Chapter 1: Q1.38P (page 42)

Express the unit vectors $r, θ, ϕ$ in terms of x, y, z (that is, deriveEq. 1.64). Check your answers several ways ( $r \cdot r \overset{?}{=} 1$ , $θ \cdot ϕ \overset{?}{= 0}$ , $r \times θ \overset{?}{= ϕ}$ ).Also work out the inverse formulas, giving x, y, z in terms of $r, θ, ϕ$ (and $θ, ϕ$ ).

Short Answer

Expert verified

The formula of $\overset{\land}{r}$ is obtained to be equal to $\overset{\land}{r} = \sin θ \cos ϕ \overset{\land}{x} + \sin θ \sin ϕ \overset{\land}{y} + \cos θ \overset{\land}{z}$ . The formula for $\overset{\land}{θ}$ is obtained as $\overset{\land}{θ} = \cos θ \cos ϕ \overset{\land}{x} + \cos θ \sin ϕ \overset{\land}{y} - \sin θ \overset{\land}{z}$ and the value of $\overset{\land}{ϕ}$ is obtained as.

$\overset{\land}{ϕ} = - \sin ϕ \overset{\land}{x} + \cos ϕ \overset{\land}{y}$

The product of $\overset{\land}{r} . \overset{\land}{r}$ , is obtained as,1 and the product $\overset{\land}{θ} . \overset{\land}{ϕ}$ is obtained as 0

The inverse formulae are obtained as $\overset{\land}{x} = \sin θ \cos ϕ \overset{\land}{r} + \cos θ \cos ϕ \overset{\land}{θ} - \sin ϕ \overset{\land}{ϕ}$ , $\overset{\land}{y} = \sin θ \sin ϕ \overset{\land}{r} + \cos θ \sin ϕ \overset{\land}{θ} + \cos ϕ \overset{\land}{ϕ}$ , $\overset{\land}{z} = \cos θ \overset{\land}{r} - \sin θ \overset{\land}{θ}$

Step by step solution

Define the spherical coordinates.

The spherical coordinates are defined in terms of $r, θ, ϕ$ , where r is the distance from origin, $θ$ is the pole angle and $ϕ$ is the azimuthal angle.

The spherical coordinates can be drawn as,

The scalar potentials is $v = \frac{\overset{\land}{r}}{r^{2}}$ and the position vector is $\vec{r} = x i + y j + z k$ . The unit vector in the direction of $\vec{r}$ , is obtained as,

$\begin{array}{rcl} \overset{\land}{r} & = & \frac{\vec{r}}{|r|} \\ = & \frac{x i + y j + z k}{|\sqrt{x^{2} + y^{2} + z^{2}}|} \end{array}$

The spherical coordinates of the system is defined as,

$x = (r \sin θ) \cos ϕ$

$y = (r \sin θ) \sin ϕ$

$z = r \cos θ$

Substitute $(r \sin θ) \cos ϕ$ for x , $(r \sin θ) \sin ϕ$ for y and $r \cos θ$ for z into

$\vec{r} = x i + y j + z k$

$\begin{array}{rcl} \vec{r} & = & x i + y j + z k \\ = & (r \sin θ) \cos ϕ i + (r \sin θ) \sin ϕ j + r \cos θ k \end{array}$

The unit vector $\overset{\land}{r}$ is obtained as $\overset{\land}{r} = \sin θ \cos ϕ \overset{\land}{x} + \sin θ \sin ϕ \overset{\land}{y} + \cos θ \overset{\land}{z}$

Step: 2 Obtain the formula for θ .

The infinitesimal displacement along the direction $θ$ , is obtained as ${\vec{d l}}_{θ} = r d θ \overset{\land}{θ}$ ……. (3)

The infinitesimal displacement along the direction $θ$ , in terms of Cartesian coordinates is written as,

${\vec{d l}}_{θ} = d x \overset{\land}{x} + d y \overset{\land}{y} + d z \overset{\land}{x}$

As $x = (r \sin θ) \cos ϕ$ , $y = (r \sin θ) \sin ϕ$ , $z = r \cos θ$ , infinitesimal displacement along the direction $θ$ , can be written as,

${\vec{d l}}_{θ} = ((r \sin θ) \cos ϕ) \overset{\land}{x} + ((r \sin θ) \sin ϕ) \overset{\land}{y} + (r \cos θ) \overset{\land}{z}$

From equation (3), ${\vec{d l}}_{θ} = r d θ \overset{\land}{θ}$

$r d θ \overset{\land}{θ} = ((r \sin θ) \cos ϕ) \overset{\land}{x} + ((r \sin θ) \sin ϕ) \overset{\land}{y} + (r \cos θ) \overset{\land}{z}$

Step: 3Obtain the formula for ϕ

The infinitesimal displacement along the direction $θ$ , is obtained as

${\vec{d l}}_{ϕ} = r \sin θ d ϕ \overset{\land}{ϕ}$ ……. (3)

The infinitesimal displacement along the direction $θ$ , in terms of Cartesian coordinates is written as,

${\vec{d l}}_{θ} = d x \overset{\land}{x} + d y \overset{\land}{y} + d z \overset{\land}{z}$

As $x = (r \sin θ) \cos ϕ$ , $y = (r \sin θ) \sin ϕ$ , $z = r \cos θ$ , infinitesimal displacement along the direction $θ$ , can be written as,

${\vec{d l}}_{θ} = ((r \sin θ) \cos ϕ) \overset{\land}{x} + ((r \sin θ) \sin ϕ) \overset{\land}{y} + (r \cos θ) \overset{\land}{z}$

From equation (3), ${\vec{d l}}_{ϕ} = r \sin θ d ϕ \overset{\land}{ϕ}$

$\begin{array}{rcl} r \sin θ d ϕ \overset{\land}{ϕ} & = & - ((r \sin θ) \cos ϕ) \overset{\land}{x} + ((r \sin θ) \sin ϕ) \overset{\land}{y} \\ \overset{\land}{ϕ} & = & - \sin ϕ \overset{\land}{x} + \cos ϕ \overset{\land}{y} \end{array}$

Step: 4 Check the products

The product of $\overset{\land}{r} . \overset{\land}{r}$ , is calculated as,

$\begin{array}{rcl} \overset{\land}{r} . \overset{\land}{r} & = & \sin^{2} θ (\cos^{2} ϕ + \sin^{2} ϕ) + \cos^{2} θ \\ = & \sin^{2} θ + \cos^{2} θ \\ = & 1 \end{array}$

Multiply the vectors $\overset{\land}{θ}$ and $\overset{\land}{ϕ}$

$\begin{array}{rcl} \overset{\land}{θ} . \overset{\land}{ϕ} & = & - \cos θ \sin ϕ \cos ϕ + \cos θ \sin ϕ \cos ϕ \\ = & 0 \end{array}$

Step: 5 Find the value of x∧,y∧,z∧

As $x = (r \sin θ) \cos ϕ$ , $y = (r \sin θ) \sin ϕ$ , $z = r \cos θ$ , the position vector

$\overset{\land}{r} = (r \sin θ) \cos \overset{\land}{x} + \sin θ \sin ϕ \overset{\land}{y} + \cos θ \overset{\land}{z}$

Multiply above equation by $\sin θ$ on both sides,

$\sin θ \overset{\land}{r} = \sin^{2} θ \cos \overset{\land}{x} + \sin^{2} θ \sin ϕ \overset{\land}{y} + \sin θ \cos θ \overset{\land}{z}$ ……. (1)

Now the theta vector is $\overset{\land}{θ} = \cos θ \cos ϕ \overset{\land}{x} + \cos θ \sin ϕ \overset{\land}{y} - \sin θ \overset{\land}{z}$

Multiply above equation by $\cos ϕ$ on both sides,

$\cos θ \overset{\land}{θ} = \cos^{2} θ \cos ϕ \overset{\land}{x} + \cos^{2} θ \sin ϕ \overset{\land}{y} - \sin θ \cos θ \overset{\land}{z}$ ……. (2)

Add equations (1) and (2) as,

$\begin{array}{rcl} \sin θ \overset{\land}{r} + \cos θ \overset{\land}{θ} & = & \sin^{2} θ \cos ϕ \overset{\land}{x} + \sin^{2} θ \sin ϕ \overset{\land}{y} + \sin θ \cos θ \overset{\land}{z} + \cos^{2} θ \cos ϕ \overset{\land}{x} + \cos^{2} θ \sin ϕ \overset{\land}{y} - \sin θ \cos θ \overset{\land}{z} \\ = & \sin^{2} θ \cos ϕ \overset{\land}{x} + \sin^{2} θ \sin ϕ \overset{\land}{y} + \cos^{2} θ \cos ϕ \overset{\land}{x} + \cos^{2} θ \sin ϕ \overset{\land}{y} \\ = & (\sin^{2} θ + \cos^{2} θ \overset{\land}{x}) \cos ϕ \overset{}{x + (\sin^{2} θ + \cos^{2} θ) \sin ϕ \overset{\land}{y}} \\ = & \cos ϕ \overset{\land}{x} + \sin ϕ \overset{\land}{y} \end{array}$

solve further as,

$\overset{\land}{ϕ} = - \sin ϕ \overset{\land}{x} + \cos ϕ \overset{\land}{y}$

Multiply $\sin θ \overset{\land}{r} + \cos θ \overset{\land}{θ} = \cos ϕ \overset{\land}{x} + \sin ϕ \overset{\land}{y}$ by $\cos ϕ$ on both sides,

$\sin θ \cos ϕ \overset{\land}{r} + \cos θ \cos ϕ \overset{\land}{θ} = \cos^{2} ϕ \overset{\land}{x} + \sin ϕ \cos ϕ \overset{\land}{y}$ ……. (3)

Multiply $\overset{\land}{ϕ} = - \sin ϕ \overset{\land}{x} + \cos ϕ \overset{\land}{y}$ by $\sin ϕ$ on both sides,

$\sin ϕ \overset{\land}{ϕ} = - \sin^{2} ϕ \overset{\land}{x} + \cos ϕ \sin ϕ \overset{\land}{y}$ ……. (4)

Subtract equation (4) from equation (3).

$\begin{array}{rcl} \sin θ \cos ϕ \overset{\land}{r} + \cos θ \cos ϕ \overset{\land}{θ} - \sin ϕ \overset{\land}{ϕ} & = & \cos^{2} ϕ \overset{\land}{x} + \sin ϕ \cos ϕ \overset{\land}{y} - \sin ϕ \cos ϕ \overset{\land}{y} + \sin^{2} ϕ \overset{\land}{x} \\ = & \cos^{2} ϕ \overset{\land}{x} + \sin^{2} ϕ \overset{\land}{x} \\ = & \overset{\land}{x} \end{array}$

Thus, $\overset{\land}{x} = \sin θ \cos ϕ \overset{\land}{r} + \cos θ \cos ϕ \overset{\land}{θ} - \sin ϕ \overset{\land}{ϕ}$

Multiply $\sin θ \overset{\land}{r} + \cos θ \overset{\land}{θ} = \cos ϕ \overset{\land}{x} + \sin ϕ \overset{\land}{y}$ by $\sin ϕ$ on both sides,

$\sin θ \sin ϕ \overset{\land}{r} + \cos θ \sin ϕ \overset{\land}{θ} = \cos ϕ \sin ϕ \overset{\land}{x} + \sin^{2} ϕ \overset{\land}{y}$ ……. (5)

Multiply $\overset{\land}{ϕ} = - \sin ϕ \overset{\land}{x} + \cos ϕ \overset{\land}{y}$ by $\cos ϕ$ on both sides,

$\cos ϕ \overset{\land}{ϕ} = - \sin ϕ \cos ϕ \overset{\land}{x} + \cos^{2} ϕ \overset{\land}{y}$ ……. (5)

Add equation (5) and (6).

$\begin{array}{rcl} \sin θ \sin ϕ \overset{\land}{r} + \cos θ \sin ϕ \overset{\land}{θ} + \cos ϕ \overset{\land}{ϕ} & = & \cos ϕ \sin ϕ \overset{\land}{x} + \sin^{2} ϕ \overset{\land}{y} - \sin ϕ \cos ϕ \overset{\land}{x} + \cos^{2} ϕ \overset{\land}{y} \\ = & \sin^{2} ϕ \overset{\land}{y} + \cos^{2} ϕ \overset{\land}{y} \\ = & \overset{\land}{y} \end{array}$

Thus, $\overset{\land}{y} = \begin{array}{rcl} \sin \end{array} \begin{array}{rcl} θ \end{array} \begin{array}{rcl} \sin \end{array} \begin{array}{rcl} ϕ \end{array} \begin{array}{rcl} \overset{\land}{r} \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \cos \end{array} \begin{array}{rcl} θ \end{array} \begin{array}{rcl} \sin \end{array} \begin{array}{rcl} ϕ \end{array} \begin{array}{rcl} \overset{\land}{θ} \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \cos \end{array} \begin{array}{rcl} ϕ \end{array} \begin{array}{rcl} \overset{\land}{ϕ} \end{array}$

As $x = (r \sin θ) \cos ϕ$ , $y = (r \sin θ) \sin ϕ$ , $z = r \cos θ$ , the position vector is

$\overset{}{r} = (r \sin θ) \cos \overset{\land}{x} + \sin θ \sin ϕ \overset{\land}{y} + \cos θ \overset{\land}{z}$

Multiply above equation by $\cos θ$ on both sides,

$\cos θ \overset{\land}{r} = \sin θ \cos θ \cos \overset{\land}{x} + \sin θ \cos θ \sin ϕ \overset{\land}{y} + \cos^{2} θ \overset{\land}{z}$ ……. (6)

Now the theta vector is $\overset{\land}{θ} = = \cos θ \cos ϕ \overset{\land}{x} + \cos θ \sin ϕ \overset{\land}{y} - \sin θ \overset{\land}{z}$

Multiply above equation by $\sin θ$ on both sides,

$\sin θ \overset{\land}{θ} = \sin θ \cos θ \cos ϕ \overset{\land}{x} + \sin θ \cos θ \sin ϕ \overset{\land}{y} - \sin^{2} θ \overset{\land}{z}$ ……. (7)

Subtract equation (7) from equation (8) as,

$\begin{array}{rcl} \cos θ \overset{\land}{r} - \sin θ \overset{\land}{θ} & = & [\sin θ \cos θ \cos \overset{\land}{x} + \sin θ \cos θ \sin ϕ \overset{\land}{y} + \cos^{2} θ \overset{\land}{z} - \sin θ \cos θ \cos ϕ \overset{\land}{x} - \sin θ \cos θ \sin ϕ \overset{\land}{y} + \sin^{2} θ \overset{\land}{z}] \\ = & \cos^{2} θ \overset{\land}{z} + \sin^{2} θ \overset{\land}{z} \\ = & \overset{\land}{z} \end{array}$

Thus, $\overset{\land}{z} = \cos θ \overset{\land}{r} - \sin θ \overset{\land}{θ}$

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Short Answer

Step by step solution

Define the spherical coordinates.

Step: 2 Obtain the formula for θ .

Step: 3Obtain the formula for ϕ

Step: 4 Check the products

Step: 5 Find the value of x∧,y∧,z∧

One App. One Place for Learning.

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Express the unit vectors r, θ, ϕin terms of x, y, z (that is, deriveEq. 1.64). Check your answers several ways ( r·r=?1, θ·ϕ=0?, r×θ=ϕ?).Also work out the inverse formulas, giving x, y, z in terms of r, θ, ϕ (and θ, ϕ).

Short Answer

Step by step solution

Define the spherical coordinates.

Step: 2 Obtain the formula for θ .

Step: 3Obtain the formula for ϕ

Step: 4 Check the products

Step: 5 Find the value of x∧,y∧,z∧

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Rotational Dynamics

Geometrical and Physical Optics

Nuclear Physics

Particle Model of Matter

Electricity and Magnetism

Study anywhere. Anytime. Across all devices.