Answer on Question #51712 - Math - Vector Calculus
V ( x / r 3 ) = ? , V ( r ⃗ / r 3 ) = ? \mathcal {V} (x / r ^ {3}) = ?, \qquad \mathcal {V} (\vec {r} / r ^ {3}) = ? V ( x / r 3 ) = ? , V ( r / r 3 ) = ?
where
r ⃗ = x i ⃗ + y j ⃗ + z k ⃗ \vec {r} = x \vec {i} + y \vec {j} + z \vec {k} r = x i + y j + z k
**Solution:**
V ( x / r 3 ) \mathcal{V}(x / r^3) V ( x / r 3 )
V ( x r 3 ) = [ ∂ ∂ x ( x r 3 ) ] i ⃗ + [ ∂ ∂ y ( x r 3 ) ] j ⃗ + [ ∂ ∂ z ( x r 3 ) ] k ⃗ \mathcal {V} \left(\frac {x}{r ^ {3}}\right) = \left[ \frac {\partial}{\partial x} \left(\frac {x}{r ^ {3}}\right) \right] \vec {i} + \left[ \frac {\partial}{\partial y} \left(\frac {x}{r ^ {3}}\right) \right] \vec {j} + \left[ \frac {\partial}{\partial z} \left(\frac {x}{r ^ {3}}\right) \right] \vec {k} V ( r 3 x ) = [ ∂ x ∂ ( r 3 x ) ] i + [ ∂ y ∂ ( r 3 x ) ] j + [ ∂ z ∂ ( r 3 x ) ] k
Since r = x 2 + y 2 + z 2 r = \sqrt{x^2 + y^2 + z^2} r = x 2 + y 2 + z 2
∂ ∂ x ( x r 3 ) = ∂ ∂ x ( x ) r 3 + x ∂ ∂ x ( 1 r 3 ) = = 1 r 3 + x ∂ ∂ x ( 1 ( x 2 + y 2 + z 2 ) 3 ) = = 1 r 3 − x 3 2 ∂ ∂ x ( x 2 + y 2 + z 2 ) ( x 2 + y 2 + z 2 ) 5 = 1 r 3 − 3 x 2 ( x 2 + y 2 + z 2 ) 5 = 1 r 3 − 3 x 2 r 5 \begin{array}{l}
\frac {\partial}{\partial x} \left(\frac {x}{r ^ {3}}\right) = \frac {\frac {\partial}{\partial x} (x)}{r ^ {3}} + x \frac {\partial}{\partial x} \left(\frac {1}{r ^ {3}}\right) = \\
= \frac {1}{r ^ {3}} + x \frac {\partial}{\partial x} \left(\frac {1}{\left(\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}\right) ^ {3}}\right) = \\
= \frac {1}{r ^ {3}} - x \frac {3}{2} \frac {\frac {\partial}{\partial x} \left(x ^ {2} + y ^ {2} + z ^ {2}\right)}{\left(\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}\right) ^ {5}} = \frac {1}{r ^ {3}} - \frac {3 x ^ {2}}{\left(\sqrt {x ^ {2} + y ^ {2} + z ^ {2}}\right) ^ {5}} = \frac {1}{r ^ {3}} - \frac {3 x ^ {2}}{r ^ {5}} \\
\end{array} ∂ x ∂ ( r 3 x ) = r 3 ∂ x ∂ ( x ) + x ∂ x ∂ ( r 3 1 ) = = r 3 1 + x ∂ x ∂ ( ( x 2 + y 2 + z 2 ) 3 1 ) = = r 3 1 − x 2 3 ( x 2 + y 2 + z 2 ) 5 ∂ x ∂ ( x 2 + y 2 + z 2 ) = r 3 1 − ( x 2 + y 2 + z 2 ) 5 3 x 2 = r 3 1 − r 5 3 x 2
Similarly
∂ ∂ y ( x r 3 ) = x ∂ ∂ y ( 1 r 3 ) = − 3 y x r 5 \frac {\partial}{\partial y} \left(\frac {x}{r ^ {3}}\right) = x \frac {\partial}{\partial y} \left(\frac {1}{r ^ {3}}\right) = - \frac {3 y x}{r ^ {5}} ∂ y ∂ ( r 3 x ) = x ∂ y ∂ ( r 3 1 ) = − r 5 3 y x ∂ ∂ z ( x r 3 ) = x ∂ ∂ z ( 1 r 3 ) = − 3 z x r 5 \frac {\partial}{\partial z} \left(\frac {x}{r ^ {3}}\right) = x \frac {\partial}{\partial z} \left(\frac {1}{r ^ {3}}\right) = - \frac {3 z x}{r ^ {5}} ∂ z ∂ ( r 3 x ) = x ∂ z ∂ ( r 3 1 ) = − r 5 3 z x
Substituting these into the first equation we obtain
V ( x r 3 ) = [ ∂ ∂ x ( x r 3 ) ] i ⃗ + [ ∂ ∂ y ( x r 3 ) ] j ⃗ + [ ∂ ∂ z ( x r 3 ) ] k ⃗ = ( 1 r 3 − 3 x 2 r 5 ) i ⃗ − 3 y x r 5 j ⃗ − 3 z x r 5 k ⃗ = = i ⃗ r 3 − 3 x r ⃗ r 5 \begin{array}{l}
\mathcal {V} \left(\frac {x}{r ^ {3}}\right) = \left[ \frac {\partial}{\partial x} \left(\frac {x}{r ^ {3}}\right) \right] \vec {i} + \left[ \frac {\partial}{\partial y} \left(\frac {x}{r ^ {3}}\right) \right] \vec {j} + \left[ \frac {\partial}{\partial z} \left(\frac {x}{r ^ {3}}\right) \right] \vec {k} = \left(\frac {1}{r ^ {3}} - \frac {3 x ^ {2}}{r ^ {5}}\right) \vec {i} - \frac {3 y x}{r ^ {5}} \vec {j} - \frac {3 z x}{r ^ {5}} \vec {k} = \\
= \frac {\vec {i}}{r ^ {3}} - 3 x \frac {\vec {r}}{r ^ {5}} \\
\end{array} V ( r 3 x ) = [ ∂ x ∂ ( r 3 x ) ] i + [ ∂ y ∂ ( r 3 x ) ] j + [ ∂ z ∂ ( r 3 x ) ] k = ( r 3 1 − r 5 3 x 2 ) i − r 5 3 y x j − r 5 3 z x k = = r 3 i − 3 x r 5 r V ( r ⃗ / r 3 ) \mathcal{V}(\vec{r} /r^3) V ( r / r 3 )
V ( r ⃗ r 3 ) = [ V ( x r 3 ) ] i ⃗ + [ V ( y r 3 ) ] j ⃗ + [ V ( z r 3 ) ] k ⃗ \mathcal {V} \left(\frac {\vec {r}}{r ^ {3}}\right) = \left[ \mathcal {V} \left(\frac {x}{r ^ {3}}\right) \right] \vec {i} + \left[ \mathcal {V} \left(\frac {y}{r ^ {3}}\right) \right] \vec {j} + \left[ \mathcal {V} \left(\frac {z}{r ^ {3}}\right) \right] \vec {k} V ( r 3 r ) = [ V ( r 3 x ) ] i + [ V ( r 3 y ) ] j + [ V ( r 3 z ) ] k
Since
∇ ( x r 3 ) = i ⃗ r 3 − 3 x r ⃗ r 5 ∇ ( y r 3 ) = j ⃗ r 3 − 3 y r ⃗ r 5 ∇ ( z r 3 ) = k ⃗ r 3 − 3 z r ⃗ r 5 \begin{array}{l}
\nabla \left(\frac {x}{r ^ {3}}\right) = \frac {\vec {i}}{r ^ {3}} - 3 x \frac {\vec {r}}{r ^ {5}} \\
\nabla \left(\frac {y}{r ^ {3}}\right) = \frac {\vec {j}}{r ^ {3}} - 3 y \frac {\vec {r}}{r ^ {5}} \\
\nabla \left(\frac {z}{r ^ {3}}\right) = \frac {\vec {k}}{r ^ {3}} - 3 z \frac {\vec {r}}{r ^ {5}} \\
\end{array} ∇ ( r 3 x ) = r 3 i − 3 x r 5 r ∇ ( r 3 y ) = r 3 j − 3 y r 5 r ∇ ( r 3 z ) = r 3 k − 3 z r 5 r
Therefore
∇ ( r ⃗ r 3 ) = [ ∇ ( x r 3 ) ] i ⃗ + [ ∇ ( y r 3 ) ] j ⃗ + [ ∇ ( z r 3 ) ] k ⃗ = = [ i ⃗ r 3 − 3 x r ⃗ r 5 ] i ⃗ + [ j ⃗ r 3 − 3 y r ⃗ r 5 ] j ⃗ + [ k ⃗ r 3 − 3 z r ⃗ r 5 ] k ⃗ = = 1 r 3 − 3 x x r 5 + 1 r 3 − 3 y y r 5 + 1 r 3 − 3 z z r 5 = 3 r 3 − 3 x 2 + y 2 + z 2 r 5 = = 3 r 3 − 3 r 2 r 5 = 3 r 3 − 3 r 3 = 0 \begin{array}{l}
\nabla \left(\frac {\vec {r}}{r ^ {3}}\right) = \left[ \nabla \left(\frac {x}{r ^ {3}}\right) \right] \vec {i} + \left[ \nabla \left(\frac {y}{r ^ {3}}\right) \right] \vec {j} + \left[ \nabla \left(\frac {z}{r ^ {3}}\right) \right] \vec {k} = \\
= \left[ \frac {\vec {i}}{r ^ {3}} - 3 x \frac {\vec {r}}{r ^ {5}} \right] \vec {i} + \left[ \frac {\vec {j}}{r ^ {3}} - 3 y \frac {\vec {r}}{r ^ {5}} \right] \vec {j} + \left[ \frac {\vec {k}}{r ^ {3}} - 3 z \frac {\vec {r}}{r ^ {5}} \right] \vec {k} = \\
= \frac {1}{r ^ {3}} - 3 x \frac {x}{r ^ {5}} + \frac {1}{r ^ {3}} - 3 y \frac {y}{r ^ {5}} + \frac {1}{r ^ {3}} - 3 z \frac {z}{r ^ {5}} = \frac {3}{r ^ {3}} - 3 \frac {x ^ {2} + y ^ {2} + z ^ {2}}{r ^ {5}} = \\
= \frac {3}{r ^ {3}} - 3 \frac {r ^ {2}}{r ^ {5}} = \frac {3}{r ^ {3}} - \frac {3}{r ^ {3}} = 0 \\
\end{array} ∇ ( r 3 r ) = [ ∇ ( r 3 x ) ] i + [ ∇ ( r 3 y ) ] j + [ ∇ ( r 3 z ) ] k = = [ r 3 i − 3 x r 5 r ] i + [ r 3 j − 3 y r 5 r ] j + [ r 3 k − 3 z r 5 r ] k = = r 3 1 − 3 x r 5 x + r 3 1 − 3 y r 5 y + r 3 1 − 3 z r 5 z = r 3 3 − 3 r 5 x 2 + y 2 + z 2 = = r 3 3 − 3 r 5 r 2 = r 3 3 − r 3 3 = 0
Answer: ∇ ( x r 3 ) = i ⃗ r 3 − 3 x r ⃗ r 5 , ∇ ( r ⃗ r 3 ) = 0. \nabla \left(\frac{x}{r^3}\right) = \frac{\vec{i}}{r^3} - 3x\frac{\vec{r}}{r^5},\quad \nabla \left(\frac{\vec{r}}{r^3}\right) = 0. ∇ ( r 3 x ) = r 3 i − 3 x r 5 r , ∇ ( r 3 r ) = 0.
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#340153 on Dec 2023