Geometric algebra cross product

[krcmd1]

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

 

[Nate810]

Homework Equations d/dt (fsubk) = omega X fsubk omega being angular velocityomega 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.The Attempt at a Solution The first equation is just the derivative of the vector fsubk with respect to time. The second equation is saying that the vector omega X fsubk is equal to the dot product of (-I omega) and fsubk and also equal to the dot product of fsubk and (I omega). This is just the result of a few algebraic manipulations, using the fact that for any vectors a and b, a X b = -I(a wedge b) and a dot b = I(a dot b).

 

[Moetasim]

The cross product is a mathematical operation that takes two vectors and produces a third vector that is perpendicular to both of the original vectors. In geometric algebra, this operation is represented by the wedge product, denoted by the symbol "∧". In this case, the cross product of two vectors a and b is written as a ∧ b.

In the given context, we are dealing with a rotating frame {fsubk}, which is a set of three vectors that define the axes of rotation. The angular velocity of the rotation is denoted by the symbol omega and is a vector. The notation d/dt (fsubk) represents the time derivative of the vector fsubk.

Now, the first step in the algebraic process is to use the definition of the cross product, which is a ∧ b = -I(a ∧ b), where I is the pseudoscalar. This means that the cross product of two vectors is equal to the negative of the wedge product of those same two vectors multiplied by the pseudoscalar. So, in this case, we can write omega ∧ fsubk = -I(omega ∧ fsubk).

Next, we can use the property of the pseudoscalar, which states that I² = -1. This means that we can rewrite the expression as omega ∧ fsubk = (-1)(I²)(omega ∧ fsubk).

Now, using the associative property of the wedge product, we can write this as omega ∧ fsubk = (I(-I))(omega ∧ fsubk).

Finally, we can use the definition of the geometric product, which is denoted by the symbol "·", to write this as omega ∧ fsubk = (I(omega · fsubk)). This is the same as writing omega ∧ fsubk = (I omega) · fsubk. So, the final expression is omega ∧ fsubk = (I omega) · fsubk = (-I omega) · fsubk = fsubk · (I omega).

In summary, the algebraic steps that were taken are based on the definitions and properties of the cross product, wedge product, pseudoscalar, and geometric product. I hope this explanation helps to clarify the process for you.