Question

Show that \(\nabla_{2}\left(1 / R_{12}\right)=-\nabla_{1}\left(1 / R_{12}\right)=\mathbf{R}_{21} / R_{12}^{3}\).

Short answer

Question: Show that the gradient of 1/R_12 with respect to the second coordinate and the negative gradient with respect to the first coordinate are equal to R_21/R_12^3. Answer: The gradient of 1/R_12 with respect to the second coordinate is -R_21/R_12^3, and the gradient of 1/R_12 with respect to the first coordinate is R_21/R_12^3.

Step-by-step solution

Define R_12 and R_21

We first define R_12 and R_21, where R_12 is the distance between points R_1 and R_2, and R_21 is their difference vector: \(R_{12}=|R_{2}-R_{1}|\) and \(R_{21}=R_{1}-R_{2}\). Since the magnitude of R_21 is the same as the magnitude of R_12, we can write \(R_{12}=|R_{21}|\).

Compute gradient of 1/R_12 with respect to R_2

To compute the gradient of 1/R_12 with respect to R_2, we'll first find the gradient of R_12 with respect to R_2 and use the chain rule. We have: \(\nabla_{2} R_{12} = \nabla_{2} |R_2 - R_1| = \frac{R_2 - R_1}{R_{12}}\) Now, using the chain rule, we compute the gradient of 1/R_12 with respect to R_2: \(\nabla_{2}\left(\frac{1}{R_{12}}\right)=-\frac{1}{R_{12}^{2}}\nabla_{2} R_{12}=-\frac{R_2 - R_1}{R_{12}^3}\)

Compute gradient of 1/R_12 with respect to R_1

Similarly, we compute the gradient of R_12 with respect to R_1: \(\nabla_{1} R_{12} = \nabla_{1} |R_2 - R_1| = -\frac{R_2 - R_1}{R_{12}}\) Notice that the negative of the gradient of R_12 with respect to R_1 is the same as the gradient with respect to R_2, as found in step 2: \(\nabla_{1} R_{12} = - \nabla_{2} R_{12}\) Now, using the chain rule, we compute the gradient of 1/R_12 with respect to R_1: \(\nabla_{1}\left(\frac{1}{R_{12}}\right)=-\frac{1}{R_{12}^{2}}\nabla_{1} R_{12}=\frac{R_2 - R_1}{R_{12}^3}\)

Compare with the given expressions

We can now compare the results from steps 2 and 3 to the given expressions. We found that: \(\nabla_{2}\left(\frac{1}{R_{12}}\right)=-\frac{R_2 - R_1}{R_{12}^3}\) and \(\nabla_{1}\left(\frac{1}{R_{12}}\right)=\frac{R_2 - R_1}{R_{12}^3}\). Since \(R_{21}=R_{1}-R_{2}=-(R_2 - R_1)\), the expressions can be rewritten as: \(\nabla_{2}\left(\frac{1}{R_{12}}\right)=-\frac{R_{21}}{R_{12}^3}\) and \(\nabla_{1}\left(\frac{1}{R_{12}}\right)=\frac{R_{21}}{R_{12}^3}\). We can see that the expressions match the ones we want to show, completing the proof.

Key concepts

Gradient Calculation

Gradient calculation is a fundamental technique in vector calculus, especially useful when dealing with scalar fields. When you find the gradient of a scalar function, you are essentially computing a vector field that points in the direction of the greatest rate of increase of the function.

This concept is pivotal in understanding the behavior of multi-variable functions. For our specific exercise, we are interested in the gradient of the reciprocal of a distance between two points, denoted as \(1/R_{12}\).

The gradient, \(abla\), acts as an operator that, when applied to a scalar field, returns a vector field. In the context of our problem, we need to determine how \(1/R_{12}\) changes with respect to either \(R_1\) or \(R_2\).

First, the gradient of \(R_{12}\) with respect to \(R_2\) is \(\frac{R_2 - R_1}{R_{12}}\).

• This tells you how the distance itself changes.
• Using the chain rule, the gradient of \(1/R_{12}\) becomes \(-\frac{R_2 - R_1}{R_{12}^3}\).

Understanding these calculations avoids common pitfalls where vector and scalar fields might mistakenly be swapped.

Chain Rule

The chain rule is a crucial tool in calculus, allowing us to differentiate composite functions. This rule is extremely helpful when functions are nested, that is, when one function is inside another.

For our vector calculus problem, applying the chain rule involves understanding how multi-variable functions interrelate. In simpler terms, if we have a function that is composed of an outer and an inner function, like \(1/R_{12}\), we use the chain rule to differentiate accordingly.

The chain rule states that if you want to find the derivative of \(f(g(x))\), you first differentiate \(f\) with respect to its input, and then multiply this by the derivative of \(g(x)\) with respect to \(x\).

This rule quickly translates in our case, as follows:

• The outer function is \(1/x\), which has a derivative of \(-1/x^2\).
• The inner function is \(R_{12}\).

By combining these results, we find the necessary gradients: \(abla\left(\frac{1}{R_{12}}\right) = - \frac{1}{R_{12}^2} \cdot \frac{Partial\;derivative\;of \;R_{12}}{Partial\;variable}\).

This clarity in the chain rule helps establish a clear pathway from the simple to the compound.

Distance Vector

The distance vector is key in spatial problems related to physics and engineering. It connects two points in space, providing not just the scalar distance but the full vector necessary to describe direction and separation.

For our purposes, consider \(R_{12}\), the magnitude of the vector \(R_2 - R_1\). This represents the distance between points \(R_1\) and \(R_2\). Meanwhile, \(R_{21}\) is the reverse vector, \(R_1 - R_2\), pointing in the opposite direction.

Here’s why it’s important:

• The magnitude, \(|R_{12}|\), gives you an idea of how far the points are from each other without considering direction - a pure scalar value.
• The direction comes from \(R_{21}\), showing which way the separation runs.
• In our work, we focus on using \(R_{21}\) in the gradient calculations.

The distance vector simplifies many aspects of vector calculus, weaving directionality into the concept of simple distance.

Magnitude

Magnitude in vector calculus refers to the size or length of a vector, but not its direction. Imagine it as the absolute "distance" from the vector's tail to its tip, measured in the scale of the space it occupies.

For vectors like \(R_{12}\) and \(R_{21}\), magnitude informs us about their length in physical terms. This is calculated as \(|R_{21}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\) in three dimensions.

Magnitudes play pivotal roles:

• They are scalar numbers useful in comparing vectors in terms of size.
• For expressing unit vectors from given vectors, use the formula \(\frac{Vector}{Magnitude}\).
• Magnitudes help normalize vectors, making computations such as the problem at hand more manageable.

Understanding magnitude helps manage components of calculus problems by concentrating on size and keeping direction separate. This component allows diverse applications in fields ranging from physics to computer graphics.