Recent content by quZz

Hi everyone, I don't fully understand what is the regular method to state and solve problems in GR when no handy hints like spherical symmetry or homogeneity of time are assumed. If I find myself in arbitrary reference frame with coordinates x^0, x^1, x^2, x^3 the meaning of which is unknown...

Actually, vacuum is filled with virtual particles even when there are no protons or electrons. In the presence of charged particles configuration of the e/m field changes, so you shouldn't be surprised that energy may actually decrease.

Well, you'll get the same answer because f(z) is analytic, it follows from definition. You can use Cauchy-Riemann equations to get different forms of df/dz.

Ok, first you notice that r^2-2mr = (r-m)^2-m^2 and then you substitute x\equiv r-m, and dr=dx. The integral then becomes \int\frac{dx}{\sqrt{x^2-m^2}} = \ln(\sqrt{x^2-m^2}+x) Second, on the right hand side you have \int\frac{d\rho}{\rho}=\ln\rho + C_1 Let's write C_1=\ln C so that...

try wolframalpha.com... http://www.wolframalpha.com/input/?_=1325716343605&i=d^2x%2fdy^2%3d-1%2fx^2&fp=1&incTime=true

so you just need an answer, right?

how did you get this? Seems you should have had \ln(\sqrt{(r-m)^2-m^2} + r-m) = \ln(C \rho) choosing C=2 you get your answer

Hi, the equation does not contain explicit dependency on y, so try multiplying both sides on dx/dy. You'll get a sort of "conservation of energy" for the problem: (dx/dy)^2/2 + 1/x = const const depends on the initial conditions. This way you are left with solving 1st order diff equation

Hi, thanks for a quick reply. In this paper it only proves that if energy of a body at rest is changed by dE, its mass also changes by dm=dE/c^2. It doesn't prove, however, a much stronger statement that all its energy equals its mass times c squared...

Hi, if f(z) is analytical function, you can take derivative in any direction on complex plane of z, e.g. take it along real axis dz = dx.

Hello everyone Can some one please provide (a link to) the most rigorous proof of E_0 = mc^2? Actually, I got stuck on this question: why doesn't E_0 = mc^2 + \rm{const}? How to prove that \rm{const} = 0?

It comes right from the connection between hamiltonian and lagrangian. If you have for lagrangian L = L1 + L2 and L2 does not depend on derivatives, then for hamiltonian you have H = H1 + H2, where H1 corresponds to L1 and H2 = -L2. Even more, if you have infinitesimal addition (that can...

em... which book? =)

No... momentum and energy are measured in different units =) F.dt = dp, but F.ds = -dU. dU = -dK if dE = 0 (total energy is conserved = elastic collision)

It is the sum over all particles. Density of a point particle is (as you said from the beginning) its' mass times Dirac delta function. In macroscopic physics (e.g. fluid dynamics) you average this microscopic density over a small finite volume (small comparative to all volumes under...