Problem 8 - Classical Dynamics of Particles and Systems

Question

Consider a region of space divided by a plane. The potential energy of a particle in region 1 is \(U_{1}\) and in region 2 it is \(U_{2} .\) If a particle of mass \(m\) and with speed \(v_{1}\) in region 1 passes from region 1 to region 2 such that its path in region 1 makes an angle \(\theta_{1}\) with the normal to the plane of separation and an angle \(\theta_{2}\) with the normal when in region \(2,\) show that $$\frac{\sin \theta_{1}}{\sin \theta_{2}}=\left(1+\frac{U_{1}-U_{2}}{T_{1}}\right)^{1 / 2}$$ where \(T_{1}=\frac{1}{2} m v_{1}^{2} .\) What is the optical analog of this problem?

Step-by-Step Solution

Answer

In this problem, we find that the ratio of the sine of angles \(\theta_1\) in region 1 and \(\theta_2\) in region 2 can be expressed as: \(\frac{\sin\theta_1}{\sin\theta_2} =\left(1+\frac{U_{1}-U_{2}}{T_{1}}\right)^{1 / 2}\), where \(T_1 = \frac{1}{2}mv_1^2\) is the kinetic energy of the particle in region 1, and \(U_1\) and \(U_2\) are potential energies in regions 1 and 2, respectively. The optical analog of this problem is a light ray passing from one medium to another with different refractive indices, where the angles are related through Snell's Law with the inversely proportional ratio of refractive indices: \(\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1}\).

Step 1: Calculate the final velocity of the particle in region 2

To calculate the final velocity (v2) in region 2, we can use conservation of energy. The total energy in both regions must be conserved as follows: Total energy in region 1 = Total energy in region 2 \(T_1 + U_1 = T_2 + U_2\) We have \(T_1 = \frac{1}{2}mv_1^2\) and we need to find \(T_2\), which can be expressed as \(T_2 = \frac{1}{2}mv_2^2\) Rearranging the conservation of energy equation to solve for v2, we get: \(v_2^2 = v_1^2 + \frac{2(U_1 - U_2)}{m}\)

Step 2: Apply Snell's Law

Snell's Law states that for a propagative wave or particle crossing the boundary of two media (in this case, the plane separating regions 1 and 2), the following relationship holds: \(\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}\) We have all variables except for v2, which we can find from Step 1. Substitute the expression we derived above into Snell's Law: \(\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{\sqrt{v_1^2 + \frac{2(U_1 - U_2)}{m}}}\)

Step 3: Simplify the expression

Now, we will simplify the expression derived in Step 2. To do this, we first divide the numerator and denominator of the fraction by \(v_1\): \(\frac{\sin\theta_1}{\sin\theta_2} = \frac{1}{\sqrt{1 + \frac{2(U_1 - U_2)}{mv_1^2}}}\) Recall that \(T_1 = \frac{1}{2}mv_1^2\). We can rewrite the expression as follows: \(\frac{\sin\theta_1}{\sin\theta_2} = \frac{1}{\sqrt{1 + \frac{U_1 - U_2}{T_1}}}\) Square both sides of the equation to find the final expression: \(\frac{\sin\theta_1}{\sin\theta_2} =\left(1+\frac{U_{1}-U_{2}}{T_{1}}\right)^{1 / 2}\)

Step 4: Optical Analog of the Problem

The optical analog of this problem is that of a light ray passing from one medium to another with different refractive indices (n1 and n2). In this case, the angles would be related through Snell's Law, where the ratio of velocities is replaced by the inversely proportional ratio of refractive indices: \(\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1}\) The energy of the light (photon) can be related to the refractive indices of the media, and the relationship would be analogous to what we have derived above.

Key Concepts

Conservation of Energy

The concept of conservation of energy is fundamental in classical dynamics. It asserts that energy cannot be created or destroyed, only transformed from one form to another. To better understand this, imagine a roller coaster at the peak of its track. It possesses potential energy due to its elevated position above the ground. As it descends, this potential energy is converted into kinetic energy, which is the energy of motion, causing the roller coaster to accelerate.

In the textbook problem, we observed this principle by examining a particle moving across two regions of space with different potential energies. By equating the total energy in both regions, \(T_1 + U_1 = T_2 + U_2\), and solving for \(v_2\), we ensure that the energy is conserved during the transition. This demonstrates a practical application of the conservation of energy, which is crucial for solving various problems in physics and engineering.

An interesting note is that while energy is conserved within a closed system, the forms of energy we observe may change, such as potential energy changing to kinetic energy or vice versa.

Potential Energy and Kinetic Energy

The distinction between potential energy and kinetic energy is key to analyzing and understanding many physical systems. Potential energy refers to the energy stored within an object due to its position, condition, or configuration. For instance, a book held above the ground has gravitational potential energy resulting from Earth's gravity. On the other hand, kinetic energy is the energy an object possesses due to its motion.

In the context of the particle in our exercise, potential energy is represented by \(U_1\) and \(U_2\) in the two different regions. The kinetic energy, denoted as \(T_1 = \frac{1}{2} m v_1^2\), transforms as the particle moves between these regions due to the differences in potential energy, illustrating the seamless conversion between these two forms of energy.

Understanding that the sum of potential and kinetic energy remains constant in an isolated system (ignoring other forms of energy, such as thermal or sound) allows us to predict the behavior of objects. For example, when a pendulum swings, its kinetic energy reaches a maximum at the lowest point of its swing, where its potential energy is at its minimum, and vice versa.

Refractive Indices in Optics

In optics, the refractive index of a medium quantifies how much light is bent, or refracted, when entering the medium from a vacuum or another medium. When light passes from one material into another, changes in speed occur based on the refractive indices of the materials. The higher the refractive index, the slower the light travels through the material, resulting in a greater bending of the light beam.

Snell's Law, a formula used to describe the relationship between the angles of incidence and refraction, is defined by the equation \(\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1}\), where \(\theta_1\) is the angle of incidence, \(\theta_2\) is the angle of refraction, and \(n_1\) and \(n_2\) are the refractive indices of the first and second medium, respectively. The optical analog of our textbook problem showcases this relationship when viewing the particle's transition between regions with different potential energies as analogous to light being refracted at the boundary between materials with different refractive indices.

Understanding how refractive indices affect light propagation is crucial in many applications, such as designing lenses for cameras, eyeglasses, and various optical instruments. It is an excellent example of how a concept from classical dynamics is mirrored in the behavior of light—a phenomenon governed by the principles of optics.