Radiation field of a scattering experiment moves faster than light

Question

Consider an electron with 4-momentum $p$ which receives a sudden "kick" at $t=0$. After the collision, it has momentum $p^\prime$. Now I am interested in the radiation, that the particle is emitting. According to Peskin and Schroeder, this field is: \begin{equation} A_\rho(x) = \text{Re} \int \frac{d^3k}{(2\pi)^3} \; e^{-ik\cdot x} \frac{-e}{\left|\mathbf{k} \right|} \left(\frac{p^\prime_\rho}{k \cdot p^\prime}- \frac{p_\rho}{k \cdot p} \right) \end{equation} Where $k^0 = \left|\mathbf{k}\right|$ is implicit. I was interested in the potential in real space, so I wanted to evaluate the integral. Using spherical coordinates I obtained: \begin{equation} A_\rho(x) = \int_0^\infty d\left|\mathbf{k}\right| \int_{-1}^1 d\cos \theta \int_0^{2\pi} d\phi \; \cos \left(-\left|\mathbf{k}\right| t +\left| \mathbf{k}\right| \left|\mathbf{x}\right| \cos \theta \right) \left[... \right] \end{equation} Where the expression in the brackets does not depend on $\left|\mathbf{k} \right|$. Since the integrand is symmetric in $\left|\mathbf{k}\right|$, the corresponding integral can be evaluated from $-\infty$ to $\infty$ as well, if an additional factor $1/2$ is multiplied. Then the $\left|\mathbf{k}\right|$-integration should yield a delta distribution $\delta (-t + \left| \mathbf{x} \right| \cos \theta)$. But this implies, that the solution is only non-zero, if $\left|\mathbf{x}\right| > t$, i.e. the solution is outside the lightcone. Hence the field travelled faster than light or the field appears instantanous at $t=0$ in the entire space.

How can I make sense of this? Where did I make a mistake?

The integral can be analyzed further: Let $\mathbf{x} = \left|\mathbf{x}\right| e_z$ and $\mathbf{k} = \left| \mathbf{k} \right| (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ and $\mathbf{p} = \left| \mathbf{p} \right| (\sin \theta_p \cos \phi_p, \sin \theta_p \sin \phi_p, \cos \theta_p)$. Then the integral becomes (essentially): \begin{equation} A_\rho (x) = \text{Re} \int_{-\infty}^\infty d \left|\mathbf{k}\right| \int_{-1}^1 d\cos \theta \int_0^{2\pi} d\phi \frac{e^{-i\left|\mathbf{k}\right| (x^0 - \left|\mathbf{x}\right| \cos \theta)}}{p^0 - \left| \mathbf{p} \right| (\sin \theta \sin \theta_p \cos\phi - \phi_p + \cos \theta \cos\theta_p)} \end{equation}
As mentioned, the $\left|\mathbf{k}\right|$ integration yields a delta distribution. Then $\cos \theta$ integration can be executed. (From now on $ \cos \theta = \frac{x^0}{\left|\mathbf{x}\right|}$ is implicit) What remains is: \begin{equation} \frac{1}{\left|\mathbf{x}\right|} \Theta(|\mathbf{x}| - x^0) \int_0^{2\pi} d\phi \; \frac{2\pi}{a - b \cos \theta} \end{equation} Where $a= p^0 - \left|\mathbf{p}\right| \cos\theta \cos \theta_p$ and $b = \left|\mathbf{p}\right| \sin\theta \sin\theta_p$. One can easily show that $a>b$. This integral can then be solved using the substitution $z = e^{i\phi}$. Then the integral is a contour integral with $\left|z \right| = 1$ and the poles are at $\frac{a}{b} \pm \sqrt{\left(\frac{a}{b}\right]^2 - 1}$ (here the restriction $a>b$ is important). So only one of the poles lies within the interior of the contour and we obtain the result: \begin{equation} = \frac{1}{\left|\mathbf{x}\right|} \Theta (\left|\mathbf{x} \right| - x^0 ) \frac{(2\pi)^2}{\sqrt{a^2 - b^2}} \end{equation} So now the entire solution for $t>0$ is: \begin{equation} A^\mu = \frac{ep^{\prime \mu}}{4\pi m |\mathbf{x} - \frac{\mathbf{p}^\prime}{m}t|} + \frac{e}{4\pi |\mathbf{x}|} \left( \frac{p^\mu}{\sqrt{a^2 -b^2}} - \frac{p^{\prime \mu}}{\sqrt{a^{\prime 2} - b^{\prime 2}}} \right) \Theta(|\mathbf{x}| - x^0) \end{equation} And for $t<0$ the solution is: \begin{equation} A^\mu = \frac{ep^{ \mu}}{4\pi m |\mathbf{x}- \frac{\mathbf{p}}{m}t|} \end{equation} At time $t = 0$ these solutions should be equal. For this it would be required that $a^2 - b^2 = m^2$. However we imidietly see that: \begin{equation} a^2 - b^2 = (p^{0})^2 - \left| \mathbf{p} \right|^2 \sin^2 \theta_p \end{equation} So it almost works but the $\sin \theta_p$ destroys it.

I also plotted the solution to get a better overview. [1]: https://i.stack.imgur.com/RmkEI.png

We clearly see that from the second to the third picture the distribution is much slimmer.

Answer 1

The first equation in the question is from equations (6.6)-(6.7) on page 179 in Peskin and Schroeder's An Introduction to Quantum Field Theory. According to page 177, the kick occurs at $$ \newcommand{\bfA}{\mathbf{A}} \newcommand{\bfk}{\mathbf{k}} \newcommand{\bfp}{\mathbf{p}} \newcommand{\bfv}{\mathbf{v}} \newcommand{\bfx}{\mathbf{x}} t=0 \hskip2cm \bfx=\mathbf{0}. \tag{1} $$ The charged particle is non-accelerating for all $t\neq 0$, so it does not produce any radiation at those times. All radiation is produced at $t=0$, so we should expect the radiation to be restricted to the surface of lightcone whose apex is at the point (1). Not inside or outside the lightcone, just precisely on the lightcone.

The argument shown in the question already confirms that we don't get any radiation inside the lightcone. We should not get any radiation outside the lightcone, either, but the integral shown in the question is non-zero outside the light-cone!

To see how this paradox is resolved, return to the original integral on page 178 in the cited book, which is manifesly Lorentz-covariant: $$ A_\rho(x) \sim \int\frac{d^4k}{(2\pi)^4} \frac{e^{-ik\cdot x}}{k^2} \left(\frac{p'_\rho}{k\cdot p'}-\frac{p_\rho}{k\cdot p}\right). \tag{2} $$ The quantities $p$ and $p'$ are the particle's 4-momentum before and after the kick, respectively. I wrote "$\sim$" because the integrand has four poles, which must be shifted off the real $k_0$-axis into either the upper- or lower-half complex $k_0$ plane. Shifting the poles at $k_0=\pm|\bfk|$ into the lower-half $k_0$ plane selects the solution of Maxwell's equations that doesn't have any incoming radiation. These poles don't contribute anything to the integral when $t<0$, because then the contour can be closed in the upper-half $k_0$ plane. When $t>0$, the contour is closed in the lower-half $k_0$ plane, and the contribution of these two poles is the integral shown in the question.

The paradox is resolved by thinking carefully about what the various poles represent physically, based on how (2) was derived. Peskin and Schroeder state that the two poles $k^2=0$ are "completely responsible for the radiation field," which is correct, but these two poles are also responsible for something else. To see this, let $F$ be the field we would get for a particle moving forever with constant momentum $p$, and let $F'$ be the field we would get for a particle moving forever with constant momentum $p'$. Neither $F$ nor $F'$ has any radiation. In the scenario with the kick, the field should be $F$ everywhere outside the future light-cone of the kick-event, and the field should be $F'$ everywhere inside that future light-cone. The radiation is restricted to the surface of the light-cone, the boundary between where the field is $F$ and where it is $F'$. Here's the key:

• The pole at $k\cdot p=0$ gives the field $F$ at $t<0$ (because this pole was shifted into the upper-half $k_0$ plane).

• The pole at $k\cdot p'=0$ gives the field $F'$ at $t>0$ (because this pole was shifted into the lower-half $k_0$ plane).

• The poles at $k^2=0$ are "completely responsible for the radiation field," as Peskin and Schroeder state, but they are also responsible for replacing $F'$ with $F$ everywhere outside the light-cone for $t>0$ (because these poles were shifted into the lower-half $k_0$ plane).

The integral shown in the question is the contribution to (2) of the poles at $k^2=0$, so it should be zero everywhere inside the future light-cone (because the pole at $k\cdot p'=0$ already gives the correct field there), but outside the future light-cone, where $t>0$, it should be $F-F'\neq 0$. This is not radiation, but it is needed in order to make sure that the full solution (2) remains equal to $F$ everywhere outside the future light-cone. In other words, it compensates for the contribution of the pole at $k\cdot p'=0$ outside the future light-cone at $t>0$, which would otherwise be $F'$ as though the particle had always had momentum $p'$. The radiation itself is restricted to the surface of the future light-cone, the boundary between where the field is $F$ and where the field is $F'$.