Recent content by FrankPlanck

It sounds a little bit strange to me. Anyway probably f(t=0)=infinite it's the Big Bang singularity, a point with infinite density (so the metric), but it's a very raw treatment.

As Jonathan said, it's wrong. Your galaxy isn't spherical, you can't use your (wrong) formulas. Fix them and find the rotation curve of the bulge. If you want to find the disk rotation curve you should write your potential considering a cylindrical distribution (hint: Green's functions), then...

It depends on your tool...

Black holes without spin curve spacetime, black holes with spin also distort it. This causes spacetime vibration which in turn causes thermal emission modulation (oscillations) from the inner part of the accretion disk, for more information see thirring-lense effect. Observing these...

Yes but how can we realize that quantum effects are not so important? Does a (or more) formula exist or something similar?

something similar? :) I don't know, maybe the de broglie wavelength of a graviton? I'm not saying this is the answer, but this is a nice problem, I would like to know the solution.

I'm not an expert, but I think you should use de broglie wavelength (or something similar) to estimate quantum effects, therefore amplitude is not so important, wavelength is.

I think you can use both, because (if I'm not wrong) in stars M_V \simeq M_{bol} ; from wikipedia: "The bolometric magnitude can be computed from the visual magnitude plus a bolometric correction, M_{bol} = M_V + BC . This correction is needed because very hot stars radiate mostly ultraviolet...

Thank you for all the answers. I'm sorry but in the problem R, that is the total radius, is not given. I think it's sufficient to write the formula, not the real value. So this is ok M_{tot}= \int^R_0 \rho (r) dV but it's very strange to me because, I know this...

Ok, so R is the radius of the sphere (galaxy). Integral means the sum of each contribution along the differential (it's a rough answer I know...) Maybe M_{tot}= \int^R_0 \rho (r) dV ?

Yes, if the observer is in free-fall he doesn't see emission (although the particle is accelerated)

No the v(r) formula is correct, I made a mistake considering c the speed of light, c is the constant I wrote above (cm^3/2 sec^(-1)) (I didn't follow out the text). The radius of the galaxy is the radius of the sphere that contains matter :) , I'm sorry but I don't understand what you are...

Ok mate, so there is no problem :) This problem looks like this: does an electron in free-fall in a gravitational field emit?

You are obviously right, I was trying to explain the process in the simplest way (that is wrong, but it helps to understand) :)

@tms no problem :) @turin thank you, you are obviously right so maybe I have to do this: M(r)= \int\rho (r) dV(r) so \rho (r) = \frac{dM}{dV} = \frac{1}{4\pi r^{2}} \frac{dM}{dr} hence \rho(r) = \frac{c^2}{2\pi G} \frac{r_{c}^2}{r(r_{c}^2+r^2)^2} Is this right? And the total mass? I find...