91影视

Write the expression for the radiated power (using equation 11.75).

$\mathit{P}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{6}}}{\mathbf{6}\mathbf{螤}\mathbf{c}}$ 鈥︹�� (1)

Here,${渭}_0$ is the permeability of free space, q is the charge, a is the retardation, and c is the speed of light.

Here, the value$纬$ is given as:

$纬=\sqrt{\frac{1}{1-{\displaystyle \frac{v^2}{c^2}}}}$ 鈥︹�� (2)

(a)

As the particle is in hyperbolic motion along the x-axis, write the expression for the angular velocity of a particle.

$蝇\left(t\right)=\sqrt{b^2+c^2{t}^2}$

Write the expression for the linear velocity of a particle.

$$\begin{gathered} v=\stackrel{藱}{蝇}\left(t\right) \\ v=\frac{d蝇\left(t\right)}{dt} \end{gathered}$$

Substitute the value of$蝇\left(t\right)$in the above expression.

$$\begin{gathered} v=\frac{d}{dt}\left(\sqrt{b^2+c^2{t}^2}\right) \\ v=\frac{1}{2\sqrt{b^2+c^2{t}^2}}脳2c^{2}t \\ v=\frac{c^{2}t}{\sqrt{b^2+c^2{t}^2}} \end{gathered}$$

Calculate the acceleration of a particle.

$$\begin{gathered} a=\frac{dv}{dt} \\ a=\frac{d}{dt}\left(\frac{c^{2}t}{\sqrt{b^2+c^2{t}^2}}\right) \\ a=\frac{\sqrt{b^2+c^2{t}^2}脳c^2-c^2\left(2t^2\right)\left({\displaystyle \frac{1}{2\sqrt{b^2+c^2{t}^2}}}\right)}{{\left(\sqrt{b^2+c^2{t}^2}\right)}^2} \\ a=\frac{\left(b^2+c^2{t}^2\right)c^2-c^2\left(2t^2\right)\left({\displaystyle \frac{1}{2\sqrt{b^2+c^2{t}^2}}}\right)}{\left(b^2+c^2{t}^2\right)\left(\sqrt{b^2+c^2{t}^2}\right)} \end{gathered}$$

On further solving,

$$\begin{gathered} a=\frac{c^2}{{\left(b^2+c^2{t}^2\right)}^{3/2}}\left[b^2+c^2{t}^2-c^2{t}^2\right] \\ a=\frac{b^2{c}^2}{{\left(b^2+c^2{t}^2\right)}^{3/2}} \end{gathered}$$

Substitute the value of vin equation (2).

$$\begin{gathered} 纬=\sqrt{\frac{1}{1-{\displaystyle \frac{{\left({\displaystyle \frac{c^{2}t}{\sqrt{b^2+c^2{t}^2}}}\right)}^2}{c^2}}}} \\ {纬}^2=\frac{1}{1-{\displaystyle \frac{c^4{t}^2}{\left(b^2+c^2{t}^2\right)c^2}}} \\ {纬}^2=\frac{b^2+c^2{t}^2}{b^2} \end{gathered}$$

Substitute the value of a and $纬$in equation (1).

$$\begin{gathered} P=\frac{{渭}_0{q}^2}{6蟺c}{\left[\frac{b^2{c}^2}{{\left(b^2+c^2{t}^2\right)}^{3/2}}\right]}^2{\left[\frac{b^2+c^2{t}^2}{b^2}\right]}^3 \\ P=\frac{{渭}_0{q}^2{c}^3}{6蟺b^2}\left(鈭�{渭}_0=\frac{1}{{蔚}_0{c}^2}\right) \\ P=\frac{q^2{c}^3}{6蟺b^2{蔚}_0{c}^2} \\ P=\frac{q^{2}c}{6蟺{蔚}_0{b}^2} \end{gathered}$$

Since, the terms q, c, b, and${蔚}_0$ are constant, the power is radiated at a constant rate.

Therefore, yes, the power radiated at a constant rate.

(b)

Write the expression for the force due to radiation acting on a particle.

$F_{rad}=\frac{{渭}_0{q}^2{纬}^4}{6蟺c}\left[\stackrel{藱}{a}+\frac{3{纬}^2{a}^{2}v}{c^2}\right]$ 鈥︹��. (3)

Here,$\stackrel{藱}{a}$ is the rate of acceleration which is calculated as:

$$\begin{gathered} \stackrel{藱}{a}=\frac{da}{dt} \\ \stackrel{藱}{a}=\frac{d}{dt}\left(\frac{b^2{c}^2}{{\left(b^2+c^2{t}^2\right)}^{3/2}}\right) \\ \stackrel{藱}{a}=\frac{{\left(b^2+c^2{t}^2\right)}^{3/2}\left(0\right)-b^2{c}^2\left(-{\displaystyle \frac{3}{2}}\right){\left(b^2+c^2{t}^2\right)}^{1/2}\left(2c^{2}t\right)}{{\left(b^2+c^2{t}^2\right)}^3} \\ \stackrel{藱}{a}=\frac{-b^2{c}^2\left(-{\displaystyle \frac{3}{2}}\right){\left(b^2+c^2{t}^2\right)}^{1/2}}{{\left(b^2+c^2{t}^2\right)}^3{\left(b^2+c^2{t}^2\right)}^{-1/2}} \end{gathered}$$

On further solving,

$\stackrel{藱}{a}=\frac{-3b^2{c}^{4}t}{{\left(b^2+c^2{t}^2\right)}^{5/2}}$

Substitute the value of $\stackrel{藱}{a}$,$纬$ and v in equation (3).

role="math" localid="1654061292314" $$\begin{gathered} F_{rad}=\frac{{渭}_0{q}^2{纬}^4}{6蟺c}\left[\frac{-3b^2{c}^{4}t}{{\left(b^2+c^2{t}^2\right)}^{5/2}}+\frac{3{纬}^2{a}^{2}v}{c^2}\right] \\ {F}_{rad}=\frac{{渭}_0{q}^2{纬}^4}{6蟺c}\left[\frac{-3b^2{c}^{4}t}{{\left(b^2+c^2{t}^2\right)}^{5/2}}+\frac{3{\left({\displaystyle \frac{b^2+c^2{t}^2}{b^2}}\right)}^2{a}^2\left({\displaystyle \frac{c^{2}t}{\sqrt{b^2+c^2{t}^2}}}\right)}{c^2}\right] \\ {F}_{rad}=\frac{{渭}_0{q}^2{纬}^4}{6蟺c}\left[\frac{-3b^2{c}^{4}t}{{\left(b^2+c^2{t}^2\right)}^{5/2}}+\frac{3}{c^2}\left(\frac{b^2{c}^{6}t}{{\left(b^2+c^{2}t\right)}^{5/2}}\right)\right] \\ {F}_{rad}=0 \end{gathered}$$

Hence, there is no radiation reaction experienced by a particle in hyperbolic motion.

Therefore, the particle in hyperbolic motion does not experience a radiation reaction.