Derivation of Kirchhoff's Formula in Gravitational Radiation

[jstrunk]

I am trying to understand the formula's for Gravitational Radiation. At one point in the derivation, every source I have
seen comes up with Kirchhoff's Formula but they never derive it. They always just say its a well known result from
Electromagnetic Radiation. I have been able to determine that they use a Green's function to solve the PDQ. I have
a basic knowledge of Green's functions but I can't quite get the solution in the book because the book isn't very
clear about the physical situation that is being modeled and you need that to come up with the Green's function.
Anyway, can someone recommend free or inexpensive source that carefully derives this formula?
Thanks.

Incidentally, I spent several hours trying to get Kirchhoff's Formula to format in the Preview and had to give up. The Latex was produced by MathType. Maybe it will format when I actually post it.

$h'_{jk,ii} = - 2\kappa T_{jk}$

Kirchhoff's Formula:

$h'_{jk} \left( x_0 ,t_0 \right) = \frac{\kappa }{2\pi} \int_V \frac{1}{r} T_{jk} \left( x, t_0 - \frac{r}{c} \right) dV$

[Moderator's note: edited the formula so it formats. However, it still may not be quite right; jstrunk, you may want to edit further.]

 

[PeterDonis]

every source I have seen comes up with Kirchhoff's Formula

Can you give some references?

 

[jstrunk]

Introduction to Tensor Calculus, Relativity and Cosmology by D. F. Lawden, Page 161:
"The solution of the wave equation (58.9) with source term $- 2\kappa T_{jk}$ is well known to be given by
Kirchhoff's Formula (Bateman, 1952)".

https://en.wikipedia.org/wiki/Gravitational_wave
Under the heading "Relation to the source", there is a statement "we can write the solution in terms of the tensorial
Green's function for the d'Alembertian operator". It then goes on to give a formula which I think is the Kirchhoff's Formula,
although that is obscured by the fact that every imaginable convention is different from the one used in Lawden.

A first course in general relativity by Bernard F. Schutz, page 233:
Under the heading 'Exact solution of the wave equation" there is a statement "Readers who have studied the wave equation,
Eq. (9.55), will know that its outgoing-wave solution for arbitrary $T_{\mu \nu }$ is given by the retarded
integral $\bar h_{\mu \nu } \left( {t,x^i } \right) = 4\int {\frac{{T_{\mu \nu } \left( {t - R,y^i } \right)}}{R}d^3 y}$.

I spent hours Googling various combinations of "Gravitational Waves", "Kirchhoff's Formula", "Green's Function", etc and saw
other similarly unhelpful items. It is very tedious to try to relocate them all.

 

[bobm]

This is OT for Kirchhoff's Formula, but I wanted to address the use of MathType here. Being OT, I won't be offended if a moderator deletes it.

To use MathType on Physics Forums, don't paste LaTeX from MathType into the "Write LaTeX Code" block below. Instead, follow the process here: http://www.dessci.com/en/support/MathType/works_with.asp#!target=physics_forums. If you paste the equation here into the editing space, and it will show up as LaTeX until you Post Reply. The Preview works as well.

 

[jstrunk]

Still working on this. My current question is
$$\begin{array}{l}<br /> {\rm{In solving the following equation for Electromagnetic Potential using a Green's Function approach}}\\<br /> {\nabla ^2}\psi - \frac{1}{{{c^2}}}\frac{{{\partial ^2}\psi }}{{\partial {t^2}}} = - 4\pi f\left( {x,t} \right)\\<br /> {\rm{one takes Fourier Transforms of }}\\<br /> {\delta ^3}\left( {\vec x - \vec x'} \right)\delta \left( {t - t'} \right)\\<br /> {\rm{and gets}}\\<br /> \delta \left( {t - t'} \right){\rm{ = }}\frac{1}{{2\pi }}\int {{e^{ - iw\left( {t - t'} \right)}}dw} {\rm{ }}\\<br /> \delta \left( {\vec x - \vec x'} \right){\rm{ = }}\frac{1}{{{{\left( {2\pi } \right)}^3}}}\int {{e^{i\vec k\left( {\vec x - \vec x'} \right)}}{d^3}k} \\<br /> {\rm{Can someone explain why the different signs on the exponent here?}}\\<br /> {\rm{This can be seen in context here:}}\\<br /> {\rm{https://archive}}{\rm{.org/stream/ClassicalElectrodynamics/Jackson - ClassicalElectrodynamics\# page/n201/mode/2up/search/contour}}<br /> \end{array}$$

 

[jstrunk]

I bought an upgrade to MathType so I could get the required Cut and Paste option and followed the instructions to the best of my ability but it still doesn't work.
I get very frustrated with this and go to extreme lengths to avoid having to ask questions in the forum because it is usually harder to enter math into the forum than it is to find the answer some other way.

 

[PeterDonis]

I bought an upgrade to MathType so I could get the required Cut and Paste option and followed the instructions to the best of my ability but it still doesn't work.

I'm not familiar with MathType, but from looking at your post #5 it doesn't look like its output is suitable for forum posts. It appears to be trying to format your entire post using LaTeX, instead of just the equations. It's a lot easier if you just type ordinary text the way you usually would, and only use the LaTeX features to format the actual math. It does take some practice to get used to.

The PF LaTeX Primer is here:

https://www.physicsforums.com/help/latexhelp/

 

[dextercioby]

The Kirchhoff's formula applies to find the solution of a 3-D inhomogenous wave equation. D'Alembertian something = - constant times known function (= source of radiation).

There's no particular derivation of gravitational waves. One uses the well-known derivation for the e-m waves and adjust 4-vectors to 2nd rank 4-tensors in the final solution.

 

[jstrunk]

In solving the following equation for Electromagnetic Potential using a Green's Function approach
$${\nabla ^2}\psi - \frac{1}{{{c^2}}}\frac{{{\partial ^2}\psi }}{{\partial {t^2}}} = - 4\pi f\left( {x,t} \right)$$
one takes Fourier Transforms of
$${\delta ^3}\left( {\vec x - \vec x'} \right)\delta \left( {t - t'} \right)$$
and gets
$$\delta \left( {t - t'} \right){\rm{ = }}\frac{1}{{2\pi }}\int {{e^{ - iw\left( {t - t'} \right)}}dw}$$
$$\delta \left( {\vec x - \vec x'} \right){\rm{ = }}\frac{1}{{{{\left( {2\pi } \right)}^3}}}\int {{e^{i\vec k\left( {\vec x - \vec x'} \right)}}{d^3}k}$$
Note: there is a dot product between K and (x-xprime) which is lost when I paste it into the forum.
Can someone explain why the different signs on the exponent here?
This can be seen in context here:
https://archive.org/stream/ClassicalElectrodynamics/Jackson - ClassicalElectrodynamics\# page/n201/mode/2up/search/contour

 

[samalkhaiat]

In solving the following equation for Electromagnetic Potential using a Green's Function approach
$${\nabla ^2}\psi - \frac{1}{{{c^2}}}\frac{{{\partial ^2}\psi }}{{\partial {t^2}}} = - 4\pi f\left( {x,t} \right)$$
one takes Fourier Transforms of
$${\delta ^3}\left( {\vec x - \vec x'} \right)\delta \left( {t - t'} \right)$$
and gets
$$\delta \left( {t - t'} \right){\rm{ = }}\frac{1}{{2\pi }}\int {{e^{ - iw\left( {t - t'} \right)}}dw}$$
$$\delta \left( {\vec x - \vec x'} \right){\rm{ = }}\frac{1}{{{{\left( {2\pi } \right)}^3}}}\int {{e^{i\vec k\left( {\vec x - \vec x'} \right)}}{d^3}k}$$
Note: there is a dot product between K and (x-xprime) which is lost when I paste it into the forum.
Can someone explain why the different signs on the exponent here?
This can be seen in context here:
https://archive.org/stream/ClassicalElectrodynamics/Jackson - ClassicalElectrodynamics\# page/n201/mode/2up/search/contour

You are free to use any sign. However, Lorentz invariance demands opposite signs, i.e., $kx = \omega t - \mathbf{k} \cdot \mathbf{x}$.
Below, I will solve that equation using three (basic, intermediate and advance) methods. Hopefully, I will get the same answer in each method.
1) Causal Solution from Action-at-a-distance Solution
If the velocity of propagation is infinitely large, then your equation reduces to the Poisson’s equation
$$\nabla^{2}\psi ( \mathbf{x}, t ) = - 4 \pi \ f( \mathbf{x}, t ) = - 4 \pi \int d^{3} \bar{\mathbf{x}} \ f(\bar{\mathbf{x}},t) \ \delta^{3}(\mathbf{x} - \bar{\mathbf{x}}) .$$ Using the identity $$\nabla^{2} \frac{1}{|\mathbf{x} - \bar{\mathbf{x}}|} = - 4 \pi \ \delta^{3}(\mathbf{x} - \bar{\mathbf{x}}) ,$$ we obtain the following solution to Poisson’s equation $$\psi(\mathbf{x},t) = \int d^{3}\bar{\mathbf{x}} \ \frac{f(\bar{\mathbf{x}},t)}{|\mathbf{x} - \bar{\mathbf{x}}|} . \ \ \ \ \ (1.1)$$ From this, we can infer the result for causal (i.e., finite speed) propagation and hence a “solution” to the full inhomogeneous field equation: The field at time $t$ depends not on what the source is doing at $t$, but rather on that at a time earlier than $t$ by the time required to propagate signal from $\mathbf{x}$ to $\bar{\mathbf{x}}$, that is $|\mathbf{x} - \bar{\mathbf{x}}| / c$. So, to wish for a solution to the full inhomogeneous equation, we must let $$t \to t - \frac{|\mathbf{x} - \bar{\mathbf{x}}|}{c} ,$$ on the right hand side of Eq(1.1). That is
$$\psi(\mathbf{x},t) = \int d^{3}\bar{\mathbf{x}} \ \frac{1}{|\mathbf{x} - \bar{\mathbf{x}}|} \ f\left(\bar{\mathbf{x}} , t - \frac{|\mathbf{x} - \bar{\mathbf{x}}|}{c} \right) . \ \ \ \ (1.2)$$ We can rewrite this as
$$\psi(\mathbf{x},t) = \int_{\bar{t} < t } d^{3}\bar{\mathbf{x}} \ d \bar{t} \ \frac{f(\bar{\mathbf{x}},\bar{t})}{|\mathbf{x} - \bar{\mathbf{x}}|} \ \delta \left( t - \bar{t} - \frac{|\mathbf{x} - \bar{\mathbf{x}}|}{c} \right) .$$ Indeed, we will try to show later that the causal Green’s function is actually given by
$$G^{+}(|\mathbf{x}-\bar{\mathbf{x}}| ; t - \bar{t}) = \frac{1}{|\mathbf{x} - \bar{\mathbf{x}}|} \ \delta \left( t - \bar{t} - \frac{|\mathbf{x} - \bar{\mathbf{x}}|}{c} \right) . \ \ \ \ \ (1.3)$$
2) Combination of Fourier and Laplace transforms
Let us consider your wave equation and apply to it the Laplace operator $\mathcal{L}$, where
$$\Psi (\mathbf{x} , s) = \mathcal{L} \left( \psi(\mathbf{x} , t) \right) = \int_{0}^{\infty} dt \ \psi(\mathbf{x},t) \ e^{-st} , \ \ \ \ (2.1)$$ $$F(\mathbf{x},s) = \mathcal{L}\left( f(\mathbf{x},t) \right) = \int_{0}^{\infty} dt \ f(\mathbf{x},t) \ e^{-st} . \ \ \ \ (2.2)$$ Using the initial conditions $\psi(\mathbf{x},0) = 0$, $\partial_{t}\psi(\mathbf{x},0) = 0$, we obtain
$$\nabla^{2}\Psi (\mathbf{x},s) - \frac{s^{2}}{c^{2}} \Psi (\mathbf{x},s) = - 4 \pi \ F(\mathbf{x},s) . \ \ \ \ (2.3)$$ To this equation, (2.3), we apply the Fourier operator $\mathcal{F}$, which we define by
$$\hat{\Psi}(\mathbf{k},s) = \mathcal{F} \left( \Psi(\mathbf{x},s) \right) = ( \frac{1}{2 \pi} )^{3} \int d^{3}\mathbf{x} \ \Psi(\mathbf{x},s) \ e^{- i \mathbf{k} \cdot \mathbf{x}} , \ \ \ (2.4)$$
$$\hat{F}(\mathbf{k},s) = \mathcal{F}\left( F(\mathbf{x},s) \right) = ( \frac{1}{2 \pi} )^{3} \int d^{3}\mathbf{x} \ F(\mathbf{x},s) \ e^{- i \mathbf{k} \cdot \mathbf{x}} . \ \ \ (2.5)$$ And, using the relation
$$\mathcal{F}\left( \nabla^{2} \Psi (\mathbf{x},s) \right) = - \mathbf{k}^{2} \ \hat{\Psi}(\mathbf{k},s) ,$$ equation (2.3) becomes
$$\hat{\Psi}(\mathbf{k},s) = \frac{4 \pi}{\mathbf{k}^{2} + (\frac{s}{c})^{2}} \ \hat{F}(\mathbf{k},s) . \ \ \ \ \ \ \ (2.6)$$ But, $$\mathcal{F}\left( \frac{e^{- \frac{s}{c}|\mathbf{x}|}}{|\mathbf{x}|} \right) = \frac{4 \pi}{\mathbf{k}^{2} + ( \frac{s}{c} )^{2}} ,$$ thus
$$\mathcal{F} \left( \Psi(\mathbf{x},s) \right) = \mathcal{F}\left( \frac{e^{- \frac{s}{c}|\mathbf{x}|}}{|\mathbf{x}|} \right) \ \mathcal{F}\left( F(\mathbf{x},s) \right) . \ \ \ \ \ (2.7)$$ Applying the convolution theorem to the right-hand-side, we obtain
$$\Psi(\mathbf{x},s) = \int d^{3} \bar{\mathbf{x}} \ \frac{ e^{- \frac{s}{c} \ | \mathbf{x} - \bar{\mathbf{x}} |}}{| \mathbf{x} - \bar{\mathbf{x}} |} \ F(\bar{\mathbf{x}},s) .$$ Using (2.1) and (2.2) and taking the inverse Laplace transform, we get
$$\psi(\mathbf{x},t) = \int d^{3}\bar{\mathbf{x}} \ \frac{1}{| \mathbf{x} - \bar{\mathbf{x}} |} \ \mathcal{L}^{-1} \left( e^{- \frac{s}{c}| \mathbf{x} - \bar{\mathbf{x}} |} \ \mathcal{L} \left( f( \bar{ \mathbf{x}}, t ) \right) \right) . \ \ \ \ (2.8)$$ Finally, using the relation
$$e^{- \frac{s}{c}| \mathbf{x} - \bar{\mathbf{x}} |} \ \mathcal{L}\left( f(\bar{\mathbf{x}},t) \right) = \mathcal{L} \left( f\left(\bar{\mathbf{x}}, t - \frac{| \mathbf{x} - \bar{\mathbf{x}} |}{c} \right) \right) , \ \ \ \ \ (2.9)$$ we arrive at our solution (1.2).
3) The Causal Green’s Function: Again, we consider the inhomogeneous field equation
$$\partial^{2} \Psi (x) = 4 \ \pi \ J(x) = 4 \pi \ \int d^{4}\bar{x} \ J(\bar{x}) \ \delta^{4}(x - \bar{x}) , \ \ \ (3.1)$$ where the field $\Psi$ and the source $J$ are tensors of the same rank, and $\partial^{2} = \frac{\partial^{2}}{\partial t^{2}} - \nabla^{2}$. The most general formal solution of (3.1) is given by
$$\Psi(x) = \Psi_{0}(x) + 4 \pi \ \int d^{4}\bar{x} \ J(\bar{x}) \ \partial^{-2}\delta^{4}(x - \bar{x}) , \ \ \ (3.2)$$ where $\Psi_{0}$ is the general solution to the free field equation $\partial^{2}\Psi_{0} = 0$. If we define $$\partial^{2}G( x ) = 4 \pi \ \delta^{4}(x) , \ \ \ \ \ \ (3.3)$$ then (3.2) becomes
$$\Psi(x) = \Psi_{0}(x) + \int d^{4} \bar{x} \ G(x - \bar{x}) \ J( \bar{x} ) . \ \ \ \ (3.4)$$

So, in order to solve (3.1) we need to find the Green function $G(x)$. Let us write ($k = (\omega , \mathbf{k})$)
$$G(x) = ( \frac{1}{2 \pi} )^{4} \ \int d^{4} k \ e^{- i kx} \ \hat{G}(k) , \ \ \ \ \ (3.4)$$
$$\delta^{4}(x) = ( \frac{1}{2 \pi} )^{4} \ \int d^{4} k \ e^{- i kx} . \ \ \ \ \ \ \ (3.5)$$ Substituting (3.4) and (3.5) in (3.3), we get
$$\hat{G}(k) = \frac{4 \pi}{\mathbf{k}^{2} - \omega^{2}} . \ \ \ \ \ \ \ \ (3.6)$$ Using this, equation (3.4) can now be written as
$$G(x) = \frac{4 \pi}{(2 \pi)^{4}} \lim_{\epsilon \to 0} \int d^{3}\mathbf{k} \ d \omega \ \frac{e^{ i ( \mathbf{k} \cdot \mathbf{x} - \omega t )}}{\mathbf{k}^{2} - \omega^{2} - i \epsilon \omega} . \ \ \ \ \ (3.7)$$ For $t < 0$, the $\omega$ integral can be calculated over a contour consisting of the real axis and an infinite semi-circle in the upper half-plane, on which the integrand tend to zero. Since there are no poles there, we find $G = 0$ for $t < 0$, which is nothing but the statement of causality. For $t > 0$, we close the contour in the lower half-plane where there are two poles. By calculating the residues, we find
$$G( \mathbf{x} , t) = \frac{1}{2 \pi^{2}} \int d^{3} \mathbf{k} \ e^{ i \mathbf{k} \cdot \mathbf{x} } (1 / | \mathbf{k}| ) \ \sin ( |\mathbf{k}| t ) .$$ Doing the angular integrals with $\mathbf{x}$ as a polar axis, we obtain
$$G( \mathbf{x} , t) = \frac{2}{\pi | \mathbf{x} |} \int_{0}^{\infty} d |\mathbf{k}| \ \sin ( | \mathbf{k}| | \mathbf{x} | ) \ \sin ( | \mathbf{k} | t ) ,$$ which can be rewritten as
$$G( \mathbf{x} , t) = \frac{1}{2 \pi | \mathbf{x} |} \int_{- \infty}^{\infty} dq \ [ \cos q ( |\mathbf{x}| - t ) - \cos q ( |\mathbf{x} + t ) ] ,$$ or
$$G( \mathbf{x} , t) = \frac{1}{| \mathbf{x} |} [ \delta ( | \mathbf{x} | - t ) - \delta ( |\mathbf{x} | + t ) ] .$$ Since both $|\mathbf{x}| \geq 0$ and $t \geq 0$, the second $\delta$ vanishes and we left with
$$G( \mathbf{x} , t) = \frac{1}{| \mathbf{x} |} \ \delta ( t - | \mathbf{x}| ) .$$ Restoring the speed of light, we obtain what we have expected in (1.3)
$$G( | \mathbf{x}-\bar{\mathbf{x}}| ; t - \bar{t}) = \frac{1}{ |\mathbf{x} - \bar{\mathbf{x}}|} \ \delta \left( t - \bar{t} - \frac{|\mathbf{x} - \bar{\mathbf{x}}|}{c} \right) .$$

 

[jstrunk]

Thanks for the effort you put into this.
3) The Causal Green’s Function was most similar to the derivation I am trying to understand and your explanation filled in the missing piece.