Geometric algebra cross product

In summary, the conversation discussed the use of geometric algebra in understanding physics, specifically in relation to rotating frames and angular velocity. The concept of the dual of an outer product being equal to the dot product of the dual was also explained. The conversation also mentioned the usefulness of geometric algebra in various areas of physics and recommended some books for further study.
  • #1
krcmd1
62
0
geometric algebra cross product

--------------------------------------------------------------------------------

Homework Statement [/b]

my text (Geometric Algebra for Physicists, by Doran and Lasenby), p. 69, deals with rotating frame {fsubk} (I assume in 3D)

d/dt (fsubk) = omega X fsubk omega being angular velocity

then

omega X fsubk = (-I omega) dot fsubk = fsubk dot (I omega), where I is pseudoscalar and (I omega) is I think the geometric product of I and omega.

I don't understand these algebraic steps. Can someone explain?

I did follow earlier explanations of why with vectors a,b that a X b = -I(a wedge b)

If I have posted this to the wrong forum, would the moderators kindly forward it to a more appropriate one?


Thank you all very much!

Ken Cohen
 
Physics news on Phys.org
  • #2
[tex]\omega \times f = -I (\omega \wedge f) = -(I\omega)\cdot f [/tex]

The first equality is just the definition of the cross product as the dual of
a wedge product (outer product). The next
step invokes a duality relationship between the dot product and the outer
product: the dual of the outer product of a blade A with blade B is
just the dot product of the dual of A with B (Do you know about this?).
On the left of the equals
sign we see the dual of an outer product and on the right we have the
dual of one of the factors dotted with the second factor. For the final
step, use the general relationship

[tex]A_r \cdot B_s = (-)^{r(s-1)} B_s \cdot A_r[/tex] for [tex]r \leq s[/tex]

Identify the vector f with Ar and the other factor with
Bs, so that s=2 (bivector!) and r=1 The sign is then -1:

[tex] -(I\omega)\cdot f = f \cdot (I\omega)[/tex].
 
  • #3
Thank you very, very much.

I am struggling to teach my self math to understand physics, and it's a lonely stumbling effort.

what program do you use to enter notation?

the equality of the dual of outer product equalling the dot of the dual was not presented as far as I can tell up to that point in the text I'm using.

do you think I'm barking up the wrong tree with geometric algebra entirely?

thanks, again.
 
  • #4
krcmd1 said:
what program do you use to enter notation?
This forum allows an implementation of TeX. You just sandwich the TeX expression
between two markup symbols: (tex) ab = a \cdot b + a \wedge b (/tex) to get

[tex] ab = a \cdot b + a \wedge b [/tex].
But use square brackets [] instead of round brackets (). You can also click on an equation and a window pops up
showing you how it is done. For more info: https://www.physicsforums.com/misc/howtolatex.pdf

krcmd1 said:
the equality of the dual of outer product equaling the dot of the dual was not presented as far as I can tell up to that point in the text I'm using.
Look for section 4.1.4 on pages 96 to 97 of your book. Indeed, skip to chapter 4 and
read it right away. I think it will answer a lot of your questions. You can always come
back to where you are. Chapter 4 is pretty well self-contained.

krcmd1 said:
do you think I'm barking up the wrong tree with geometric algebra entirely?

Not at all! For all of physics that has to do with vectors (mechanics, spacetime,
special relativity, electrodynamics) and for other topics (point groups, lie algebra,
conformal geometry, computer graphics, robotics), geometric algebra offers
a better understanding and provides better tools for attacking problems. Anything to do with rotations is best done with GA (quaternions are just the even-grade sub-algebra
of 3D space). Once you have the GA perspective on something like electrodynamics,
you will never look back. There is also a lot of baggage that you can forget about: e.g.
the vector cross product, axial as opposed to polar vectors. On the other hand, some
topics, such as quantum mechanics, although they can be approached by GA, would be
very difficult to learn entirely from the GA point of view, in my opinion. Also, although
GA is very valuable, you will not readily find others willing to talk physics in this language.
Which brings me to your remark:

krcmd1 said:
I am struggling to teach my self math to understand physics, and it's a lonely stumbling effort.

Essentially that is what I am doing, and I find it lonely as well. There are three other
very good books that I can recommend but they are very expensive; however, I don't know what I would do without them:

"Clifford Algebra to GA" by David Hestenes and Garret Sobczyk
(Kluwer, 1984, reprinted in 2002)
This has it all but is not an easy read.

"New Foundations for Classical Mechanics" (second edition)
David Hestenes, (Kluwer, 1999, reprinted in 2003)
A great deal on GA in addition to a wonderful treatment of mechanics.

"GA for Computer Science" by
Leo Dorst, Daniel Fontijne, and Stephen Mann
(Morgan Kaufmann, Elsevier, 2007)
A very good introduction with many things spelled out. It
includes many exercises and an internet web-site with software
for viewing many of the figures dynamically and interactively:
http://www.geometricalgebra.net/

Good luck in your endeavours. If you have specific questions to me, why not
send me a private communication within physics forums---I think this is possible.
 
Last edited by a moderator:
  • #5
You are very generous. I will!
 

Related to Geometric algebra cross product

What is the definition of the geometric algebra cross product?

The geometric algebra cross product is a mathematical operation that takes two vectors as inputs and returns a third vector that is perpendicular to both of the input vectors. It is also known as the exterior product or wedge product.

How is the geometric algebra cross product calculated?

The cross product of two vectors, A and B, is calculated as A ∧ B = |A||B|sin(θ)n, where |A| and |B| are the magnitudes of the vectors, θ is the angle between them, and n is the unit vector perpendicular to both A and B. This can also be represented using the cross product matrix: A ∧ B = [A]×[B] = [a1b1, a1b2, a1b3; a2b1, a2b2, a2b3; a3b1, a3b2, a3b3].

What are the properties of the geometric algebra cross product?

The geometric algebra cross product has the following properties: distributivity, associativity, anticommutativity, and bilinearity. This means that it follows the same rules as regular vector multiplication, but also has some unique properties such as being anticommutative (A ∧ B = -B ∧ A) and allowing for the cross product of a vector with itself (A ∧ A = 0).

What is the significance of the geometric algebra cross product?

The geometric algebra cross product is significant in many fields of science and engineering, including physics, computer graphics, robotics, and mechanics. It is used to calculate torque, angular momentum, and magnetic fields, and is also used in 3D graphics to determine the orientation of objects and to perform rotations.

Are there any alternative methods for calculating the cross product?

Yes, there are alternative methods for calculating the cross product, such as using the determinant of a matrix or using the vector triple product. However, the geometric algebra cross product offers a more elegant and general approach that can be extended to higher dimensions and is also compatible with other algebraic operations, making it a useful tool in many applications.

Similar threads

Replies
14
Views
2K
  • Linear and Abstract Algebra
Replies
32
Views
4K
  • Linear and Abstract Algebra
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
6
Views
2K
Replies
14
Views
1K
Replies
10
Views
786
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Math Proof Training and Practice
Replies
25
Views
2K
  • Linear and Abstract Algebra
Replies
6
Views
4K
Back
Top