# Tensor Notation epsilon ijk

1. Oct 3, 2012

### bbolddaslove

1. The problem statement, all variables and given/known data

Prove that (AxB) is perpendicular to A
*We know that it is in the definition but this requires an actual proof. This is what I did on the exam because it was quicker than writing out the vectors and crossing and dotting them.

2. Relevant equations

X dot Y = 0 when they are perpendicular
AxB= epsilonijk(AjBk)i

3. The attempt at a solution

AxB= epsilonijk(AjBk)i
so (AxB) dot A = AxB= epsilonijk(AjBk)iAi = AxB= epsilonijkAiAjBk which means that i=j because we have both Ai and Aj. If i=j then epsilon=0, so the value of the dot product is 0, which means the angle between them is 90.

2. Oct 3, 2012

### dextercioby

Your solution is awfully written, uninteligible. However, indeed the contraction between the $\epsilon_{ijk}$ and either $A_{i}A_{j}$ or $B_{i}B_{j}$ is always 0.

P.S. You should learn the LaTex code. It's really useful knowledge.

3. Oct 3, 2012

### D H

Staff Emeritus
The highlighted text is very wrong.

Let's do it right, using the Levi-Cevita symbol (which is not a tensor).
$$(A\times B)_i = \epsilon_{ijk} A_j b_k$$
Note that the subscript i belongs with the cross product. What you wrote, AxB= epsilonijk(AjBk)i, doesn't make sense.

The inner product of A and AxB using Einstein sum notation is
$$A\cdot (A\times B) = A_i (A\times B)_i = A_i \epsilon_{ijk} A_j B_k$$
There are two is, two js, and two ks on the right. This is shorthand for a triple sum. It does not mean that i=j. So let's write out that triple sum explicitly:
$$A\cdot (A\times B) = \sum_i \sum_j \sum_k A_i \epsilon_{ijk} A_j B_k$$
As these are finite sums, rearranging things is valid (rearranging sums is trickier with infinite sums).
$$A\cdot (A\times B) = \sum_k B_k \left( \sum_i \sum_j \epsilon_{ijk} A_i A_j \right)$$
I'll let you finish it off. Hint: What's in the parentheses?

If you are going to use tensor notation, particular the shorthand Einstein implied sum, you need to learn how to use it correctly.

4. Oct 3, 2012

### D H

Staff Emeritus
But not for the reason bbolddaslove thinks. Happening to get the right answer for the completely wrong reason is not a good way to proceed.

Besides, bbolddaslove didn't prove the conjecture. You can't prove a definition. What you can do is prove that a specific mathematical notation is consistent with the definition. bbolddaslove started with (A×B)i = εijkAjBk as the definition of the cross product. Perhaps this expression of the cross product was shown in class or in the text, but I doubt it. If it wasn't, claiming that that expression is the cross product is an unproven assertion. It would need to be proven that this is consistent with the definition given in class or in the text.

5. Oct 3, 2012

### bbolddaslove

If i=1, j=2, k=3, we have the ith component of A multiplied by the jth component of A? But this will always equal 0 because unit vectors i and j when multiplied = 0. Any variation other than that when i=j will yield the same response. And any variation where i=j means that epsilon=0. Is this right?

We did learn the cross product definition with this value of epsilon, and it is in our book. In fact, we were taught to do other proofs such as equivalent triple products using this notation for cross product. This is a higher level class and we are very familiar with what a cross product means, and I have known for a long time that a cross product is orthogonal to each of the vectors operated on, I used this notation for its simplicity. I am just trying to gauge the accuracy of the work using that definition.

Thank you all for the constructive criticism! I already realize some mistakes in the notation and how to correct them in the future. Is my new explanation better?

Last edited by a moderator: Oct 3, 2012
6. Oct 3, 2012

### D H

Staff Emeritus
Correct.
Incorrect.

You apparently don't quite understand the concept of contraction. εijkAiAj is shorthand for a sum of nine terms. Consider the case i=1, k=2. That AiAj with i=1, j=2 is just the product of the x and y (or more generically, the e1 and e2) components of the vector A. If A is, for example [3,4,5], A1A2 is 12. It is not zero.

Maybe it will help to expand out the sum, no Einstein notation, not even a sum notation. Just brute force expansion.

εijkAiAj =
ε11kA1A1 +
ε12kA1A2 +
ε13kA1A3 +
ε21kA2A1 +
ε22kA2A2 +
ε23kA2A3 +
ε31kA3A1 +
ε32kA3A2 +
ε33kA3A3

Three of those terms obviously vanish, those involving ε11k, ε22k, and ε33k. Rearranging those that don't obviously vanish,
εijkAiAj =
12k + ε21k)A1A2 +
23k + ε32k)A2A3 +
31k + ε13k)A3A1

Look at the first expression, (ε12k + ε21k)A1A2. Given some k, either both ε12k and ε21k will be zero, or ε12k = -ε21k, and the sum is zero. Either way, the sum is zero. The same goes for the other two terms.

That's the brute force approach. A better way to look at it is that εijkAiAj is zero because the Levi Civita symbol is skew symmetric.

7. Oct 4, 2012

### bbolddaslove

So the reason the answer is 0 is because every term in which epsilon does not equal 0 has an equal and opposite term somewhere else.

8. Oct 4, 2012

### bbolddaslove

i.e. if there is an A1A2 there is also a -A2A1

9. Oct 4, 2012

### D H

Staff Emeritus
Exactly.