Recent content by felicja

OK, I sorry. The result is the same. ##2\ln(\sqrt{x}+\sqrt{x-1})=\ln(\sqrt{x}+\sqrt{x-1})^2=\ln(2\sqrt{x}\sqrt{x-1}+x+x-1)## But look at this. The distance with the Schwarzschild metric is equal to: ##s=\sqrt{r(r-a)} + a.arcosh(\sqrt{r/a})## thus for the case of very big distances: ##r >> a##...

Say me, what is incorrect? We observe the angles directly - in the space! There no time element has any impact. The observation is instant: t = const, thus any angular shift must be a pure and static geometry relation, nothing more. Oh! I discovered an additional fantastic question: the time...

The question: 'what is a distance...', and with a given metric is rather very precise - there is no room for any more specification.

A spatial curvature is in the spatial dimensions, thus it's: ##K_r\phi=-m/r^3## These other curvatures, the mixed like: t-r,, are irrelevant because the GR claims: the space is curved, thus we observe the planetary precession... and the light deflection too. We observe these all effects in the...

Try to compute some definite integral using these 'alternative solutions' then You get it. For example: what is a correct distance, means: according to the GR, to the Sun from the Earth?

I don't ask about any particular planetary motion model, but about the angular deficit in the GR only! If the model can't reproduce the correct angle deficit along the circle, then the rest is nothing but another improvised numerology only! df = ?

I showed this already: #3. And there are much more idiotic versions in the net!

So, I don't what is going on. The proposed solutions are quite nonsensical - what it the reason? ##arcosh(x)=\ln(x+\sqrt{x^2-1})##

I easily compute this integral - just make a simple substitute: ##x = a\cosh^2(t)## then: ##dx = 2a\cosh(t)\sinh(t)## and ##x-a = a\sinh^2(t)## so, the integral now is: ##\int cosh^2(t)dt = \int(\cosh(2t)+1)dt = \frac{1}{2}\sinh(2t)+t+C=\sinh(t)\cosh(t)+t+C## finally: ##I = \sqrt{x(x-a)}+a\cdot...

I think a time has nothing to the geometric curvature of the space alone. So, the orbital precession must be equal to the angle deficit along any closed loop in the curved space, and exactly, not other.

I have a problem with this integral, because there are several versions in the net, so, I'm very surprised... what is going on? for example: http://www.physicspages.com/2013/04/05/schwarzschild-metric-radial-coordinate/ and the proposed solution 6) is rather strange. I can use the wolfram...

##\int\sqrt{\frac{x}{x-a}}dx=?##

So, a curvature radius is equal to: ##R=1/\sqrt{|K|}##

And what is the Gaussian curvature of the space in the GR: ##K=m/r^3## ?

I think this should be equal to the famous precession angle, but with a negative sign: ##d\phi = 6\pi m/r## correct?