Advances in our understanding of perturbation theory suggest the existence of a correspondence between classical general relativity and Yang-Mills theory. A concrete example of this correspondence, which is known as the double copy, was recently introduced for the case of stationary Kerr-Schild spacetimes. Building on this foundation, we examine the simple time-dependent case of an accelerating, radiating point source. The gravitational solution, which generalises the Schwarzschild solution, includes a non-trivial stress-energy tensor. This stress-energy tensor corresponds to a gauge theoretic current in the double copy. We interpret both of these sources as representing the radiative part of the field. Furthermore, in the simple example of Bremsstrahlung, we determine a scattering amplitude describing the radiation, maintaining the double copy throughout. Our results provide the strongest evidence yet that the classical double copy is directly related to the BCJ double copy for scattering amplitudes.
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\epsilon \end{array} \right. \label{frange} \end{equation} and, in the interval $(-\epsilon,\epsilon)$, $f(\tau)$ is smooth but otherwise arbitrary. This describes a particle which moves with constant velocity $\lambda^\mu=u^\mu$ for $\tau<-\epsilon$, while for $\tau>\epsilon$ the particle moves with a different constant velocity $\lambda^\mu = u'^\mu$. Thus, the particle undergoes a rapid change of direction around $\tau=0$, assuming $\epsilon$ to be small. The form of $f(\tau)$ acts as a regulator needed to avoid pathologies in the calculation that follows. However, dependence on this regulator cancels out, so that an explicit form for $f(\tau)$ will not be needed. Owing to the constant nature of $u$ and $u'$, the acceleration is given by \begin{equation} \dot{\lambda}^\mu=\dot{f}(\tau)\left({u'}^{\mu}-u^\mu\right). \label{lambdadot} \end{equation} The acceleration vanishes for $\tau<-\epsilon$ and $\tau>\epsilon$, but is potentially large in the interval $(-\epsilon,\epsilon)$. Without loss of generality, we may choose the spatial origin to be the place at which the particle changes direction, so that $y^\mu(0)=0$. ]]>
0$: \begin{align} {\cal F}\left\{\frac{1}{u\cdot x}\right\}&=\frac{2\pi i}{u^0} \int d^3x \, e^{-i\vec{x}\cdot\left[\vec{q}-\frac{q^0}{u^0}\vec{u}\right]}\notag\\ &=\frac{i(2\pi)^4}{u^0}\delta^{(3)}\left(\vec{q}-\frac{q^0}{u^0}\vec{u}\right). \label{jfourier2} \end{align} It is possible to regain a covariant form for this expression by introducing a mass variable $m$, such that \begin{align} {\cal F}\left\{\frac{1}{u\cdot x}\right\}&=\frac{i(2\pi)^4}{u^0} \int_0^\infty dm\, \delta\left(m-\frac{q^0}{u^0}\right)\delta^{(3)}(\vec{q} -m\vec{u})\notag\\ &=i(2\pi)^4\int_0^\infty dm \, \delta^{(4)}(q-mu), \label{jfourier3} \end{align} where the integral is over non-negative values of $m$ only, given that $q^0>0$. Given that $\theta(x^0)\delta(x^2)$ is a retarded propagator,\footnote{The retarded nature of the propagator is implemented by the prescription $\frac{1}{(p^0+i\varepsilon)^2-\vec{p}^2}$, where $\varepsilon$ ensures convergence of the integrals in what follows.} one may also note the transform \begin{equation} {\cal F}\left\{\theta(x^0)\delta(x^2)\right\}=-\frac{2\pi}{q^2}. \label{proptrans} \end{equation} We then use the convolution theorem to obtain the Fourier transform of the current from eq.~(\ref{janu4}). The theorem states that the Fourier transform of a product is equal to the convolution of the transforms of each term. That is, \begin{eqnarray} \mathcal{F}\{f\cdot g\}=\mathcal{F}\{f\}\ast\mathcal{F}\{g\}, \end{eqnarray} where the convolution operation in four dimensions takes the form \begin{eqnarray} (F\ast G)(k)=\frac{1}{(2\pi)^4}\int d^4qF(q)G(k-q). \label{conv} \end{eqnarray} Then, we can compute the Fourier transform of the current \begin{align} \nonumber\tilde{j}^{\nu}(k)&=\mathcal{F}\{j_\textrm{\tiny{KS}}^\nu(x)\}\\ &=\frac{2g}{4\pi} \frac{\partial}{\partial u'_\nu}\left[\mathcal{F}\{\theta(x^0)\delta(x^2)\}\ast\mathcal{F}\left\lbrace\frac{1}{x\cdot u'}\right\rbrace\right] -(u\leftrightarrow u'), \end{align} so inserting eqs.~(\ref{proptrans}) and~(\ref{jfourier3}), and using the convolution definition eq.~(\ref{conv}) we obtain the expression \begin{align} \nonumber\tilde{j}^{\nu}(k)&=\frac{2g}{4\pi} \frac{\partial}{\partial u'_\nu}\left[\frac{1}{(2\pi)^4}\int d^4q\left(-\frac{2\pi}{q^2}\right)\left(i(2\pi)^4\int_0^\infty dm \, \delta^{(4)}(k-q-mu')\right)\right] -(u\leftrightarrow u')\\ &=-ig\int_0^\infty dm\left(\frac{\partial}{\partial u'_\nu} \left[\frac{1}{(k-mu')^2}\right]-(u\leftrightarrow u')\right). \label{jfourier4} \end{align} where we have carried out the integral over $q$ in the last line. The derivative in the $m$ integral can be carried out to give \begin{equation} \int_0^\infty dm \frac{2m(k-mu')^\nu}{(k-mu')^4}=-\int_0^\infty dm \frac{2m^2 {u'}^\nu}{(m^2-2mu'\cdot k)^2}, \label{mint1} \end{equation} where, on the right-hand side, we have used the onshellness condition $k^2=0$, and also neglected terms $\sim k^\mu$, which vanish upon contraction of the current with a physical polarisation vector. The remaining integral over $m$ is easily carried out, and leads directly to the result of eq.~(\ref{jafourier}). Similar steps to those leading to eq.~(\ref{jfourier4}) can be used to rewrite eq.~(\ref{Tmunurad2}) in the form \begin{equation} T^{\mu\nu}_\textrm{\tiny{KS}}=\frac{iM}{2} \int_0^\infty dm\left(\frac{\partial}{\partial u'_\mu}\frac{\partial}{\partial u'_\nu}\left[\frac{1}{(k-mu')^2}\right] -(u\leftrightarrow u')\right). \label{Tmunufourier1} \end{equation} Carrying out the double derivative gives \begin{align} \frac{\partial}{\partial u'_\mu}\frac{\partial}{\partial u'_\nu}\left[\frac{1}{(k-mu')^2}\right] &=-\frac{2m^2\eta^{\mu\nu}}{(m^2-2mu'\cdot k)^4}+\frac{8m^2(k-mu')^\mu (k-mu')^\nu}{(m^2-2mu'\cdot k)^3}\notag\\ &\simeq \frac{8m^4{u'}^\mu{u'}^\nu}{(m^2-2mu'\cdot k)^3}, \label{Tmunufourier2} \end{align} where in the second line we have again used onshellness ($k^2=0$), and ignored terms which vanish when contracted with the graviton polarisation tensor. Substituting eq.~(\ref{Tmunufourier2}) into eq.~(\ref{Tmunufourier1}), the $m$ integral is straightforward, and one obtains the result of eq.~(\ref{Tmunufourier}). ]]>