# Homework Help: Vector Calculus identity

1. Jan 9, 2010

### hhhmortal

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

I'm a bit confused as to the following vector calculus identity:

[∇ (∇.A)]_i = (δ/δx_i )( δA_j/δx_j)

Shouldn’t it be = (δ/δx_i )( δA_i/δx_i) why is it ‘j’ if we are taking it over ‘i’ ?

Thanks.

2. Jan 9, 2010

### phsopher

j is a "dummy" index, i.e. it is summed over. You could name it whatever you want. Remember that ∇.A is a scalar so it can't have any indices.

$$\nabla \cdot \mathbf A = \partial_1 A_1 + \partial_2 A_2 + \partial_3 A_3 = \sum_{j=1}^3 \partial_j A_j \equiv \partial_j A_j$$

3. Jan 9, 2010

### hhhmortal

Ok. But why are we summing over 'j'? This is where Im getting confused. Shouldn't it be 'i'

4. Jan 10, 2010

### Altabeh

Because j is the repeated index (it appears as an upper and lower index simultaneously) and due to Einstein summation convention, the repeated index must be summed over all possible values for that index.

AB

5. Jan 10, 2010

### hhhmortal

Ok I'm missing something out here. I'm summing over 'i' and 'i' appears as the repeated index..how does 'j' come into this. I can't see how it appears as a lower and upper index simultaneously.

6. Jan 10, 2010

### phsopher

Let's start from the beginning. First consider ∇.A. This is

$\nabla \cdot \mathbf A = \partial_1 A_1 + \partial_2 A_2 + \partial_3 A_3 = \sum_{j=1}^3 \partial_j A_j$

Do you agree so far? The j is just a summation index, i.e. it is not a fixed number but rather it gets all values from 1 to 3. We could have used any other letter instead of j, it doesn't matter. It's just a label.

If we use the Einstein's summation convention that means we drop the summation sign and agree that whenever an index appears twice it is summed over. Thus we write $\nabla \cdot \mathbf A = \partial_j A_j$

Now consider ∇(∇.A). This is a vector and we decide to only consider one component - the i'th component. Now i is a fixed index - it is not summed over. i.e. it refers to a particular component we are considering. So using our earlier expression for ∇.A we write $[\nabla(\nabla \cdot \mathbf A)]_i = \partial_i (\partial_j A_j)$

Does that make it more clear?