Chapter 15, Problem 15.84P

To determine

The velocity of mass $m$ as a function of time $t$ and check the trajectory in the non-relativistic limit.

Answer to Problem 15.84P

The velocity of mass $m$ as a function of time $t$ is $\underset{\_}{\frac{p_0}{m\sqrt{1+k{t}^2}}+\frac{F_{0}t}{m\sqrt{1+k{t}^2}}}$ and checked the trajectory in the non-relativistic limit as parabolic nature.

Explanation of Solution

Write the expression for the momentum of a particle having mass $m$ along $x$ axis.

    $p=Ft$        (I)

Here, $p$ is the relativistic momentum, $F$ is the constant force on the particle along $x$ axis, $t$ is the time.

Write the expression for the relativistic momentum of the mass.

    $p=\gamma mv$        (II)

Here, $m$ is the mass, $v$ is the velocity of the mass as a function of time, $\gamma$ is the relativistic factor.

Write the expression for the relativistic total energy.

    $E^2={\left(m{c}^2\right)}^2+{\left(pc\right)}^2$        (III)

Write the general expression for the relativistic energy.

    $E=\gamma m{c}^2$        (IV)

Write the expression for $p$.

    ${\left(p_0\right)}_x=F_{0}t$        (V)

Use equation (IV) and (V) in (III) to solve for $\gamma$.

    $\begin{array}{c}{\left(\gamma m{c}^2\right)}^2={\left(m{c}^2\right)}^2+{\left(pc\right)}^2\\ {\gamma }^2=1+{\left(\frac{p}{mc}\right)}^2\\ =1+{\left(\frac{F_{0}t}{mc}\right)}^2\\ \gamma =\sqrt{1+\left(\frac{F_{0}t}{mc}\right)}\end{array}$        (VI)

It is provided that the particle has initial three momentum as $p_0$ in $y$ direction and a constant force $F_0$ in $x$ direction.

Write the expression for the relativistic momentum.

    $\begin{array}{c}p={\left(p_0\right)}_y+{\left(p_0\right)}_x\\ =p_0+F_{0}t\end{array}$        (VII)

Write the general expression for the relativistic momentum.

    $p=\gamma mv$        (VIII)

Here, $p$ is the relativistic momentum.

Use equation (VIII) and (VI) in (VII) to solve for $v\left(t\right)$.

    $\begin{array}{c}m\gamma v=p_0+F_{0}t\\ m\left(\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}\right)v=p_0+F_{0}t\\ v\left(t\right)=\frac{p_0}{m\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}}+\frac{F_{0}t}{m\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}}\end{array}$        (IX)

Consider ${\left(\frac{F_0}{mc}\right)}^2=k$ and the equation (IX) becomes,

    $v\left(t\right)=\frac{p_0}{m\sqrt{1+k{t}^2}}+\frac{F_{0}t}{m\sqrt{1+k{t}^2}}$        (X)

Write the expression for the trajectory of the particle.

    $r\left(t\right)={\displaystyle \int v\left(t\right)dt}$        (XI)

Here, $r\left(t\right)$ is the trajectory of the particle.

Use equation (X) in (XI) to solve for $r\left(t\right)$.

    $\begin{array}{c}r\left(t\right)={\displaystyle \int \left[\frac{p_0}{m\sqrt{1+k{t}^2}}+\frac{F_{0}t}{m\sqrt{1+k{t}^2}}\right]dt}\\ =\frac{p_0}{m}{\displaystyle \int \frac{dt}{m\sqrt{1+k{t}^2}}}+\frac{F_0}{m}{\displaystyle \int \frac{dt}{m\sqrt{1+k{t}^2}}}\\ =\frac{p_0}{m}\left(\frac{{\sinh }^{-1}\left(\sqrt{kt}\right)}{\sqrt{k}}\right)+\frac{F_0}{m}\left(\frac{\sqrt{1+k{t}^2}}{k}\right)+\text{C}\end{array}$        (XII)

Here, $\text{C}$ is an integration constant.

Use ${\left(\frac{F_0}{mc}\right)}^2=k$ in equation (XII) to solve for $r\left(t\right)$.

    $\begin{array}{c}r\left(t\right)=\left(\frac{p_0}{m}\right)\frac{{\sinh }^{-1}\left(\sqrt{{\left(\frac{F_0}{mc}\right)}^{2}t}\right)}{\sqrt{{\left(\frac{F_0}{mc}\right)}^2}}+\frac{F_0}{m}\left[\frac{\sqrt{1+\left({\left(\frac{F_0}{mc}\right)}^2\right)t^2}}{{\left(\frac{F_0}{mc}\right)}^2}\right]+\text{C}\\ =\frac{p_{0}c}{F_0}{\sinh }^{-1}\left(\frac{F_0}{mc}t\right)+\frac{m{c}^2\left(F_0\right)}{F_0^2}\left[\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}\right]+\text{C}\end{array}$        (XIII)

Applying the boundary conditions $t=0$ and $r=0$ to solve for $\text{C}$.

    $\begin{array}{c}0=\frac{p_{0}c}{F_0}{\sinh }^{-1}\left(\frac{F_0\left(0\right)}{mc}t\right)+\frac{m{c}^2\left(F_0\right)}{F_0^2}\left[\sqrt{1+{\left(\frac{F_0\left(0\right)t}{mc}\right)}^2}\right]+\text{C}\\ =\text{0+}\frac{m{c}^2\left(F_0\right)}{F_0^2}+\text{C}\\ \text{C}=-\frac{m{c}^2\left(F_0\right)}{F_0^2}\end{array}$        (XIV)

Use equation (XIV) in (XIII) to solve for $r\left(t\right)$.

    $\begin{array}{c}r\left(t\right)=\frac{p_{0}c}{F_0}{\sinh }^{-1}\left(\frac{F_0}{mc}t\right)+\frac{m{c}^2\left(F_0\right)}{F_0^2}\left[\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}\right]-\frac{m{c}^2\left(F_0\right)}{F_0^2}\\ =\frac{p_{0}c}{F_0}{\sinh }^{-1}\left(\frac{F_0}{mc}t\right)+\frac{m{c}^2\left(F_0\right)}{F_0^2}\left[\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}-1\right]\end{array}$        (XV)

The constant force $F_0$ is along the $x$ axis and $p_0$ is along the $y$ axis. So, the term containing $F_0$ represents the displacement along $x$ axis and the term containing $p_0$ represents the displacement along $y$ axis.

Write the expression for the displacement along the $y$ axis.

    $y\left(t\right)=\frac{p_{0}c}{m{F}_0}{\sinh }^{-1}\left(\frac{F_0}{mc}t\right)$        (XVI)

Write the expression for the displacement along the $x$ direction.

    $\begin{array}{c}x\left(t\right)=\frac{m{c}^2\left(F_0\right)}{F_0^2}\left(\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}-1\right)\\ =\frac{m{c}^2}{F_0}\left(\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}-1\right)\end{array}$        (XVII)

Replace $x\left(t\right)$ with $x$ and $y\left(t\right)$ with $y$ in equation (XVI) and (XVII) for solving.

Use equation (XVII) to solve for $t$.

    $\begin{array}{c}t=\left(\frac{mc}{F_0}\right)\sqrt{\left({\left(\frac{x{F}_0}{m{c}^2}+1\right)}^2-1\right)}\\ =\sqrt{{\left(\frac{m_{0}c}{F_0}\right)}^2\left({\left(\frac{x{F}_0}{m{c}^2}\right)}^2+\frac{2x{F}_0}{m{c}^2}+1-1\right)}\\ =\sqrt{\left(\frac{x}{c^2}+\frac{2mx}{F_0}\right)}\end{array}$        (XVIII)

Use equation (XVIII) in (XVI) to solve for $y$.

    $\begin{array}{c}y\left(t\right)=\frac{p_{0}c}{m{F}_0}{\sinh }^{-1}\left(\frac{F_0}{mc}\sqrt{\left(\frac{x}{c^2}+\frac{2mx}{F_0}\right)}\right)\\ =\frac{p_{0}c}{m{F}_0}{\sinh }^{-1}\left(\sqrt{{\left(\frac{F_0}{mc}\right)}^2\left(\frac{x}{c^2}+\frac{2mx}{F_0}\right)}\right)\\ =\frac{p_{0}c}{m{F}_0}{\sinh }^{-1}\left(\sqrt{\left({\left(\frac{F_0}{mc}\right)}^{2}x+\frac{2F_{0}x}{m{c}^2}\right)}\right)\end{array}$        (XIX)

If the motion of the mass is considered as non-relativistic, then use the condition $c\to \infty$ in equation (IX) to solve for $v\left(t\right)$.

    $\begin{array}{c}\underset{c\to \infty }{\lim }v\left(t\right)=\underset{c\to \infty }{\lim }\frac{p_0}{m\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}}+\frac{F_{0}t}{m\sqrt{1+{\left(\frac{F_{0}t}{mc}\right)}^2}}\\ =\frac{p_0}{m}+\frac{F_0}{m}t\end{array}$        (XX)

Integrate the equation (XIX) to solve for $r\left(t\right)$.

    $\begin{array}{c}r\left(t\right)={\displaystyle \int v\left(t\right)dt}\\ ={\displaystyle \int \left(\frac{p_0}{m}+\frac{F_0}{m}t\right)dt}\\ =\frac{p_{0}t}{m}+\frac{F_0}{2m}{t}^2+\text{C}\end{array}$        (XXI)

Apply the boundary conditions in equation (XXI) to solve for $r\left(t\right)$.

    $\begin{array}{c}0=0+0+\text{C}\\ \text{C}=0\end{array}$        (XXII)

Use equation (XXII) to solve for the displacement along the $y$ axis.

    $y=\frac{p_{0}t}{m}$        (XXIII)

Use equation (XXIII) to solve for $t$.

    $t=\frac{my}{p_0}$        (XXIV)

Use equation (XXIV) to solve for the displacement along the $x$ direction.

    $\begin{array}{c}x=\frac{F_0}{2m}{t}^2\\ =\frac{F_0}{2m}{\left(\frac{my}{p_0}\right)}^2\\ =\left(\frac{m{F}_0}{2p_0}\right)y^2\end{array}$        (XXV)

The trajectory of the particle in the case of non-relativistic motion is parabola.

Conclusion:

Therefore, the velocity of mass $m$ as a function of time $t$ is $\underset{\_}{\frac{p_0}{m\sqrt{1+k{t}^2}}+\frac{F_{0}t}{m\sqrt{1+k{t}^2}}}$ and checked the trajectory in the non-relativistic limit as parabolic nature.