How Does the Cross Product Transform Under Inversion in Vector Analysis?

[noblegas]

Homework Statement

How does the cross product (i.e. A X B) of two vectors transform under inversion? [The cross product of two vectors is properly called a pseudovector because of this "anomalous behavior]. Is the cross product of two pseudo vectors, a vector, and a pseudovector? Name two psuedovector quantities in classical mechanics.

Homework Equations

A=A_x*x-hat+A_y*y_hat+A_z*z_hat

B=B_x*x-hat+B_y*y_hat+B_z*z_hat

The Attempt at a Solution

should I show/prove that
A X B=-B X A

 

[gabbagabbahey]

Do you understand what inversion means in this context?

If so, what happens to an actual vector $\textbf{C}$ under inversion?

 

[noblegas]

Do you understand what inversion means in this context?

If so, what happens to an actual vector $\textbf{C}$ under inversion?

Under inversion , C=C^-1

 

[gabbagabbahey]

Under inversion , C=C^-1

Huh?! How exactly does one compute $\textbf{C}^{-1}$?

No. In this context the word inverse has nothing to do with compositional/functional inverses (such as $\tan^{-1}x$). Rather, when you talk about vectors and inversion, you want to find the effect of inverting the coordinate axes (i.e. $x\yo-x$, $y\to-y$ and $z\to-z$)...What effect does this kind of inversion have on $\textbf{C}$?

 

[noblegas]

Huh?! How exactly does one compute $\textbf{C}^{-1}$?

No. In this context the word inverse has nothing to do with compositional/functional inverses (such as $\tan^{-1}x$). Rather, when you talk about vectors and inversion, you want to find the effect of inverting the coordinate axes (i.e. $x\yo-x$, $y\to-y$ and $z\to-z$)...What effect does this kind of inversion have on $\textbf{C}$?

I was thinking that C was a matrix rather than a vector because I assumed the C u were referring to was C=A XB; If C is a vector, then C=-C

 

[gabbagabbahey]

I was thinking that C was a matrix rather than a vector because I assumed the C u were referring to was C=A XB;

Why do you think $\textbf{A}\times\textbf{B}$ is a matrix and not a vector?

If C is a vector, then C=-C

Yes, $\textbf{C}\to-\textbf{C}$ under inversion (Strictly speaking, the equal sign is incorrect, a transformation like this is represented by an arrow sign))...so $\textbf{A}\to$___? And $\textbf{B}\to$?...Which makes $\textbf{A}\times\textbf{B}\to$____?

 

[noblegas]

Why do you think $\textbf{A}\times\textbf{B}$ is a matrix and not a vector?

Because A X B can be written in matrix formA X B =(A_y*B_z-B_y*A_z)x-hat+(A_x*B_z-B_x*A_z)y-hat+(A_x*B_y-A_y*B_x)z-hat

Yes, $\textbf{C}\to-\textbf{C}$ under inversion (Strictly speaking, the equal sign is incorrect, a transformation like this is represented by an arrow sign))...so $\textbf{A}\to$___? And $\textbf{B}\to$?...Which makes $\textbf{A}\times\textbf{B}\to$____?

A-> -A, B-> -B => A X B => -A X -B

 

[gabbagabbahey]

Because A X B can be written in matrix form


A X B =(A_y*B_z-B_y*A_z)x-hat+(A_x*B_z-B_x*A_z)y-hat+(A_x*B_y-A_y*B_x)z-hat

But that's not matrix form, that's just component form for a vector...matrix form is more like $C_{xx}\hat{\textbf{x}}\wedge\hat{\textbf{x}}+C_{xy}\hat{\textbf{x}}\wedge\hat{\textbf{y}}+C_{xz}\hat{\textbf{x}}\wedge\hat{\textbf{z}}+C_{yx}\hat{\textbf{y}}\wedge\hat{\textbf{x}}\ldots$

You can write the cross product as a determinant of a matrix

$$\textbf{A}\times\textbf{B}=\begin{vmatrix}\hat{\textbf{x}} & \hat{\textbf{y}} & \hat{\textbf{z}} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$$

But, a determinant is not a matrix.

A-> -A, B-> -B => A X B => -A X -B

Keep going...the product of two negative numbers is a _______ number?

 

[noblegas]

but that's not matrix form, that's just component form for a vector...matrix form is more like $c_{xx}\hat{\textbf{x}}\wedge\hat{\textbf{x}}+c_{xy}\hat{\textbf{x}}\wedge\hat{\textbf{y}}+c_{xz}\hat{\textbf{x}}\wedge\hat{\textbf{z}}+c_{yx}\hat{\textbf{y}}\wedge\hat{\textbf{x}}\ldots$

you can write the cross product as a determinant of a matrix

$$\textbf{a}\times\textbf{b}=\begin{vmatrix}\hat{\textbf{x}} & \hat{\textbf{y}} & \hat{\textbf{z}} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}$$

but, a determinant is not a matrix.

Keep going...the product of two negative numbers is a _______ number?

-Ax -B=AxB

 

[gabbagabbahey]

Right (again, you need to use an arrow instead of an equal sign, otherwise you are telling whoever is marking your paper that the two vectors are equal, when you want to tell the marker that the first vector transforms into the second), but if $\textbf{A}\times\textbf{B}$ was an actual vector (instead of a pseudo vector) then you would expect it to transform as $\textbf{A}\times\textbf{B}\to-(\textbf{A}\times\textbf{B})$ wouldn't you?...Therefor, $\textbf{A}\times\textbf{B}$ is a _________.

 

[noblegas]

Right (again, you need to use an arrow instead of an equal sign, otherwise you are telling whoever is marking your paper that the two vectors are equal, when you want to tell the marker that the first vector transforms into the second), but if $\textbf{A}\times\textbf{B}$ was an actual vector (instead of a pseudo vector) then you would expect it to transform as $\textbf{A}\times\textbf{B}\to-(\textbf{A}\times\textbf{B})$ wouldn't you?...Therefor, $\textbf{A}\times\textbf{B}$ is a _________.

a pseudovector since the expression AXB does not change when AXB -> -AX-B