RepeatProb. 11.19, but this time let the external force be a Dirac delta function:$F\left(t\right)=k\partial \left(t\right)$(for some constant k)[Note that the acceleration is now discontinuous at$t=0$(though the velocity must still be continuous); use the method ofProb. 11.19(a) to show that$\mathbf{\Delta a}=-\mathbf{k}/\mathbf{m\tau }$. In this problem there are only two intervals to consider: (i)$\mathbf{t}<\mathbf{0}$and (ii)$\mathbf{t}>\mathbf{0}$.]

(b) As inProb. 11.30, check that energy is conserved in this process.

Dirac Delta Function is a function that exists only at t=0, and before and after that time instant, it has no existence. Its height will be based upon the constant written before the delta function.

(a)

By using the force as the Dirac Delta Function, using the problem (11.19).

$a=\tau \dot{a}+\frac{k\text{}\delta \left(t\right)}{m}$ ….. (i)

Here $\dot{a}=\frac{da}{dt}$is the rate of change of acceleration with respect to time.

Now, by integrating and solving equation (i) solve as:

$\begin{array}{c}\underset{-\epsilon }{\overset{+\epsilon }{\int }}a\left(t\right)dt=v(+\epsilon )-v(-\epsilon )\\ =\tau \underset{-\epsilon }{\overset{+\epsilon }{\int }}\frac{da}{dt}dt+\frac{k}{m}\underset{-\epsilon }{\overset{+\epsilon }{\int }}\delta \left(t\right)dt\\ =\tau [a\left(\epsilon \right)-a(-\epsilon )]+\frac{k}{m}\text{}\end{array}$

For further calculation,considering the velocity as constant as mentioned in the question, then we can write:

$v(+\epsilon )=v(-\epsilon )$

Therefore from equation (ii), we get:

$\begin{array}{l}\tau \left[a\right(\epsilon )-a(-\epsilon \left)\right]+\frac{k}{m}=0\\ \tau \left[a\right(\epsilon )-a(-\epsilon \left)\right]=-\frac{k}{m}\\ \left[a\right(\epsilon )-a(-\epsilon \left)\right]=-\frac{k}{m\tau }\\ \Delta a=-\frac{k}{m\tau }\end{array}$ …… (iii)

From step (i), we know that the Dirac Delta Function does not exist less than $t=0$,

therefore:

For,$t<0$in equation (i):

$\begin{array}{l}a=\tau \dot{a}+\frac{k\text{}\delta \left(t\right)}{m}\\ a=\tau \dot{a}\\ a=\tau \frac{da}{dt}\\ a\left(t\right)=A{e}^{\frac{t}{\tau }}\end{array}$

Here, A is constant.

From step (i), it's known that the Dirac Delta Function does not exist greater than $t=0$,.

therefore:

Using the condition$t>0$in equation (i), we get:

$\begin{array}{l}a=\tau \dot{a}+\frac{k\text{}\delta \left(t\right)}{m}\\ a=\tau \dot{a}\\ a=\tau \frac{da}{dt}\\ a\left(t\right)=B{e}^{\frac{t}{\tau }}\end{array}$ ...... (iv)

Here, B is constant.

Therefore, from equation (iii), the change of acceleration can be written as:

$\begin{array}{l}\Delta a=-\frac{k}{m\tau }\\ B-A=-\frac{k}{m\tau }\\ B=A-\frac{k}{m\tau }\end{array}$

$a\left(t\right)=A{e}^{\frac{t}{\tau }}$

(b)

The Work done (W) can be written as:

$\begin{array}{c}{W}_{ext}=\int F\cdot dx\\ =\int F\cdot vdt\\ =k\int \delta \left(t\right)v\left(t\right)dt\end{array}$

Here $W_{ext}$is the external Work done.

As the delta function exists only at $t=0$, therefore, from the above equation and solve to get:

$W=k\int v\left(0\right)dt$ …… (v)

Now, as we know that,

$\begin{array}{c}v\left(t\right)=\text{}\underset{-\infty }{\overset{t}{\int }}a\left(t\right)dt\\ =\frac{k}{m\tau }\text{}\underset{-\infty }{\overset{t}{\int }}e{}^{\frac{t}{\tau }}dt\\ =\frac{k}{m}{e}^{\frac{t}{\tau }}\text{}\end{array}$

Therefore, at $t=0$, the above equation becomes,

$v\left(0\right)=\frac{k}{m}$ …... (vi)

Hence, putting the value of equation (vi) in (v), and solve.

$W=\frac{k^2}{m}$

The Kinetic Energy can be calculated as:

$\begin{array}{c}{W}_{kin}=\frac{1}{2}m{v}^2\\ =\frac{1}{2}m\times \frac{k^2}{m^2}\\ =\frac{1}{2}\frac{k^2}{m}\end{array}$

Here$W_{kin}$is the Work done due to kinetic energy.

The Work done due to radiation can be calculated as follows:

$\begin{array}{c}{W}_{rad}\text{}=\int P_{rad}\text{}dt\\ =\frac{{\mu }_0{q}^2}{6\pi c}\int {\left[a\left(t\right)\right]}^2\text{}dt\\ =\tau m{\left(\frac{k}{m\tau }\right)}^2\underset{-\infty }{\overset{0}{\int }}\left(e^{\frac{2t}{\tau }}\right)dt\\ =\frac{k^2}{2m}\end{array}$

Here,$W_{rad}$is the Work done due to radiation.

Here, q is the charged particle.

Here, $c$is the velocity of light in a vacuum.

It is observed thatenergy is conserved when the Work done due to the external effect becomes equal to the sum of the Work done due to radiation and Kinetic energy.

Therefore,

$\begin{array}{c}{W}_{rad}+W_{ext}=\frac{k^2}{2m}+\frac{k^2}{2m}\\ =2\times \frac{k^2}{2m}\\ =\frac{k^2}{m}\\ =W_{ext}\end{array}$

Hence, energy has been conserved.