# “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!

• For those interested in context, this is problem 4 from chapter 6 of Wald. His style is very noticeable. – user10851 Mar 19 '13 at 0:40

First, as you wrote the energy $E$ can be written as \begin{equation} E=m(-\xi_\mu\xi^\mu)^{1/2}=mV. \end{equation} Because $E$ is conserved, $V$ is also conserved.
Second, we are now in asymptotically flat spacetime. This means that at infinity you get $V=1$. Therefore at infinity you get \begin{equation} F_\infty=m(\nabla_\mu V\nabla^\mu V)^{1/2}. \end{equation} Here $(\nabla_\mu V\nabla^\mu V)^{1/2}$ has same value everywhere, because $V$ is conserved.
My GR is not very good but if $E$ is the energy as measured at infinity, as you mention, then it is reasonable that $\nabla_b E= \partial_b E$ be interpreted as the force that acts on the particle as measured at infinity because it is the gradient of the energy. Thus its magnitude should be equal to the magnitude of the force exerted at the other end of the string which is $F_\infty$.
• Hi alex, thank you so much for the response! What I'm confused by is why does $(F_{\infty })_b = -\triangledown _{b}E$ necessarily? I tried to look in references when I did that but I couldn't find anything on why this relationship would hold true i.e. why the force applied by the observer on that end of the rope would be related to the 4 - gradient of the total energy of the particle, on the other end, as measured at infinity. And I'm also not sure how it is related to conservation of energy, which the book says to use. Thanks again! – WannabeNewton Mar 18 '13 at 23:55
• It's related to conservation of energy by virtue of the fact that we related the force to energy flux, i.e. in much the same way as $-\nabla U=F$ in 1st year mechanics is related to conservation of energy. As to your first question that is a bit of a handwave on my part right now :$– alexarvanitakis Mar 19 '13 at 0:35 • I've only seen force defined in Wald as$F^{a} = ma^{a}$where$m$is the rest mass. In newtonian mechanics we derive the relation above by using conservation of energy by using as a consequence that the work done by the external force itself, to which we attribute the potential energy, is independent of path etc. but does that make sense in GR? All we know is the particle's total energy is constant along it's path but I'm not seeing how that is related to the force applied by the observer at infinity and the other problem is that$\triangledown _{b}\$ is a 4 - gradient unlike newtonian theory. – WannabeNewton Mar 19 '13 at 0:58