# Newtonian limit of kinetic energy in general relativity

Rindler's Relativity (2nd ed.) implies on p.239, Eq. 11.34, that the kinetic energy of a body falling radially into a black hole is the square of the derivative of $r$ with respect to $s$. But in the Newtonian limit this becomes $v^2$, not half of that, as it should be. What is the reason for this apparent contradiction?

• If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Mar 19 at 21:27

Start with the Schwarzschild metric $$ds^2~=~\left(1~-~\frac{2GM}{r}\right)dt^2~-~\left(1~-~\frac{2GM}{r}\right)^{-1}dr^2~-~r^2d\Omega^2,$$ where $c~=~1$. Now multiply by the particle mass or $m^2$ and divide by ds^2 $$m^2~=~m^2\left(1~-~\frac{2GM}{r}\right)\left(\frac{dt}{ds}\right)^2~-~m^2\left(1~-~\frac{2GM}{r}\right)^{-1}\left(\frac{dr}{ds}\right)^2~-~m^2r^2\left(\frac{d\Omega}{ds}\right)^2.$$ We have a radial moving particle that $$m^2~=~\left(1~-~\frac{2GM}{r}\right)p_t^2~-~\left(1~-~\frac{2GM}{r}\right)^{-1}p_r^2.$$ We now explore the weak field limit for $1~>>~\frac{2GM}{rc^2}$ so that $$m^2~=~\left(1~-~\frac{2GM}{r}\right)p_t^2~-~\left(1~+~\frac{2GM}{r}\right)p_r^2.$$ In the limit that $r~\rightarrow~\infty$ or $M~=~0$ this is the momentum invariant in special relativity $$m^2~=~p_t^2~-~p_r^2,$$ where $p_t~=~E$.

This limit is different from the limit that recovers nonrelativistic Newtonian mechanics. The energy is $$E~=~\sqrt{m^2~+~p_r^2}.$$ The nonrelativistc limit is then where $m^2~>>~p_r$ and so with binomial theorem the energy is $$E~=~m\left(1~+~\frac{p_2}{2m^2}\right)~=~m~+~\frac{1}{2}mv_r^2,$$ where the momentum $p_r~=~mv_r$. This is the standard Newtonian kinetic energy with the $E~=~mc^2$ added in.

I was always in agreement with your calculation. But you haven't addressed my question: what is the formula for KE in GR? Rindler says, which wd seem incorrect in my view, that alpha=1-2m/r (he uses natural units) is the PE, which makes the PE nonzero at infinity. He calls alpha + p^2 sub r the total energy, and p^2 sub r the KE. He includes the rest energy in the PE, but even with this unusual formulation, his figures don't add up properly to the law of conservation of energy. He is off by a factor of 2. p^2 sub r gives mv^2 as Newtonian limit.

Using E=mc^2 =(m sub 0) gamma c^2, I finally figured out that, if alpha=(1 − 2GM/r) the formula shd be, for c=1:

                                        KE=1/2 p^2 sub r)=(m sub 0)(gamma-1)
PE=(alpha-1)/2=M/r
E sub 0 = 1


Now KE + E sub 0 + PE = total energy, but it is important to be clear in what frame one is measuring v and KE.

and these qties reduce correctly in the Newtonian limit. By the way, where are the instructions for the markup language?