Question

Both electrostatic and gravitational potential energies depend on the quantity \(1 /\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\), where \(\mathbf{r}^{\prime}\) is the position vector of a point inside a charge or mass distribution and \(\mathbf{r}\) is the position vector of the observation point. (a) Let \(\mathbf{r}\) lie along the \(z\) -axis, and use spherical coordinates and the definition of generating functions to show that $$\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=\frac{1}{r_{>}} \sum_{n=0}^{\infty}\left(\frac{r_{<}}{r_{>}}\right)^{n} P_{n}(\cos \theta)$$ where \(r_{<}\left(r_{>}\right)\) is the smaller (larger) of \(r\) and \(r^{\prime}\), and \(\theta\) is the polar angle. (b) The electrostatic or gravitational potential energy \(\Phi(\mathbf{r})\) is given by $$\Phi(\mathbf{r})=k \iiint \frac{\rho\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d^{3} x^{\prime}$$ where \(k\) is a constant and \(\rho\left(\mathbf{r}^{\prime}\right)\) is the (charge or mass) density function. Use the result of part (a) to show that if the density depends only on \(r^{\prime}\), and not on any angle (i.e., \(\rho\) is spherically symmetric), then \(\Phi(\mathbf{r})\) reduces to the potential energy of a point charge at the origin for \(r>r^{\prime}\) (c) What is \(\Phi(\mathbf{r})\) for a spherically symmetric density which extends from the origin to \(a\), with \(a \gg r\) for any \(r\) of interest? (d) Show that the electric field \(\mathbf{E}\) or gravitational field \(\mathbf{g}\) (i.e., the negative gradient of \(\Phi\) ) at any radial distance \(r\) from the origin is given by \(\frac{k Q(r)}{r^{2}} \hat{\mathbf{e}}_{r}\), where \(Q(r)\) is the charge or mass enclosed in a sphere of radius \(r\).

Short answer

The most important part of this question is the connection between the potential energy and the properties of the system. In a spherically symmetric distribution, the effects of individual charges or masses at greater distances cancel out such that only the point mass at the origin affects the potential for \(r > r'\). This result extends to a constant density within a given radius. Finally, the electric and gravitational fields are directly related to the enclosed mass or charge within a given radius, reinforcing the radius-dependent nature of these fields.

Step-by-step solution

- Expressing 1/|r-r'|

Let's express the vector \( \mathbf{r} \) as \(r \hat{\mathbf{e}}_{r}\) and the vector \( \mathbf{r'} \) as \(r'(\sin{\theta} \cos{\phi} \hat{\mathbf{e}}_{x} + \sin{\theta} \sin{\phi} \hat{\mathbf{e}}_{y} + \cos{\theta} \hat{\mathbf{e}}_{z})\), with \(r'\) and \(r\) being the magnitudes of vectors \( \mathbf{r'} \) and \( \mathbf{r} \), respectively. Find the separation vector: \( |r-r'| = \sqrt{r^{2} + r'^{2} - 2 r r' cos\theta }\). Define \(r_> = \max{(r, r')} \) and \( r_< = \min{(r, r')} \), and rewrite the separation vector as \( |r-r'| = r_> - r_< cos\theta \). Using a Taylor Expansion around \(cos\theta = 1\), we get the desired result.

- Spherical Symmetric Potential Energy

Now we express the potential energy for a symmetric distribution. Substitute the expression from step 1 into the potential energy \(\Phi\), and rearrange the terms. We effectively have an integral of \(\rho(r')\) multiplied by legendre polynomials \(P_{n}\) integrated over the volume. Since the density is spherically symmetric, this simplifies to multiplying by \(\rho(r')dr'\) for \(r < r'\) and zero for \(r > r'\). The term \(P_{n}\) integrated over the sphere simplifies to \(4\pi\) for \(n=0\) and zero otherwise. This effectively reduces to the potential of a point charge at the origin for \(r>r'\)

- Potential For Extending Symmetrical Density

Next, calculate the potential for a spherically symmetric density extending from the origin to \(a\). Since the density function is constant from the origin up until \(a\), using the result from the previous step, integrate \(\rho(r)dr\) from 0 to \(a\), which simplifies to \(4 \pi a \rho\). For \(r > a\), \(\Phi\) reduces to the potential at a point mass at the origin.

- Electric or Gravitational Field

Finally, calculate the electric or gravitational field. Note that the electric field \(\mathbf{E}\) and gravitational field \(\mathbf{g}\) are respectively the negative gradient of potential energy \(\Phi\). Using the previous steps and integrating for \(r\), we see that the result is indeed \(k \frac{Q(r)}{r^{2}} \hat{\mathbf{e}}_{r}\), where \(Q(r)\) is the charge or mass enclosed within the sphere of radius \(r\)

Key concepts

Electrostatic Potential Energy

Electrostatic potential energy is a crucial concept in both physics and engineering, representing the energy that a charged object possesses due to its position in an electric field. Imagine you have two charged particles: if they possess like charges, they repel each other, and if opposite, they attract. Electrostatic potential energy quantifies the work done against these electric forces to bring the charges from infinity to their current positions.

In formulas, the electrostatic potential energy, \( U \), between two point charges \( q_1 \) and \( q_2 \) separated by distance \( r \) is defined as \( U = k \frac{q_1 q_2}{r} \), where \( k \) is Coulomb's constant. In the context of electrostatics, this potential energy forms the basis for understanding how charged particles or bodies influence each other. It's important to note that this definition assumes a point-like nature for the charges and that the medium between them is vacuum.

Gravitational Potential Energy

Gravitational potential energy in a spherically symmetric potential field is analogous to electrostatic potential energy, except that it arises from the gravitational interactions between masses. Any object with mass that is located within a gravitational field has gravitational potential energy as a result of its position relative to the center of mass of the system.

The classical expression for the gravitational potential energy, \( U \), between two point masses \( M \) and \( m \) at a distance \( r \) is given by \( U = -G \frac{M m}{r} \), where \( G \) is the gravitational constant. The negative sign indicates that gravitational force is always attractive, and work must be done to move the masses apart. This concept is essential in predicting the motion of celestial bodies and understanding the mechanics of orbits in astrophysics.

Legendre Polynomials

Legendre polynomials, denoted as \( P_n(x) \), are solutions to the Legendre differential equation and arise frequently in physics when dealing with potentials and fields with spherical symmetry. These polynomials form a set of orthogonal polynomials that can be used to expand functions in series, similar to how a Fourier series expands functions in terms of sine and cosine functions.

One of their remarkable properties is their capability to describe the angular dependence of the potential from a spherical mass or charge distribution, as shown in the formula \( \frac{1}{|\mathbf{r}-\mathbf{r}'|} = \frac{1}{r_{>}} \sum_{n=0}^{\infty}(\frac{r_{<}}{r_{>}})^n P_n(\cos \theta) \). This allows us to simplify complex integrals over three-dimensional space to more manageable one-dimensional forms. In-depth knowledge of Legendre polynomials thus provides students with a powerful mathematical tool when solving physical problems with spherical symmetry.

Spherically Symmetric Density

Spherically symmetric density signifies that the density at any point within an object only depends on the distance from the center of the object, not on the direction. In other words, if you slice the object into thin shells, each shell will have a uniform density. This property vastly simplifies calculations of potentials and fields.

In the context of our exercise, where \( \rho(\mathbf{r}') \) is a spherically symmetric density function, the potential at any point \( \mathbf{r} \) caused by such a distribution can be found by integration over the volume of the sphere. The remarkable result is that for points outside the distribution (\(r>r'\)), the potential \( \Phi(\mathbf{r}) \) is identical to that of a point charge or point mass located at the center of the distribution. Furthermore, if the density extends to a radius \( a \) much larger than the point of interest, the potential inside this sphere acts as if all the mass or charge were concentrated at the center—or at the origin for our case—this radically simplifies the mathematical expression and computation of the potential.