Perturbation

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Copenhagen is too reliant on the observer, and some aspects are just plain wrong in my opinion. Everett seems pretty superfluous. So neither.
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The speed of light doesn't depend on alpha anyway. In fact, we already know that alpha is not constant  it changes with energy scale.
Damn it, I was going to say that.
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I doubt I'd be able to get there any way. Nevermind.
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Do you have to be involved in research or actually be qualified for them to let you into these sorts of things? I'd expect there's some restriction. I've spent the past three or four years teaching myself mathematics and physics and the thought of being neglected from things like this because I'm not formally educated, or too young, irritates me.
York's not too far from me, it's about an hour or so away, If I could get there I'd go.
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Oh, yeah, axiom of choice. Ignore what I said about separability.
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It's only an analogy. I don't know any better way of explaining it without going into the mathematics. Spacetime is some abstract thing that has nonzero curvature around bodies with massenergy. If you really hate using the trampoline with gravity just say it's a naturally positively curved trampoline. Maybe some grossly obese person has fallen asleep on it and left an indentation.
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It depends if the Hilbert space is separable, that is that any statevector can be written as a sum of elements of a countable basis set. And whether you can expand it into a basis of an operator's eigenvectors depends whether the eigenvectors are linearally independant, i.e. orthogonal. For the eigenvectors of a Hermitian operator this is always the case, it's a simple property of Hermitian operators, at least if they're nondegenerate.
It's usually assumed that the Hilbert space is separable, but I do believe the guys working on early quantum gravity came across problems with separability when they tried to build a Hilbert space, and was one reason why they moved to using spin networks.
If you wanted to use a set of eigenvectors of some operator as an expansion consisting of purely diagonal terms for some other operator, the two operators would have to commute.
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Schrodinger quantum mechanics has the time dependence in two things: the statevector and consequently the state operator. There is no time dependance in the operators that represent dynamical variables. The Schrodinger equation expresses the time dependence of statevectors as a differential equation, i.e. as a dynamical equation.
Start with your statevector [imath]\psi (0)\rangle[/imath] defined at initial time t=0 and perform a unitary evolution
[math]\psi (t)\rangle =e^{iHt/\hbar}\psi (0)\rangle[/math]
Differentiate with respect to time
[math]\frac{\partial}{\partial t}\psi (t)\rangle =\frac{iH}{\hbar}e^{iHt/\hbar}\psi (0)\rangle =\frac{iH}{\hbar}\psi (t)\rangle[/math]
Putting this in the usual form gives
[math]H\psi (t)\rangle =i\hbar\frac{\partial}{\partial t}\psi (t)\rangle[/math]
This is the Schrodinger equation.
For a nonrelativistic particle the Hamiltonian H is
[imath] H=\frac{P^2}{2M}+V [/imath]
And in coordinate representation the wavefunction [math]\psi (\vec{x}, t)=\langle\vec{x} \psi\rangle[/math] obeys
[math]\left(\frac{\hbar^2\nabla^2}{2M}+V(\vec{x}, t)\right) \psi (\vec{x}, t)=i\hbar\frac{\partial}{\partial t}\psi (\vec{x}, t)[/math]
The momentum operator [imath]\vec{P}[/imath] is [imath]i\hbar\vec{\nabla}[/imath] in coordinate representation.
For a relativistic particle we have the Dirac equation
[math]\left(i\gamma^{\mu}\partial_{\mu}m\right)\psi (\vec{x}, t)=0[/math]
Heinsenberg QM differs in that the statevectors and state operators have no explicit time depedence; instead the time dependence is in the dynamical operators, such as the Hamiltonian or Momenta.
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So basicially what hes saying is that the HUP really just helps people used to thinking in classical terms grasp QM right? And again... I do not know any mathematics so all of that is gibberish to me!
I din't expect you'd get some of it, having already said you don't really know the maths. I was just bored and went for a bit of a ramble. You should be alright with the bit about on and offshell if you ignore the mention of fourmomentum, which I needn't really have included, and take the equation I gave for the square of the mass as given. I'd edit the reply but it's too late to edit it now.
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Im sorry but I dont know what that means.. Im very new to this and dont know what onshell or offshell is.. nor do I understand the mathematics. Maybe I also need a more thorough understanding of HUP.. When I read that it just sort of struck me as odd because from what Ive read HUP is one of the bases of QM.
It's not the basis of, it's a result of the approach of QM. The more general relation is between two selfadjoint operators A and B such that [imath][A, B]=iC[/imath], where C is another selfadjoint operator, then their variances [imath]\Delta_A[/imath] and [imath]\Delta_B[/imath] satisfy
[math]\Delta_A\Delta_B\geq\tfrac{1}{2}\langle C\rangle [/math]
Here the angle brackets denote the mean of C in some state ([math]=Tr(\rho C)/Tr(\rho )[/math], for some state operator [imath]\rho[/imath]).
In the case of position and momentum the commutation relation between the two is [imath][\vec{P},\vec{Q}] =i\hbar[/imath] giving the familiar Heisenberg uncertainty relation between position and momentum
[math]\Delta\vec{P}\Delta\vec{Q}\geq\tfrac{1}{2}\hbar[/math]
For the uncertainty relation between energy and time, the derivation is not the same as this, for there is no "time operator" per se. The time in this uncertainty relation is the "characteristic time" associated with the variance of some dynamical variable, say that of some measuring apparatus.
[math]\Delta E\Delta\tau\geq\tfrac{1}{2}\hbar[/math]
The deal with on/off mass shell is to do with fourmomentum, a fourdimensional vector that has the energy and spatial momentum as components. The definition of fourmomentum is
[math]p^{\mu}p_{\mu}=E^2\vec{p}^2=m_0^2[/math] (units with c=1 for simplicity)
Here the p's on the left are the fourmomentum, E is the energy, [imath]\vec{p}[/imath] is the spatial momentum and [imath]m_0[/imath] is the rest mass/energy (rest mass if we weren't using units with c=1).
In quantum field theory it is feasable for this equality to be violated by virtual particles. If it is the case that the equality holds the particle is said to be onmassshell. If the identity does not hold then the particle is said to be offmassshell. As Severian said, the violation of the equation he gave for energy indicates a particle being offshell. It's usually called being offmassshell so I gave a definition where the relation to mass is more explicit, the two are obviously equivalent.
What Severian was saying is that particles being of shell can be seen to imply Heisenberg uncertainty: a particle could have too much energy by being offshell than it should have, so it has sort of borrowed energy from the uncertainty it is allowed in its energy from Heisenberg. So the norm [imath]p^{\mu}p_{\mu}[/imath] needn't be equal to the square of the mass because of this uncertainty.
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I've heard that they are entropic' date=' but it doesn't seem that way, have scientists twisted the idea of entropy so it seems to apply here.
Here's how I understand entropy: Light from the sun hits a surface on earth, gets reradiated at room temperature. Visable light is degraded into less useful, less powerful infrared.
Now to our black hole: After a septillion years, background radition is radio waves, that get sucked into a black hole, which explodes into gamma rays.
How is entropy preserved?[/quote']
Because their emission is essentially that of a perfect black body, a thermodynamic system. Its radiation of energy is pretty much spot on with thermodynamics, so it radiates like any ideal thermodynamic system will do, entropically. I can't remember the equations and things off hand to figure out the specific mechanics of it though, but I do remember that a consequence of the Hawking area theorem is that the change in entropy must be strictly positive.
If stuff keeps falling into the black hole to increase its massenergy it ain't gonna explode. It will "explode" due to a kind of runaway in Hawking radiation as its mass gets increasingly smaller. The derivation of Hawking radiation does not consider the CMBR, the calculation takes place in spacetime that is a vacuum apart from the black hole itself.
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I KNEW IT !!! Nobody was responding to my original post because it's ROCK SOLID. I'm getting good. Now I suspect the insults will start flowing, as per historical forum bahaviour patterns.... bring it.
Yes, a revolution is at hand.
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Original post deleted.
Edit: Actually, stuff it. I'm just going to be labelled a dogmatist by this cretin.
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I know nothing of the mathematics involved with QM.. Im trying to break into that slowly.
Go for it. I'm 16, well nearly 17, and in college (UK equivalent of Highschool) and I taught myself quantum mechanics/field theory and all the prerequisite stuff. Reading popular science is definitely the thing to do first. If you need any help any time just give me a shout.
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Ok.. first off Im no physicist.. hell Im not even in school. I have a deep interest in Quantum Mechanics.. Ive been reading alot about it. I want to get a better grasp on both QED and QCD. What books would you guys/girls recommend?
Would these texts be mathematically orientated or purely qualitative, popular science and what not?
Popular science wise, "QEDThe Strange Theory of Light and Matter"Feynman is a must, but you've probably read that, it's a popular book. Popular science wise for particle physics I don't really know any more than that, I don't read much popular science any more.
The only QED books I own and have really read a great deal of are field theoretic, chief amongst which is Schroeder and Peskin's, where the first part is devoted entirely to QED and Feynman diagrams. The second part focuses on the systematics of renormalisation and the final part on nonabelian field theories and the Standard Model of particle physics, i.e. QCD, GSW (electroweak), the Higgs etc. But it's very good, though it requires a fair amount of mathematics/physics as a prerequisite. It would probably be much better if you've been introduced to QED through relativistic quantum mechanics, I'd have thought. Though I never did much relativistic QM before doing field theory, and it didn't take me long to get the hang of the relativistic formulation.
I have read a bit of Halzen and Martin, and it doesn't seem too bad.
I teach myself so I have to make do with what I can get hold of, so my recommendations aren't as substantial as Severian's or ajb's.
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They're the same thing.
[math]\mathbf{g}=\frac{\mathbf{F}}{m}[/math]
The force is just
[math]G\frac{mM}{r^2}\hat{\mathbf{r}}[/math]
To get the absolute value just whack  around the things your taking abs of, there should just be a key on your keyboard for it.
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let's make it
[math]\frac{dy}{dt}=\sqrt{a^2y^2}[/math]
[math]\int\frac{1}{\sqrt{a^2y^2}}dy=\int dt = sin^1 \frac{y}{a} = t[/math]
so [math]y=a.sin(t)[/math]
But this is still periodic' date=' why is y not staying constant when it reaches a?[/quote']
That should be root a by the way in the argument of the inverse sine, as
[math]\frac{d(arcsin\tfrac{x}{a})}{dx}=\frac{1/a}{\sqrt{1x^2/a^2}}=\frac{1}{\sqrt{a^2x^2}}[/math]
Whatever it is that you've used to solve and graph it is just extending the range of the sine function to be the reals. Your y(t) is only defined on the range [90, 90].
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It's seperable
[math]\frac{dy}{dx}=\sqrt{ay^2}[/math]
[math]\frac{1}{\sqrt{ay^2}}\frac{dy}{dx}=1[/math]
[math]\int\frac{1}{\sqrt{ay^2}}\frac{dy}{dx}dx=\int\frac{1}{\sqrt{ay^2}}dy=\int dx[/math]
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well, 1st of all, yur substitutin z from elements fo teh same equation...i dunt think it will yield anythin that way.(ex: y+r = 2 > r= 2y > y+(2y) = 2 > 0=0?) So i dunt know if yur method is valid. Does the problem give any other information?
What he's done is used pythag to get the distance from the origin to (x, y, z) and then sub'ed in what z is from the equation for the surface.
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Its a philosophy movie, not a science movie. As long as people don't confuse it as such, its fine. What bothers me is that people will inevitably confuse it. It should come with a warning "5 out of 5 real physicists agree that the movie is complete crap, on a mathematical level."
They use accepted, well established science to sell pseudoscience under the guise of philosophy.
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Damn it, stupid accidental post.
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You're equation for energy, and thus the de' Broglie wavelength, won't work for light, as you've multiplied it by [imath]m_0\gamma[/imath] and [imath]m_0=0[/imath] for real photons.
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I'm not even going to start on how much I despise "What the Bleep Do We Know".
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No, I said we don't just whack a Coloumb potential into Schrodinger's equation to get QED. We can get ourselves potentials from QFT if we apply the nonrelativistic limit and use the Born approximation, which relates Smatrices to the classical potential (that in the Schrodinger equation). The Coloumb potential comes quite nicely out of QED.
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Between Casimir plates
in Quantum Theory
Posted
[math]\langle q_2, t_2q_1, t_1\rangle =\int_{Q(t_1)=q_1}_{Q(t_2)=q_2}\mathcal{Q(t)}\int\mathcal{P(t)}\exp\left[i\int dt\left(\sum_iP_i\dot{Q}^iH(Q, P)\right)\right][/math]