- #1
- 5,844
- 551
"Redshifting" of forces in stationary space - times
I'm trying to solve the following:
Let [itex](M,g_{ab})[/itex] be a stationary spacetime with timelike killing field [itex]\xi ^{a}[/itex]. Let [itex]V^{2} = -\xi _{a}\xi ^{a}[/itex] ([itex]V[/itex] is called the redshift factor).
(a) Show that the acceleration [itex]a^{b} = u^{a}\triangledown _{a}u^{b}[/itex] of a stationary observer is given by [itex]a^{b} = \triangledown^{b}\ln V[/itex].
(b) Suppose in addition that [itex](M,g_{ab})[/itex] is asymptotically flat. Then, the "energy as measured at infinity" of a particle of mass [itex]m[/itex] and 4 - velocity [itex]u^{a}[/itex] is [itex]E = - m\xi _{a}u^{a}[/itex]. Suppose a particle of mass [itex]m[/itex] is held stationary by a (massless) string, with the other end of the string being held by a stationary observer at infinity. Let [itex]F[/itex] denote the magnitude of the local force exerted by the string on on the particle. According to part (a), we have [itex]F = mV^{-1}(\triangledown ^{a}V\triangledown _{a}V)^{1/2}[/itex]. 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 [itex]F_{\infty } = VF[/itex].
So here's what I got:
(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, [itex]u^{a} = \frac{\xi ^{a}}{(-\xi ^{c}\xi _{c})^{1/2}}[/itex]. Now we compute, [itex]\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}}[/itex] so [itex]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})}[/itex] 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, [itex]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[/itex] as desired.
(b) This is where I'm totally stuck. As far as conservation of energy goes, we know that [itex]E[/itex], as defined above, is a conserved quantity along the worldline of the stationary particle; physically [itex]E[/itex] 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 [itex]F_{\infty }[/itex] at event [itex]P_{1}[/itex] and the particle feels the local force [itex]F[/itex] at event [itex]P_{2}[/itex]. As far as I can tell, all we know is that in between these events, [itex]E[/itex] is constant. But how do I relate [itex]E[/itex] to [itex]F_{\infty}[/itex] 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 [itex]\triangledown _{b}E[/itex]. We know that for the stationary particle hanging from the string, for which this total energy is attributed, [itex]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}[/itex] so if we compute the derivative we get [itex]\triangledown _{b}E = mV\triangledown _{b}\ln V[/itex] thus [itex](\triangledown ^{b}E\triangledown _{b}E)^{1/2} = mV(\triangledown ^{b}\ln V\triangledown _{b}\ln V)^{1/2} = VF[/itex] but I have NO idea how this quantity is related to [itex]F_{\infty }[/itex], 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!
I'm trying to solve the following:
Let [itex](M,g_{ab})[/itex] be a stationary spacetime with timelike killing field [itex]\xi ^{a}[/itex]. Let [itex]V^{2} = -\xi _{a}\xi ^{a}[/itex] ([itex]V[/itex] is called the redshift factor).
(a) Show that the acceleration [itex]a^{b} = u^{a}\triangledown _{a}u^{b}[/itex] of a stationary observer is given by [itex]a^{b} = \triangledown^{b}\ln V[/itex].
(b) Suppose in addition that [itex](M,g_{ab})[/itex] is asymptotically flat. Then, the "energy as measured at infinity" of a particle of mass [itex]m[/itex] and 4 - velocity [itex]u^{a}[/itex] is [itex]E = - m\xi _{a}u^{a}[/itex]. Suppose a particle of mass [itex]m[/itex] is held stationary by a (massless) string, with the other end of the string being held by a stationary observer at infinity. Let [itex]F[/itex] denote the magnitude of the local force exerted by the string on on the particle. According to part (a), we have [itex]F = mV^{-1}(\triangledown ^{a}V\triangledown _{a}V)^{1/2}[/itex]. 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 [itex]F_{\infty } = VF[/itex].
So here's what I got:
(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, [itex]u^{a} = \frac{\xi ^{a}}{(-\xi ^{c}\xi _{c})^{1/2}}[/itex]. Now we compute, [itex]\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}}[/itex] so [itex]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})}[/itex] 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, [itex]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[/itex] as desired.
(b) This is where I'm totally stuck. As far as conservation of energy goes, we know that [itex]E[/itex], as defined above, is a conserved quantity along the worldline of the stationary particle; physically [itex]E[/itex] 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 [itex]F_{\infty }[/itex] at event [itex]P_{1}[/itex] and the particle feels the local force [itex]F[/itex] at event [itex]P_{2}[/itex]. As far as I can tell, all we know is that in between these events, [itex]E[/itex] is constant. But how do I relate [itex]E[/itex] to [itex]F_{\infty}[/itex] 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 [itex]\triangledown _{b}E[/itex]. We know that for the stationary particle hanging from the string, for which this total energy is attributed, [itex]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}[/itex] so if we compute the derivative we get [itex]\triangledown _{b}E = mV\triangledown _{b}\ln V[/itex] thus [itex](\triangledown ^{b}E\triangledown _{b}E)^{1/2} = mV(\triangledown ^{b}\ln V\triangledown _{b}\ln V)^{1/2} = VF[/itex] but I have NO idea how this quantity is related to [itex]F_{\infty }[/itex], 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!
Last edited by a moderator: