# Index Notation, taking derivative

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1. Sep 14, 2015

### hellomrrobot

Can anyone explain how to take the derivative of (Aδij),j? I know that since there is a repeating subscript I have to do the summation then take the derivative, but I am not sure how to go about that process because there are two subscripts (i and j) and that it is the Kronecker's Delta (not sure if this would change anything).

2. Sep 15, 2015

### Orodruin

Staff Emeritus
You really should be using sub and super scripts because it becomes difficult to imagine what you intend, I am going to assume that you are meaning $(A\delta_{ij})_{,j} = \partial_j A\delta_{ij}$.

The wonderful thing about index notation is that you can treat each term as if it was just a number and in the end you sum over repeated indices. So what you need to think about is what is the partial derivative $\partial_k (A \delta_{ij})$. Once you have done that you can let $k = j$ and perform the sum.

3. Sep 15, 2015

### hellomrrobot

How would the partial derivative look like? This is what I got $(dp/dx_{j}) \delta_{ij} + p (d \delta_{ij}/dx_{j})$

4. Sep 15, 2015

### Avodyne

The second term is zero, since $\delta_{ij}$ is a constant.

5. Sep 15, 2015

### hellomrrobot

Okay so after I do the summation, I would get $(dp/dx_{1})A\delta_{i1} + (dp/dx_{2})A\delta_{i2} + (dp/dx_{3})A\delta_{i3}$

Then if I set i=1, I would get just $dp/dx_{1}$ ?

6. Sep 15, 2015

### Orodruin

Staff Emeritus
In general you can use the property $\delta_{ij}A_j = A_i$ as long as $A_i$ is any indexed expression.

7. Sep 15, 2015

### Avodyne

OP, I'm not sure what you're doing with the $p$ and the $A$, so let's go back to basics:

Suppose $A$ is a scalar and $T_{ij}$ is a two-index tensor. Consider $(AT_{ij})_{,k}$. (Later we will set $T_{ij}=\delta_{ij}$ and $k=j$.) Then, using the product rule for derivatives, we have $(AT_{ij})_{,k}=A_{,k}T_{ij}+A(T_{ij})_{,k}$. (As a side comment, the standard notation would be to write $T_{ij,k}$ instead of $(T_{ij})_{,k}$.)

In the case of interest, we have $T_{ij}=\delta_{ij}$. This is constant, and so its derivatives vanish: $(\delta_{ij})_{,k}=0$. We are left with $(A\delta_{ij})_{,k}=A_{,k}\delta_{ij}$.

Now let $k=j$, with an implicit sum over the repeated $j$ index. Then we have $(A\delta_{ij})_{,j}=A_{,j}\delta_{ij}$. We can perform the sum over $j$ using the general rule given by Orodruin: $A_{,j}\delta_{ij}=A_{,i}$. So the final result is $(A\delta_{ij})_{,j}=A_{,i}$.

Now you can set $i$ to a particular value (1, 2, or 3) if you like.