□h=-16πT

You've got [math](9-x^2)^{\tfrac{1}{2}}[/math] Then you use the chain rule.

No it doesn't. You can get stable orbits about black holes.

Your derivative is wrong by the way. [math]y=-\sqrt{9-x^2} [/math] [math]y'=-\frac{1}{2}(9-x^2)^{-\frac{1}{2}}\frac{d(-x^2)}{dx}=\frac{x}{\sqrt{9-x^2}}[/math] 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.

Well for a weak field we can consider the metric as a small perturbation from flat space-time metric [math]g_{\alpha\beta}=\eta_{\alpha\beta}+h_{\alpha\beta}[/math] 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) [math]\left(\frac{1}{c^2}\frac{\partial}{\partial t}-\nabla^2\right)h_{\alpha\beta}=-16\pi T_{\alpha\beta}[/math] 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.

Jesus, he polluted myspace with this as well.

Officially, no. Off the recod, yes.

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

Or have you done something that gets rid of that factorial? Cheers, man.

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.

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.

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

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.

*shameless bump* Anyone?

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 [math]\vec{G}=\frac{d\vec{J}}{dt}[/math]