A point charge q, of mass m, is attached to a spring of constant $\mathit{k}\mathbf{.}{\mathit{Y}}^{\mathbf{2}}\mathbf{<}\mathbf{<}{\mathit{\omega }}_{\mathbf{0}}\mathbf{}\mathit{A}\mathit{t}\mathbf{}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathbf{}\mathit{t}\mathbf{=}\mathbf{0}$it is given a kick, so its initial energy is ${\mathit{U}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{0}}^{\mathbf{2}}$. Now it oscillates, gradually radiating away this energy.

(a) Confirm that the total energy radiated is equal to U₀. Assume the radiation damping is small, so you can write the equation of motion as and the solution as

role="math" localid="1658840767865" $\mathit{x}\mathbf{+}\mathit{y}\mathbf{+}\mathit{x}\mathbf{+}{\mathit{\omega }}_{\mathbf{0}}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}$

and the solution as

$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{0}}}{{\mathbf{\omega }}_{\mathbf{0}}}{\mathit{e}}^{\mathbf{-}\mathbf{y}\mathbf{t}\mathbf{/}\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\left({\omega }_{0}t\right)$

with ${\mathit{\omega }}_{\mathbf{0}}\mathbf{\equiv }\sqrt{\mathbf{k}\mathbf{/}\mathbf{m}}\mathbf{,}\mathit{Y}\mathbf{=}{\mathit{\omega }}_{\mathbf{0}}^{\mathbf{2}}\mathit{T}$, and ${\mathit{Y}}^{\mathbf{2}}\mathbf{<}\mathbf{<}{\mathit{\omega }}_{\mathbf{0}}$ (drop Y²in comparison to ${\mathit{\omega }}_{\mathbf{0}}^{\mathbf{2}}$, and when you average over a complete cycle, ignore the change in ${\mathit{e}}^{\mathbf{-}\mathbf{y}\mathbf{\tau }}$).

(b) Suppose now we have two such oscillators, and we start them off with identical kicks. Regardless of their relative positions and orientations, the total energy radiated must be 2U₀. But what if they are right on top of each other, so it's equivalent to a single oscillator with twice the charge; the Larmor formula says that the power radiated is four times as great, suggesting that the total will be 4U₀. Find the error in this reasoning, and show that the total is actually2U₀, as it should be.

The point charge is q.

The mass is m.

The spring constant is k.

The initial energy is$U_0=\frac{1}{2}m{v}_0^2$

The natural frequency,${\omega }_0=\sqrt{\frac{k}{m}}$

The equation of the motion,

$x+y+x+{\omega }_0^{2}x=0,$

The expression to calculate the total radiated energy is given as follows.

${\mathit{U}}_{\mathbf{0}}\mathbf{=}\mathbf{\int }<P>\mathbf{·}\mathit{d}\mathit{a}$ …… (1)

The expression for the force is given by

$F=mx$

The radiated force is given by,

role="math" localid="1658841100943" $F_{\mathrm{rad}}=m\tau x$

The spring force is given by,

$F_{spring}=-kx$

The force on the system is given by,

$F=F_{spring}+F_{rad}$

Substitute mx for F, -kx for F_spring and $m\tau x$ for F_rad into above equation.

$mx=-kx+m\tau x$

Divide the above equation by m,

$x=-\frac{k}{m}x+\tau x$

Substitute ${\omega }_0^2$ for k/m into above equation.

$\begin{array}{rcl}x& =& -{\omega }_0^{2}x+\tau x\\ x+{\omega }_0^{2}x+\tau x& =& 0\end{array}$ …… (2)

Since the damping is small, therefore it oscillates at the natural frequency.

Thus,

$x=-{\omega }_0^{2}x$

Substitute ${\omega }_0^{2}x$for -x into equation (2).

$$\begin{gathered} x+{\omega }_0^{2}x+\tau {\omega }_0^{2}x=0 \\ x+\tau {\omega }_0^{2}x+{\omega }_0^{2}x=0 \end{gathered}$$ …… (3)

The damping factor is given by,

$\gamma ={\omega }_0^2\tau$

Substitute ${\omega }_0^2\tau$for $\gamma$into equation (3).

$x+\gamma x+{\omega }_0^{2}x=0$

The solution of the equation to motion is given by,

$x\left(t\right)=A{e}^{-\gamma t/2}\sin {\omega }_{0}t$ ……. (4)

Differentiate the above equation with respect to time.

role="math" localid="1658842249857" $v\left(t\right)=-\frac{\gamma }{2}A{e}^{-\gamma t/2}\sin \left({\omega }_{0}t\right)+{\omega }_{0}A{e}^{-\gamma t/2}\cos {\omega }_{0}t$

At $t=0,v\left(0\right)=v_0$

$\begin{array}{rcl}v\left(t\right)& =& -\frac{\gamma }{2}A{e}^{-\gamma t/2}\sin {\omega }_0\left(0\right)+{\omega }_{0}A{e}^{-\gamma t/2}\cos {\omega }_0\left(0\right)\\ v_0& =& {\omega }_{0}A\\ A& =& \frac{v_0}{{\omega }_0}\end{array}$

Substitute $\frac{v_0}{{\omega }_0}$ for A into equation (4).

role="math" localid="1658842771258" $x\left(t\right)=\frac{v_0}{{\omega }_0}{e}^{-\gamma t/2}\sin {\omega }_{0}t$

The expression for the power is given by,

$P=\frac{{\mu }_0{q}^2{a}^2}{6\pi c}$

Substitute $-{\omega }_0^{2}x$for a into above equation.

$$\begin{gathered} P=\frac{{\mu }_0{q}^2}{6\pi c}{\left(-{\omega }_0^{2}x\right)}^2 \\ P=\frac{{\mu }_0{q}^2}{6\pi c}-{\omega }_0^4{x}^2 \end{gathered}$$

Substitute $\frac{v_0}{{\omega }_0}{e}^{-\gamma t/2}\sin {\omega }_{0}t$for x into above equation.

role="math" localid="1658843158125" $$\begin{gathered} P=-\frac{{\mu }_0{q}^2}{6\pi c}{\omega }_0^4{\left(\frac{v_0}{{\omega }_0}{e}^{-\gamma t/2}\sin {\omega }_{0}t\right)}^2 \\ P=\frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{6\pi c}{e}^{-\gamma t}{\sin }^2{\omega }_{0}t \end{gathered}$$

Average power over the full cycle keeping the $e^{-\gamma t}$constant.

role="math" localid="1658843227983" $$\begin{gathered} P=\frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{6\pi c}{e}^{-\gamma t}\frac{1}{2} \\ P=\frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{12\pi c}{e}^{-\gamma t} \end{gathered}$$

Calculate the total energy,

Substitute $\frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{12\pi c}{e}^{-\gamma t}$ for P into equation (1).

role="math" localid="1658843599437" $\begin{array}{rcl}{U}_0& =& {\int }_0^{\infty }\frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{12\pi c}{e}^{-\gamma t}dt\\ U_0& =& \frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{12\pi c}{\int }_0^{\infty }{e}^{-\gamma t}dt\\ U_0& =& \frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{12\pi c}{\left(\frac{1}{\gamma }{e}^{-\gamma t}dt\right)}_0^{\infty }\\ U_0& =& \frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{12\pi c}\end{array}$

Substitute ${\omega }_0^2\tau$for $\gamma$into above equation.

role="math" localid="1658843809745" $\begin{array}{rcl}{U}_0& =& \frac{{\mu }_0{q}^2\left({\omega }_0^2{v}_0^2\right)}{12\pi c{\omega }_0^2\tau }\\ U_0& =& \frac{{\mu }_0{q}^2{v}_0^2}{12\pi c}\frac{1}{\tau }\end{array}$ ……. (5)

The reaction torque is given by,

$\tau =\frac{{\mu }_0{q}^2}{6\pi mc}$

Substitute$\frac{{\mu }_0{q}^2}{6\pi mc}$ for $\tau$into equation (5).

$\begin{array}{rcl}{U}_0& =& \frac{{\mu }_0{q}^2{v}_0^2}{12\pi c}\frac{1}{{\displaystyle \frac{{\mu }_0{q}^2}{6\pi mc}}}\\ U_0& =& \frac{1}{2}m{v}_0^2\end{array}$

Hence it is proved that the total radiated energy is equal to U₀.

The reaction torque is given by,

$\tau =\frac{{\mu }_0{q}^2}{6\pi mc}$ …… (6)

The single oscillator has twice charge and twice mass.

Substitute 2m for m and 2q fir q into equation (6).

$$\begin{gathered} {\tau }_1=\frac{{\mu }_0{\left(2q\right)}^2}{6\pi \left(2m\right)c} \\ {\tau }_1=2\tau \end{gathered}$$

From the above the torque is double. Since$\int Pdt$ goes $q^2/\gamma$, it is also double.

The power radiated is indeed four times, but the oscillator died ways faster, therefore the total energy radiated is just a twice as one oscillator.