# Permutation Index

1. Feb 3, 2008

### EugP

1. The problem statement, all variables and given/known data
I'm taking a Magnetic Fields class, and the professor taught us doing cross and dot products using the permutation index. But I don't quite understand how it works completely.

I have these problems:

Given:

$$\vec A=\hat x + 2\hat y - 3\hat z$$
$$\vec B=3\hat x - 4\hat y$$
$$\vec C=3\hat y - 4\hat z$$

Find:

1) $$\vec A \times \vec C$$

2) $$\hat x \times \vec B$$

2. Relevant equations

3. The attempt at a solution

1) Using what I know about the permitivity constant:

$$(\vec A \times \vec C)=$$

$$\varepsilon_{xyz}\vec A_y \vec C_z=$$

But I don't know where to go from here. All I know is that $$\varepsilon_{xyz} = 1[tex] because indices are a cyclic permutation, but I don't know what to do next. 2) For this one I don't even know where to begin. Please someone help, any help at all would be great. 2. Feb 3, 2008 ### Avodyne You can do these with rules for the cross products of the unit vectors: [tex]\hat x\times\hat x = 0\quad\quad\hat x\times\hat y = \hat z\quad\quad\hat x\times\hat z = -\hat y$$

$$\hat y\times\hat x = -\hat z\quad\quad\hat y\times\hat y = 0\quad\quad\hat y\times\hat z = \hat x$$

$$\hat z\times\hat x = \hat y\quad\quad\hat z\times\hat y = -\hat x\quad\quad\hat z\times\hat z = 0$$

3. Feb 4, 2008

### Nevetsman

Hi!

I'm new to the forum so I'm still getting used to the formatting techniques... instead of using the fancy mathematical symbols, I'm going to use easier assignments. In this case, E will be my permutation symbol.

Remember that the Einstein notation for vectors is incredibly useful, but it operates with an implied summation. E(xyz) is summed as x goes from 0 to 3, y goes from 0 to 3, and z goes from 0 to 3. These represent nothing more than component indexing numbers, with 1=i'th component, 2=j'th component, and 3=k'th component.

Your representation of the cross product is accurate, but don't include the vector line above each letter. E(xyz)A(y)C(z)=AxC. Using the summation, let's expand this, we get...

=E(123)A(2)C(3)1+E(132)A(3)C(2)1+E(213)A(1)C(3)2+E(231)A(3)C(1)2+
E(312)A(1)C(3)3+E(321)A(2)C(1)3

Then, for all cyclic arrangements of E -> E(123)=E(312)=E(231)=1
For all non-clyclic arrangements of E -> E(132)=E(213)=E(321)=-1

Now simply plug in the corresponding values for each component index.

Part B in your problem is solved in almost the exact same way, just remember that the x-unit vector does not have a y or z component, so all those components will be zero.

Hope this helps!

Steve

4. Feb 4, 2008

### HallsofIvy

Staff Emeritus
As nevetsman said, this should be
$$\varepsilon_{xyz} A_y C_z=$$ where Ax and Cz are the x component of A and the z component of C, not vectors themselves.
For any set of indices $\varepsilon_{ijklm}$ is defined to be "1 if ijklm is an even permutation of 12345, -1 if an odd permutation, 0 otherwise". There are 3!= 6 permutions of 123. 3 are even: 123, 231, and 312, 3 are odd: 132, 213, and 321.So $\varepsilon_{123}= \varepsilon_{231}= \varepsilon_{321}= 1$ while $\varepsilon_{132}= \varepsilon_{213}= \varepsilon_{321}= -1$.

Do you know how to find a 3 by 3 determinant? That's another mnenonic that might be simpler.

Last edited: Feb 4, 2008
5. Feb 4, 2008

### EugP

Thanks for the replies. I accidentally left A and C as vectors, didn't mean to. And now I see what I have to do, thank you very much!