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 ).

The formula of r∧ is obtained to be equal to r∧=sinθcosϕ x∧+sinθsinϕy∧+cosθz∧ . The formula for θ∧ is obtained as θ∧=cosθcosϕx∧+cosθsinϕ y∧ -sin&#952...

Q1.38P - Introduction to Electrodynamics

Q1.38P

Question

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 $θ, ϕ$ ).

Step-by-Step Solution

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Answer

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

1Step 1: 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 $\overset{\land}{r} = \sin θ \cos ϕ \overset{\land}{x} + \sin θ \sin ϕ \overset{\land}{y} + \cos θ \overset{\land}{z}$

2Step: 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), ${\vec{d l}}_{θ} = r d θ \overset{\land}{θ}$

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

3Step: 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), ${\vec{d l}}_{ϕ} = r \sin θ d ϕ \overset{\land}{ϕ}$

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

4Step: 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 $\overset{\land}{ϕ}$

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

5Step: 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, $\overset{\land}{z} = \cos θ \overset{\land}{r} - \sin θ \overset{\land}{θ}$

1.38P

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