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By integrating the element of solid angle \((14.15), d \Omega=\sin \theta d \theta d \phi\), over all directions, verify that the solid angle corresponding to all directions is 4 \(\pi\) steradians.

Short Answer

Expert verified
The solid angle encompassing all directions is 4\(\pi\) steradians.

Step by step solution

01

Understand the Element of Solid Angle

The element of solid angle in spherical coordinates is given by \( d\Omega = \sin \theta \, d\theta \, d\phi \). This describes a small patch of area on the unit sphere, which is determined by angles \( \theta \) (polar angle) and \( \phi \) (azimuthal angle). We are tasked to integrate over all possible directions to find the total solid angle.
02

Set Integration Limits

To find the total solid angle, we need to integrate over all possible values of \( \theta \) and \( \phi \). For \( \theta \), which is the angle from the positive z-axis, it ranges from \( 0 \) to \( \pi \). For \( \phi \), which is the angle in the xy-plane from the positive x-axis, it ranges from \( 0 \) to \( 2\pi \).
03

Integrate Over \( \phi \)

The integral over \( \phi \) from 0 to \( 2\pi \) is straightforward as \( \phi \) does not appear in the expression for \( d\Omega \). Evaluating the integral we have: \[ \int_0^{2\pi} d\phi = 2\pi\]
04

Integrate Over \( \theta \)

Next, integrate the remaining part over \( \theta \): \[ \int_0^{\pi} \sin\theta \, d\theta \] The antiderivative of \( \sin\theta \) is \( -\cos\theta \). Therefore, substituting the limits, we get: \[ \left[ -\cos\theta \right]_0^{\pi} = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2\]
05

Multiply Results from Step 3 and Step 4

Now, multiply the results of the two integrals to find the total solid angle: \[ \text{Total solid angle} = (2\pi) \times (2) = 4\pi \, \text{steradians}\]
06

Conclusion: Verification Complete

The calculated total solid angle of \( 4\pi \) steradians confirms the known fact that the solid angle encompassing all directions in space is indeed \( 4\pi \) steradians.

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Key Concepts

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

Spherical Coordinates
Spherical coordinates are an essential tool in dealing with 3D problems, especially when spherical symmetry is involved. In contrast to Cartesian coordinates which use \(x, y, z\) representations, spherical coordinates describe a point in space using three values: the radial distance \(r\), the polar angle \(\theta\), and the azimuthal angle \(\phi\). This is particularly useful in physics and engineering when analyzing phenomena involving spheres, cylinders, cones, etc.
  • The radial distance \(r\) measures how far a point is from the origin, similar to the radius of a sphere.
  • The polar angle \(\theta\) measures the angle from the positive z-axis down towards the xy-plane.
  • The azimuthal angle \(\phi\) represents the angle in the xy-plane from the positive x-axis.

Spherical coordinates simplify the mathematics of certain integrals, like the one you're exploring for the solid angle integration.
Integration Limits
When integrating to find total solid angles, it's crucial to set the correct integration limits. In spherical coordinates, these limits correspond to full sweeps of the angles \(\theta\) and \(\phi\).
  • The polar angle \(\theta\) varies from 0 to \(\pi\). This range covers from the top of the sphere (the positive z-axis) to the bottom (negative z-axis).
  • The azimuthal angle \(\phi\) changes from 0 to \(2\pi\), completing a full circle in the xy-plane.

These specific limits ensure that all possible directions on a unit sphere are considered, thereby covering an entire solid angle as required.
Unit Sphere
The concept of a unit sphere is pivotal in understanding the representation of solid angles in spherical coordinates. A unit sphere is simply a sphere with a radius of one unit. This makes calculations simpler and results more generalizable.
Here's why it's important:
  • All points on a unit sphere are equidistant from the origin, simplifying distance-related calculations.
  • On a unit sphere, any area element can directly be described with the angular measurements only, thus the element of solid angle is linked to the angles alone.
  • This concept is utilized frequently in physics to normalize equations and quantify directional quantities like radiation or field intensities using steradians.

Utilizing a unit sphere helps in creating standard solutions that are easily scalable and applicable across various scenarios.
Polar Angle
The polar angle \(\theta\) is a crucial component of spherical coordinates, as it describes the angle from the positive z-axis of the sphere. Understanding \(\theta\) is necessary for integrating areas on a sphere accurately.
  • It varies between \(0\) (pointing directly at the positive z-axis) to \(\pi\) (directly down the negative z-axis).
  • As \(\theta\) increases, points are angled progressively downwards away from the z-axis towards the xy-plane.
  • This measurement is critical when determining areas and volumes involved in 3D space revolving around a center point.

In the context of integrating the solid angle, the correct limits of \(\theta\) are essential for encompassing the entire sphere.
Azimuthal Angle
The azimuthal angle \(\phi\) is another essential angle in spherical coordinates, representing the direction in the xy-plane. This angle facilitates a full rotational perspective around the z-axis.
  • This angle ranges from 0 to \(2\pi\), allowing for a complete 360-degree rotation around the z-axis.
  • At \(\phi = 0\), the direction is aligned with the positive x-axis, and progressive increases in \(\phi\) rotate the point counterclockwise in the xy-plane.
  • The azimuthal angle helps determine positions and directions in a plane, making it vital for tasks involving full spatial orientation.
    The limits for \(\phi\) ensure that integration covers all horizontal directions, completing the calculation for the total solid angle.

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Most popular questions from this chapter

[Computer] Consider a point projectile moving in a fixed, spherical force whose potential energy is $$U(r)=\left\\{\begin{array}{ll} -U_{0} & (0 \leq r \leq R) \\ 0 & (R < r) \end{array}\right.\quad\quad\quad\quad\quad\quad(14.60)$$ where \(U_{\mathrm{o}}\) is a positive constant. This so-called spherical well represents a projectile which moves freely in either of the regions \(r < R\) and \(R < r_{1}\) but, when it crosses the boundary \(r=R_{1}\) receives a radially inward impulse that changes its kinetic energy by \(\pm U_{\mathrm{o}}\left(+U_{\mathrm{o}}\right.\) going inward, \(-U_{\circ}\) going outward). (a) Sketch the orbit of a projectile that approaches the well with momentum \(p_{\mathrm{o}}\) and impact parameter \(b < R .\) (b) Use conservation of energy to find the momentum \(p\) of the projectile inside the well \((r < R)\). Let \(\zeta\) denote the momentum ratio \(\zeta=p_{\mathrm{o}} / p\) and let \(d\) denote the projectile's distance of closest approach to the origin. Use conservation of angular momentum to show that \(d=\zeta b\). (c) Use your sketch to prove that the scattering angle \(\theta\) is $$\theta=2\left(\arcsin \frac{b}{R}-\arcsin \frac{\zeta b}{R}\right).\quad\quad\quad\quad\quad\quad(14.60)$$ This gives \(\theta\) as a function of \(b\), which is what you need to get the cross section. The relation depends on the momentum ratio \(\zeta,\) which in turn depends on the incoming momentum \(p_{\mathrm{o}}\) and the well depth \(U_{\mathrm{o}} .\) Plot \(\theta\) as a function of \(b\) for the case that \(\zeta=0.5 .\) ( \(\mathbf{d}\) ) By differentiating \(\theta\) with respect to \(b\), find an expression for the differential cross section as a function of \(b\), and make a plot of \(d \sigma / d \Omega\) against \(\theta\) for the case that \(\zeta=0.5 .\) Comment. [Hint: To plot as a function of \(\theta\) you don't need to solve for \(b\) in terms of \(\theta ;\) instead, you can make a parametric plot of the point \((\theta, d \sigma / d \Omega)\) as a function of the parameter b running from 0 to \(R\).] (e) By integrating \(d \sigma / d \Omega\) over all directions, find the total cross section.

Our definition of the scattering cross section, \(N_{\mathrm{sc}}=N_{\text {inc }} n_{\text {tar }} \sigma,\) applies to an experiment using a narrow beam of projectiles all of which pass through a wide target assembly. Experimenters sometimes use a wide incident beam, which completely engulfs a small target assembly (the beam of photons from a car's headlamp, directed at a small piece of plastic, for example). Show that in this case \(N_{\mathrm{sc}}=n_{\text {inc }} N_{\mathrm{tar}} \sigma\) where \(n_{\text {inc }}\) is the density (number/area) of the incident beam, viewed head-on, and \(N_{\text {tar }}\) is the total number of targets in the target assembly.

In their famous experiment, Rutherford's assistants, Geiger and Marsden, detected the scattered alpha particles using a zinc sulphide screen, which produced a tiny flash of light when struck by an alpha particle. If their screen had area \(1 \mathrm{mm}^{2}\) and was \(1 \mathrm{cm}\) from the target, what solid angle did it subtend?

The cross section for scattering a certain nuclear particle by a nitrogen nucleus is 0.5 barns. If \(10^{11}\) of these particles are fired through a cloud chamber of length \(10 \mathrm{cm},\) containing nitrogen at STP, how many particles are scattered? (Use the ideal gas law and remember that each nitrogen molecule has two atoms. The scattering by any atomic electrons is completely negligible.)

A beam of particles is directed through a tank of liquid hydrogen. If the tank's length is \(50 \mathrm{cm}\) and the liquid density is 0.07 gram/cm \(^{3}\), what is the target density (number/area) of hydrogen atoms seen by the incident particles?

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