Prove Perpendicularity of (AxB) and A Using Tensor Notation

[bbolddaslove]

Homework Statement

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.

Homework Equations

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

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.

 

[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.

 

[D H]

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.

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.

 

[D H]

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

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 = ε_ijkA_jB_k 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.

 

[bbolddaslove]

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 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?

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 = ε_ijkA_jB_k 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.

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?

 

[D H]

If i=1, j=2, k=3, we have the ith component of A multiplied by the jth component of A?

Correct.

But this will always equal 0 because unit vectors i and j when multiplied = 0.

Incorrect.

You apparently don't quite understand the concept of contraction. ε_ijkA_iA_j is shorthand for a sum of nine terms. Consider the case i=1, k=2. That A_iA_j 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.

ε_ijkA_iA_j =
ε_11kA₁A₁ +
ε_12kA₁A₂ +
ε_13kA₁A₃ +
ε_21kA₂A₁ +
ε_22kA₂A₂ +
ε_23kA₂A₃ +
ε_31kA₃A₁ +
ε_32kA₃A₂ +
ε_33kA₃A₃

Three of those terms obviously vanish, those involving ε_11k, ε_22k, and ε_33k. Rearranging those that don't obviously vanish,
ε_ijkA_iA_j =
(ε_12k + ε_21k)A₁A₂ +
(ε_23k + ε_32k)A₂A₃ +
(ε_31k + ε_13k)A₃A₁

Look at the first expression, (ε_12k + ε_21k)A₁A₂. Given some k, either both ε_12k and ε_21k will be zero, or ε_12k = -ε_21k, 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 ε_ijkA_iA_j is zero because the Levi Civita symbol is skew symmetric.

 

[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.

 

[bbolddaslove]

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

 

[D H]

Exactly.