What Do These Einstein Summation Convention Expressions Represent?

[bon]

sure.. so 3x^2

and 2x^2 + 2yx dx/dy + 2zx dx/dz?

how can i solve my question now using Einstein convention? thanks :)

 

[Cyosis]

The first one is correct, the second one is not.

$$(\vec{r} \cdot \nabla)x^2=(x \frac{d}{dx}+y\frac{d}{dy}+z\frac{d}{dz})x^2=2x^2$$

The dx^2/dy and dx^2z terms are 0. The reason we are doing this is so that you understand what the operator does, without that understanding you cannot make the exercise properly. The dot product of a vector and a differential operator does not commute.

Back to the original problem

d/dx_i e_i . (a_j x_j b_k e_k)

This is correct

why can't i just rewrite this as a_i x_i d/dx_j bj?

This is not, can you see why now?

 

[Cyosis]

The first one is correct, the second one is not.

$$(\vec{r} \cdot \nabla)x^2=(x \frac{d}{dx}+y\frac{d}{dy}+z\frac{d}{dz})x^2=2x^2$$

The dx^2/dy and dx^2z terms are 0. The reason we are doing this is so that you understand what the operator does, without that understanding you cannot make the exercise properly. The dot product of a vector and a differential operator does not commute.

Back to the original problem

d/dx_i e_i . (a_j x_j b_k e_k)

This is correct

why can't i just rewrite this as a_i x_i d/dx_j bj?

This is not, can you see why now?

 

[bon]

ok thanks, yes..so is it d/dx_i(a_j x_j) b_k e_i.e_k

=d/dx_i (a_j x_j) b_i

how do i proceed? thanks

 

[Cyosis]

You know that a is a constant vector therefore a_j is just some constant which you can take in front of the differential operator. Now look at the expression d/dx_i x_j, what can you say about that? If you don't see it let i and j run through the values 1 2 3 and see what this operation yields for various combinations.

 

[bon]

ahh so its just the identity dij? so the answer is aibi?

 

[Cyosis]

Correct.

 

[bon]

for curl (a.r)b i get axb is this right too? (here's hoping!)

 

[Cyosis]

Looks good.

 

[bon]

haha..excellent..just solved a harder one also curl (a x r) = 2athanks so much for your great tutoring..understand it now..
:)

 

[Cyosis]

You're welcome and that last one is correct as well.