1.38P

Question

Express the unit vectors in terms of x, y, z (that is, derive Eq. 1.64). Check your answers several ways ( $r . r =$ ^?1, $θ . ϕ =$ ^? $r x θ =$ ^? $ϕ$ ), .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 r is obtained to be equal to

r = $\sin θ \cos ϕ x^{\land} + \sin θ \sin ϕ y^{\land} + \cos θ z^{\land} .$ The formula $θ^{\land}$ is obtain as

$θ^{\land} = \cos θ \cos ϕ x^{\land} + \cos θ \sin ϕ y^{\land} - θ z^{\land}$ and the value of $θ^{\land}$ is obtain as

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

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

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

1Define the spherical coordinates.

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

The spherical coordinates can be drawn as,

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

$r^{\land} = {\frac{\overset{⇀}{r}}{|r|}}_{} = \frac{x i + y j + z k}{|\sqrt{x^{2} + y^{2} + z^{2}}|}$

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

$\overset{⇀}{r} = x i + y j + z k . \overset{⇀}{r} = x i + y j + z k = (r \sin θ) \cos ϕ + (r \sin θ) \sin ϕ + r \cos θ k$

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

2Obtain the formula for θ .

The infinitesimal displacement along the direction, is obtained as

$\overset{⇀}{d I}_{θ_{}} = r d θ θ^{\land}$ ……. (3)

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

${\overset{⇀}{d I}}_{θ} = d x x^{\land} + d y y^{\land} + d z z^{\land}$

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,

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

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

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

3Obtain the formula for ϕ

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

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

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

${\vec{d I}}_{θ} = d x x^{\land} + d y y^{\land} + d z z^{\land}$

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,

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

From equation (3), ${\vec{d l}}_{θ} = r \sin θ d ϕ ϕ^{\land} r \sin θ d ϕ ϕ^{\land} = ((r \sin θ) \cos ϕ) x^{\land} + ((r \sin θ) \sin ϕ) y^{\land} ϕ^{\land} = - \sin ϕ x^{\land} + \cos ϕ y^{\land}$

4Check the products

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

$r^{\land} . r^{\land} = \sin^{2} θ (\cos^{2} ϕ + \sin^{2} ϕ) + \cos^{2} θ = \sin^{2} θ + \cos^{2} θ = 1$

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

$θ^{\land} . ϕ^{\land} = - \cos θ \sin ϕ \cos ϕ + \cos θ \sin ϕ \cos ϕ = 0$

5Find 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

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

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

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

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

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

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

Add equations (1) and (2) as,

$\sin θ r^{\land} + \cos θ θ^{\land} = \sin^{2} θ \cos ϕ x^{\land} + \sin^{2} θ \sin ϕ y^{\land} + \sin θ \cos θ z^{\land} + \cos^{2} θ \cos ϕ x^{\land} + \cos^{2} θ \sin ϕ y^{\land} - \sin θ \cos ϕ = \sin^{2} θ \cos ϕ x^{\land} + \sin^{2} θ \sin ϕ y^{\land} + \cos^{2} θ \cos ϕ x^{\land} + \cos^{2} θ \sin ϕ y^{\land} = [\begin{matrix} \sin^{2} θ + & \cos^{2} θ x^{\land} \end{matrix}] \cos θ x^{\land} + (\sin^{2} θ + c o s^{2} θ) \sin θ y^{\land} = \cos θ x^{\land} + \sin θ y^{\land}$

solve further as,

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

Multiply $\sin θ r^{\land} + \cos θ θ^{\land} = \cos ϕ x^{\land} + \sin ϕ y^{\land}$ by $\sin ϕ$ on both sides, $\sin θ \cos φ r^{\land} + \cos θ \cos ϕ θ^{\land} = \cos^{2} ϕ x^{\land} + \sin ϕ \cos ϕ y^{\land}$ ……. (3)

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

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

Subtract equation (4) from equation (3).

$\sin θ \cos ϕ r^{\land} + \cos θ \cos ϕ θ^{\land} - \sin ϕ ϕ^{\land} = \cos^{2} ϕ x^{\land} + \sin ϕ \cos ϕ y^{\land} - \sin ϕ \cos ϕ y^{\land} + \sin^{2} ϕ x^{\land} = \cos^{2} ϕ x^{\land} + \sin^{2} ϕ x^{\land} = x^{\land}$

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

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

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

Multiply $ϕ^{\land} = - \sin ϕ \cos ϕ x^{\land} + \cos^{2} ϕ y^{\land}$ by $\cos ϕ$ on both sides, $\cos ϕ ϕ^{\land} = - \sin ϕ \cos ϕ x^{\land} + \cos^{2} ϕ y^{\land}$ ……. (5)

Add equation (5) and (6).

$\sin θ \sin ϕ r^{\land} + \cos θ \sin ϕ θ^{\land} + \cos ϕ ϕ^{\land} = \cos ϕ ϕ^{\land} \sin ϕ x^{\land} + \sin^{2} ϕ y^{\land} - \sin ϕ \cos ϕ x^{\land} + \cos^{2} ϕ y^{\land} = \sin^{2} ϕ y^{\land} + \cos^{2} ϕ y^{\land} = y^{\land}$