Calculate Exterior Differentials: Step-by-Step Guide

In summary, the conversation discusses the calculation of exterior differentials and the use of the wedge product and anticommutativity. The simplified expression for the given problem is provided, as well as the full calculation for exterior differentiating r^3cos(4θ)dθ. The concept of decreasing work by ignoring partials with respect to differentials already present in a form is also mentioned.
  • #1
Zurtex
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Hey, having a little problem with a question because I am a little to unsure how to do it, I've been asked to calculate some exterior differentials, would this be write:

[tex]d\left(r^3 (\cos (4 \theta) dr - r \sin (4\theta) d\theta)\right)[/tex]

Simplified:

[tex]d\left(r^3 \cos (4\theta) dr - r^4 \sin (4\theta) d\theta\right)[/tex]

Goes to:

[tex]2r^3 dr \left(-4 \sin (4\theta) d\theta dr - 4r^3 dr \cos (4\theta) d\theta d\theta \right) = 0[/tex]

Am I even slightly right?
 
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  • #2
Note that if p and q are k-forms, then [itex]dp\wedge dq = -dq\wedge dp[/itex] which leads to [itex]dp\wedge dp = 0[/itex] and the differential follows the product rule. Thus we get:
[tex]d\left(r^3 \cos (4\theta) dr - r^4 \sin (4\theta) d\theta\right)[/tex]
[tex]=-4r^3\sin(4\theta)d\theta\wedge dr - 4r^3\sin(4\theta)dr\wedge d\theta[/tex]
= 0.
You could have saved a bit of work by noting that your expression is just
[tex]d(\frac{r^4\cos(4\theta)}{4})[/tex]
since the differential of a differential is zero.
 
  • #3
When say exterior differentiating [itex]r^3 \cos (4\theta) d\theta[/itex] Does only dr get generated and not an extra [itex]d\theta[/itex]?
 
  • #4
Zurtex said:
When say exterior differentiating [itex]r^3 \cos (4\theta) d\theta[/itex] Does only dr get generated and not an extra [itex]d\theta[/itex]?
That's due to the anticommutativity of the wedge product. The full calculation is:
[tex]d(r^3 \cos (4\theta) d\theta) = d(r^3\cos(4\theta))\wedge d\theta + (-1)^{1}r^3\cos(4\theta)\wedge d(d\theta))[/tex]
[tex] = (3r^2 \cos(4\theta) dr - 4r^3 \sin(4\theta )d\theta ) \wedge d\theta - 0[/tex]
[tex] = (3r^2 \cos(4\theta)) dr\wedge d\theta - (4r^3 \sin(4\theta )) d\theta\wedge d\theta[/tex]
[tex] = (3r^2 \cos(4\theta)) dr\wedge d\theta - 0[/tex]
Just too much to type out. :wink: if you have a k-form where k is greater than or equal to one, you can decrease your drudge work by just ignoring partials with respect to differentials already present in the form and noticing existing differentials.
All you really need is the anticommutativity, from which can be derived [itex]dp\wedge dp = 0[/itex], the generalized product rule, and d(dp)=0 for any form p.
 
Last edited:
  • #5
Thanks a lot :smile:
 

1. What is the purpose of calculating exterior differentials?

The purpose of calculating exterior differentials is to find the change in a function with respect to changes in its independent variables. This can be used to approximate the values of a function at a given point, or to optimize a function to find its maximum or minimum value.

2. How do I calculate exterior differentials?

To calculate exterior differentials, you will need to take the partial derivative of the function with respect to each independent variable. Then, multiply each partial derivative by its corresponding differential (dx, dy, etc.). Finally, add all of these terms together to get the exterior differential.

3. What is the difference between exterior differential and total differential?

The exterior differential only considers changes in the independent variables, while the total differential takes into account changes in both the independent and dependent variables. The total differential is used to find the exact change in a function, while the exterior differential is an approximation.

4. When should I use exterior differentials?

Exterior differentials are commonly used in physics, engineering, and economics to approximate the values of functions and optimize them. They are also useful in multivariable calculus for finding gradients and directional derivatives.

5. Are there any limitations to using exterior differentials?

Yes, exterior differentials are only accurate for small changes in the independent variables. If the changes are too large, the approximation may not be accurate. Additionally, exterior differentials may not work for all types of functions and may require other mathematical techniques to solve.

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