jstrunk said:
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) .
