# Simple question on Gravito - Magnetism inside Shell

1. Mar 1, 2013

### WannabeNewton

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 $\omega$ about the z - axis.

The metric perturbation, in these coordinates, turns out to be $h_{0x} = \frac{4M}{3R}\omega y, h_{0y} = -\frac{4M}{3R}\omega x$ with all other components zero.
We define a 4 - potential, similar to the one of EM, by $A_{a} = -\frac{1}{4}h_{ab}t^{b}$ where $t^{a} = (\frac{\partial }{\partial t})^{a}$. Thus we get, for the vector potential $\vec{A}$, $A_{z} = 0, A_{x} = -\frac{M}{3R}\omega y, A_{y} = \frac{M}{3R}\omega x$. The gravito - magnetic fields is defined the same was as in EM i.e. $B^{i} = \epsilon ^{ijk}\partial _{j}A_{k}$ (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, $\vec{E} = 0$). Anyways, we can calculate the gravito - magnetic field and we find $B^{z} = \frac{2M}{3R}\omega , B^{x} = B^{y} = 0$.

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 $S^{a}$ that is orthogonal to his 4 - velocity i.e. $S^{a}u_{a} = 0$. The "lorentz force" law for the gravito - magnetic field turns out to be $\vec{a} = -\frac{8M}{3R}\vec{v}\times \vec{\omega }$ (here $\vec{v}$ 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 $u^{a} = (\frac{\partial }{\partial t})^{a} = (1,0,0,0)$. We are also told $u^{a}\triangledown _{a}S^{b} = 0$. The book asks for the precession of the spatial components of the spin so in particular
$\frac{\mathrm{d} S^{i}}{\mathrm{d} t} = -\Gamma ^{i}_{00}S^{0} - \Gamma ^{i}_{j0}S^{j} = \frac{1}{2}\partial _{i}h_{00} - \partial _{0}h_{0i} - \Gamma ^{i}_{j0}S^{j} = -\Gamma ^{i}_{j0}S^{j}$. So we find that $\frac{\mathrm{d} S^{i}}{\mathrm{d} t} = \frac{1}{2}S^{j}(\partial _{i}h_{0j} - \partial _{j}h_{0i})$. This tells us that, after calculating the components of the spatial parts of the spin vector, $\frac{\mathrm{d} \vec{S}}{\mathrm{d} t} = \frac{4M}{3R}\vec{\omega }\times \vec{S}$ 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 $S^{a}u_{a} = 0$ (which in this case, if the observer is at rest in these coordinates, just tells us that $S^{0} = 0$ 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 $\frac{\mathrm{d} S^{\mu }}{\mathrm{d} \tau } = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} \tau } = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}u^{0} = \frac{\mathrm{d} S^{\mu }}{\mathrm{d} t}$ (this is legal correct? =D)

Last edited: Mar 1, 2013
2. Mar 2, 2013

### micromass

Staff Emeritus
This question is very interesting!

3. Mar 2, 2013

### WannabeNewton

yeah...>.< lol

4. Mar 2, 2013

### Bill_K

Your method and results look fine to me. Even though it doesn't matter where the particle is located inside the sphere, placing it at the center makes sense. Also, u·S = 0 is true for a spin vector, it doesn't hurt to state it.

5. Mar 2, 2013

### WannabeNewton

Thank you so much Bill! Just a few questions:
- So the lorentz force argument leading to the statement that the observer placed at rest anywhere inside the shell will remain at rest for all time is in fact correct?

- Why do you say that placing it at the center makes sense? I'm just having trouble seeing what is particularly special about it for this problem.

- So we technically didn't need to use that u.S = 0 for the solution itself?

-Also, in this case since the observer was at rest, S turned out to have no time component and only a space component which physically makes sense to me since the spin orientation intuitively should be a spatial orientation because one precess's in space. However if the observer was moving hence making u having non zero spatial components then S can have a non zero time component right? I can't really interpret what it means for a spin orientation to have a time component.

Again, sorry for all the questions and thank you so much for the help.

6. Mar 2, 2013

### Bill_K

Spin is a strange quantity. It's really an antisymmetric rank-two tensor, Jμν. For example for a Dirac particle, spin is represented by the matrix σμν. (Actually ½ σμν)

In three dimensions, in the rest frame of the particle, we can define a spin 3-vector to be Si = ½ εijkJjk. Whereas in four dimensions we define a 4-vector Sμ = ½ εμνστpνJστ and this gives us a vector that's automatically perpendicular to pν. Even for a particle that's accelerating, we write down equations of motion that maintain the orthogonality.

7. Mar 2, 2013

### WannabeNewton

Ah ok. So there isn't really much of a physical interpretation to the spin possibly having a time component? Also how do the other 3 questions asked above fare? Thank you so much again Bill!

8. Mar 3, 2013

### WannabeNewton

Also, when one says the given observer is at rest with respect to the global inertial coordinates of the background minkowski metric does that mean the same thing as it does in SR, that is there exists a global inertial frame of another observer fixed in the background in which the given observer is at rest?

9. Mar 3, 2013

### pervect

Staff Emeritus
The way I look at spin is similar to Bill K. It's really a bivector, a rank 2 antisymmetric tensor.

If you define some vector $\hat{t}$ that is the time vector in some local inertial frame , you can express the spin bivector as the wedge product of t and a number of different vectors. I.e. so that S_{ij} = $\hat{t}$ ^ x.

There should be one unique x that's perpendicular to $\hat{t}$. (Intuitively, I haven't done a formal proof of this). This vector x is what I call the "spin axis".

If you have some complex spin vector y, I believe it still satisfies $S_{ij} = \hat{t}$ ^ y. However, I don't give this complex vector the same physical interpretation as the spin axis.

The spin axis I think of, physically, as representing the axis around which some gyroscope rotates. If you've got a tube or telescope on this spin axis, given an inertial frame you can say it points in one specific direction. If you look through the telescope, you'll see some particular star, for instance. This interpretation requires that the vector have no time component in the specified inertial frame.

I dont think I"ve ever seen a textbook define things this way, so unfortunately the idea of the 'spin axis' may not be a widely recognized entity. Also, it doesn't transform via the Lorentz transform. I still find the idea useful, and I hope it generates more illumination than confusion.

[add]The other point,is that you can always define spin as a bivector, but you need to single out "the direction of time" to be able to express spin as a vector. Again, I don't have an explicit textbook statement reference for this statement, but I beleive it to be correct.

Last edited: Mar 3, 2013
10. Mar 3, 2013

### Staff: Mentor

I think the reason for this is that spin as an antisymmetric 2nd rank tensor represents rotation in a plane that is "purely spacelike", so to speak; in other words, there is change in two of the four spacetime dimensions. The bivector represents the two dimensions in which there is not change; but that alone is not enough to pin down a spacelike vector within those two dimensions. To do that, you have to specify a timelike vector, which is what picking a "direction of time" does.

11. Mar 3, 2013

### WannabeNewton

Thanks guys. So in the case of the observer placed at rest anywhere inside the shell could we view $S^{a}$ as just being a choice of a normal / orientation vector along the spin axis for the observer (normal to the time direction in this case) and then we can look at $dS^{a} / dt$ as describing the rate at which this normal/ orientation vector is precessing? So basically, physically will the observer just remain at rest for all time when initially placed at rest anywhere inside the shell due to the nature of the gravito magnetic lorentz force discussed above (by rest I mean that in the global inertial frame fixed to the background minkowski space - time, the 4 - velocity of said observer is just (1,0,0,0)), but start precessing in terms of his/her orientation?

I'm not really sure how to rigorously describe this "spin axis" because the precession of the spin vector is being measured in the fixed background global inertial frame but the spin vector itself is assigned and parallel propagated by the observer in the shell. As mentioned before, the observer has no spatial velocity so this isn't any kind of orbital angular momentum but is more akin to a spin angular momentum but this would only make sense for extended bodies so does frame dragging not apply to point particles or does that get into intrinsic spin and QM related stuff? Sorry for all the questions :[

Last edited: Mar 3, 2013
12. Mar 3, 2013

### pervect

Staff Emeritus
Something else I should mention - I'm most likely using the dual (hodges dual) of the usual spin bivector.

One form of the bivector is radius ^ linear momentum. The (hodges) dual of that will be time ^ spin axis.

^ represents the wedge product as before.

13. Mar 3, 2013

### WannabeNewton

I guess i'm just having a hard time visualized what would happen to the observer as seen in the background global inertial frame. I am relatively sure that as measured in this frame the 4 - velocity of the observer inside the shell would just be $u^{a} = (1,0,0,0)$ because the gravito magnetic lorentz force $\vec{F}\propto \vec{v}\times \vec{\omega }$ and as such if the observer is initially placed at rest anywhere inside the shell (not just the center), with respect to the background inertial frame, he will remain at rest for all time because this force cannot change the speed. We could then talk about a spin axis for the observer along the spin vector $S^{a}$ he parallel propagates along his worldine and see how it starts to precess due to the frame dragging. So what is physically happening? Is the observer twirling around in space while remaining fixed to the point in the shell he was placed in because the frame dragging force can't change his position in the shell if he was at rest to begin with? (of course, all these things regarding the analysis of the spin vector and its precession are being seen and measured in the background global inertial frame as soon as the observer in the shell actually defines and parallel propagates it along his worldline IF I'm not mistaken)

Last edited: Mar 3, 2013
14. Mar 6, 2013

### pervect

Staff Emeritus
Thinking about what measurements one would actually take, what springs to mind is having a fixed gyroscope with a telescope down the center and observing distant "fixed stars".

This sort of measurement would work most directly at the center, as gravitational lensing effects would cause additonal shifts in the angles elsewhere, as only at the center would the geodesics be radial (by the spherical symmetry).

I think that you could measure it anywhere if what you measured was the time it took to make a complete precession circuit - assuming that the precession makes the gyroscope return to its starting point (i think it should).