Solving poisson's equation slowly rotating spherical shell of mass

[WannabeNewton]

Solving "poisson's equation" slowly rotating spherical shell of mass

Homework Statement

We have that \partial ^{\alpha}\partial _{\alpha}\bar{\gamma _{0\mu}} = -16\pi T_{0\mu} which is very similar to Poisson's equation if we treat each component of the metric tensor as a scalar field (the reason we only care about the time - space components is because the to be described mass - energy distribution has negligible pressure and rotates very slowly). The solution, in general, is given by \bar{\gamma _{0\mu}} = 4\int \frac{T_{0\mu}(x')dS(x')}{|x' - x|}. I have a thin hollow slowly rotating spherical shell of uniform mass density \rho, radius R, and constant angular velocity \boldsymbol{\omega } about the z - axis and I must solve for the components for such a case.

The Attempt at a Solution

The stress energy tensor for a perfect fluid is (I only care about the 0i components here which is where the rotational effects come into play anyways and I have assumed the time component of 4 - velocity is approximately 1 to first order which is fine because this is a slow motion approximation) T_{0i} = -\rho u{^{i}} = -\rho (\boldsymbol{\omega } \times r')^{i} = -\frac{M}{4\pi R^{2}}\delta(r' - R)(\boldsymbol{\omega } \times r')^{i} so in particular T_{0x} = -\frac{M}{4\pi R^{2}}\delta(r' - R)\omega y' = -\frac{M}{4\pi R^{2}}\delta(r' - R)\omega r'sin\theta 'sin\varphi ' and T_{0y} = \frac{M}{4\pi R^{2}}\delta(r' - R)\omega x' = \frac{M}{4\pi R^{2}}\delta(r' - R)\omega r'sin\theta 'cos\varphi ' (the \omega that shows up here is just the magnitude of \boldsymbol{\omega })

Basically I'm solving for these functions \bar{\gamma _{0x}}, \bar{\gamma _{0y}} that are pretty much like the scalar potential in EM and the solution is given by the usual solution to poisson's equation for a scalar potential as noted above. So, for example, \bar{\gamma _{0x}} = -4\int\frac{ \frac{M}{4\pi R^{2}}\delta(r' - R)\omega r'sin\theta 'sin\varphi 'dS'}{|r' - r|} = -4\int\frac{ \frac{M}{4\pi R^{2}}\delta(r' - R)\omega r'sin\theta 'sin\varphi 'r'^{2}sin\theta 'dr'd\theta 'd\varphi '}{|r' - r|} and I have to evaluate this over the entire sphere. I'm not sure if I even set this up correctly and even if I did set it up correctly I'm also not sure how to go about actually evaluating that thing. |r' - r| = \sqrt{r'^{2} + r^{2} - 2rr'cos\beta _{r'r}} but I don't know how to relate the angle between them, \beta _{r'r} to the spherical coordinate angles so that I could get started off trying to integrate. Again, still not sure if I even set up the whole thing right. Thanks a ton guys!

 

[TSny]

|r' - r| = \sqrt{r'^{2} + r^{2} - 2rr'cos\beta _{r'r}} but I don't know how to relate the angle between them, \beta _{r'r} to the spherical coordinate angles so that I could get started off trying to integrate. Again, still not sure if I even set up the whole thing right.

I think you have it set up right although I haven't checked all the numerical factors. To deal with the angle \beta _{r'r} issue, you can set up the problem initially in a coordinate system where the field point (observation point) lies on the z axis and the angular velocity vector lies in the x-z plane. Then \beta _{r'r} is just the angle θ' between the source point r' on the sphere and the z-axis. Later you can convert back to the coordinate system where the angular velocity is along the z axis. Here is a tutorial for a similar calculation of the vector potential of a rotating charged spherical shell.

 

[WannabeNewton]

I will try that thanks and thank you so much for the tutorial. I will see what I can do and come back with further questions if that is ok. I appreciate the help!

 

[WannabeNewton]

Hi TSny. So first off can I start off by saying
-4\int\frac{ \frac{M}{4\pi R^{2}}\delta(r' - R)\omega r'sin\theta 'sin\varphi 'r'^{2}sin\theta 'dr'd\theta 'd\varphi '}{|r' - r|} = -4\int\frac{ \frac{M}{4\pi R^{2}}\omega R^{3}sin\theta 'sin\varphi 'sin\theta 'dr'd\theta 'd\varphi '}{|R - r|} = -\frac{MR}{\pi}\omega\int\frac{ sin^{2}\theta 'sin\varphi 'dr'd\theta 'd\varphi '}{|R - r|}?
Secondly, so if I pick my coordinate system so that the field point is on the z - axis I would just have |R - r| = \sqrt{R^{2} + r^{2} - 2Rrcos\theta '} correct? I feel like I'm missing one thing: if I do make this choice of coordinates how exactly will it affect the angular velocity vector? It will no longer be perfectly aligned along the z - axis so I have to go back to the start of my set up and redo the v^{i} = (\boldsymbol{\omega} \times r')^{i} stuff right (so that my above expression for the integral would have to be changed to accommodate this if I'm not mistaken)? Thanks a ton!

 

[TSny]

You have two factors of sinθ' in your integrand. But one of these is incorrect. Instead, you should have sinψ' as in equation 11 of the tutorial. To see where this comes from, you have to evaluate the cross product ω x r' where ω makes the angle ψ to the z-axis. ω x r' will have x, y, and z components, but you can show that only the y-component will contribute to the integral.

 

[WannabeNewton]

I got the first sin factor from replacing the y coordinate from the cross product with it's spherical coordinate representation. The other came from the volume element. Was that not the right procedure? I'll quote it here: T_{0i} = -\rho u{^{i}} = -\rho (\boldsymbol{\omega } \times r')^{i} = -\frac{M}{4\pi R^{2}}\delta(r' - R)(\boldsymbol{\omega } \times r')^{i} so in particular T_{0x} = -\frac{M}{4\pi R^{2}}\delta(r' - R)\omega y' = -\frac{M}{4\pi R^{2}}\delta(r' - R)\omega r'sin\theta 'sin\varphi '. This was for when the angular velocity was aligned with the z - axis. Is what you are talking about when you make it lie on the x - z plane so that v = \omega \times r' = (-\omega _{z}y',\omega _{z}x' - \omega _{x}z', \omega _{x}y')? Thanks.

 

[TSny]

Is what you are talking about when you make it lie on the x - z plane so that v = \omega \times r' = (-\omega _{z}y',\omega _{z}x' - \omega _{x}z', \omega _{x}y')?

Yes, that's right. Note that x' and y' have factors of cos\phi' and sin\phi' respectively. When you integrate over \phi' what happens to these terms?

 

[WannabeNewton]

Well integrating the cos\phi terms will make them vanish and integrating the sin\phi term will also make them vanish, and the only term without any of these is the -\omega _{x}z' in the y - component, so the only surviving integral should be
\bar{\gamma _{0y}} = 4\int \frac{T_{0y}dS'}{|r' - r|} = -4\int \frac{\frac{M}{4\pi R^{2}}\delta (r' - R)\omega _{x}z'r'^{2}dr'd\Omega '}{|r' - r|} = -4\int \frac{\frac{M}{4\pi R^{2}}\delta (r' - R)\omega _{x}r'^{3}cos\theta 'sin\theta 'dr'd\theta '}{\sqrt{r'^{2} + r^{2} - 2r'rcos\theta '}}
I think.

I am also unsure about where the angle \Psi comes in here. I didn't seem to be able to extract any such angle from the cross product. Thank you!

 

[TSny]

Well integrating the cos\phi terms will make them vanish and integrating the sin\phi term will also make them vanish, and the only term without any of these is the -\omega _{x}z' in the y - component, so the only surviving integral should be
\bar{\gamma _{0y}} = 4\int \frac{T_{0y}dS'}{|r' - r|} = -4\int \frac{\frac{M}{4\pi R^{2}}\delta (r' - R)\omega _{x}z'r'^{2}dr'd\Omega '}{|r' - r|} = -4\int \frac{\frac{M}{4\pi R^{2}}\delta (r' - R)\omega _{x}r'^{3}cos\theta 'sin\theta 'dr'd\theta '}{\sqrt{r'^{2} + r^{2} - 2r'rcos\theta '}}

Yes, that looks good.

I am also unsure about where the angle \Psi comes in here.

The angular velocity ω lies in the x-z plane and makes an angle ψ to the z-axis. You need ω_x. See the figure in the tutorial.

 

[WannabeNewton]

Oh I see what you mean. So yes \omega _{x} = \omega sin\Psi (where \omega is the same angular velocity magnitude as from when it was aligned on the z axis as rotations preserve norm) I think. Also sorry but I think I forgot a factor of \pi in front of the 4 that is outside the integral otherwise I have a leftover pi. I ran the integral through wolfram and then evaluated the bounds and hopefully if I didn't make a mistake, I got \bar\gamma {_{0y}} = -\frac{2M}{3R}r\omega sin\Psi. Am I supposed to rotate back now?

EDIT: I should add that I chose the signs for the square roots resulting from the integration in accordance with r <= R because the problem only asked for the field when within the shell.

 

[TSny]

OK. I think that's basically it, although if I'm not mistaken the factor of 2/3 should be 4/3. Anyway, note that you can write $r\omega sin\psi$ (in the negative y-direction) as $\vec{\omega}\times\vec{r}$. This is now in a form that is independent of coordinate system. So, you can think of it as applying to a coordinate system where ω is along the z-axis and the observation point r is at angle θ = ψ to the z-axis. See equations 15 and 16 in the tutorial.

 

[WannabeNewton]

I'll check my integration again to make sure I didn't make any arithmetic mistakes regarding that factor of 2. However going along with you, so we can now write \vec{{\gamma }} = (\bar{\gamma _{0x}}, \bar{\gamma _{0y}}, \bar{\gamma _{0z}}) = (0,-\frac{4M}{3R}\omega rsin\Psi , 0) = \frac{4M}{3R}(\vec{\omega} \times \vec{r})? And that last expression is coordinate independent so do I have the freedom to then re - express it in coordinates it by shifting r back to its original observation angle and shift the angular velocity back to the z - axis? Thanks!

 

[TSny]

Yes. That's right.

 

[WannabeNewton]

So I can rewrite it like this: \vec{r} = (x,y,z), \vec{\omega } = (0,0,\omega ) so that \vec{\gamma } = \frac{4M}{3R}(\vec{\omega} \times \vec{r}) = \frac{4M}{3R}(-\omega y,\omega x,0) thus in particular \bar{\gamma} _{0x} = -\frac{4M}{3R}\omega y,\bar{\gamma} _{0y} = \frac{4M}{3R}\omega x ,\bar{\gamma} _{0z} = 0. The vector potential for the gravito - magnetic field is defined as A_{i} = -\frac{1}{4}\bar{\gamma }_{0i} therefore A_{x} = \frac{M}{3R}\omega y, A_{y} = -\frac{M}{3R}\omega x, A_{z} = 0. The gravito - magnetic itself is B^{i} = \epsilon ^{ijk}\partial _{j}A_{k} which gives B^{z} = -\frac{2M}{3R}\omega ,B^{x} = B^{y} = 0 so \vec{B} = -\frac{2M}{3R}\vec{\omega}. I'm off by a sign from what the book says is the final answer but that isn't an issue I can just work back and see where I made the mistake. Can you just check if me re - expressing the angular velocity and field point position vector as I did above is a valid move? Thank you so much TSny!

 

[TSny]

That looks correct to me.

 

[WannabeNewton]

Ok one more thing then: I thought I understood it in the beginning but it seems I don't - why is it that aligning the arbitrary field point with the z - axis will cause the angular velocity, which was originally along the z - axis, to now be on only the x - z plane as opposed to some general direction?

 

[WannabeNewton]

If I'm not mistaken, is it because, starting with the field point vector in an arbitrary direction and the angular velocity vector along the z - axis, we can first rotate about the z - axis until the field point vector is on the x - z plane, and this will still keep the angular velocity vector fixed along the z -axis because it is a rotation about the z - axis, and then rotate about the origin of the x - z plane until the field point vector is along the z - axis and now the angular velocity vector will lie at some angle on the x - z plane?

 

[TSny]

That would be one way to get from the one coordinate system to the other. The main point is that we can choose the coordinate system anyway we wish. We just chose it so that ω would be in the x-z plane. We could just as well have chosen it such that ω was in the y-z plane. What made the problem easier was to choose the z-axis through the field point. That way, the angle between r and r' was just the spherical coordinate angle θ. Since the answer in this system could be expressed as a vector equation, the answer must be valid as a vector equation in any coordinate system. So, we write it out in any coordinate system we choose. We don't really need to worry about how to get from one system to the other.

 

[WannabeNewton]

Yeah I was just about to ask why we couldn't have chosen to do it on the y - z plane and get the same answer. You answered that already however :). Well thank you so much, again, for all the help. Would you mind if I tried to summarize what we did just to make sure I got the procedure down?

 

[TSny]

Not at all. Please do.

 

[WannabeNewton]

Cool thanks! So we start off by re - orienting our axes, as described above, so that the field point vector is along the z - axis and the angular velocity is on the x - z plane (although we could have just as easily chosen the y - z plane). From here we expressed the components of the stress energy tensor in terms of this coordinate system, eliminated the components that would integrate to zero, and just kept the one surviving integral. We calculated the integral whichever way and found our result was \vec{\gamma } = -\frac{4M}{3R}\omega rsin\Psi \hat{y} where \Psi is the angle \vec{\omega }, which is on the x - z plane, makes with the z - axis.

Now this is the part I want to make sure I understand: we know from the way we oriented our coordinate system that the field point vector \vec{r} is along the z - axis, and \vec{\omega } is on the x - z plane, therefore -\omega rsin\Psi \hat{y} is simply their cross product by the usual right hand rule argument (if \Psi is negative then the right hand rule gives a positive y direction but sin\Psi will be negative so that gives a negative y direction again and if \Psi is positive then the right hand rule immediately gives a negative y direction correct?). Hence we can write -\omega rsin\Psi \hat{y} = \vec{\omega }\times \vec{{r}} and \vec{\gamma } = \frac{4M}{3R}(\vec{\omega }\times \vec{{r}}). This is of course coordinate indepedent so we have the freedom to re - express it any coordinate system we choose, in particular the original one we started with where the angular velocity vector was along the z - axis and the field point vector was off in some arbitrary direction.

Would that be accurate to say (especially that last paragraph)? Thank you TSny!