Index Notation, taking derivative

[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).

 

[Orodruin]

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.

 

[hellomrrobot]

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.

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

 

[Avodyne]

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

 

[hellomrrobot]

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

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 $dp/dx_{1}$ ?

 

[Orodruin]

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

 

[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.