□h=-16πT

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• Rank
Baryon
• Birthday 06/30/1989

Profile Information

• Location
NE Lincs, England
• Interests
All sorts of shizzle
• College Major/Degree
None, but I'm pretty much teaching myself one.
• Favorite Area of Science
Relativity/Astrophysics
• Occupation
Student

2. Calculus with Physics question

You've got $(9-x^2)^{\tfrac{1}{2}}$ Then you use the chain rule.
3. Blackholes and EM waves

No it doesn't. You can get stable orbits about black holes.
4. Calculus with Physics question

Your derivative is wrong by the way. $y=-\sqrt{9-x^2}$ $y'=-\frac{1}{2}(9-x^2)^{-\frac{1}{2}}\frac{d(-x^2)}{dx}=\frac{x}{\sqrt{9-x^2}}$ I can't remember how you do coefficient of friction and all the gubbins, but the problem without coefficients of friction just requires that the kid's kinetic energy at the bottom of the pipe be the same as its potential energy at its initial position, height=3. Can you not integrate friction coefficients into conservation of energy.
5. Consequences of general relativity - gravitational waves

Well for a weak field we can consider the metric as a small perturbation from flat space-time metric $g_{\alpha\beta}=\eta_{\alpha\beta}+h_{\alpha\beta}$ Where the field h is not necessarilly a tensor, but a function added to each component of the Lorentzian metric. However, if we consider Minkowski space-time as a background, the field h will transform as a (0 2) tensor, so we can consider it as a tensor in such circumstances. If we now give ourselves an expression for the Einstein tensor in terms of our perturbed flat metric we can impose certain simplifying restrictions on h. These are the traceless-transverse gauge and the Lorentz (or de Donder) gauge, the latter being similar to that in classical electrodynamics. If we do this we end up with the weak field version of the field equations of general relativity which serves the basis of linearised field theory (linearised because we only one the terms linear in h) $\left(\frac{1}{c^2}\frac{\partial}{\partial t}-\nabla^2\right)h_{\alpha\beta}=-16\pi T_{\alpha\beta}$ Where T is the usual symmetric stress energy tensor of the generating field. "A First Course In General Relativity"-B. Schutz has a good discussion on the treatment of gravitational waves in linearised field theory, including its detection and the energy radiated from bodies by their gravitational wave emission. Note that spherically symmetric bodies do not radiate gravitational radiation, as their quadrupole and higher moments vanish due to the symmetry.
6. A new view for the universe

Jesus, he polluted myspace with this as well.
7. Getting rid of time once and for all

Officially, no. Off the recod, yes.
8. Expectation of the Wilson Loop

Danke, I wish it did work though . I hate functional integrals, as elegant as the whole procedure is for deriving Feynman rules.
9. Expectation of the Wilson Loop

Or have you done something that gets rid of that factorial? Cheers, man.
10. Getting rid of time once and for all

It allows one to take some state defined at some particular time and find the state at a further point in time. So one can remove time dependance, which can then be replaced by evolution operators, and makes things a bit easier in places.
11. Getting rid of time once and for all

The Hamiltonian is one of the most important elementary quantities in quantum mechanics. It arises because of time, and gives the energy of the system, in a round about way.
12. Getting rid of time once and for all

Oh, really? Yeah, I guess you're right, after all Hamiltonians and time evolutions opreators play a minor, insignificant role.
13. Expectation of the Wilson Loop

I was trying to approach it, as the question suggested, by constructing it out of Wilson lines and using a functional integral. There should be factorials in the exponent that you've missed out, which will muck up the end result.
14. Expectation of the Wilson Loop

*shameless bump* Anyone?
15. A.M. Conservation?

Angular momentum is conserved for central systems without external forces. That this results in angular momentum conservation arises from the following relationship between turning moment G and angular momentum J in an analogous manner to the conservation of linear momentum from Newton's second law $\vec{G}=\frac{d\vec{J}}{dt}$
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