Chapter 15, Problem 15.74P

(a)

To determine

The speed of either $b$ particle in the rest frame of particle $a$.

Answer to Problem 15.74P

The speed of the newly formed particles in the rest frame of particle $a$ are $\underset{\_}{0.8401c}$ and $\underset{\_}{0.1122c}$ or $\underset{\_}{0.3912c}$ and $\underset{\_}{0.7122c}$.

Explanation of Solution

Write the expression for the relativistic momentum of particle $a$.

    $p=\gamma mv$        (I)

Here, $p$ is the relativistic momentum, $\gamma$ is the relativistic factor, $m$ is the mass of the particle, $v$ is the velocity of the particle.

Write the expression for $\gamma$.

    $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$        (II)

Here, $c$ is the speed of light.

Consider a particle $a$ is moving along positive $x$ axis of $\text{S}$ frame. This particle decayed into two identical particle $\left(b\text{and }b\right)$. These two new formed particles are also moving along the $x$ axis.

    $a\to b+b$        (III)

Write the expression for the conservation of momentum for the reaction in equation (III).

    $p_a=p_{b1}+p_{b2}$        (IV)

Here, $p_{b1}$ is the momentum of the first new particle $b$, $p_{b2}$ is the momentum of the second new particle $b$.

Use equation (I) and (II) to solve for the relativistic momentum of particle $a$.

    $\begin{array}{c}{p}_a={\gamma }_a{m}_a{v}_a\\ =\frac{m_a{v}_a}{\sqrt{1-\frac{v_a^2}{c^2}}}\end{array}$        (V)

Substitute $2.5m_b$ for $m_a$, $0.5c$ for $v_a$ in equation (V) to solve for $p_a$.

    $\begin{array}{c}{p}_a=\frac{\left(2.5m_b\right)\left(0.5c\right)}{\sqrt{1-\frac{{\left(0.5c\right)}^2}{c^2}}}\\ =1.44m_{b}c\end{array}$        (VI)

Substitute $1.44m_{b}c$ for $p_a$ in equation (IV) to solve for $p_{b1}$.

    $\begin{array}{c}1.44m_{b}c=p_{b1}+p_{b2}\\ p_{b1}=144m_{b}c-p_{b2}\end{array}$        (V)

The total energy of the particle $a$ is equal to the sum of the total energy of the newly formed particles.

Write the expression for the conservation of energy.

    $m_a{c}^2+T_a=m_b{c}^2+T_{b1}+m_b{c}^2+T_{b2}$        (VI)

Here, $T_i$ is the kinetic energy of $i^{\text{th}}$ particle.

Write the expression for the relativistic kinetic energy.

    $\begin{array}{c}T=\frac{1}{2}\gamma m{v}^2\\ =\frac{1}{2}\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}{v}^2\end{array}$        (VII)

Use equation (VII) in (VI) to solve for the conservation of energy.

    $m_a{c}^2+\frac{1}{2}\frac{m_a}{\sqrt{1-\frac{v_a^2}{c^2}}}{v}_a^2=2m_b{c}^2+T_{b1}+T_{b2}$        (VIII)

Substitute $2.5m_b$ for $m_a$, $0.5c$ for $v_a$ in equation (VIII) to solve for the conservation of energy.

    $\begin{array}{c}2.5m_b{c}^2+\frac{1}{2}\frac{2.5m_b}{\sqrt{1-\frac{{\left(0.5c\right)}^2}{c^2}}}{\left(0.5c\right)}^2=2m_b{c}^2+T_{b1}+T_{b2}\\ 0.5m_b{c}^2+0.3608m_b{c}^2=T_{b1}+T_{b2}\\ 0.8608m_b{c}^2=T_{b1}+T_{b2}\end{array}$        (IX)

Write the expression for the kinetic energy of first $b$ particle.

    $T_{b1}=\sqrt{{\left(p_{b1}c\right)}^2+{\left(m_b{c}^2\right)}^2}-m_b{c}^2$        (X)

Write the expression for the kinetic energy of second $b$ particle.

    $T_{b2}=\sqrt{{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2}-m_b{c}^2$        (XI)

Substitute $\sqrt{{\left(p_{b1}c\right)}^2+{\left(m_b{c}^2\right)}^2}-m_b{c}^2$ for $T_{b1}$, $\sqrt{{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2}-m_b{c}^2$ for $T_{b2}$ in equation (IX) and it becomes,

    $\begin{array}{c}0.8608m_b{c}^2=\left(\sqrt{{\left(p_{b1}c\right)}^2+{\left(m_b{c}^2\right)}^2}-m_b{c}^2\right)+\left(\sqrt{{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2}-m_b{c}^2\right)\\ \sqrt{{\left(p_{b1}c\right)}^2+{\left(m_b{c}^2\right)}^2}=2.8068m_b{c}^2-\sqrt{{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2}\end{array}$        (XII)

Use equation (V) in (XII) and squaring on both side to solve for $p_{b2}$.

    $\begin{array}{c}{\left(p_{b1}c\right)}^2+{\left(m_b{c}^2\right)}^2=8.1841m_b^2{c}^4+\left[{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2\right]-5.7216m_b{c}^2\sqrt{{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2}\\ {\left[\left(1.44m_{b}c-p_{b2}\right)c\right]}^2+{\left(m_b{c}^2\right)}^2=\left\{\begin{array}{c}8.1841m_b^2{c}^4+\left[{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2\right]\\ -5.7216m_b{c}^2\sqrt{{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2}\end{array}\right\}\\ 2.0736m_b^2{c}^2+{\left(p_{b2}c\right)}^2-2.88m_b{c}^2{p}_{b2}c=\left\{\begin{array}{c}8.1841m_b^2{c}^4+\left[{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2\right]\\ -5.7216m_b{c}^2\sqrt{{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2}\end{array}\right\}\\ 6.1105m_b{c}^2+2.88p_{b2}c=5.7216\sqrt{{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2}\end{array}$        (XIII)

Squaring the equation (XIII) on both sides to solve for $p_{b2}$.

    $\begin{array}{c}37.34m_b^2{c}^4+8.2944{\left(p_{b2}c\right)}^2+17.59824m_b{c}^2\left(p_{b2}c\right)=32.74\left[{\left(p_{b2}c\right)}^2+{\left(m_b{c}^2\right)}^2\right]\\ 4.6m_b^2{c}^4-24.4406{\left(p_{b2}c\right)}^2+17.59824m_b{c}^2\left(p_{b2}c\right)=0\\ 24.4406\left(p_{b2}c\right)-17.59824m_b{c}^2\left(p_{b2}c\right)-4.6m_b^2{c}^4=0\end{array}$        (XIV)

Use the general solution of quadratic equation in (XIV) to solve for $p_{b2}$.

    $\begin{array}{c}{p}_{b2}=\frac{-\left(17.5982m_b{c}^2\right)\pm \sqrt{{\left(-17.5982m_b{c}^2\right)}^2-4\left(24.4406\right)\left(-4.6m_b^2{c}^4\right)}}{2\left(24.4406\right)}\\ =\frac{48.35m_b{c}^2\pm 27.5572m_b{c}^2}{2\left(24.4406\right)}\\ =1.533m_b{c}^2\text{or }0.4253m_b{c}^2\end{array}$        (XV)

Write the expression for the relativistic momentum of the second newly formed particle.

    $p_{b2}=\frac{m_b{v}_{b2}}{\sqrt{1-\frac{{\left(v_{b2}\right)}^2}{c^2}}}$        (XVI)

Substitute $1.533m_b{c}^2$ for $p_{b2}$ in equation (XVI) to solve for $v_{b2}$.

    $\begin{array}{c}1.533m_b{c}^2=\frac{m_b{v}_{b2}}{\sqrt{1-\frac{{\left(v_{b2}\right)}^2}{c^2}}}\\ {\left(v_{b2}\right)}^2=2.4c^2\left(1-\frac{{\left(v_{b2}\right)}^2}{c^2}\right)\\ 3.4{\left(v_{b2}\right)}^2=2.4c^2\\ v_{b2}=\sqrt{\left(\frac{2.4}{3.4}\right)c}\\ =0.8401c\end{array}$        (XVII)

Substitute $0.4253m_b{c}^2$ for $p_{b2}$ in equation (XVI) to solve for $v_{b2}$.

    $\begin{array}{c}0.4253m_b{c}^2=\frac{m_b{v}_{b2}}{\sqrt{1-\frac{{\left(v_{b2}\right)}^2}{c^2}}}\\ {\left(v_{b2}\right)}^2=0.18088c^2\left(1-\frac{{\left(v_{b2}\right)}^2}{c^2}\right)\\ 1.18088{\left(v_{b2}\right)}^2=0.18088c^2\\ {\left(v_{b2}\right)}^2=\frac{0.18088}{1.18088}{c}^2\\ v_{b2}=\sqrt{0.1531}c\\ =0.3912c\end{array}$        (XVIII)

Substitute $1.553m_{b}c$ for $p_{b2}$ in equation (V) to solve for $p_{b1}$.

    $\begin{array}{c}{p}_{b1}=1.44m_{b}c-1.553m_{b}c\\ =-0.113m_{b}c\end{array}$        (XIX)

Write the expression for the relativistic momentum of the first newly formed particle.

    $p_{b1}=\frac{m_b{v}_{b1}}{\sqrt{1-\frac{{\left(v_{b1}\right)}^2}{c^2}}}$        (XX)

Substitute $-0.113m_{b}c$ for $p_{b1}$ in equation (XX) to solve for $v_{b1}$.

    $\begin{array}{c}-0.113m_{b}c=\frac{m_b{v}_{b1}}{\sqrt{1-\frac{{\left(v_{b1}\right)}^2}{c^2}}}\\ {\left(v_{b1}\right)}^2=0.012769c^2\left(1-\frac{{\left(v_{b1}\right)}^2}{c^2}\right)\\ v_{b1}=\sqrt{\left(\frac{0.012769}{1.012769}\right)}c\\ =0.1122c\end{array}$        (XXI)

Substitute $0.4253m_{b}c$ for $p_{b2}$ in equation (V) to solve for $p_{b1}$.

    $\begin{array}{c}{p}_{b1}=1.44m_{b}c-\left(0.4253m_{b}c\right)\\ =1.0147m_{b}c\end{array}$        (XXII)

Substitute $1.0147m_{b}c$ for $p_{b1}$ in equation (XX) to solve for $v_{b1}$.

    $\begin{array}{c}1.0147m_{b}c=\frac{m_b{v}_{b1}}{\sqrt{1-\frac{{\left(v_{b1}\right)}^2}{c^2}}}\\ {\left(v_{b1}\right)}^2=1.0296c^2\left(1-\frac{{\left(v_{b1}\right)}^2}{c^2}\right)\\ v_{b1}=\sqrt{\left(\frac{1.0296}{2.0296}\right)c^2}\\ =0.7122c\end{array}$        (XXIII)

Conclusion:

Therefore, the speed of the newly formed particles in the rest frame of particle $a$ are $\underset{\_}{0.8401c}$ and $\underset{\_}{0.1122c}$ or $\underset{\_}{0.3912c}$ and $\underset{\_}{0.7122c}$.

(b)

To determine

The velocities of the two $b$ particles in the original frame $\text{S}$.

Answer to Problem 15.74P

The velocities of the newly formed particles in the frame of reference $\text{S}$ are $\underset{\_}{0.9436c}$ and $\underset{\_}{0.5797c}$ or $\underset{\_}{0.7454c}$ and $\underset{\_}{0.8939c}$ respectively.

Explanation of Solution

The relative velocity of the newly formed particles with respect to the frame of reference $\text{S}$.

Write the expression for the velocity of the newly formed particle in the $S$ frame.

    $v^{\prime }=\frac{v+V}{1+\frac{vV}{c^2}}$        (XXIV)

Here, $v^{\prime }$ is the velocity of the particle in the original frame $\text{S}$, $v$ is the speed of the moving frame, $V$ is the speed of the newly formed particle.

Substitute $0.8401c$ for $v$, $0.5c$ for $V$ in equation (XXIV) to solve for $v_{b2}{}^{\prime }$.

    $\begin{array}{c}{v}_{b2}{}^{\prime }=\frac{\left(0.8401c\right)+\left(0.5c\right)}{1+\frac{\left(0.8401c\right)\left(0.5c\right)}{c^2}}\\ =0.9436c\end{array}$        (XXV)

Here, $v_{b2}{}^{\prime }$ is the speed of the second particle $b$ in the frame of reference $\text{S}$.

Substitute $0.3912c$ for $v$, $0.5c$ for $V$ in equation (XXIV) to solve for $v_{b1}{}^{\prime }$.

    $\begin{array}{c}{v}_{b1}{}^{\prime }=\frac{\left(0.3912c\right)+\left(0.5c\right)}{1+\frac{\left(0.3912c\right)\left(0.5c\right)}{c^2}}\\ =0.7454c\end{array}$        (XXVI)

Here, $v_{b1}{}^{\prime }$ is the speed of the first particle $b$ in the $\text{S}$ frame.

Substitute $0.1122c$ for $v$, $0.5c$ for $V$ in equation (XXIV) to solve for $v_{b2}{}^{\prime }$.

    $\begin{array}{c}{v}_{b2}{}^{\prime }=\frac{\left(0.1122c\right)+\left(0.5c\right)}{1+\frac{\left(0.1122c\right)\left(0.5c\right)}{c^2}}\\ =0.5797c\end{array}$        (XXVII)

Substitute $0.7122c$ for $v$, $0.5c$ for $V$ in equation (XXIV) to solve for $v_{b1}{}^{\prime }$.

    $\begin{array}{c}{v}_{b1}{}^{\prime }=\frac{\left(0.7122c\right)+\left(0.5c\right)}{1+\frac{\left(0.7122c\right)\left(0.5c\right)}{c^2}}\\ =0.8939c\end{array}$        (XXVIII)

Conclusion:

Therefore, the velocities of the newly formed particles in the frame of reference $\text{S}$ are $\underset{\_}{0.9436c}$ and $\underset{\_}{0.5797c}$ or $\underset{\_}{0.7454c}$ and $\underset{\_}{0.8939c}$ respectively.