I Linearised gravity approach to Lense Thirring metric

ergospherical
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Doing some revision and getting confused. It's under GR but may as well be under electromagnetism or calculus because that is where the problem is. Taking a shell of mass ##\rho = M\delta(r-R)/(4\pi R^2)## and four velocity corresponding to rotation about ##z## axis i.e. ##U = (1, -\omega y, \omega x, 0)##. Looking at the ##\bar{h}_{i0}## component, say, the Green's function solution if I remember right,
$$\bar{h}_{i0} = -4 \int \frac{T_{i0}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} d^3 \mathbf{x}'$$And I can take a multipole expansion
$$\frac{1}{|\mathbf{x} - \mathbf{x}'|} = \frac{1}{r} + \frac{\mathbf{x} \cdot \mathbf{x}'}{r^3} + \dots$$
Start with ##i = x##, where ##U_x = \omega y##

$$\bar{h}_{x0} = -\frac{4M \omega}{r}\int \frac{\delta(r'-R)}{4\pi R^2} y' \ d^3 \mathbf{x}' - \frac{4M \omega}{r^3} \mathbf{x} \cdot \int \mathbf{x}' \frac{\delta(r'-R)}{4\pi R^2} y' \ d^3\mathbf{x}' + \dots$$
So look at the first term. With ##d^3 \mathbf{x}' = r'^2 \sin{\theta'} dr' d\theta' d\phi'## and putting also ##y' = r' \sin{\phi'}##, then the integral over ##\phi'## from ##0 \rightarrow 2\pi## is going to zero.

With the second term, look at the part$$\mathbf{I} = \int \mathbf{x}' \frac{\delta(r'-R)}{4\pi R^2} y' \ d^3\mathbf{x}'$$Then $$I_{x'} = \int x' y' \frac{\delta(r'-R)}{4\pi R^2} d^3 \mathbf{x}$$and ##x'y' = r'^2 \sin{\phi'} \cos{\phi'}## is also going to go to zero when integrated over ##\phi'##, same story for the other two components. What's the issue?
 
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ergospherical said:
four velocity corresponding to rotation about ##z## axis i.e. ##U = (1, -\omega y, \omega x, 0)##.
The ##t## component of 4-velocity can't be ##1## if any of the other components are nonzero; the 4-velocity has to be a unit vector.
 
ergospherical said:
With the second term, look at the part$$\mathbf{I} = \int \mathbf{x}' \frac{\delta(r'-R)}{4\pi R^2} y' \ d^3\mathbf{x}'$$Then $$I_{x'} = \int x' y' \frac{\delta(r'-R)}{4\pi R^2} d^3 \mathbf{x}$$and ##x'y' = r'^2 \sin{\phi'} \cos{\phi'}## is also going to go to zero when integrated over ##\phi'##, same story for the other two components. What's the issue?
Typo: You should have ##x'y' = r'^2 \sin^2 \theta' \sin{\phi'} \cos{\phi'}##. But, still, the ##\phi'## integration for ##I_{x'}## will be zero .

But for ##I_y'##, you will deal with ##y'y' = r'^2 \sin^2 \theta' \sin ^2 {\phi'} ##. Integrating over ##\phi'## will no longer give zero.
 
Thanks for spotting the error. I get ##I_y = R^2/3## which gives $$h_{x0} = \bar{h}_{x0} = -\frac{4}{3} \frac{MR^2 \omega y}{r^3}$$and I think this makes sense as the result because if ##L \sim MR^2 \omega## then I remember roughly that the vector potential should go like ##\mathbf{L} \times \mathbf{x}/r^3##. You'd also get $$h_{y0}=\bar{h}_{y0} = \frac{4}{3} \frac{MR^2 \omega x}{r^3}$$Also, close to the shell, ##r \sim R##, the size of the perturbation looks like ##\sim (4/3) \phi \omega x^i## with ##\phi = GM/R##. But taking ##r \rightarrow 0## you get divergence of the metric, which can't be right, although I'd still have expected it to be weak field there?
 
ergospherical said:
Doing some revision and getting confused. It's under GR but may as well be under electromagnetism or calculus because that is where the problem is. Taking a shell of mass ##\rho = M\delta(r-R)/(4\pi R^2)## and four velocity corresponding to rotation about ##z## axis i.e. ##U = (1, -\omega y, \omega x, 0)##. Looking at the ##\bar{h}_{i0}## component, say, the Green's function solution if I remember right,
$$\bar{h}_{i0} = -4 \int \frac{T_{i0}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} d^3 \mathbf{x}'$$And I can take a multipole expansion
$$\frac{1}{|\mathbf{x} - \mathbf{x}'|} = \frac{1}{r} + \frac{\mathbf{x} \cdot \mathbf{x}'}{r^3} + \dots$$
The form of the expansion depends on whether ##r > r'## or ##r < r'##. Grabbing my copy of Jackson's E&M text, I find $$\frac{1}{|\mathbf{x} - \mathbf{x}'|} = \sum_{l = 0}^{\infty} \frac{r_<^l}{r_>^{l+1}}P_l(\cos \gamma)$$ where ##r_< \, (r_>)## is the smaller (larger) of ##r## and ##r'##. ##P_l## are the Legendre polynomials and ##\gamma## is the angle between ##\mathbf{x}## and ##\mathbf{x}'##. Your expression for the expansion would be for ##r > r'## (corresponding to ##\mathbf{x}## being outside the shell).

There's also the need to justify dropping the higher-order terms in the expansion.

The integral $$\bar{h}_{i0} = -4 \int \frac{T_{i0}(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|} d^3 \mathbf{x}'$$ can be done without a series expansion. With ##T_{i0} = \rho \;\boldsymbol{\omega} \times \mathbf{x}'##, the key part is the evaluation of $$\int \rho(r') \frac{\mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|} d^3x'$$ Choose spherical coordinates with the z-axis along ##\mathbf{x}## (not along ##\boldsymbol{\omega}##). So, in this integration, ##\theta '## is the angle between ##\mathbf x'## and ##\mathbf x##.The angular part of the integration is $$ \mathbf{I} = \int \frac{\mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|} d\Omega'= \int_0^\pi d\theta' \sin \theta' \int_0^{2 \pi} d \phi' \frac{\mathbf x'}{\sqrt{r^2-2rr'\cos\theta' + r'^2}}$$ You can show that only the z-component survives the ##\phi'## integration. I get $$I_z = \frac {4 \pi} 3 \frac {r'^2}{r^2} \,\,\,\, \text{for } r > r'$$ $$I_z = \frac {4 \pi} 3 \frac {r}{r'} \,\,\,\, \text{for } r < r'$$
 
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TSny said:
I get $$I_z = \frac {4 \pi} 3 \frac {r'^2}{r^2} \,\,\,\, \text{for } r > r'$$ $$I_z = \frac {4 \pi} 3 \frac {r}{r'} \,\,\,\, \text{for } r < r'$$
This result shows that the integral ## \mathbf{I} = \int \dfrac{\mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|} d\Omega'## has a direction along the z-axis that was chosen to compute the integral. This z-axis was chosen along ##\mathbf{x}##. So, ##\mathbf{I}## is along ##\mathbf{x}##. The result for ##\mathbf{I}## can be written $$\mathbf{I} = \frac {4 \pi r'^2} {3 r^2} \hat {\mathbf{x}}\,\,\,\, \text{for } r > r'$$ $$\mathbf{I}= \frac {4 \pi r} {3 r'} \hat {\mathbf{x}}\,\,\,\, \text{for } r < r'$$
 
@TSny I've come across quite a cool way to solve this. Maybe you've see it before... it's based on defining a complex variable ##H = h_{01} + ih_{02}##. (If we're only looking at the contribution sourced by ##T_{0i}## then ##\bar{h} = 0## and ##\bar{h}_{\mu \nu} = h_{\mu \nu}##, so we can just work with no bars...).

Instead of going via. the Greens function solution, just try to solve directly:$$\nabla^2 H = \nabla^2 (h_{01} + ih_{02}) = -16\pi(T_{01} + iT_{02})$$The nice thing is that ##T_{0k} = \rho u_0 u_k = -\rho (-\Omega y, \Omega x, 0)_k = -\rho \Omega R \sin{\theta} (-\sin{\phi}, \ cos{\phi}, 0)_k##
So we could define another complex variable ##T \equiv T_{01} + iT_{02}## which comes out as$$T = -i \frac{\Omega M}{4\pi R} \delta(r-R) \sin{\theta} e^{i \phi}$$and the equation to solve is ##\nabla^2 H = -16\pi T##. You can take a trial solution ##H = \mathcal{R}(r) \sin{\theta} e^{i\phi}##. And writing the Laplacian as ##\nabla^2 = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r} \right) + \hat{L}##, with ##\hat{L}## the angular Laplacian, gives\begin{align*}
\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \mathcal{R}}{\partial r}\right) \sin{\theta} e^{i\phi} + (\hat{L} \sin{\theta} e^{i\phi})\mathcal{R} &= i \frac{4\Omega M}{ R} \delta(r-R) \sin{\theta} e^{i \phi} \\
\frac{\partial}{\partial r} \left( r^2 \frac{\partial \mathcal{R}}{\partial r} \right) - 2r^2 \mathcal{R} &= i \frac{4\Omega M}{ R} \delta(r-R)
\end{align*}Given that ##\hat{L} \sin{\theta} e^{i\phi} = -2\sin{\theta} e^{i\phi}##.
You can then solve the homogeneous part in for ##r<R## and ##r>R## respectively, then join them together with continuity at ##r=R## and by integrating the jump condition between ##R-\epsilon## and ##R+\epsilon##. This gets the same result as the methods you described.
 
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