Riemann surface
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A particle moves in one dimension along the $x$-axis, bouncing between two perfectly reflecting walls at $x=0$ and $x=\ell$. In between collisions with the walls no forces act on the particle. Suppose there is an uncertainty $\Delta v_0$ in the initial velocity $v_0$. Determine the corresponding uncertainty $\Delta x$ in the position of the particle after a time $t$.
Solution
In between the instants of reflection, the particle moves with constant velocity equal to the initial value. Thus, if the initial velocity is $v_0$ then the distance moved by the particle in a time $t$ is $v_0 t$, whereas if the initial velocity is $v_0+\Delta v_0$ the distance moved is $\left(v_0+\Delta v_0\right) t$. Therefore, the uncertainty in position after a time $t$ is
$$
\begin{aligned}
\Delta x & =\left(v_0+\Delta v_0\right) t-v_0 t \
& =\left(\Delta v_0\right) t
\end{aligned}
$$
Question 2.3
The active gravitational mass $\left(m^{\mathrm{A}}\right)$ of a particle is an attribute that enables it to establish a gravitational field in space, whereas the passive gravitational mass $\left(m^{\mathrm{P}}\right)$ is an attribute that enables the particle to respond to this field.
(a) Write Newton’s law of universal gravitation in terms of the relevant active and passive gravitational masses.
(b) Show that the third law of motion makes it unnecessary to distinguish between active and passive gravitational mass.
Solution
(a) The gravitational force $\mathbf{F}{12}$ that particle 1 exerts on particle 2 is proportional to the product of the active gravitational mass $m_1^A$ of particle 1 and the passive gravitational mass $m_2^{\mathrm{P}}$ of particle 2. Thus, the inverse-square law of gravitation is $$ \mathbf{F}{12}=-G \frac{m_1^A m_2^{\mathrm{P}}}{r^2} \hat{\mathbf{r}}
$$
where $G$ is the universal constant of gravitation, $r$ is the distance between the particles and $\hat{\mathbf{r}}$ is a unit vector directed from particle 1 to particle 2 . By the same token, the force $\mathbf{F}{21}$ which particle 2 exerts on particle 1 is $$ \mathbf{F}{21}=G \frac{m_2^A m_1^{\mathrm{P}}}{r^2} \hat{\mathbf{r}} \text {. }
$$
(b) According to Newton’s third law, $\mathbf{F}{12}=-\mathbf{F}{21}$. It therefore follows from (1) and (2) that
$$
\frac{m_2^{\mathrm{A}}}{m_2^{\mathrm{P}}}=\frac{m_1^{\mathrm{A}}}{m_1^{\mathrm{P}}}
$$
We conclude from (3) that the ratio of the active to the passive gravitational mass of a particle is a universal constant. Furthermore, this constant can be
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