Expression for energy of a massive radially moving particle in a static spacetime

Question

I have become utterly confused about times and velocities measured by different observers. Let us take the specific case of a Schwarzschild spacetime. When we say that there is a particle of mass m moving with 4-velocity $u^a = dx^a/d\tau$, $\tau$ is the proper time of the particle (i.e. the time registered by a clock sitting on the particle). But who is measuring the coordinate $x^a$? Is it measured by the asymptotic observer in Schwarzschild coordinates?

My next confusion is about the expression for the energy E of the particle as given by Frolov and Novikov in the book `Black Hole Physics' (p19, footnote). I am giving their argument below:

If $u^a$ is the 4-velocity of a particle of mass m moving freely in a stationary gravitational field with metric $\bf g$ (having signature $(-+++)$) then its energy \begin{align} E = - g_{00} ~c^2 ~m u^0 . \end{align} Then they claim that the energy in a static spacetime can be rewritten as \begin{align} E = \frac{\sqrt{-g_{00}}~m c^2}{\sqrt{1-v^2/c^2}} , \end{align} where $v$ is the physical three velocity s.t. \begin{align} v^2 = \frac{g_{\alpha\beta}}{-g_{00}}\frac{dx^\alpha dx^\beta}{dt^2} . \end{align} For example, for radial motion in Schwarzschild spacetime, \begin{align} E= mc^2 \frac{\sqrt{1-r_s/r}}{\sqrt{1-\dot r^2/(1-r_s/r)}} . \end{align} I do not understand how they arrived at the second expression for E in a static spacetime. I think they have used the result $$u^0 = \frac{1}{\sqrt{-g_{00}\left(1-v^2/c^2\right)}}$$ but I cannot see how it comes about.

Here, the indices $a,b,...$ run over spacetime coordinate values while $\alpha,\beta,...$ run over spatial ones. Any help would be greatly appreciated.

Answer 1

Then they claim that the energy in a static spacetime can be rewritten as \begin{align} E = \frac{\sqrt{-g_{00}}~m c^2}{\sqrt{1-v^2/c^2}} , \end{align}

It seems to be incorrect. Somewhat simpler version of already given answer is to start from $$c^2d\tau^2\equiv-g_{00}c^2 dt^2+g_{rr}dr^2+r^2d\Omega^2. \tag{1}$$ Using definitions and relations $$u^{0}\equiv dt/d\tau,~~~u^{r}\equiv dr/d\tau =dr/dt\cdot dt/d\tau\equiv v\cdot u^{0},~~~g_{rr}=g_{00}^{-1},~~~d\Omega/d\tau=0, \tag{2}$$ the equation (1) reads $$1=(u^{0})^{2}\cdot\Big[-g_{00}+(v/c)^{2} g_{00}^{-1}\Big]. \tag{3}$$ Deriving the $ u^{0}$ from equation (3) and inserting it into your first equation, defining the energy, one receives $$E=\frac{\sqrt{-g_{00}}~m c^2}{\sqrt{1-v^2/c^2~ g_{00}^{-2}}}.$$

Answer 2

Oops: for the second formula,we have :$$E=-g_{00}mc^{2}u^{0}=-g_{00}mc^{2}\frac{dx^{0}}{ds}=\frac{-g_{00}mc^{2}dx^{0}}{\sqrt{-g_{00}(dx^{0})^{2}+g_{\alpha\beta}dx^{\alpha}dx^{\beta}}}$$

we know that : $dl^{2}=g_{\alpha\beta}dx^{\alpha}dx^{\beta}$ and $\;\;c^{2}d\tau^{2}=-g_{00}(dx^{0})^{2}$

$$=\frac{-g_{00}mc^{2}dx^{0}}{\sqrt{-g_{00}(dx^{0})^{2}+dl^{2}}}$$ $$=\frac{-g_{00}mc^{2}dx^{0}}{\sqrt{-g_{00}}(dx^{0})\sqrt{1- ( \frac{dl}{g_{00}dx^{0}}} )^{2}}$$ $$=\frac{\sqrt{-g_{00}}mc^{2}}{\sqrt{1- \frac{(\frac{dl}{ d\tau})^{2}}{c^{2}}}}$$ $$E=\frac{\sqrt{-g_{00}}mc^{2}}{\sqrt{1- \frac{v^{2}}{c^{2}}}}$$ the other equation is obtained by replacing $ds^{2}=-g_{00}(dx^{0})^{2}+g_{\alpha\beta}dx^{\alpha}dx^{\beta}$

by Schwarzschild metric (with $\theta=\phi=0, \;-g_{00}=1/g_{rr})$: $$\;\;ds^{2}=-g_{00}(dx^{0})^{2}+g_{rr}dr^{2}=-g_{00}(dx^{0})^{2}\left(1-\frac{g_{rr}dr^{2}}{g_{00}(dx^{0})^{2}}\right)$$ $$=-g_{00}(dx^{0})^{2}\left(1+\frac{g_{rr}dr^{2}}{c^{2}d\tau^{2}}\right)$$ $$=-g_{00}(dx^{0})^{2}\left(1-\frac{dr^{2}}{g_{00}\,c^{2}d\tau^{2}}\right)$$ $$ds=\sqrt{-g_{00}}(dx^{0})\sqrt{1-\frac{\dot{r}^{2}}{g_{00}\,c^{2}}}$$ $$ds=\sqrt{-g_{00}}(dx^{0})\sqrt{1-\frac{\dot{r}^{2}/c^{2}}{1-\frac{r_{s}}{r}}}$$ $$E=-g_{00}mc^{2}\frac{dx^{0}}{ds}=\frac{-g_{00}mc^{2}dx^{0}}{\sqrt{-g_{00}(dx^{0})^{2}+g_{rr}dr^{2}}}$$ $$= \frac{-g_{00}mc^{2}dx^{0}}{\sqrt{-g_{00}}(dx^{0})\sqrt{1-\frac{\dot{r}^{2}/c^{2}}{1-\frac{r_{s}}{r}}}}$$ $$= \frac{\sqrt{-g_{00}}mc^{2}}{\sqrt{1-\frac{\dot{r}^{2}/c^{2}}{1-\frac{r_{s}}{r}}}}$$ $$E= mc^{2}\frac{\sqrt{1-\frac{r_{s}}{r}}}{\sqrt{1-\frac{\dot{r}^{2}/c^{2}}{1-\frac{r_{s}}{r}}}}$$