# Einstein Summation Convention

1. Feb 28, 2010

### joe:)

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

Basically need to use einstein's summation convention to find the grad of (mod r)^n and a.r where a is a vector and r = (x,y,z)

2. Relevant equations

3. The attempt at a solution

Not sure where to begin really.. :S

grad (mod r)^n= (d/dx, d/dy, d/dz) of root (X1^2 + x2^2 + x3^2)^n..

Just not sure what to do now..
thanks

2. Feb 28, 2010

### gabbagabbahey

If you write it out using unit vector notation instead of ordered triplets, you will see a sum of three terms. The Einstein summation convention can be used to reduce that sum to just one term with an implied summation over an index.

3. Feb 28, 2010

### joe:)

sorry not sure what you mean here..

How can i write (mod r)^n in unit vector notation?

4. Feb 28, 2010

### gabbagabbahey

You can't, it's a scalar. However, $\textbf{r}$ and $\mathbf{\nabla}(r^n)$ are vectors (Are you comfortable with using boldface type to denote vectors, and normal font for scalars?...If so, you can simply write $r$ to represent the modulus of the position vector $\textbf{r}$ as I have done).

5. Feb 28, 2010

### joe:)

Sorry..I'm not really very familiar with the eistein notation and still cant see how to do this :(

6. Feb 28, 2010

### gabbagabbahey

Start by writing $\textbf{r}$ in unit vector notation...

7. Feb 28, 2010

### joe:)

so r=r'r where r' is a unit vector

8. Feb 28, 2010

### gabbagabbahey

I should have been more specific, try writing it in terms of Cartesian unit vectors.

9. Feb 28, 2010

### joe:)

r=(x1i, x2j, x3k)?

10. Feb 28, 2010

### gabbagabbahey

Why do you have an ordered triplet with unit vector inside?

I would say $\textbf{r}=x\textbf{i}+y\textbf{j}+z\textbf{k}$....does this look familiar to you?

You can rewrite this by defining $x_1\equiv x$, $x_2\equiv y$ and $x_3\equiv z$ as well as $\textbf{e}_1\equiv \textbf{i}$, $\textbf{e}_2\equiv \textbf{j}$, and $\textbf{e}_3\equiv \textbf{k}$ to get;

$$\textbf{r}=x\textbf{i}+y\textbf{j}+z\textbf{k}=x_1\textbf{e}_1+x_2\textbf{e}_2+x_3\textbf{e}_3$$

Using the Einstein summation convention, this can be written as $\textbf{r}=x_i\textbf{e}_i$

Now, try rewriting the gradient operator,

$$\mathbf{\nabla}=\frac{\partial}{\partial x}\textbf{i}+\frac{\partial}{\partial y}\textbf{j}+\frac{\partial}{\partial z}\textbf{k}$$ using the same definitions...

11. Feb 28, 2010

### joe:)

Ahh this helps. Thanks

so del is eidi??

12. Feb 28, 2010

### joe:)

how do i apply this on the r^n now?

13. Feb 28, 2010

### gabbagabbahey

Sure, if you define $\partial_i\equiv\frac{\partial}{\partial x_i}$, you get $\mathbf{\nabla}=\textbf{e}_i\partial_i$

And how about $r$, the modulus of $\textbf{r}$...what do you get for that in index notation?

14. Feb 28, 2010

### joe:)

um xi2??

15. Feb 28, 2010

### gabbagabbahey

That's not very good notationally (there is no repeated index in your expression, so no summation is implied); I would say $r^2=x_ix_i$, and hence $r^n=(x_ix_i)^{n/2}$.

So, what is $\mathbf{\nabla}(r^n)$ in index notation?

16. Feb 28, 2010

### joe:)

Thank you..

Sorry I imagine that this is painful for you..sorry :(

so i think it is eidi(xixi)^n/2

But i don't know how to simplify this? ahhhhhh :S

Is there some key concept im missing..

17. Feb 28, 2010

### gabbagabbahey

Your welcome, and don't worry; a lot of students struggle with this stuff when they are first introduced to it.

Another rule when using index notation is that the same index should not be used more than twice in a single term of an expression (If it's used once, it is a free index. If it's used twice, it's a repeated index and a summation is implied. If it is used 3 or 4 or more times, it's just nonsense)

For that reason, this should be written as $\mathbf{\nabla}r^n=\textbf{e}_i\partial_i(x_jx_j)^{n/2}$. To simplify this, just use the product and chain rules to calculate the derivatives involved.

18. Feb 28, 2010

### joe:)

So now im just thinking about how the differentiation rules might apply here.

Is it basically a product rule i.e. d/dx of x^n/2 x^n/2 so it simplifies to nei? but thats nonsensical since there needs to be two indices to convey a summation

arghhhhhhhhh :( :(

19. Feb 28, 2010

### gabbagabbahey

Start by applying the chain rule:

$$\partial_i(x_jx_j)^{n/2}=(x_kx_k)^{n/2-1}\partial_i(x_jx_j)$$

(I switched the indices from $j$ to $k$ for the first sum in order to comply with the index notation rule I previously stated)

Now, use the product rule to calculate [itex]\partial_i(x_jx_j)[/tex]

20. Feb 28, 2010

### joe:)

Thanks..

Is that 2(dixj)xj?

What now:S?!