Hi dreamchaser, welcome to PFs!
I'm not sure just what you're asking here... do you want a proof that such a relation exists, a recipe to get U(R), what?
The theorem you're talking about is true in the context of quantum mechanics. In QM we represent the state of a system by a ray in...
Nope. It's true that spin-up and spin-down particles have different energies in a homogenous B field. (They will differ by \Delta E = 2 \mu B.) But since this is a constant difference, there will be no force -- remember F=-\nabla E.
Not really. There's a really good article about this, probably aimed at just your level, in Scientific American here:
http://www.sciam.com/article.cfm?chanID=sa006&colID=1&articleID=0009F0CA-C523-1213-852383414B7F0147
As other posters have said, the Big Bang is an expansion of space, not in...
^^^ Yes, that's correct about the recession speed. But the universe only began accelerating relatively recently; before that it was decelerating. (Sort of like a 'bounce.') The 'dark energy' causing the expansion could have been there all along, but the mutual gravitational attraction of...
You're thinking of the big bang as an explosion from a center point, but this is entirely inaccurate. There is no apparent 'center' to the universe: as far as anyone can tell, the universe is the same everywhere and in all directions (homogenous & isotropic).
I do not know how that estimate...
What makes you think the energy stored in a spring (for a given force) is constant? If we write the energy stored as a function of F and k, we get
x = F/k
W = 1/2*kx^2 = 1/2*F^2/k.
Intuitively, it might help to remember that energy is the integral of force (w.r.t. displacement.) Under...
Don't confuse energy with force! They're different concepts. Let's draw out our toy model a bit more. If each end of the molecule has charge q, and the imposed field is E (ignoring other molecules for the moment), then the work done on the molecule (neglecting the electric attraction of the...
No. Or rather, it depends on which diagonal you're talking about. One will allow |H>+|V> through and completely block |H>-|V> ; the other will do the reverse. You can pretty much see what happens here just by looking at things: the lower-left to upper-right diagonal clearly corresponds to...
^^^ You're right in that when we impose a field on a dielectric, work is done on the dielectric itself. But we need to be careful about terminology here. When we talk about imposing a field, that means we set up charges in order to create the specified field (or in this case, what would be the...
If you want to learn how to do particle physics calculations in the Standard Model, then you want to learn, well, basic particle physics. :) There are sort of two paths you can take: a QFT book will teach you how QFT works, why the theory is the way it is, etc., and in the process show you how...
Hi MV,
You got it, though I wouldn't talk about dielectrics etc. "absorbing" electric fields. When we apply an electric field, we put that field in and that field is there -- that's what we mean. It doesn't get "absorbed" or anything like that. Rather, this field causes the dielectric...
Hi regent,
That's a good question. In general, if you really do only test a small region of position space, you will *not* collapse the wavefunction into a position eigenstate. In other words, your view of the situation from "the observer's perspective" is the correct one.
In detail, what...
To add to what others have said, you can think about the direction of the acceleration thusly:
When we say the universe is expanding at all, we mean it's just "stretching out" (like the black dots here: http://www.astro.ucla.edu/~wright/cphotons.gif ). All galaxies are getting further apart...
^^^ It does give you a better approximation. Here, maybe this will make it clear. The exact energy of the perturbed system is
n0 + n1 + n2 + n3 + n4 + ...
where n0 is the energy of the unperturbed system, and n1, n2, etc are the corrections you're talking about. So we need to add them...
Sure. There's a bit of a difficulty in talking about direct experiments involving the vacuum, of course, because it's the vacuum... it's not like we can turn it on and off to see what changes! But we can calculate its expected effects from some theory and look for those. jal mentioned the...