Gradient of a Scalar Function: \(\phi = \ln|\vec{r}|\)

To find the gradient of the given scalar function \(\phi = \ln \left| {\vec r} \right|\), we need to understand the definitions of a position vector, its norm, and the gradient operator. The question asks us to compute \(\nabla\phi\), where \(\vec r = x\hat i + y\hat j + z\hat k\) is a position vector.

Position Vector and Norm Definition

A position vector \(\vec r\) in three-dimensional space is given by:

\[ \vec r = x\hat i + y\hat j + z\hat k \]

The norm of the vector \(\vec r\), denoted as \(\left| {\vec r} \right|\), represents its magnitude or length. It is defined as:

\[ \left| {\vec r} \right| = \sqrt {{x^2} + {y^2} + {z^2}} \]

Function Transformation for Gradient Calculation

The given scalar function is \(\phi = \ln \left| {\vec r} \right|\). To compute its gradient, it's often easier to express \(\phi\) in terms of the Cartesian coordinates \(x, y, z\).

Substitute the definition of \(\left| {\vec r} \right|\) into the function \(\phi\):

\[ \phi = \ln \left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right) \]

Using the logarithm property \(\ln(a^b) = b \ln(a)\), we can simplify this expression:

\[ \phi = \ln \left( {({x^2} + {y^2} + {z^2})^{1/2}} \right) \]

\[ \phi = \frac{1}{2}\ln \left( {{x^2} + {y^2} + {z^2}} \right) \]

Gradient Operator Definition

The gradient of a scalar function \(\phi(x,y,z)\) is a vector quantity that represents the direction of the greatest rate of increase of the function. It is denoted by \(\nabla\phi\) and defined as:

\[ \nabla\phi = \frac{{\partial\phi}}{{\partial x}}\hat i + \frac{{\partial\phi}}{{\partial y}}\hat j + \frac{{\partial\phi}}{{\partial z}}\hat k \]

Here, \(\frac{{\partial\phi}}{{\partial x}}\), \(\frac{{\partial\phi}}{{\partial y}}\), and \(\frac{{\partial\phi}}{{\partial z}}\) are the partial derivatives of \(\phi\) with respect to \(x, y,\) and \(z\), respectively.

Partial Derivatives Calculation

Now, we will calculate each partial derivative of \(\phi = \frac{1}{2}\ln \left( {{x^2} + {y^2} + {z^2}} \right)\).

Partial Derivative with respect to \(x\):

\[ \frac{{\partial\phi}}{{\partial x}} = \frac{\partial}{{\partial x}}\left( {\frac{1}{2}\ln \left( {{x^2} + {y^2} + {z^2}} \right)} \right) \]

Using the chain rule, \(\frac{d}{du}\ln(u) = \frac{1}{u}\) and \(\frac{\partial}{\partial x}(x^2+y^2+z^2) = 2x\):

\[ \frac{{\partial\phi}}{{\partial x}} = \frac{1}{2} \cdot \frac{1}{{{x^2} + {y^2} + {z^2}}} \cdot (2x) \]

\[ \frac{{\partial\phi}}{{\partial x}} = \frac{x}{{{x^2} + {y^2} + {z^2}}} \]

Partial Derivative with respect to \(y\):

Similarly, for \(\frac{{\partial\phi}}{{\partial y}}\), we treat \(x\) and \(z\) as constants:

\[ \frac{{\partial\phi}}{{\partial y}} = \frac{\partial}{{\partial y}}\left( {\frac{1}{2}\ln \left( {{x^2} + {y^2} + {z^2}} \right)} \right) \]

\[ \frac{{\partial\phi}}{{\partial y}} = \frac{1}{2} \cdot \frac{1}{{{x^2} + {y^2} + {z^2}}} \cdot (2y) \]

\[ \frac{{\partial\phi}}{{\partial y}} = \frac{y}{{{x^2} + {y^2} + {z^2}}} \]

Partial Derivative with respect to \(z\):

And for \(\frac{{\partial\phi}}{{\partial z}}\), we treat \(x\) and \(y\) as constants:

\[ \frac{{\partial\phi}}{{\partial z}} = \frac{\partial}{{\partial z}}\left( {\frac{1}{2}\ln \left( {{x^2} + {y^2} + {z^2}} \right)} \right) \]

\[ \frac{{\partial\phi}}{{\partial z}} = \frac{1}{2} \cdot \frac{1}{{{x^2} + {y^2} + {z^2}}} \cdot (2z) \]

\[ \frac{{\partial\phi}}{{\partial z}} = \frac{z}{{{x^2} + {y^2} + {z^2}}} \]

Constructing the Gradient Vector

Now, substitute these partial derivatives back into the gradient formula:

\[ \nabla\phi = \frac{x}{{{x^2} + {y^2} + {z^2}}}\hat i + \frac{y}{{{x^2} + {y^2} + {z^2}}}\hat j + \frac{z}{{{x^2} + {y^2} + {z^2}}}\hat k \]

We can factor out the common denominator:

\[ \nabla\phi = \frac{1}{{{x^2} + {y^2} + {z^2}}}(x\hat i + y\hat j + z\hat k) \]

Relating to Position Vector Notation

Recall the definition of the position vector \(\vec r = x\hat i + y\hat j + z\hat k\). So, the term \((x\hat i + y\hat j + z\hat k)\) can be replaced by \(\vec r\).

Also, remember the norm of the vector: \(\left| {\vec r} \right| = \sqrt {{x^2} + {y^2} + {z^2}}\). Squaring both sides gives us \({{\left| {\vec r} \right|}^2} = {x^2} + {y^2} + {z^2}\).

Another important vector identity is the dot product of a vector with itself: \(\vec r \cdot \vec r = (x\hat i + y\hat j + z\hat k) \cdot (x\hat i + y\hat j + z\hat k) = x^2 + y^2 + z^2\). Therefore, we have \(\vec r \cdot \vec r = {{\left| {\vec r} \right|}^2}\).

Substituting these into the expression for \(\nabla\phi\):

\[ \nabla\phi = \frac{1}{{{{\left| {\vec r} \right|}^2}}}\vec r \]

Or, equivalently:

\[ \nabla\phi = \frac{{\vec r}}{{{{\left| {\vec r} \right|}^2}}} \]

And since \({{\left| {\vec r} \right|}^2} = \vec r \cdot \vec r\), we can also write:

\[ \nabla\phi = \frac{{\vec r}}{{\vec r \cdot \vec r}} \]

Comparing this result with the given options, we find that it matches option 3.

Therefore, the gradient of the function \(\phi = \ln \left| {\vec r} \right|\) is \(\frac{{\vec r}}{{\vec r \cdot \vec r}}\).