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Another way to evaluate the potential \(U\) of the example in Section \(5.16\) is to use the following information: a) \(\nabla^{2} U=-2 \pi b\) for \(Ra\), c) \(U=M \log \frac{1}{R}+\frac{p(x, y)}{R}\) for large \(R\), as in Eq. (5.172), d) \(U\) and \(\operatorname{grad} U\) are continuous for all \((x, y)\), e) \(U\) depends on \(R\) alone, by symmetry. Since \(\nabla^{2} U=\partial^{2} U / \partial R^{2}+(1 / R) \partial U / \partial R\) in polar coordinates, for a function depending only on \(R\) (Section 2.17), (a) and (b) give differential equations for \(U\). By (c) and (d) the arbitrary constants in the solution can be determined. Carry out the process suggested to find \(U\).

Short Answer

Expert verified
Question: Find the potential U using the given information and boundary conditions. Answer: The potential U has two expressions depending on the region: 1) For \(Ra\): \(U(R) = -M\ln R\). The constants \(C_1\) and \(C_2\) depend on the specific problem and its boundary conditions.

Step by step solution

01

Write the Laplacian in polar coordinates for a function depending only on R

As the problem states, the function depends only on \(R\), so we will use the Laplacian in polar coordinates for a function depending only on \(R\): \(\nabla^2 U = \frac{\partial^2 U}{\partial R^2} + \frac{1}{R} \frac{\partial U}{\partial R}\).
02

Write the differential equations for U

We have two cases to consider for the differential equations: a) For \(Ra\), the Laplacian is given as \(\nabla^2 U = 0\). So, the differential equation is: \(\frac{\partial^2 U}{\partial R^2} + \frac{1}{R} \frac{\partial U}{\partial R} = 0\).
03

Solve the differential equations for U

For \(Ra\), the equation is: \(\frac{\partial^2 U}{\partial R^2} + \frac{1}{R} \frac{\partial U}{\partial R} = 0\). Integrating with respect to \(R\), we get: \(\frac{\partial U}{\partial R} = \frac{C_3}{R}\). Integrating again, we obtain: \(U(R) = C_3\ln R + C_4\) for \(R>a\).
04

Determine the arbitrary constants using boundary conditions

We know that \(U\) and its gradient are continuous, so we have two boundary conditions: 1) \(U(R)\) must be continuous at \(R = a\). 2) \(\frac{\partial U}{\partial R}\) must be continuous at \(R = a\). Using these conditions, we can find the constants as follows: 1) \(-\pi ba^2 + C_1\ln a + C_2 = C_3\ln a + C_4\). 2) \(-2\pi ba + C_1 = \frac{C_3}{a}\). Finally, we have the behavior of \(U\) at large \(R\): \(U = M \log \frac{1}{R} + \frac{p(x, y)}{R}\). Comparing this with our general solution for \(R>a\), we can determine \(C_3 = -M\) and \(C_4 = 0\) since \(p(x, y)\) does not depend on \(R\). Now, we can substitute these values back into our boundary condition equations to solve for \(C_1\) and \(C_2\).
05

Final expression for U

Therefore, the potential \(U\) has the following expressions: 1) For \(Ra\): \(U(R) = -M\ln R\). Determining the constants \(C_1\) and \(C_2\) depends on the specific problem and its boundary conditions.

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Key Concepts

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

Potential Function
A potential function, in this context, describes a scalar field that gives rise to a particular vector field when the gradient is taken. This potential function is key in physics, especially in gravitational and electric fields, where it represents potential energy per unit charge or mass.
One interesting thing about potential functions is their relationship with Laplace's equation, which is a second-order partial differential equation. The solution to this equation often helps in determining the potential in a given field, which, in turn, helps us understand how forces will act within that field.
In our case, we're considering the potential function in polar coordinates. Here, potential depends solely on the radial coordinate, which simplifies the problem significantly. The radial symmetry is due to the nature of the problem or the object causing the potential, often resulting in equations that are easier to solve. These are expressed in terms of the Laplacian operator, found frequently in physics and engineering problems concerning fields.
Differential Equations
Differential equations are vital to understanding continuous change. They describe an unknown function in terms of its derivatives and are hugely important in modeling real-world phenomena.
In this exercise, we deal with second-order differential equations derived from Laplace’s equation in polar coordinates. These equations are:
  • For \(R
  • For \(R>a\): \(\frac{\partial^2 U}{\partial R^2} + \frac{1}{R} \frac{\partial U}{\partial R} = 0\)
These equations capture the essence of how the potential evolves within and outside a given region. Once solved, these differential equations will reveal expressions that define our potential function, which can then describe the system's behavior.
Boundary Conditions
Boundary conditions are the constraints or limits that solutions to differential equations must satisfy. Many physical problems require these conditions to be defined, as they ensure the problem is well-posed and has a unique solution.
In our particular example, the boundary conditions are crucial. We have conditions that relate to both the continuity of the potential function itself and its gradient:
  • For the potential function to be continuous across the boundary at \(R = a\).
  • For its gradient, or rate of change, to be continuous across the same boundary.
These conditions allow us to determine the arbitrary constants that arise during the integration process of solving the differential equations. Effectively, they ensure that our solution smoothly transitions across the regions, matching both the potential and its rate of change.
Continuity in Calculus
Continuity in calculus refers to a function being smoothly connected, with no breaks, jumps, or abrupt changes at any point within its domain. It's a critical property, especially in physical problems described by differential equations, as it ensures that solutions are realistic and physically meaningful.
In the context of our problem, we're focused on both the potential function and its gradient being continuous. A continuous potential ensures a smooth transition of values across different regions, which is significant when interpreting the physical phenomena represented by the potential, such as force fields.
Additionally, ensuring the continuity of the gradient means that the rate at which the potential changes is equally smooth. This smoothness is essential in many fields, as it can impact how calculations related to changes in energy, force, or field behavior are performed.
Continuity is not just a mathematical concept but a foundation of how we model real-world phenomena accurately, ensuring models align with observable reality when describing continuous systems.

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Most popular questions from this chapter

(The solid angle) Let \(S\) be a plane surface, oriented in accordance with a unit normal \(\mathbf{n}\). The solid angle \(\Omega\) of \(S\) with respect to a point \(O\) not in \(S\) is defined as $$ \Omega(O, S)=\pm \text { area of projection of } S \text { on } S_{1}, $$ where \(S_{1}\) is the sphere of radius 1 about \(O\) and the \(+\) or - sign is chosen according to whether \(\mathbf{n}\) points away from or toward the side of \(S\) on which \(O\) lies. This is suggested in Fig. 5.37. a) Show that if \(O\) lies in the plane of \(S\) but not in \(S\), then \(\Omega(O, S)=0\). it b) Show that if \(S\) is a complete (that is, infinite) plane, then \(\Omega(O, S)=\pm 2 \pi\). c) For a general oriented surface \(S\) the surface can be thought of as made up of small elements, each of which is approximately planar and has a normal \(\mathbf{n}\). Justify the following definition of element of solid angle for such a surface element: $$ d \Omega=\frac{\mathbf{r} \cdot \mathbf{n}}{r^{3}} d \sigma $$ where \(\mathbf{r}\) is the vector from \(O\) to the element. d) On the basis of the formula of (c), one obtains as solid angle for a general oriented surface \(S\) the integral $$ \Omega(O, S)=\iint_{S} \frac{\mathbf{r} \cdot \mathbf{n}}{r^{3}} d \sigma . $$ Show that for surfaces in parametric form, if \(O\) is the origin, $$ \Omega(O, S)=\iint_{R_{x r}}\left|\begin{array}{ccc} x & y & z \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{array}\right| \frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}} d u d v . $$ This formula permits one to define a solid angle for complicated surfaces that interac themselves. e) Show that if the normal of \(S_{1}\) is the outer one, then \(\Omega(O, S)=4 \pi\). f) Show that if \(S\) forms the boundary of a bounded, closed, simply connected region \(R\). then \(\Omega(O, S), \pm 4 \pi\), when \(O\) is inside \(S\) and \(\Omega(O, S)=0\) when \(O\) is outside \(S\). g) If \(S\) is a fixed circular disk and \(O\) is variable, show that \(-2 \pi \leq \Omega(O, S) \leq 2 \pi\) and that \(\Omega(O, S)\) jumps by \(4 \pi\) as \(O\) crosses \(S\).

Evaluate the following line integrals: a) \(\int_{(0,-1)}^{(0,1)} y^{2} d x+x^{2} d y\), where \(C\) is the semicircle \(x=\sqrt{1-y^{2}}\); b) \(\int_{(0,0)}^{(2,4)} y d x+x d y\), where \(C\) is the parabola \(y=x^{2}\); c) \(\int_{(1,0)}^{(0,1)} \frac{y d x-x d y}{x^{2}+y^{2}}\), where \(C\) is the curve \(x=\cos ^{3} t, y=\sin ^{3} t, 0 \leq t \leq \frac{\pi}{2}\). \(C\)

Let \(D\) be a domain that has a finite number of "holes" at points \(A_{1}, A_{2}, \ldots, A_{k}\), so that \(D\) is ( \(k+1)\)-tuply connected; cf. Fig. 5.23. Let \(P\) and \(Q\) be continuous and have continuous derivatives in \(D\), with \(\partial P / \partial y=\partial Q / \partial x\) in \(D\). Let \(C_{1}\) denote a circle about \(A_{1}\) in \(D\), enclosing none of the other \(A^{\text {'s }}\). Let \(C_{2}\) be chosen similarly for \(A_{2}\), and so on. Let $$ \oint_{C_{1}} P d x+Q d y=\alpha_{1}, \oint_{C_{2}} P d x+Q d y=\alpha_{2} \ldots, \oint_{C_{k}} P d x+Q d y=\alpha_{k} $$ a) Show that if \(C\) is an arbitrary simple closed path in \(D\) enclosing \(A_{1}, A_{2} \ldots, A_{k}\), then $$ \oint_{C} P d x+Q d y=\alpha_{1}+\alpha_{2}+\cdots+\dot{u}_{k} $$ b) Determine all possible values of the integral $$ \int_{\left(x_{1}, y_{1}\right)}^{\left(x_{2}, y_{2}\right)} P d x+Q d y $$ between two fixed points of \(D\), if it is known that this integral has the value \(K\) for one particular path. 9\. Let \(P\) and \(Q\) be continuous and have continuous derivatives, with \(\partial P / \partial y=\partial Q / \partial x\), except at the points \((4,0),(0,0),(-4,0)\). Let \(C_{1}\) denote the circle \((x-2)^{2}+y^{2}=9\); let \(C_{2}\) denote the circle \((x+2)^{2}+y^{2}=9\); let \(C_{3}\) denote the circle \(x^{2}+y^{2}=25\). Given that $$ \oint_{C_{1}} P d x+Q d y=11, \quad \oint_{C_{2}} P d x+Q d y=9, \quad \oint_{C_{3}} P d x+Q d y=13 $$ find $$ \int_{C_{4}} P d x+Q d y $$ «s where \(C_{4}\) is the circle \(x^{2}+y^{2}=1\). [Hint: Use the result of Problem \(8(\) a).]

(Degree of mapping of one surface into another) Let \(S_{w v w}\) and \(S_{x y z}\) be surfaces forming the boundaries of regions \(R_{u v w}\) and \(R_{x y z}\) respectively: it is assumed that \(R_{u v w}\) and \(R_{x y}\). are bounded and closed and that \(R_{x y z}\) is simply connected. Let \(S_{u v w}\) and \(S_{x y z}\) be oriented by the outer normal. Let \(s, t\) be parameters for \(S_{u v w}\) : $$ u=u(s, t), \quad v=v(s, t), \quad w=w(s, t), $$ the normal having the direction of $$ \left(u_{s} \mathbf{i}+v_{s} \mathbf{j}+w_{s} \mathbf{k}\right) \times\left(u_{\mathbf{i}} \mathbf{i}+v_{t} \mathbf{j}+w_{t} \mathbf{k}\right) $$ Let $$ x=x(u, v, w), \quad y=y(u, v, w), \quad z=z(u, v, w) $$ be functions defined and having continuous derivatives in a domain containing \(S_{u v w}\), an let these equations define a mapping of \(S_{u v w}\) into \(S_{x y z}\). The degree \(\delta\) of this mapping i defined as \(1 / 4 \pi\) times the solid angle \(\Omega(O, S)\) of the image \(S\) of \(S_{u v w}\) with respect t a point \(O\) interior to \(S_{x y z}\). If \(O\) is the origin, the degree is hence given by the integra (Kronecker integral) $$ \delta=\frac{1}{4 \pi} \iint_{R_{u t}}\left|\begin{array}{ccc} x & y & z \\ \frac{\partial x}{\partial s} & \frac{\partial y}{\partial s} & \frac{\partial z}{\partial s} \\ \frac{\partial x}{\partial t} & \frac{\partial y}{\partial t} & \frac{\partial z}{\partial t} \end{array}\right| \frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}} d s d t $$ where \(x, y, z\) are expressed in terms of \(s, t\) by (a) and (b). It can be shown that \(\delta\), as thu defined, is independent of the choice of the interior point \(O\), that \(\delta\) is a positive or negativ integer or zero, and that \(\delta\) does measure the effective number of times that \(S_{x y z}\) is covered Let \(S_{k v w}\) be the sphere \(u=\sin s \cos t, v=\sin s \sin t, w=\cos s, 0 \leq s \leq \pi, 0\) \(t \leq 2 \pi\). Let \(S_{x y z}\) be the sphere \(x^{2}+y^{2}+z^{2}=1\). Evaluate the degree for the followin; mappings of \(S_{u v w}\) into \(S_{x y z}\); a) \(x=v, y=-w, z=u\) b) \(x=u^{2}-v^{2}, y=2 u v, z=w \sqrt{2-w^{2}}\)

By showing that the integrand is an exact differential, evaluate a) \(\int_{(1,1,2)}^{(3.5 .0)} y z d x+x z d y+x y d z\) on any path; b) \(\int_{z=t}^{(1.0 .2 \pi)} \sin y z d x+x z \cos y z d y+x y \cos y z d z\). on the helix \(x=\cos t . y=\sin t\).

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