensions.

http://www.sciencemag.org/cgi/content/full/278/5336/226

Lets just say that the model has alreadybeen proposed an analysed in some detail.

Speicla relativty can handle accelarted frames, it's a bit of a myth that it can't. Anyway look at this site as it explains why the spatial slice can no longer be Euclidean: http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

8, because if z = 3x + 5y, then z + 1 = 3(x +2) + 5(y - 1) = 3(x - 3) + 5(y + 2). and any z greater than or equal to 8 must contain at least 1 y term or at least 3 x terms.

By using relativty what you are infact doing is changing the geometry. In Euclidean geomerty the ratio of a circles diameter to it's circumference is a constant.

Yes, but the matter coming from them, in the form of stellar winds for example will meet.

There is a thoey of galaxies and antigalaxies, but the problem is that there is no way to prevnt the matter from galaxies and antigalxies meeting; there would always be a zone between galaxy and an adjacent antigalaxy where matter from the two would meet and anihilate leading to the emisson of gamma rays in huge quantities. No such zones are observed between galaxies.

Yep, but you're not measuring weight tho', which is what the original question asked.

1) We have certainly seen firm evidnece for quarks in scattering experiments, so it's safe to say they exist. 2) The speed of a quark, is within the usual relativsstic limitations for tardyons.

Weight is not the net force on the balloon it it is the force due to the gravity of the Earth, so even if the helium balloon moves upwards it's weight is still in the same direction.

It's rubbish: even helium balloons have 'postive weight' as weight is defined as mg and 'zero' is ther trivial solution and should be ignored.

Rubbish Dr.Belfrey: 1) you're being irrelevant by talking about whether infinity is real 2) If you mean pi isn't irrational you are very much mistaken and should be pummeled. Pi has been proven to be irrational, this is nothing to do with coputers calculating the nth digit of pi.

27kg + 3kg on one side 1kg + the object you wnat to weigh. For 2kg, 3kg on one side and 1kg plus the object you want to weigh on the other.

indeed, infact it's usually introduced as a method of solving s. linear equations.

Use Gaussian elimination to determine the rank.

I believ you can weigh any weight between 1kg-40kg, though I haven't checked.

5614, did I not tell you before that quantum teleportation, provides no useful means of transferring information as you still need a classical commuincations channel.

Anyone predicting a few decades is going to be way off, we haven't even buitl one that is remotely useful: Take the example quote facorizing 15: to give you some perspective on this, for numbers as small as 15 the quantum algorithm is actually about 150 times slower than the classical algorithm!!!! The quantum algorithm only comes into it's own when when factorizing very large numbers (even for a number like 7,857,567,342 the classical algorithm is faster). QUantum computing has many, many hurdles to overcome before it is viable, inafct Dr. Kaku's prediction is probably on the optimistic side.

The relativstic version of this I guess would be: [math]\hat{H}\Psi = \sqrt{\hat{P}^2 c^2 + m^2c^4}\Psi[/math] Which is not actually a very useful equation and is only half-way to the deriving the Dirac equation as the Dirac equation can't be derived directly from the Schroedinger equation and the Lorentz transformation, you've got to do a little extra work. Infact it looks more like the Klein gordon equation which is: [math]\hbar^2\frac{\partial^2\psi}{\partial t^2} + [c^2\hat{P}^2 + m^2c^4]\psi = 0[/math]

I can see a couple of fudges: You've change the energy operator into the Hamiltonian operator in the second equation(For historical reasons the operator whose eigenvalues are the energy of the system is called the Hamiltonian operator represented as a 'H' plus the usual notation to indicate that it is a Hermitian operator) which describes the total energy of the system. You've also tacitly assumed that the Hamiltonian is time-independent, i.e. that the total energy of the system is conserved (time-depenednt Hamiltonians are extermely difficult to deal with in general and solutions are usually obtained perturbatively). Also anohter very minor fudge is that your delaing with the one-diemsnional case, but the extension to 3 dimensions is obvious. You've also forgetten to square the momentum operator in your restaemnt of the Shroedinger equation, but I assume that's just a typing error. edited to add: and I've missed the most obvious thing, as F.K points out you've lost a minus sign!

That is probably what those who originally formulated the Copenhagen interpretatipn would of said, but it has been the orthodox view in QM fdor many years now that QM applies at all scales (we do not see quantum effects at the macroscopic level mainly due to statistical effects).

The derivitaion does not come from considering electrons alone, but 'matter waves' (i.e. massive particles obeying the De Broglie relations as all masive particle should do) in general. What I'd guess you'd get is this (if you look at the derivation you find for light waves you can stop before you have to actually postualte anything new): [math]\frac{\partial^2 \Psi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2\Psi}{\partial t^2}[/math] (note: this is a partial differential equation as psi is a function of both x and t) which is just the equation for a classical plane wave (and obeys the relationship, from De Broglie, E = pc) which has the solution: [math]\Psi(x,t) = \Psi_0 e^{i(kx - \omega t)[/math] and it's only the real part of Psi that we're interested in to give us th dispalcement (classically). edited to add: I've just looked up the equation and does indeed take the form of the classical wave equation above.

Wrong. You simply cannot talk about elemnatry particles cretaing singularities as not enough is known Clearly the particle will not have a defintie postion most of the time, but you still cannot use this to assign it a radius. We can be certain that our value for the elctrons mass is correc within experimentla error. Though experimentally we can only ever put an upper limit on the radius of an electron, in the standard model electrons are point particles, infact the standard model gets into trouble if they are not. This is because if electrons are not point particles they will be subject to Lorentz-Fitzgerald contraction which will have a definite and noticeable effect and also I beleive it would violate gauge invariance.

Schroedinger's equations describes any particle, all you need to do is simply feed the informtation such as the potential and mass into the Schroedinger equation and solve it in the usual manner. rememebr tho' make sure the wavefunction meets it's boundary conditions.