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A particle of mass \(m\) moves along one of two "paths" through space and time connecting the points \((x, t)=(0,0)\) and \((x, t)=(D, T)\). One path is quadratic in time, i.e. \(x_{1}(t)=\frac{1}{2} a t^{2}\) where \(a\) is a constant. The second path is linear in time, i.e. \(x_{2}(t)=v t\) where \(v\) is a constant. The correct classical path is the quadratic path, that is \(x_{1}(t)\). a. Find the acceleration \(a\) for the correct classical path. Use freshman physics to find the force \(F=m a=-d V / d x\) and then the potential energy function \(V(x)\) in terms of \(m, D\), and \(T\). Also find the velocity \(v\) for the linear (i.e. incorrect classical) path. b. Calculate the classical action \(S[x(t)]=\int_{0}^{T}\left[\frac{1}{2} m \dot{x}^{2}-V(x)\right] d t\) for each of the two paths \(x_{1}(t)\) and \(x_{2}(t)\). Confirm that \(S_{1} \equiv S\left[x_{1}(t)\right]

Step by step solution

01

Establish Equations of Motion

For the quadratic path, we know `x_1(t)` is given as \(x_1(t) = \frac{1}{2} a t^2\). For the linear path, \(x_2(t) = v t\). Here, our aim is to find the acceleration \(a\) for the quadratic path and the velocity \(v\) for the linear path.

02

Calculate Acceleration for Quadratic Path

Since the final position at time \(T\) is \(D\), for \(x_1(t) = \frac{1}{2} a t^2\), we have \(D = \frac{1}{2} a T^2\). Solving for \(a\), we get \(a = \frac{2D}{T^2}\).

03

Calculate Velocity for Linear Path

For the linear path, the final position is \(D = v T\). Solving for \(v\), we get \(v = \frac{D}{T}\).

04

Determine the Force and Potential Energy

The force is given by \(F = ma = m \left(\frac{2D}{T^2}\right)\). Since \(F = -\frac{dV}{dx}\), by integrating \(F\) with respect to \(x\), the potential energy function \(V(x)\) is given by \(V(x) = -m \left(\frac{2D}{T^2} \right) x + C\). By setting \(V(0) = 0\), we find that \(C = 0\).

05

Classical Action for Quadratic Path

For \(x_1(t)\), we compute the kinetic energy as \(\frac{1}{2} m \left( a t \right)^2 = \frac{1}{2} m \left( \frac{2D}{T^2} t \right)^2\). The classical action \(S_1\) is \(\int_0^T \left[ \frac{1}{2} m \left( \frac{2D}{T^2} t \right)^2 - \left( -m \frac{2D}{T^2} x \right) \right] dt\). Solve this integral for \(S_1\).

06

Classical Action for Linear Path

For \(x_2(t)\), compute \(\frac{1}{2} m v^2 = \frac{1}{2} m \left( \frac{D}{T} \right)^2\) and \(V(x) = -m \left(\frac{2D}{T^2} \right) xt\). Solve the integral \(\int_0^T \left[ \frac{1}{2} m \left( \frac{D}{T} \right)^2 - V(x) \right] dt\) to find \(S_2\).

07

Compute \(\Delta S\) and \(\Delta S/ \hbar\)

Calculate the difference \(\Delta S = S_2 - S_1\). For a nanoparticle and electron moving \(1 \text{ mm}\) in \(1 \text{ ms}\), use components from each particle, calculate \(\frac{\Delta S}{\hbar}\), and determine whether the motion should be considered quantum mechanical or classical based on the action's scale.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Classical Path

In classical mechanics, the concept of a "Classical Path" refers to the trajectory that a particle follows through space and time based on the principles established by Isaac Newton. The problem here looks at two hypothetical paths—a quadratic path and a linear path—to illustrate this concept.
A classical path is not just any path; it is the path that minimizes the action, a principle attributed to Pierre Maupertuis and championed by later scientists such as Lagrange and Hamilton. For two endpoints in space and time, the correct classical path, according to the principle of least action, is one that takes into consideration forces acting on the particle and results in a specified acceleration.

• Quadratic Path: This problem cuts to the heart of classical dynamics by introducing a path described by a quadratic equation, where the position is a function of time squared, as in \( x_1(t) = \frac{1}{2} a t^2 \). This path signifies a particle experiencing constant acceleration.
• Determining Forces: The acceleration \( a = \frac{2D}{T^2} \) is key to understanding the forces at play. Recognizing the force gives insight into potential energy and enables us to verify the correctness of the classical path using Newton's second law: \( F = ma \).

By solving for the correct classical path, students grasp how certain physical conditions derive the trajectory a particle must follow classically.

Action Calculation

The action, a pivotal concept in classical mechanics, is quantified as the integral of the Lagrangian over time. It helps students understand which path out of potential candidates is physically plausible. In this exercise, we compare the classical action calculated for both the quadratic and linear paths to confirm which one requires less action. This involves using the Lagrangian, \( L = T - V \), where \( T \) is kinetic energy and \( V \) is potential energy, and then integrating it over time.
For the quadratic path, the action is:

• \( S_1 = \int_0^T \left[ \frac{1}{2} m \left( \frac{2D}{T^2} t \right)^2 - \left(-m \frac{2D}{T^2} x \right) \right] dt \)

For the linear (incorrect) path, the action is different:

• \( S_2 = \int_0^T \left[ \frac{1}{2} m \left( \frac{D}{T} \right)^2 - V(x) \right] dt\)

The classical path is, by definition, the one with the smaller action (i.e., \( S_1 < S_2 \)). This disciplinary approach ingrains a deeper understanding of why nature "prefers" one path over another—adhering to the principle that action is minimized.

Quantum Mechanical Motion

When investigating whether a particle's motion is quantum mechanical, the calculation of \( \Delta S / \hbar \) becomes essential. \( \hbar \) is the reduced Planck's constant, which is notably small. Thus, only when \( \Delta S \) is comparable to or smaller than \( \hbar \) does quantum mechanical behavior emerge.
For example, in the textbook exercise, we account for a particle that moves 1 mm in 1 ms and compute this ratio for a nanoparticle and an electron. These two entities yield different results due to their substantial difference in masses and sizes:

• Nano-Particle: Given its considerable mass, the action difference \( \Delta S \) divided by \( \hbar \) usually indicates more classical behavior.
• Electron: With a much smaller mass, the action difference is often comparable to \( \hbar \), indicating more significant quantum effects.

Understanding which case depicts quantum mechanical motion enriches the student's grasp of how scale and environment could dictate the nature of motion.