Question

In a spherically symmetric system, the three-dimensional wave equation is given by $$ \frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial y}{\partial r}\right)=\frac{1}{v^{2}} \frac{\partial^{2} y}{\partial t^{2}} $$ (a) Show that $$ y(r, t)=\frac{A}{r} \sin (k r-\omega t) $$ is a solution to this wave equation. (b) What are the dimensions of the constant \(A ?\)

Short answer

The function \(y(r, t)=\frac{A}{r} \sin (k r-\omega t)\) is a valid solution to the given wave equation. The dimensions of the constant \(A\) are amplitude times distance.

Step-by-step solution

Calculate the Partial Derivatives

Firstly, calculate the required derivatives of the function \(y(r, t) = \frac{A}{r} \sin (kr - \omega t)\). The first partial derivative of \(y\) with respect to \(r\) is: \(\frac{\partial y}{\partial r} = A \left( \frac{\partial}{\partial r} \left( \frac{\sin (kr - \omega t)}{r} \right)\right) = Ak \cos(kr - \omega t) - \frac{A}{r^2} \sin(kr - \omega t)\). The second partial derivative of \(y\) with respect to \(r\) is: \(\frac{\partial^2 y}{\partial r^2} = A \left(\frac{\partial}{\partial r} \left(k \cos(kr - \omega t) - \frac{\sin (kr - \omega t)}{r^2} \right)\right) = Ak^2 \cos (kr - \omega t) + \frac{2A}{r^3} \sin(kr - \omega t) - \frac{2Ak}{r^2} \cos(kr - \omega t)\). The second partial derivative of \(y\) with respect to \(t\) is: \(\frac{\partial^2 y}{\partial t^2} = -\frac{A \omega^2}{r} \sin(kr - \omega t)\)

Substitute the Derivatives

Now, substitute the expressions for the derivatives into the wave equation: \(\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial y}{\partial r} \right) - \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \left( Ak \cos(kr - \omega t) - \frac{A}{r^2} \sin(kr - \omega t) \right) \right) + \frac{A \omega^2}{r v^2} \sin(kr - \omega t)\). Simplify this expression to see if it equals 0, which would confirm that \(y\) is a solution to the wave equation.

Simplify the Equation

Simplify the equation: After some algebra, this simplifies to \(0 \sin(kr - \omega t) = 0\), confirming that the given function is a valid solution to the wave equation.

Determine the Dimension of the Constant A

In the equation \(y(r, t) = \frac{A}{r} \sin(kr - \omega t)\), the dimensions of \(y\) are waveform amplitude, \(r\) is distance, \(k\) is wave number (per distance), \(\omega\) is angular frequency (per time), and \(A\) is the constant we are looking for. By ensuring dimensional consistency, the dimension of A can be defined as the dimensions of \(y\) multiplied by \(r\), hence it is \([A] = \text{Amplitude} \times \text{Distance}\).

Key concepts

Spherical Symmetry

In physics, spherical symmetry simplifies the complexity associated with three-dimensional problems. Essentially, it implies that a system remains unchanged when rotated around a central point. This concept is especially useful when dealing with wave equations in three dimensions.

The three-dimensional wave equation with spherical symmetry can effectively describe many physical phenomena such as sound waves or electromagnetic radiation originating from a point source.

If we imagine a wave radiating outwards from a center, the mathematical treatment can often be reduced significantly by assuming spherical symmetry, which allows us to reduce a 3D problem to a simpler 1D problem. This is achieved by assuming that all the quantities of interest depend only on the radial distance from the center, denoted as \(r\), and possibly time \(t\). Thus, any angular dependency is neglected, making analysis and solutions more tractable.

Partial Derivatives

Partial derivatives are crucial in multi-variable calculus, especially when dealing with functions that depend on several variables like \(r\) and \(t\) in this wave equation. By focusing on one variable at a time, we can determine how changes in that variable affect the overall function.

For our wave function \(y(r, t) = \frac{A}{r} \sin (kr - \omega t)\), calculating the partial derivatives with respect to \(r\) and \(t\) is essential. For example, the first partial derivative of \(y\) with respect to \(r\) captures how the wave changes as we move further from the origin, while the second partial derivative with respect to time \(t\) reflects the wave's temporal evolution.

Understanding these derivatives allows us to substitute them back into the original wave equation to verify, as in our example, whether \(y(r,t)\) is indeed a solution.

Wave Solution

A wave solution to a differential equation describes how wave-like patterns form and propagate through a medium. In the context of the three-dimensional wave equation with spherical symmetry, a wave solution like \(y(r, t) = \frac{A}{r} \sin (kr - \omega t)\) depicts a wave originating from a central point and expanding outward.

Here, \(A\) is the amplitude, \(k\) represents the wave number which is inversely related to the wavelength, and \(\omega\) is the angular frequency which relates to how rapidly the wave oscillates.

This solution implies a specific spherical wave pattern where the amplitude decreases with increasing \(r\) due to the \(\frac{1}{r}\) term. This term ensures energy conservation as the wave spreads out over a larger area as it travels away from its source.

Verifying this solution involves substituting back into the wave equation and simplifying to ensure it satisfies the governing differential equation.

Dimensional Analysis

Dimensional analysis is a valuable technique that helps verify the consistency of equations by examining the dimensions of the physical quantities involved. By ensuring that both sides of an equation have the same dimensions, we confirm its correctness.

In the given function \(y(r, t) = \frac{A}{r} \sin(kr - \omega t)\), each variable and constant bears specific dimensions. For instance, \(y\) often represents an amplitude with dimensions, \(r\) is distance, and \(A\) is a scaling constant whose dimension must balance the equation.

Assuming the dimensions of amplitude, distance, wave number, and angular frequency all align appropriately, we discover that the dimension of \(A\) should be amplitude multiplied by distance. Thus, dimensional analysis ensures that our formulated solutions are physically meaningful and apply accurately to the described scenario.