- #1
- 5,844
- 551
Hello there! I have a question regarding the behavior of test particles inside a slowly rotating thin spherical shell of matter under the influence of the gravito - magnetic "lorentz force" (this is all in the weak field slow motion approximation). I want to see just how analogous I can claim the behavior is to regular magnetism. The inertial coordinate system used is relative to the background flat metric and oriented so that the shell rotates with angular velocity of magnitude [itex]\omega[/itex] about the z - axis.
The metric perturbation, in these coordinates, turns out to be [itex]h_{0x} = \frac{4M}{3R}\omega y, h_{0y} = -\frac{4M}{3R}\omega x[/itex] with all other components zero.
We define a 4 - potential, similar to the one of EM, by [itex]A_{a} = -\frac{1}{4}h_{ab}t^{b}[/itex] where [itex]t^{a} = (\frac{\partial }{\partial t})^{a}[/itex]. Thus we get, for the vector potential [itex]\vec{A}[/itex], [itex]A_{z} = 0, A_{x} = -\frac{M}{3R}\omega y, A_{y} = \frac{M}{3R}\omega x[/itex]. The gravito - magnetic fields is defined the same was as in EM i.e. [itex]B^{i} = \epsilon ^{ijk}\partial _{j}A_{k}[/itex] (the gravito - electric field is also defined in the same was as in EM and it is easy to see that for the aforementioned metric describing the interior of the slowly rotating thin spherical shell, [itex]\vec{E} = 0[/itex]). Anyways, we can calculate the gravito - magnetic field and we find [itex]B^{z} = \frac{2M}{3R}\omega , B^{x} = B^{y} = 0[/itex].
The problem I am dealing with now states there is an observer who is placed at rest at the center of the shell and parallel propagates along his world line a spin orientation vector [itex]S^{a}[/itex] that is orthogonal to his 4 - velocity i.e. [itex]S^{a}u_{a} = 0[/itex]. The "lorentz force" law for the gravito - magnetic field turns out to be [itex]\vec{a} = -\frac{8M}{3R}\vec{v}\times \vec{\omega }[/itex] (here [itex]\vec{v}[/itex] is the spatial part of the 4 - velocity). According to this equation, it seems if I place an object at rest anywhere in the uniform gravito - magnetic field inside the shell, then it will remain at rest just as in the analogous case of a uniform magnetic field. But then it confuses me why the text specifically talks about placing the observer at rest at the center because it seems if I solve the decoupled form of the lorentz force equations, just as in magnetism, for the components of the 3 -velocity and use zero initial velocity for all components as my initial conditions then the observer remains at rest no matter where he is placed inside the shell. So why care particularly about the center? Where did I go wrong here? This is what I need clarified the most, thanks!
If I then do assume the observer remains at rest for all time relative to the inertial coordinates of the background flat metric, then [itex]u^{a} = (\frac{\partial }{\partial t})^{a} = (1,0,0,0)[/itex]. We are also told [itex]u^{a}\triangledown _{a}S^{b} = 0[/itex]. The book asks for the precession of the spatial components of the spin so in particular
[itex]\frac{\mathrm{d} S^{i}}{\mathrm{d} t} = -\Gamma ^{i}_{00}S^{0} - \Gamma ^{i}_{j0}S^{j} = \frac{1}{2}\partial _{i}h_{00} - \partial _{0}h_{0i} - \Gamma ^{i}_{j0}S^{j} = -\Gamma ^{i}_{j0}S^{j} [/itex]. So we find that [itex]\frac{\mathrm{d} S^{i}}{\mathrm{d} t} = \frac{1}{2}S^{j}(\partial _{i}h_{0j} - \partial _{j}h_{0i})[/itex]. This tells us that, after calculating the components of the spatial parts of the spin vector, [itex]\frac{\mathrm{d} \vec{S}}{\mathrm{d} t} = \frac{4M}{3R}\vec{\omega }\times \vec{S}[/itex] which is exactly what the book says the answer should be. As noted; however, the solution required the assumption that the observer will stay at rest at the center for all time and I am not sure if my hand wavy justification of this in the previous paragraph is actually correct. Also, I never had to use the constraint that [itex]S^{a}u_{a} = 0[/itex] (which in this case, if the observer is at rest in these coordinates, just tells us that [itex]S^{0} = 0[/itex] so the spin orientation only has spatial components), so this also worries me that my solution is wrong. Thanks for the help guys, sorry for the long post on multiple requests for clarifications!
EDIT: I should probably add that in writing the geodesic equation for the observer the way it looks in the above paragraph, I used the fact that in this coordinate system [itex]\frac{\mathrm{d} S^{\mu }}{\mathrm{d} \tau } = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} \tau } = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}u^{0} = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}[/itex] (this is legal correct? =D)
The metric perturbation, in these coordinates, turns out to be [itex]h_{0x} = \frac{4M}{3R}\omega y, h_{0y} = -\frac{4M}{3R}\omega x[/itex] with all other components zero.
We define a 4 - potential, similar to the one of EM, by [itex]A_{a} = -\frac{1}{4}h_{ab}t^{b}[/itex] where [itex]t^{a} = (\frac{\partial }{\partial t})^{a}[/itex]. Thus we get, for the vector potential [itex]\vec{A}[/itex], [itex]A_{z} = 0, A_{x} = -\frac{M}{3R}\omega y, A_{y} = \frac{M}{3R}\omega x[/itex]. The gravito - magnetic fields is defined the same was as in EM i.e. [itex]B^{i} = \epsilon ^{ijk}\partial _{j}A_{k}[/itex] (the gravito - electric field is also defined in the same was as in EM and it is easy to see that for the aforementioned metric describing the interior of the slowly rotating thin spherical shell, [itex]\vec{E} = 0[/itex]). Anyways, we can calculate the gravito - magnetic field and we find [itex]B^{z} = \frac{2M}{3R}\omega , B^{x} = B^{y} = 0[/itex].
The problem I am dealing with now states there is an observer who is placed at rest at the center of the shell and parallel propagates along his world line a spin orientation vector [itex]S^{a}[/itex] that is orthogonal to his 4 - velocity i.e. [itex]S^{a}u_{a} = 0[/itex]. The "lorentz force" law for the gravito - magnetic field turns out to be [itex]\vec{a} = -\frac{8M}{3R}\vec{v}\times \vec{\omega }[/itex] (here [itex]\vec{v}[/itex] is the spatial part of the 4 - velocity). According to this equation, it seems if I place an object at rest anywhere in the uniform gravito - magnetic field inside the shell, then it will remain at rest just as in the analogous case of a uniform magnetic field. But then it confuses me why the text specifically talks about placing the observer at rest at the center because it seems if I solve the decoupled form of the lorentz force equations, just as in magnetism, for the components of the 3 -velocity and use zero initial velocity for all components as my initial conditions then the observer remains at rest no matter where he is placed inside the shell. So why care particularly about the center? Where did I go wrong here? This is what I need clarified the most, thanks!
If I then do assume the observer remains at rest for all time relative to the inertial coordinates of the background flat metric, then [itex]u^{a} = (\frac{\partial }{\partial t})^{a} = (1,0,0,0)[/itex]. We are also told [itex]u^{a}\triangledown _{a}S^{b} = 0[/itex]. The book asks for the precession of the spatial components of the spin so in particular
[itex]\frac{\mathrm{d} S^{i}}{\mathrm{d} t} = -\Gamma ^{i}_{00}S^{0} - \Gamma ^{i}_{j0}S^{j} = \frac{1}{2}\partial _{i}h_{00} - \partial _{0}h_{0i} - \Gamma ^{i}_{j0}S^{j} = -\Gamma ^{i}_{j0}S^{j} [/itex]. So we find that [itex]\frac{\mathrm{d} S^{i}}{\mathrm{d} t} = \frac{1}{2}S^{j}(\partial _{i}h_{0j} - \partial _{j}h_{0i})[/itex]. This tells us that, after calculating the components of the spatial parts of the spin vector, [itex]\frac{\mathrm{d} \vec{S}}{\mathrm{d} t} = \frac{4M}{3R}\vec{\omega }\times \vec{S}[/itex] which is exactly what the book says the answer should be. As noted; however, the solution required the assumption that the observer will stay at rest at the center for all time and I am not sure if my hand wavy justification of this in the previous paragraph is actually correct. Also, I never had to use the constraint that [itex]S^{a}u_{a} = 0[/itex] (which in this case, if the observer is at rest in these coordinates, just tells us that [itex]S^{0} = 0[/itex] so the spin orientation only has spatial components), so this also worries me that my solution is wrong. Thanks for the help guys, sorry for the long post on multiple requests for clarifications!
EDIT: I should probably add that in writing the geodesic equation for the observer the way it looks in the above paragraph, I used the fact that in this coordinate system [itex]\frac{\mathrm{d} S^{\mu }}{\mathrm{d} \tau } = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} \tau } = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}u^{0} = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}[/itex] (this is legal correct? =D)
Last edited: