homework and exercises - "Redshifting" of forces in stationary spacetime

Here's the problem statement: Let $(M,g_{ab})$ be a stationary spacetime with timelike killing field $\xi ^{a}$. Let $V^{2} = -\xi _{a}\xi ^{a}$ ($V$ is called the redshift factor). (a) Show that the acceleration $a^{b} = u^{a}\triangledown _{a}u^{b}$ of a stationary observer is given by $a^{b} = \triangledown^{b}\ln V$.

(b) Suppose in addition that $(M,g_{ab})$ is asymptotically flat. Then, the "energy as measured at infinity" of a particle of mass $m$ and 4 - velocity $u^{a}$ is $E = - m\xi _{a}u^{a}$. Suppose a particle of mass $m$ is held stationary by a (massless) string, with the other end of the string being held by a stationary observer at infinity. Let $F$ denote the magnitude of the local force exerted by the string on on the particle. According to part (a), we have $F = mV^{-1}(\triangledown ^{a}V\triangledown _{a}V)^{1/2}$. Use conservation of energy arguments to show that the magnitude of the force exerted on the other end of the string by the observer at infinity, is $F_{\infty } = VF$.

My attempt: (a) A stationary observer's 4 - velocity must be proportional to the time - like killing vector and it must be normalized to -1 so we find that for a stationary observer,

$$u^{a} = \frac{\xi ^{a}}{(-\xi ^{c}\xi _{c})^{1/2}}$$ . Now we compute, $$\triangledown _{b}u^{a} = \frac{\triangledown _{b}\xi ^{a}}{(-\xi ^{c}\xi _{c})^{1/2}} + \frac{\xi ^{a}\xi ^{c}\triangledown _{b}\xi _{c}}{(-\xi ^{c}\xi _{c})^{3/2}}$$ so $$u^{b}\triangledown _{b}u^{a} = \frac{\xi^{b}\triangledown _{b}\xi ^{a}}{(-\xi ^{c}\xi _{c})} + \frac{\xi ^{a}\xi^{b}\xi ^{c}\triangledown _{b}\xi _{c}}{(-\xi ^{c}\xi _{c})^{5/2}} = -\frac{\xi^{b}\triangledown ^{a}\xi _{b}}{(-\xi ^{c}\xi _{c})}$$ where the second term vanishes because it is a contraction of a symmetric tensor with an anti - symmetric one and I have swapped the indices in the first expression using killing's equation. Therefore, $$a^{a} = u^{b}\triangledown _{b}u^{a} = \frac{1}{2}\frac{\triangledown ^{a}(-\xi^{b}\xi _{b})}{(-\xi ^{c}\xi _{c})} = \frac{1}{2}\triangledown ^{a}\ln V^{2} = \triangledown ^{a}\ln V$$ as desired.

(b) This is where I'm totally stuck. As far as conservation of energy goes, we know that $E$, as defined above, is a conserved quantity along the worldline of the stationary particle; physically $E$ is the energy needed to bring in the particle from infinity to its orbit. Here we have a stationary observer at infinity holding this particle stationary by a long thread. There is a tension force at the end the observer holds and at the end the particle hangs by. Let's say the observer exerts the force $F_{\infty }$ at event $P_{1}$ and the particle feels the local force $F$ at event $P_{2}$. As far as I can tell, all we know is that in between these events, $E$ is constant. But how do I relate $E$ to $F_{\infty}$ and how do I do this using the conservation of energy explained above?

I should note that I tried something on a whim and looked at $\triangledown _{b}E$. We know that for the stationary particle hanging from the string, to which this total energy is attributed,

$$E = -m\xi _{a}u^{a} = -\frac{m\xi_{a}\xi^{a}}{(-\xi _{c}\xi^{c})^{1/2}} = m(-\xi^{c}\xi_{c})^{1/2}$$ so if we compute the derivative we get $\triangledown _{b}E = mV\triangledown _{b}\ln V$ thus $(\triangledown ^{b}E\triangledown _{b}E)^{1/2} = mV(\triangledown ^{b}\ln V\triangledown _{b}\ln V)^{1/2} = VF$ but I have NO idea how this quantity is related to $F_{\infty }$, if at all. If it is related somehow I have no idea how one uses conservation of energy to arrive at the relation. Thanks in advance!