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

Short Answer

Expert verified
Question: Compute the partial derivatives of the spherical polar basis vectors \(\frac{\partial e_i}{\partial q_j}\), and verify the results obtained in Problem 21. Answer: We computed the following partial derivatives: - With respect to \(r\): $$\frac{\partial e_r}{\partial r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, \quad \frac{\partial e_\theta}{\partial r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, \quad \frac{\partial e_\phi}{\partial r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ - With respect to \(\theta\): $$\frac{\partial e_r}{\partial \theta} = \begin{pmatrix} \cos \theta \cos \phi \\ \cos \theta \sin \phi \\ -\sin \theta \end{pmatrix}, \quad \frac{\partial e_\theta}{\partial \theta} = \begin{pmatrix} -\sin \theta \cos \phi \\ -\sin \theta \sin \phi \\ -\cos \theta \end{pmatrix}, \quad \frac{\partial e_\phi}{\partial \theta} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$ - With respect to \(\phi\): $$\frac{\partial e_r}{\partial \phi} = \begin{pmatrix} -\sin \theta \sin \phi \\ \sin \theta \cos \phi \\ 0 \end{pmatrix}, \quad \frac{\partial e_\theta}{\partial \phi} = \begin{pmatrix} -\cos \theta \sin \phi \\ \cos \theta \cos \phi \\ 0 \end{pmatrix}, \quad \frac{\partial e_\phi}{\partial \phi} = \begin{pmatrix} -\cos \phi \\ -\sin \phi \\ 0 \end{pmatrix}$$ These results are in agreement with those found in Problem 21, and thus, we have successfully computed the partial derivatives of the spherical polar basis vectors and verified the results.

Step by step solution

01

Understand Spherical Polar Coordinates and Basis Vectors

Spherical polar coordinates represent a point in space as a tuple \((r, \theta, \phi)\), where \(r\) is the radial distance from the origin, \(\theta\) is the polar angle (angle from the positive \(z\)-axis), and \(\phi\) is the azimuthal angle (angle from the positive \(x\)-axis, measured in the \(xy\) plane). In this coordinate system, the basis vectors also depend on the coordinates, and are expressed as: - \(e_r\): radial basis vector, pointing outwards from the origin, parallel to the radial line. - \(e_\theta\): polar angle basis vector, tangential to the polar angle circle, in the direction of increasing \(\theta\). - \(e_\phi\): azimuthal basis vector, tangential to the circle of constant \(r\) and \(\theta\), in the direction of increasing \(\phi\).
02

Write Cartesian Representation of Spherical Polar Basis Vectors

We'll first write the Cartesian representation of the spherical polar basis vectors: $$ e_r = \begin{pmatrix} \sin \theta \cos \phi \\ \sin \theta \sin \phi \\ \cos \theta \end{pmatrix}, \quad e_\theta = \begin{pmatrix} \cos \theta \cos \phi \\ \cos \theta \sin \phi \\ -\sin \theta \end{pmatrix}, \quad e_\phi = \begin{pmatrix} -\sin \phi \\ \cos \phi \\ 0 \end{pmatrix} $$
03

Compute Partial Derivatives of \(e_i\) with respect to \(q_j\)

Now we'll compute the partial derivatives of the basis vectors \(e_i\) (\(e_r, e_\theta, e_\phi\)) with respect to the spherical polar coordinates \(q_j\) (\(r, \theta, \phi\)), using the chain rule: $$ \frac{\partial e_i}{\partial q_j} = \frac{\partial e_i}{\partial x_k} \frac{\partial x_k}{\partial q_j} $$ where \(x_k\) are the Cartesian coordinates \((x, y, z)\), and the sums are taken over the repeated index \(k\). We'll compute this for all possible combinations of \(i\) and \(j\).
04

Calculate \(\frac{\partial e_r}{\partial r}, \frac{\partial e_\theta}{\partial r}, \frac{\partial e_\phi}{\partial r}\)

First, we'll compute the derivatives with respect to \(r\): $$ \frac{\partial e_r}{\partial r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, \quad \frac{\partial e_\theta}{\partial r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, \quad \frac{\partial e_\phi}{\partial r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
05

Calculate \(\frac{\partial e_r}{\partial \theta}, \frac{\partial e_\theta}{\partial \theta}, \frac{\partial e_\phi}{\partial \theta}\)

Next, we'll compute the derivatives with respect to \(\theta\): $$ \frac{\partial e_r}{\partial \theta} = \begin{pmatrix} \cos \theta \cos \phi \\ \cos \theta \sin \phi \\ -\sin \theta \end{pmatrix}, \quad \frac{\partial e_\theta}{\partial \theta} = \begin{pmatrix} -\sin \theta \cos \phi \\ -\sin \theta \sin \phi \\ -\cos \theta \end{pmatrix}, \quad \frac{\partial e_\phi}{\partial \theta} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
06

Calculate \(\frac{\partial e_r}{\partial \phi}, \frac{\partial e_\theta}{\partial \phi}, \frac{\partial e_\phi}{\partial \phi}\)

Finally, we'll compute the derivatives with respect to \(\phi\): $$ \frac{\partial e_r}{\partial \phi} = \begin{pmatrix} -\sin \theta \sin \phi \\ \sin \theta \cos \phi \\ 0 \end{pmatrix}, \quad \frac{\partial e_\theta}{\partial \phi} = \begin{pmatrix} -\cos \theta \sin \phi \\ \cos \theta \cos \phi \\ 0 \end{pmatrix}, \quad \frac{\partial e_\phi}{\partial \phi} = \begin{pmatrix} -\cos \phi \\ -\sin \phi \\ 0 \end{pmatrix} $$
07

Verify the Results from Problem 21

Now we will verify the results obtained in Problem 21 by comparing the partial derivatives found above. If our results agree with those in Problem 21, then we have successfully verified the results.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

*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.

*A light rigid cylinder of radius \(2 a\) is able to rotate freely about its axis, which is horizontal. A particle of mass \(m\) is fixed to the cylinder at a distance \(a\) from the axis and is initially at rest at its lowest point. A light string is wound on the cylinder, and a steady tension \(F\) applied to it. Find the angular acceleration and angular velocity of the cylinder after it has turned through an angle \(\theta\). Show that there is a limiting tension \(F_{0}\) such that if \(F\) \(F_{0}\) it continues to accelerate. Estimate the value of \(F_{0}\) by numerical approximation.

*Parabolic co-ordinates \((\xi, \eta)\) in a plane are defined by \(\xi=r+x, \eta=\) \(r-x\). Find \(x\) and \(y\) in terms of \(\xi\) and \(\eta\). Show that the kinetic energy of a particle of mass \(m\) is $$ T=\frac{m}{8}(\xi+\eta)\left(\frac{\dot{\xi}^{2}}{\xi}+\frac{\dot{\eta}^{2}}{\eta}\right) $$ Hence find the equations of motion.

*Find the geodesics on a sphere of unit radius. [Hint: Use \(\theta\) as independent variable, and look for the path \(\varphi=\varphi(\theta)\). To perform the integration, use the substitution \(x=\cot \theta .]\)

A wedge-shaped block of mass \(M\) rests on a smooth horizontal table. A small block of mass \(m\) is placed on its upper face, which is also smooth and inclined at an angle \(\alpha\) to the horizontal. The system is released from rest. Write down the horizontal component of momentum, and the kinetic energy of the system, in terms of the velocity \(v\) of the wedge and the velocity \(u\) of the small block relative to it. Using conservation of momentum and the equation for the rate of change of kinetic energy, find the accelerations of the blocks. Given that \(M=1 \mathrm{~kg}\) and \(m=\) \(250 \mathrm{~g}\), find the angle \(\alpha\) that will maximize the acceleration of the wedge.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free