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You are correct. Wikipedia is wrong. It wouldn't be the first time. In general relativity a real force i.e. four-vector force can not be transformed away. The force of gravity can be locally transformed away simply by going transforming to a free fall frame so you can tell right away that it...

Neither the gravitational potential, nor the gravitational potential energy which is proportional to that are zero at the center when as in this case the zero point is taken to be at infinity.

Corrections: \int \bold g_{inc} \bullet d\bold a = -4\pi G M_{inc} \Phi(r)_{<} = -\frac{GMr^2}{2R^3} + C \Phi(r)_{<} = -\frac{GMr^2}{2R^3} -\frac{3GM}{2R}

No. The gravitational potential for the interior is given by \Phi = \frac{1}{2}\frac{GM_{tot}r^{2}}{R^{3}} - \frac{3}{2}\frac{GM_{tot}}{R} At the center this becomes \Phi = - \frac{3}{2}\frac{GM_{tot}}{R} and that is where the magnitude of the potential and gravitational time dilation is at...

No it doesn't. This is yet another example why "relativistic mass" is a bad concept.

That formula is only valid for the region exterior to the earth. Inside the Earth the corresponding formula would be t = \frac{\tau}{\sqrt{1 + \frac{2\Phi}{c^2}}} \Phi = \frac{1}{2}\frac{GM_{tot}r^{2}}{R^{3}} - \frac{3}{2}\frac{GM_{tot}}{R} where M_{tot} is the mass of the planet, R is its...

Yes. It is called the energy parameter. For a time independent metric the following is a timelike killing vector: [T^{\mu }] = \left[\begin{array}{cc}1\\0\\0\\0\end{array}\right] And the conserved energy parameter will be \frac{E_{cons}}{mc} = g_{\mu}_{\nu}T^{\mu }U^{\nu } which reduces to...

Just to be clear to you, in light of the notation having been used here for a while, it is an expression for ordinary force f, not the four-vector force F.

No, you are not talking about energy. You are talking about the energy parameter and p^{0} is trivial to work out for a Kerr-Newman spacetime of which the Schwarzschild result is a special case given p^{0} = mc(\frac{dt}{d\tau }) and that \frac{dt}{d\tau } is given by the differential time...

Inertial mass is not relativistic mass as I have mathematically proven here in the past. Relativistic mass has no place in modern relativity as it is just a redundant name for relativistic energy which is defined as the time element of the momentum four-vector of the first kind. From the...

admin edit: personal attack removed admin edit: personal attack removed Perhaps you should do the algebra right. You could also stop innappropriately subscipting the mass with a zero as he did not do that and don't claim that p = mv as it was not. It was clearly stated for this thread that m...

Thats close to ordinary force f. To be precise ordinary force f involves the limit of that as t2-t1 becomes infinitesimal dt in a calculus limit. Also, you shouldn't subscript the mass with a zero as it is invariant.

Using a relativistic mass substitution somewhere you shouldn't have is probably what led to your mistake. Do not use relativistic mass and try again. There is no place for relativistic mass in modern relativity.

Ok. Conservation of energy: mc^{2}cosh(\theta ) = (m + dm)(cosh(\theta ) + d(cosh(\theta )))c^{2} + m_{fex}\gamma _{fex}c^{2} Conservation of momentum: mcsinh(\theta ) = (m + dm)(sinh(\theta ) + d(sinh(\theta )))c + m_{fex}\gamma _{fex}ctanh(\theta _{fex}) Simplified 0 = sinh(\theta )dm...

Instead of using out of date concepts one should use modern relativity according to which this translates to saying that the conserved "energy parameter" in the absence of nongravitational fields is in some special cases equal to the time element of the covariant momentum four-vector which in a...