Problem 28 - Thomas Calculus

Question

Zero curl, yet the field is not conservative Show that the curl of $$\mathbf{F}=\frac{-y}{x^{2}+y^{2}} \mathbf{i}+\frac{x}{x^{2}+y^{2}} \mathbf{j}+z \mathbf{k}$$ is zero but that $$\oint_{C} \mathbf{F} \cdot d \mathbf{r}$$ is not zero if \(C\) is the circle \(x^{2}+y^{2}=1\) in the \(x y\) -plane. (Theorem 7 does not apply here because the domain of \(\mathbf{F}\) is not simply connected. The field \(\mathbf{F}\) is not defined along the z-axis so there is no way to contract \(C\) to a point without leaving the domain of \(\mathbf{F} .\) )

Step-by-Step Solution

Answer

The curl is zero, yet \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = 2\pi \), not zero.

Step 1: Compute the curl

The curl of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is given by \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k} \). Substitute \( P = \frac{-y}{x^2 + y^2}, Q = \frac{x}{x^2 + y^2}, R = z \). Since \( R = z \), \( \frac{\partial R}{\partial y} = 0 \) and \( \frac{\partial R}{\partial x} = 0 \). Therefore, the curl simplifies to \( \left( 0 - 0 \right)\mathbf{i} + \left( 0 - 0 \right)\mathbf{j} \). We compute \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \) and find that both equal \( \frac{x^2 - y^2}{(x^2 + y^2)^2} + \frac{y^2 - x^2}{(x^2 + y^2)^2} = 0 \). Thus, the curl is zero.

Step 2: Determine the line integral

Since the curl of \( \mathbf{F} \) is zero, the vector field seems to be conservative, but we proceed to calculate \( \oint_{C} \mathbf{F} \cdot d \mathbf{r} \) directly. Considering \( C \) as the circle \( x^2 + y^2 = 1 \), parameterize it by \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \) for \( t \) in \( [0, 2\pi] \). As \( d\mathbf{r} = (-\sin t \mathbf{i} + \cos t \mathbf{j})dt \), substitute \( \mathbf{F} \) into the integral and compute \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{2\pi} \left( \frac{-\sin t}{1}(-\sin t) + \frac{\cos t}{1}\cos t \right)dt = \int_{0}^{2\pi}(\sin^2 t + \cos^2 t)dt = \int_{0}^{2\pi} 1 dt = 2\pi \).

Step 3: Conclusion

Even though the curl of \( \mathbf{F} \) is zero, which typically indicates a conservative field, the line integral around a closed path is non-zero. This contradiction is explained by the domain's lack of simple connectivity due to the undefined region along the z-axis, making it impossible to shrink the path \( C \) to a point within the domain without crossing the singularity.

Key Concepts

Curl

Curl is a vital concept in vector calculus utilized to describe the rotational behavior of a vector field. It is particularly useful in physics, especially in describing fluid motion and electromagnetism. The curl of a vector field \( \mathbf{F} \) is denoted by \( abla \times \mathbf{F} \) and provides a measure of the field's tendency to rotate around a point.
This is calculated using the determinant of a matrix formed by the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), the partial derivative operators, and the components of the vector field, often expressed as \( P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \). For the vector field \( \mathbf{F} = \frac{-y}{x^2+y^2} \mathbf{i} + \frac{x}{x^2+y^2} \mathbf{j} + z \mathbf{k} \), the curl was computed to be zero.
A zero curl generally indicates that the vector field is irrotational. However, this does not guarantee that the field is conservative unless certain conditions, like being defined over a simply connected domain, are met.

Conservative Field

A conservative vector field is one where the line integral around any closed path is zero. This is typically true for fields where the curl is zero, meaning no rotation occurs about any point within the field.
Mathematically, a vector field \( \mathbf{F} \) is conservative if there exists a scalar potential function \( \phi \), such that \( \mathbf{F} = abla \phi \). For such fields, the line integral between two points \( A \) and \( B \) is path-independent and depends only on the endpoints.
However, the field \( \mathbf{F} = \frac{-y}{x^2+y^2} \mathbf{i} + \frac{x}{x^2+y^2} \mathbf{j} + z \mathbf{k} \) is not conservative despite having a zero curl because its domain is not simply connected, meaning there are essential paths where the integral is non-zero, indicating the presence of hidden complexities or singularities.

Line Integral

The concept of line integrals is crucial in understanding how vector fields interact with paths. A line integral along a curve \( C \) is computed as \( \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( \mathbf{F} \) is a vector field and \( d\mathbf{r} \) is an infinitesimal vector tangent to the path.
In this context, line integrals provide the accumulated effect of a vector field along a curve, which could represent quantities like work done by a force field in physics. For the given vector field and curve \( C \) as the circle \( x^2+y^2=1 \), the line integral was found to be \( 2\pi \), showing that despite the field's zero curl, it accumulates a non-zero value along this path.
This result suggests complexities like non-conservativeness can arise from the nature of the domain or vector field itself, reflecting physical issues like the presence of a vortex or other non-intuitive features.

Simple Connectedness

In mathematics, particularly in vector calculus, a domain is said to be simply connected if any loop within it can be continuously contracted to a point without exiting the domain. Simple connectedness plays a key role in determining whether a zero curl implies a vector field is conservative.
For a domain to be simply connected, it cannot contain holes, gaps, or excluded regions, which are points where the vector field is not defined, like along the z-axis for the field \( \mathbf{F} \). This lack of simple connectedness allows for the possibility of paths within the domain on which line integrals yield non-zero values, even if the curl is zero.
For the vector field in question, the presence of an undefined region along the z-axis interrupts simple connectedness. This interruption is why the line integral around the circle \( x^2+y^2=1 \) cannot be zero, revealing the intricate relationship between a field's curl, conservation properties, and the topology of its domain.