Tom Mattson

No, relativity does not imply that. Work it out explicitly (without the handwaving this time). For simplicity, take the paths to be linear, antiparallel over-and-back paths. That's an amazing conclusion since at no point did you: 1. explicitly solve the problem or, 2. ever make use of the definition of a hypersphere. I'll echo what Janus said to you: You should stop trying to explain relativity to anyone until you understand it yourself. Someone will let you know when that is.

I did explain why, and I did keep it simple: Time dilation simply does not necessitate velocity changes! It is a result derived for nonaccelerating frames. Instead of me trying to explain this to you, why don't you explain to me why you think time dilation does require velocity changes, and therefore energy?

But this is wrong. Time dilation simply does not necessitate nonconstant velocities. Time dilation is a derived result for inertial (nonaccelerating) frames. No energy is required to sustain that kind of motion.

Sure they are. Newton's laws adapt to relativity just fine, but of course the 2nd law is in 4-vector form.

Unfortunately' date=' no. 11 dimensional pictures cannot be represented on a 2 dimensional surface (like a piece of paper or a computer screen). It can be hard enough to represent 3 dimensions on a piece of paper! It's not yet clear that a quantum theory of gravity will insist that time began at the big bang. That inference (I think) is an extrapolation back to the Big Bang singularity, under the assumption that GR is rigorously correct in all its particulars. But GR is a classical field theory, so there is substantial doubt that that is the case.

SR is certainly consistent with this view' date=' but GR lends a lot more support to Leibniz, who I happen to agree with. Leibniz held to what is now called the relational view of space and time. That is, if there were no objects and no events, we would have no concept of space or time. It holds that if the universe were devoid of matter, energy, and dynamical processes that there would not be this huge empty container with a ticking clock left over. There would be literally [i']nothing[/i]. Not really. It's just a convenient mental picture. Relativity does not say that, nor does it imply it. There are invariants in relativity, which are known as Lorentz scalars.

"Yes" on all counts.

I know that. You asked if "everything is relative" is the point of QM. And I said "no, I think you're mistaking QM for SR".

No, it isn't. I think you're thinking relativity, not QM.

No, by definition you would have no momentum in your own frame of reference. Of course, that means that your position is completely uncertain.

"Probability waves" are so named because the probability densities associated with quantum mechanical systems are solutions to a wave equation. One is not obligated to think of probability waves in the same way as, say, propagating light waves.

Once I was eating super hot buffalo wings, and they were actually making my eyes water. So like an idiot I wiped the tears from my eyes...with sauce covered hands. Mama mia!

Music: What gets you rockin? Supa-Fo-Hunnit! Super 400, the band in my avatar. Mind blowing mid-60's style guitar-heavy rock. It's all their own music, but you can tell that they listen to a lot of Hendrix and Cream, which is my kind of music.

Yes, but parity and translation also preserve the norm of a vector, and neither is a rotation. The Moral: Preservation of distance is not enough to guarantee that an operation is a rotation.

By <x,y>, do you mean (x2+y2)1/2? If so, then I agree that x+iy=<x,y>eiq, and that it is the Euler formula. I originally disagreed because when I see <x,y> I normally think of the R2 vector xi+yj. And of course, x+iy does not equal (xi+yj)eiwt. Hmmm...Maybe I'm not seeing what you mean, because I still don't think so. If I make a row vector [x y], and do a transformation on it to get [x iy], then I am struggling to see how that could be called a "rotation". About what axis and through what angle was it rotated? Only if those complex quantities add up to a real quantity. If complex energy eigenvalues are allowed, then probability is not conserved. *I* wouldn't expect it. In fact I've never seen it treated that way. Can you elaborate?

No, not quite. First, x+iy does not equal (<x,y>eiwt). And second, it doesn't give a rotation. What the operator X+iY does is maps coordinates x and y into vectors (x,y) in the complex plane. In other words it's a linear transformation T:R2-->C. I can't cite a practical application of it at this time but that's OK for the purpose of this thread, which is to discuss why the inference from "complex eigenvalue" to "unphysical, unobservable" is invalid.

It appears that when the QTC core is unloaded the potential barrier to the electrons is (nearly) insurmountable. But when pressure is put on it, the "walls" of the barrier are moved closer together (thus decreasing the barrier width). When that happens, the barrier is much more easily crossed, and a non-negligible current starts flowing.

Wow. I woke up this morning and saw that I have a bright, shiny new star. Many thanks to the SFN staff. I'm honored!

A+ isn't the complex conjugate of A. It's the Hermitian conjugate. Just to highlight the difference: take the following matrix representation of some operator (doesn't matter what it represents): [1 0 1] A= [2 2 0] [1 0 2] Now if you took the complex conjuate A* of A, you would just get A again (because A is real). So A*=A. But if you took the Hermitian conjugate A+ of A, you would not get A back. That's because A+=A*T, where T denotes "transpose". So you would get for A+: [1 2 1] A[sup]+[/sup]=[0 2 0] [1 0 2] So A+ does not equal A.

I gave one in my opening post: X+iY. Going on with that' date=' if [b']R[/b]=<X,Y,Z> is the displacement operator, and |r> are the eigenkets of X, then: X|r>=x|r> Y|r>=y|r> From this it is possible to construct the normal (but not Hermitian) operator X+iY, with complex eigenvalues x+iy. The point of my opening post is to highlight the fact that X+iY is in fact an observable, and that normal, non-Hermitian operators should not be dismissed as unphysical simply because they have complex eigenvalues. That's right, and it can't be non-Hermitian. But that's not because it's an observable, it's because H is the generator of time translations. If H is not Hermitian, it will give rise to a nonunitary time evolution operator, which will violate probability conservation. Don't know. But it's so jittery here that I've given up on using it.

No, it's the Hermitian conjugate. You can't take the complex conjugate of an operator.