jason12345
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This comes from Jackson's Classical Electrodynamics 3rd edition, page 613. He finds the Green's function for the covariant form of the wave equation as:
D(z) = -1/(2\pi)^{4}\int d^{4}k\: \frac{e^{-ik\cdot z}}{k\cdot k}
Where z = x - x' the 4 vector difference, k\cdot z = k_0z_0 - \mathbf{k \cdot z}
He then performs the integral over k0 first by considering it as a complex variable to give:
D(z) = -1/(2\pi)^{4}\int d^{3}k e^{i\mathbf{k\cdot z}}\int_{-\infty }^{\infty } dk_0\: \frac{e^{-ik_0z_0}}{k^2_0 - \kappa^2}
where \kappa = |\mathbf{k}|
But then he says for z_0>0, e^{-ik_0z_0} increases without limit in the upper half plane.
Is this correct? There's a minus in the exponential so i would have thought it's the exact opposite.
He considers a contour r in the upper half of the k0 plane from +oo to -oo, closed by a semicircle also in the upper half of k0 and says that for z0 < 0, the resulting integral vanishes, whereas for z0 > 0, the integral over k0 is:
\oint_{r} dk_0\: \frac{e^{-ik_0z_0}}{k^2_0 - \kappa^2} = -2\pi i\: Res \left (\frac{e^{-ik_0z_0}}{k^2_0 - \kappa^2} \right) = \frac{-2\pi}{\kappa} sin(\kappa z_0)
The Green function is then:
D_r(z) = \frac{\theta(z_0)}{(2\pi)^3} \int d^3k \: e^{i\mathbf{k \cdot z}}\ \frac{sin(\kappa z_0)}{\kappa}
Where does the \theta(z_0) come from?
Thanks in advance for your interest.
D(z) = -1/(2\pi)^{4}\int d^{4}k\: \frac{e^{-ik\cdot z}}{k\cdot k}
Where z = x - x' the 4 vector difference, k\cdot z = k_0z_0 - \mathbf{k \cdot z}
He then performs the integral over k0 first by considering it as a complex variable to give:
D(z) = -1/(2\pi)^{4}\int d^{3}k e^{i\mathbf{k\cdot z}}\int_{-\infty }^{\infty } dk_0\: \frac{e^{-ik_0z_0}}{k^2_0 - \kappa^2}
where \kappa = |\mathbf{k}|
But then he says for z_0>0, e^{-ik_0z_0} increases without limit in the upper half plane.
Is this correct? There's a minus in the exponential so i would have thought it's the exact opposite.
He considers a contour r in the upper half of the k0 plane from +oo to -oo, closed by a semicircle also in the upper half of k0 and says that for z0 < 0, the resulting integral vanishes, whereas for z0 > 0, the integral over k0 is:
\oint_{r} dk_0\: \frac{e^{-ik_0z_0}}{k^2_0 - \kappa^2} = -2\pi i\: Res \left (\frac{e^{-ik_0z_0}}{k^2_0 - \kappa^2} \right) = \frac{-2\pi}{\kappa} sin(\kappa z_0)
The Green function is then:
D_r(z) = \frac{\theta(z_0)}{(2\pi)^3} \int d^3k \: e^{i\mathbf{k \cdot z}}\ \frac{sin(\kappa z_0)}{\kappa}
Where does the \theta(z_0) come from?
Thanks in advance for your interest.