Problem 34
Question
If \(\omega\) is the differential one-form in three dimensions given by \(\omega=A d x+B d y+C d z\), show that $$ \begin{aligned} d \omega=&\left(\frac{\partial C}{\partial y}-\frac{\partial B}{\partial z}\right) d y d z+\left(\frac{\partial A}{\partial z}-\frac{\partial C}{\partial x}\right) d z d x \\ &+\left(\frac{\partial B}{\partial x}-\frac{\partial A}{\partial y}\right) d x d y \end{aligned} $$
Step-by-Step Solution
Verified Answer
Short Answer: The exterior derivative \(d\omega\) of the given 1-form \(\omega = A dx + B dy + C dz\) in three dimensions is:$$
d\omega = \left(\frac{\partial C}{\partial y}-\frac{\partial B}{\partial z}\right) dy \wedge dz+\left(\frac{\partial A}{\partial z}-\frac{\partial C}{\partial x}\right) dz \wedge dx+\left(\frac{\partial B}{\partial x}-\frac{\partial A}{\partial y}\right) dx \wedge dy.
1Step 1: Recall the definition of exterior derivative
To find the exterior derivative \(d\omega\), we need to differentiate each component of \(\omega\) with respect to each variable (x, y, z) and introduce the corresponding wedge (\wedge) products: $$
d\omega = \left( \frac{\partial A}{\partial x} dx + \frac{\partial B}{\partial x} dy + \frac{\partial C}{\partial x}dz \right) \wedge dx
+ \left( \frac{\partial A}{\partial y} dx + \frac{\partial B}{\partial y} dy + \frac{\partial C}{\partial y}dz \right) \wedge dy
+ \left( \frac{\partial A}{\partial z} dx + \frac{\partial B}{\partial z} dy + \frac{\partial C}{\partial z}dz \right) \wedge dz
$$
2Step 2: Apply rules for wedge products
The wedge product is antisymmetric, so the product of a form with itself is 0. That is, \(dx \wedge dx = dy \wedge dy = dz \wedge dz = 0\). Therefore, we need to expand the wedge product and eliminate terms that involve the product of a form with itself:$$
d\omega = \left(\frac{\partial B}{\partial x} dy\wedge dx - \frac{\partial C}{\partial x}dz\wedge dx \right)
+ \left(-\frac{\partial A}{\partial y} dx\wedge dy + \frac{\partial C}{\partial y}dz\wedge dy \right)
+ \left(-\frac{\partial A}{\partial z} dx\wedge dz - \frac{\partial B}{\partial z} dy\wedge dz \right)
$$
3Step 3: Rearrange the terms
Since the wedge product is antisymmetric, it changes the sign of a term when we swap its orders. So, applying this property, we can rearrange the terms to the final expression:$$
\begin{aligned}
d \omega=&\left(\frac{\partial C}{\partial y}-\frac{\partial B}{\partial z}\right) dy \wedge dz+\left(\frac{\partial A}{\partial z}-\frac{\partial C}{\partial x}\right) dz \wedge dx \\
&+\left(\frac{\partial B}{\partial x}-\frac{\partial A}{\partial y}\right) dx \wedge dy
\end{aligned}
$$
Key Concepts
Differential FormsWedge ProductExterior Derivative
Differential Forms
Differential forms are a central concept in exterior calculus, enabling us to describe various geometrical and physical phenomena. At their core, differential forms are expressions that can integrate over manifolds, making them highly versatile tools.
Understanding differential forms provides a profound way to analyze changes and interactions within multidimensional spaces.
- A differential one-form, like the ones given by the expression \(\omega = A \, dx + B \, dy + C \, dz\), is an object that assigns a linear combination of differentials to functions.
- These one-forms can be thought of as generalized vectors that help us in analyzing directions and magnitudes in a geometrical space.
- They are especially useful in physics and engineering, where they represent things like electric and magnetic fields.
Understanding differential forms provides a profound way to analyze changes and interactions within multidimensional spaces.
Wedge Product
The wedge product is a special mathematical operation used in exterior calculus. It combines differential forms and results in a new differential form that expresses area, volume, or higher-dimensional spaces.
Using wedge products, you can express complex interactions and motions succinctly and consistently, proving indispensable in advanced calculus, physics, and engineering.
- It is denoted by the symbol \( \wedge \), and is key in operations involving differential forms, such as finding \(d\omega\) in our exercise.
- The wedge product is antisymmetric, which means \(dx \wedge dx = 0\), and switching the order of the forms in the product changes its sign. For example, \(dx \wedge dy = -dy \wedge dx\).
- This property is crucial in simplifying expressions and eliminating terms, as seen in the solution where terms like \(dx \wedge dx\) and \(dy \wedge dy\) vanish.
Using wedge products, you can express complex interactions and motions succinctly and consistently, proving indispensable in advanced calculus, physics, and engineering.
Exterior Derivative
The exterior derivative is an operation that extends the concept of derivative to differential forms, a way to compute derivatives that maintain the essential properties of calculus.
The exterior derivative's ability to extend the derivative to abstract geometrical contexts revolutionized the way mathematicians and scientists describe multi-dimensional systems, offering a robust framework for understanding the foundational behaviors of various physical systems.
- For a differential one-form \(\omega = A \, dx + B \, dy + C \, dz\), the exterior derivative \(d\omega\) creates a two-form.
- It's computed by differentiating each component of the differential form and applying the wedge product with differentials, as shown in the exercise solution.
- The exterior derivative is linear, and it obeys a property of nilpotency, meaning applying it twice results in zero, \(d(d\omega) = 0\).
The exterior derivative's ability to extend the derivative to abstract geometrical contexts revolutionized the way mathematicians and scientists describe multi-dimensional systems, offering a robust framework for understanding the foundational behaviors of various physical systems.
Other exercises in this chapter
Problem 32
Letting \(i, j, k\) be first-order units in three-dimensional space, determine the combinatory product of \(2 i+3 j-4 k\), \(3 i-i+k \cdot i+2 i-k\)
View solution Problem 33
Show that in Grassmann's combinatory multiplication, $$ \begin{aligned} &\left(\sum \alpha_{1 i} \epsilon_{i}\right)\left(\sum \alpha_{2 i} \epsilon_{i}\right)
View solution Problem 35
Show that the exterior derivative of \(\omega=A d y d z+\) \(B d z d x+C d x d y\) is the three-form $$ \left(\frac{\partial A}{\partial x}+\frac{\partial B}{\p
View solution Problem 36
Show that \(d(d \omega)=0\), where \(\omega\) is a differential one-form or two- form in three-dimensional space.
View solution