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I am trying to find a Green's function for the wave equation: \begin{equation} \bigg(\nabla^2 - \frac{1}{c}^2\frac{\partial^2}{\partial t^2}\bigg)G(\textbf{r},t) = \delta^3(\textbf{r})\delta(t) \end{equation} After a fairly trivial calculation, one easily finds that \begin{equation} G(\textbf{r},t) = \frac{1}{4\pi^3} \int_0^\infty dk \ c^2 k^2 \frac{\sin(kr)}{kr} \int_{-\infty}^\infty d\omega \frac{e^{-i\omega t}}{(\omega-ck)(\omega+ck)} \end{equation} I am mainly interested in the integral in $\omega$: \begin{equation} \int_{-\infty}^\infty d\omega \frac{e^{-i\omega t}}{(\omega-ck)(\omega+ck)} \end{equation} We can evaluate this integral in the complex $\omega$-plane by choosing a contour running over $\text{Re}(\omega)$ but jumping over the poles at $\omega=\pm ck$. There are several choices for such a contour, and I'm wondering why choosing one contour will give a different value for this integral than another?

For example, suppose that $t<0$ so that $e^{-i \omega t} \rightarrow 0$ as $\omega\rightarrow i\infty$. This suggests that we close our contour in the upper half plane ensuring that the integral due to the upper semi-circle does not give any contribution. This contour does not enclose either pole so by the residue theorem the integral vanishes.

Now consider $t>0$. Then $e^{-i\omega t} \rightarrow 0$ as $\omega \rightarrow -i\infty$. This suggests that we close our contour in the lower half plane ensuring that the integral due t the lower semi-circle doe snot give any contribution. This time, we enclose both poles, so by the residue theorem: \begin{equation} \int_{-\infty}^\infty d\omega \frac{e^{-i\omega t}}{(\omega-ck)(\omega+ck)} = -2\pi i\bigg(\frac{e^{-ickt}}{2ck}-\frac{e^{ickt}}{2ck}\bigg) = -\frac{2\pi}{ck} \sin (kct), \ t>0 \end{equation} where the negative sign comes from the fact that the contour runs clockwise. This gives rise to a retarded Green's function.

However suppose that instead of skipping over the poles I not skip under them. The integral now evaluates to something else (which gives an advanced green's function): \begin{equation} \int_{-\infty}^\infty d\omega \frac{e^{-i\omega t}}{(\omega-ck)(\omega+ck)} = 2\pi i\bigg(\frac{e^{-ickt}}{2ck}-\frac{e^{ickt}}{2ck}\bigg) = \frac{2\pi}{ck} \sin (kct), \ t<0 \end{equation} So which of these contours am I to choose? Why do they give different values for a definite integral?

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    $\begingroup$ The integral diverges because of the poles on the real axis. Sometimes, such an integral still be interpreted as a Cauchy principle value. If it diverges even in the principle value sense, one can try various regularization schemes. Different regularizations can yield different answers because the object wasn't well defined to begin with. $\endgroup$
    – Sal
    Commented Oct 17, 2021 at 18:13
  • $\begingroup$ Oh, so the integral is divergent? $\endgroup$ Commented Oct 17, 2021 at 18:16
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    $\begingroup$ Yes. Near $\omega= ck$, the integrand behaves like $\frac{1}{\omega-ck} $ which is a non-integrable singularity. Similarly for near $\omega=-ck$ $\endgroup$
    – Sal
    Commented Oct 17, 2021 at 18:22
  • $\begingroup$ Ok thanks. So if the integral did converge then it would not matter what contour I use, right? $\endgroup$ Commented Oct 17, 2021 at 18:24
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    $\begingroup$ Precisely. If it converged it would do so to a unique value. $\endgroup$
    – Sal
    Commented Oct 17, 2021 at 18:24

1 Answer 1

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Summary of comments: The integral diverges because of the non-integrable poles on the integration contour. Since these poles are of first order, the Cauchy principle value (PV) of the integral exists, and corresponds to avoiding the poles with semi-circular arcs. The PV is unchanged whether we make an arc above or below the pole. This is a type of regularization. Other regularizations may give different results. Even though we will be integrating above/ below the poles, this is different to the popular $\pm i \epsilon$ prescription whereby one considers a related integrand with poles shifted above or below the axis$^\dagger$ (the $\pm i\epsilon $ regularization leads to results that do depend on the contour taken).

The principle value is

$$\tag{1} \text{PV}\int_{-\infty}^\infty d\omega \frac{ e^{-i\omega t}}{\omega^2-k^2} = -\frac{\pi}{k}\sin(k|t|) $$

Consider the case $t>0$, so we close the contour in the lower half plane. I've also set $c=1$ for my convenience. Suppose we make arcs above the poles:

enter image description here

Then we have by residue theorem

$$\tag{2} I+C_-+C_++C_R=-2\pi i (R_- +R_+) $$

Where $I$ is the integral in eq (1), $C_{\pm}$ are the contributions due to integrating over the poles at $\pm k$, $C_R$ is the (vanishing) contribution due to the large arc, $R_\pm$ are the residues at $\pm k$, and the minus sign on the RHS is because the contour is CW.

Near the pole $\omega=k$ we can write

$$ \frac{ e^{-i\omega t}}{\omega^2-k^2} = \frac{e^{-ikt}}{2k(\omega-k)} + \text{finite} $$

To integrate around the pole, parameterize: $\omega=k+\epsilon e^{i\phi}$, then $d\omega=i \epsilon e^{i\phi}d\phi$. Integrating around a CCW semicircle yields

$$\tag{3} \int\limits_0^\pi \frac{i e^{-ikt}}{2k} d\phi + \mathcal{O}(\epsilon) = i\pi \left(\frac{ e^{-ikt}}{2k}\right) + \mathcal{O}(\epsilon) $$

When we take the radius of the arc, $\epsilon$ to zero, all the other terms vanish. The result is precisely $i\pi$ times the residue at the pole. This will generally be the case for simple first order poles$^\ddagger$. Substituting this and the similar calculation for the other pole into eq (2) we have

$$ I-i\pi R_- - i \pi R_+ = -2 \pi i (R_-+R_+) $$

The new minus signs on the LHS are because of the orientation of $C_\pm$. We are left with

$$ I=-i\pi(R_-+R_+)=-i\pi \left(-\frac{e^{ikt}}{2k}+\frac{e^{-ikt}}{2k} \right)=-\frac{\pi}{k}\sin(kt) $$

Suppose we instead made the arcs beneath the poles (still with $t>0$). The analogue of eq (2) is

$$ I+C'_-+C'_+=0 $$

Where $C'_\pm$ are the contributions due to CCW arcs beneath the poles (already calculated in eq (3), they differ only by a sign from the $C_\pm$). The RHS is zero as this contour encloses no poles. Substituting in

$$ I+i\pi R_-+i\pi R_+=0 \\ I=-i \pi (R_- + R_+) $$

Precisely as before. You can check that other combinations of above/ below the poles yield the same result. When $t>0$, the solution will be (slightly) different, but using the same method you can show that the answer is independent of whether your chosen contour goes above or below the poles.

$\dagger$ A fact that even some well respected textbooks forget.

$\ddagger$ You can see why this only works for first order poles, if there were any terms more singular than $1/\omega$ in eq (3), there would remain contributions proportional to $1/\epsilon$ which prevent the limit of zero radius being taken safely.

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    $\begingroup$ Thanks for the in-depth answer, cleared all my doubts! $\endgroup$ Commented Nov 3, 2021 at 19:13
  • $\begingroup$ @KoutaDagnino Great, glad it helped! $\endgroup$
    – Sal
    Commented Nov 3, 2021 at 21:04

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