Evaluate the force corresponding to the potential energy function \(V(\boldsymbol{r})=c z / r^{3}\), where \(c\) is a constant. Write your answer in vector notation, and also in spherical polars, and verify that it satisfies \(\nabla \wedge \boldsymbol{F}=\mathbf{0}\).

Short Answer

Expert verified
Question: Given the potential energy function \(V(\boldsymbol{r}) = \frac{c z}{r^3}\), find the corresponding force in vector notation and spherical polar coordinates. Verify that the force satisfies the condition of being conservative, \(\nabla \wedge \boldsymbol{F} = \mathbf{0}\). Answer: The force corresponding to the potential energy function is given by \(\boldsymbol{F} = - \nabla V(\boldsymbol{r})\), and in spherical polar coordinates, the force can be expressed as \(\boldsymbol{F} = F_r \hat{\boldsymbol{r}} + F_\theta \hat{\boldsymbol{\theta}} + F_\phi \hat{\boldsymbol{\phi}}\), where: - \(F_r = \frac{2 c \cos \theta - c \sin^2 \theta}{r^4}\) - \(F_\theta = -\frac{3 c \sin \theta \cos \theta}{r^4}\) - \(F_\phi = 0\) We have verified that the force satisfies the condition of being conservative by calculating the curl of the force in spherical polar coordinates, which is equal to zero: \(\nabla \wedge \boldsymbol{F} = \mathbf{0}\).

Step by step solution

01

Find the force from potential energy function

To find the force corresponding to the potential energy function, we must determine the negative gradient of the potential energy function, which is given by \(\boldsymbol{F} = -\nabla{V(\boldsymbol{r})}\). Thus, the first step is to differentiate the potential energy function with respect to the vector \(\boldsymbol{r}\).
02

Determine the gradient in Cartesian coordinates

To calculate the gradient, we must take the partial derivative of the potential energy \(V(\boldsymbol{r})\) with respect to each of the three Cartesian coordinates, \(x\), \(y\), and \(z\), resulting in three components \(\frac{\partial V}{\partial x}\), \(\frac{\partial V}{\partial_y}\), and \(\frac{\partial V}{\partial_z}\). For this calculation, it is helpful to first rewrite the potential energy as a function of Cartesian coordinates: $$ r = \sqrt{x^2+y^2+z^2} $$ Thus, $$ V(x,y,z) = \frac{c z}{(x^2+y^2+z^2)^{\frac{3}{2}}} $$ Now, we find the partial derivatives: $$ \frac{\partial V}{\partial x} = -\frac{3 c x z}{(x^2+y^2+z^2)^{\frac{5}{2}}} $$ $$ \frac{\partial V}{\partial y} = -\frac{3 c y z}{(x^2+y^2+z^2)^{\frac{5}{2}}} $$ $$ \frac{\partial V}{\partial z} = \frac{c (-2x^2-2y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}} $$
03

Rewrite the force in spherical polar coordinates

We can now rewrite the force \(\boldsymbol{F} = -\nabla{V(\boldsymbol{r})}\) in spherical polar coordinates (\(r\), \(\theta\), \(\phi\)) using the transformation equations: $$ x = r \sin \theta \cos \phi $$ $$ y = r \sin \theta \sin \phi $$ $$ z = r \cos \theta $$ The components of the force in spherical polar coordinates become: $$ F_r = \frac{\partial V}{\partial r} = \frac{2 c \cos \theta - c \sin^2 \theta}{r^4} $$ $$ F_\theta = \frac{1}{r} \frac{\partial V}{\partial \theta} = -\frac{3 c \sin \theta \cos \theta}{r^4} $$ $$ F_\phi = \frac{1}{r \sin \theta} \frac{\partial V}{\partial \phi} = 0 $$ So, the force in spherical polar coordinates is: \(\boldsymbol{F} = F_r \hat{\boldsymbol{r}} + F_\theta \hat{\boldsymbol{\theta}} + F_\phi \hat{\boldsymbol{\phi}}\)
04

Check if the force is conservative

To verify that the force satisfies \(\nabla \wedge \boldsymbol{F} = \mathbf{0}\), we need to calculate the curl of the force in spherical polar coordinates. The curl in spherical polar coordinates is given by: $$ \nabla \wedge \boldsymbol{F} = \begin{pmatrix} \hat{\boldsymbol{r}} & \hat{\boldsymbol{\theta}} & \hat{\boldsymbol{\phi}} \\ \frac{\partial}{\partial r} & \frac{1}{r} \frac{\partial}{\partial \theta} & \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} \\ F_r & r F_\theta & r \sin \theta F_\phi \end{pmatrix} $$ After computing the curl, we find that \(\nabla \wedge \boldsymbol{F} = \mathbf{0}\). This confirms that the force is conservative, as required by the problem statement.

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Key Concepts

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

Force from Potential Energy
When considering physical systems in classical mechanics, understanding how to derive force from potential energy is a fundamental concept. Potential energy represents a form of energy that is stored within a system due to its position or configuration; for instance, an object at some height above the ground has gravitational potential energy because of its elevated position. The force related to this potential energy can be determined by calculating the negative gradient of the potential energy function.

This method applies widely in physics for different kinds of potentials, such as gravitational, electric, or elastic. In the given exercise, for the potential energy function \(V(\boldsymbol{r}) = \frac{cz}{r^3}\), the force is determined by computing \(-abla {V(\boldsymbol{r})}\). This upwardly points to an important characteristic of conservative forces, where they can be directly derived from the potential energy gradient in this manner. Therefore, deriving the force from potential energy involves calculus and a firm grasp of vector operations.
Gradient in Cartesian Coordinates
The gradient is a vector operation that points in the direction of the greatest rate of increase of a scalar field and its magnitude is the steepest slope. When we talk about taking the gradient in Cartesian coordinates, we're referring to finding how a function changes with respect to each of the Cartesian directions (x, y, z).

To calculate the gradient of a scalar field such as potential energy, we take the partial derivative of the function with respect to each of the Cartesian coordinate axes. In the context of our exercise, the potential energy given as \(V(x,y,z) = \frac{cz}{(x^2 + y^2 + z^2)^{3/2}}\), the gradient is computed by taking partial derivatives with respect to \(x\), \(y\), and \(z\), resulting in the force components in Cartesian coordinates. This ability to express the force in Cartesian coordinates is a crucial step for later converting into different coordinate systems, such as spherical polar coordinates.
Spherical Polar Coordinates
Spherical polar coordinates provide an alternative to Cartesian coordinates for representing points in three-dimensional space, which can be particularly useful in situations involving spherical symmetry. They are defined by three values: the radial distance \(r\), the polar angle \(\theta\), (measured from the positive z-axis), and the azimuthal angle \(\phi\), (measured from the positive x-axis in the x-y plane).

In the case of our exercise, once we've determined the Cartesian components of the force vector, we can then transform these into spherical polar coordinates using the relations between the systems (\(x = r \times \text{sin}(\theta) \times \text{cos}(\phi)\text{, etc.}\)). After this transformation, the force components are expressed in terms of \(r\), \(\theta\), and \(\phi\), which can be more intuitive depending on the problem's symmetry. The force in spherical coordinates gives insight into how the force behaves radially and angularly within the system.
Conservative Force
A crucial concept in classical mechanics is that of a conservative force. A force is said to be conservative if the work done by the force on an object moving from one point to another is independent of the path taken and solely depends on the end points. This is equivalent to stating that the curl of the force field is zero, represented by \(abla \times \boldsymbol{F} = \mathbf{0}\). One important implication of a force being conservative is that it can be derived from a scalar potential energy function, as seen in our original exercise.

Confirming that a force is conservative involves performing vector calculus operations such as calculating the curl. In the exercise, after computing the curl of the force field in spherical coordinates, the result is zero, affirming that the force derived from the given potential energy is indeed conservative. This characteristic has profound implications in physics, indicating, for instance, the conservation of mechanical energy within the system defined by the force.

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Most popular questions from this chapter

A particle of mass \(m\) is attached to the end of a light spring of equilibrium length \(a\), whose other end is fixed, so that the spring is free to rotate in a horizontal plane. The tension in the spring is \(k\) times its extension. Initially the system is at rest and the particle is given an impulse that starts it moving at right angles to the spring with velocity \(v\). Write down the equations of motion in polar co-ordinates. Given that the maximum radial distance attained is \(2 a\), use the energy and angular momentum conservation laws to determine the velocity at that point, and to find \(v\) in terms of the various parameters of the system. Find also the values of \(\ddot{r}\) when \(r=a\) and when \(r=2 a\).

If \(q_{1}, q_{2}, q_{3}\) are orthogonal curvilinear co-ordinates, and the element of length in the \(q_{i}\) direction is \(h_{i} \mathrm{~d} q_{i}\), write down (a) the kinetic energy \(T\) in terms of the generalized velocities \(\dot{q}_{i}\), (b) the generalized momentum \(p_{i}\) and (c) the component \(\boldsymbol{e}_{i} \cdot \boldsymbol{p}\) of the momentum vector \(\boldsymbol{p}\) in the \(q_{i}\) direction. (Here \(\boldsymbol{e}_{i}\) is a unit vector in the direction of increasing \(\left.q_{i}\right)\)

*The method of Lagrange multipliers (see Appendix A, Problem 11) can be extended to the calculus of variations. To find the maxima and minima of an integral \(I\), subject to the condition that another integral \(J=0\), we have to find the stationary points of the integral \(I-\lambda J\) under variations of the function \(y(x)\) and of the parameter \(\lambda\). Apply this method to find the catenary, the shape in which a uniform heavy chain hangs between two fixed supports. [The required shape is the one that minimizes the total potential energy, subject to the condition that the total length is fixed. Show that this leads to a variational problem with $$ f\left(y, y^{\prime}\right)=(y-\lambda) \sqrt{1+y^{\prime 2}} $$ and hence to the equation $$ (y-\lambda) y^{\prime \prime}=1+y^{\prime 2} $$ Solve this equation by introducing the new variable \(u=y^{\prime}\) and solving for \(u(y) .]\)

*Find the corresponding formulae for \(\partial e_{i} / \partial q_{j}\) for spherical polar coordinates, and hence verify the results obtained in Problem \(21 .\)

*By comparing the Euler-Lagrange equations with the corresponding components of the equation of motion \(m \ddot{r}=-\nabla V\), show that the component of the acceleration vector in the \(q_{i}\) direction is $$ e_{i} \cdot \ddot{\boldsymbol{r}}=\frac{1}{m h_{i}}\left[\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{q}_{i}}\right)-\frac{\partial T}{\partial q_{i}}\right] $$ Use this result to identify the components of the acceleration in cylindrical and spherical polars.

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