(a) Repeat Prob. 11.19, but this time let the external force be a Dirac delta function: F(t) = kδ(t) (for some constant k) .^{25} [Note that the acceleration is now discontinuous at t = 0 (though the velocity must still be continuous); use the method of Prob. 11.19 (a) to show that Δa = −k/mτ. In this problem there are only two intervals to consider: (i) t < 0, and (ii) t > 0.]
(b) As in Prob. 11.30, check that energy is conserved in this process.
\text { (a) } a=\tau \dot{a}+\frac{k}{m} \delta(t) \Rightarrow \int_{-\epsilon}^{\epsilon} a(t) d t=v(\epsilon)-v(-\epsilon)=\tau \int_{-\epsilon}^{\epsilon} \frac{d a}{d t} d t+\frac{k}{m} \int_{-\epsilon}^{\epsilon} \delta(t) d t=\tau[a(\epsilon)-a(-\epsilon)]+\frac{k}{m}.
If the velocity is continuous, so v(\epsilon)=v(-\epsilon), \text { then } a(\epsilon)-a(-\epsilon)=-\frac{k}{m \tau}.
\text { When } t<0, a=\tau \dot{a} \Rightarrow a(t)=A e^{t / \tau} ; \text { when } t>0, a=\tau \dot{a} \Rightarrow a(t)=B e^{t / \tau} ; \quad \Delta a=B-A=-\frac{k}{m \tau}\Rightarrow B=A-\frac{k}{m \tau} , so the general solution is a(t)= \begin{cases}A e^{t / \tau}, & (t<0) \\ {[A-(k / m \tau)] e^{t / \tau},} & (t>0)\end{cases}
To eliminate the runaway we’d need A=k / m \tau ; to eliminate preacceleration we’d need A = 0. Obviously,
you can’t do both. If you choose to eliminate the runaway, then a(t)= \begin{cases}(k / m \tau) e^{t / \tau}, & (t<0) \\0,&(t>0)\end{cases}
v(t)=\int_{-\infty}^{t} a(t) d t=\frac{k}{m \tau} \int_{-\infty}^{t} e^{t / \tau} d t=\left.\frac{k}{m \tau}\left(\tau e^{t / \tau}\right)\right|_{-\infty} ^{t}=\frac{k}{m} e^{t / \tau}(\text { for } t<0);
\text { for } t>0, v(t)=v(0)+\int_{0}^{t} a(t) dt=v(0)=\frac{k}{m} . \text { So } v(t)= \begin{cases}(k / m) e^{t / \tau}, & (t<0) \\ (k / m), & (t>0)\end{cases}For an uncharged particle we would have a(t)=\frac{k}{m} \delta(t), v(t)=\int_{-\infty}^{t} a(t) d t= \begin{cases}0, & (t<0) \\ (k / m), & (t>0)\end{cases}
The graphs:
(b) W_{\text {ext }}=\int F d x=\int F v d t=k \int \delta(t) v(t) d t=k v(0)=\frac{k^{2}}{m}.
W_{ kin }=\frac{1}{2} m v_{f}^{2}=\frac{1}{2} m\left(\frac{k}{m}\right)^{2}=\frac{k^{2}}{2 m}.
W_{ ext }=\int P_{ rad } d t=\frac{\mu_{0} q^{2}}{6 \pi c} \int[a(t)]^{2} d t=\tau m\left(\frac{k}{m \tau}\right)^{2} \int_{-\infty}^{0} e^{2 t / \tau} d t=\left.\frac{k^{2}}{m \tau}\left(\frac{\tau}{2} e^{2 t / \tau}\right)\right|_{-\infty} ^{0}=\frac{k^{2}}{m \tau} \frac{\tau}{2}=\frac{k^{2}}{2 m}.
\text { Clearly, } W_{\text {ext }}=W_{\text {kin }}+W_{\text {rad }}.
