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Solving "poisson's equation" slowly rotating spherical shell of mass
We have that [itex]\partial ^{\alpha}\partial _{\alpha}\bar{\gamma _{0\mu}} = -16\pi T_{0\mu}[/itex] 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 [itex]\bar{\gamma _{0\mu}} = 4\int \frac{T_{0\mu}(x')dS(x')}{|x' - x|}[/itex]. I have a thin hollow slowly rotating spherical shell of uniform mass density [itex]\rho[/itex], radius [itex]R[/itex], and constant angular velocity [itex]\boldsymbol{\omega }[/itex] about the z - axis and I must solve for the components for such a case.
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) [itex]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}[/itex] so in particular [itex]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 '[/itex] and [itex]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 '[/itex] (the [itex]\omega[/itex] that shows up here is just the magnitude of [itex]\boldsymbol{\omega }[/itex])
Basically I'm solving for these functions [itex]\bar{\gamma _{0x}}, \bar{\gamma _{0y}}[/itex] 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, [itex]\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|}[/itex] 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. [itex]|r' - r| = \sqrt{r'^{2} + r^{2} - 2rr'cos\beta _{r'r}} [/itex] but I don't know how to relate the angle between them, [itex]\beta _{r'r}[/itex] 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!
Homework Statement
We have that [itex]\partial ^{\alpha}\partial _{\alpha}\bar{\gamma _{0\mu}} = -16\pi T_{0\mu}[/itex] 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 [itex]\bar{\gamma _{0\mu}} = 4\int \frac{T_{0\mu}(x')dS(x')}{|x' - x|}[/itex]. I have a thin hollow slowly rotating spherical shell of uniform mass density [itex]\rho[/itex], radius [itex]R[/itex], and constant angular velocity [itex]\boldsymbol{\omega }[/itex] 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) [itex]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}[/itex] so in particular [itex]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 '[/itex] and [itex]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 '[/itex] (the [itex]\omega[/itex] that shows up here is just the magnitude of [itex]\boldsymbol{\omega }[/itex])
Basically I'm solving for these functions [itex]\bar{\gamma _{0x}}, \bar{\gamma _{0y}}[/itex] 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, [itex]\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|}[/itex] 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. [itex]|r' - r| = \sqrt{r'^{2} + r^{2} - 2rr'cos\beta _{r'r}} [/itex] but I don't know how to relate the angle between them, [itex]\beta _{r'r}[/itex] 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!
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