Linearised gravity approach to Lense Thirring metric

[ergospherical]

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?

 

[PeterDonis]

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.

 

[TSny]

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.

 

[ergospherical]

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?

 

[TSny]

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 $$I_z = \frac {4 \pi} 3 \frac {r'^2}{r^2} \,\,\,\, \text{for } r > r'$$ $$I_z = \frac {4 \pi} 3 \frac {r}{r'} \,\,\,\, \text{for } r < r'$$

 

[TSny]

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 $$\mathbf{I} = \frac {4 \pi r'^2} {3 r^2} \hat {\mathbf{x}}\,\,\,\, \text{for } r > r'$$ $$\mathbf{I}= \frac {4 \pi r} {3 r'} \hat {\mathbf{x}}\,\,\,\, \text{for } r < r'$$

 

[ergospherical]

@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.

 

[TSny]

Thanks @ergospherical, I have not seen that before. Very nice!