Quick question on Einstein Notation

1. Apr 27, 2005

Autodidact

Hello all,

I have a quick question on Einstein notation. I'll write the tensors as a capital letter and the covariant indices as lower case letters (and not use anything that has contravariant indices). I'll also use != for not equal or not congruent to.

In Schaum's outline of tensor calculus, the author stipulates (page 3):

Aij (Xi + Yj) != AijXi + AijYj

but I don't understand how to interpret the right hand side, or to figure out exactly what it does equal in an Einstein-summation form without the parenthetical expression, without choosing some "n" and simply expanding it out entirely.

This becomes relevant because problem 1.28 part c on page 7 (which sets Ei = 1 for all i) asks for a proof of:

Aij (Xi + Xj) = 2 Aij Ei Xj (with Aij symmetric)

I was doing this by expanding the right hand side, but apparently I have not been able to figure out how to do that properly as I have been unable to demonstrate this equality.

I appreciate any insight!

Thanks.

2. Apr 27, 2005

Autodidact

Version with proper math

(This is a re-post now that I see that I can use LaTeX in the message.)

Hello all,

I have a quick question on Einstein notation.

In Schaum's outline of tensor calculus, the author stipulates (page 3):

$$a_{ij}(x_i + y_j) \not= a_{ij}x_i + a_{ij}y_j$$

but I don't understand how to interpret the right hand side, or to figure out exactly what it does equal in an Einstein-summation form without the parenthetical expression, without choosing some "n" and simply expanding it out entirely.

This becomes relevant because problem 1.28 part c on page 7 asks for a proof of:

Given $$a_{ij}$$ symmetric and $$\varepsilon_i = 1$$ for all $$i$$, prove:

$$a_{ij} (x_i + x_j) = 2 a_{ij} \varepsilon_i x_j$$

I was doing this by expanding the right hand side, but apparently I have not been able to figure out how to do that properly as I have been unable to demonstrate this equality.

I appreciate any insight!

Thanks.

3. Apr 27, 2005

dextercioby

You meant,expand the LHS,right...?

Daniel.

4. Apr 27, 2005

Autodidact

I did mean the RHS...

I actually did mean the RHS (the side with the parenthetical expression) but you are correct in that I could expand either side. The problem is that I don't understand the RHS while the LHS makes perfect sense to me.

Thanks!

5. Apr 28, 2005

dextercioby

Since $\varepsilon_{i}$ is 1,no matter the value of the subscript "i" and is involved in a multiplication,what is the soul purpose for its presence in the RHS...?

Daniel.

6. Apr 28, 2005

Doodle Bob

Always keep in mind that all Einstein notation is, is a convention as to when one doesn't have to write the summation signs, just for convenience. So, when in doubt, add the sigmas. The above expression becomes obvious when you do this:

$\sum_{i,j} \left(a_{ij}(x_i + y_j) \right) \not= \sum_{i} (a_{ij}x_i) + \sum_j(a_{ij}y_j)$

There is only one summation for each of the addends on the right-hand side, because in each of those expressions, there is only one repeated index.

I'll be honest with you: I've reviewed dozens of tensor analysis books and the Schuam's study book is by far NOT the best one. It's a bit heavier than most, but I would recommend Bishop and Goldberg's Tensor Analysis on Manifolds. (Note: you might interpret this as a subjective opinion, since Goldberg is my academic grandfather).

7. Apr 28, 2005

quetzalcoatl9

I agree with Doodle. I started out with the Schuam book quite frankly because it was monetarily cheap and I wound up more confused than when I started. Fortunately, a friend of mine (knowledgeable in diff geo) pointed me in the right direction: away from the physics-oriented texts and more toward the math-oriented approach to the subject. This indices business becomes nonsensical after awhile.

The Bishop text is good, as are some others that folks on here have recommended in other threads. These books take the more modern approach of viewing tensors as multilinear functionals and not just some arbitrary thing that changes in such-and-such a way with "coordinate transformation". (this begs the question: whats the point of working in coordinates anyway?)