Einstein Summation Convention

[joe:)]

Homework Statement

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)

Homework Equations

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

 

[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.

 

[joe:)]

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.

sorry not sure what you mean here..

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

 

[gabbagabbahey]

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

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).

 

[joe:)]

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).

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

 

[gabbagabbahey]

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

 

[joe:)]

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

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

 

[gabbagabbahey]

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

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

 

[joe:)]

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

r=(x₁i, x₂j, x₃k)?

 

[gabbagabbahey]

r=(x₁i, x₂j, x₃k)?

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...

 

[joe:)]

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...

Ahh this helps. Thanks

so del is e_id_i??

 

[joe:)]

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

 

[gabbagabbahey]

Ahh this helps. Thanks

so del is e_id_i??

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?

 

[joe:)]

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?

um x_i²??

 

[gabbagabbahey]

um x_i²??

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?

 

[joe:)]

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?

Thank you..

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

so i think it is e_id_i(x_ix_i)^n/2

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

Is there some key concept I am missing..

 

[gabbagabbahey]

Thank you..

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

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

so i think it is e_id_i(x_ix_i)^n/2

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.

 

[joe:)]

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.

Thanks for your patience.

So now I am 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 that's nonsensical since there needs to be two indices to convey a summation


arghhhhhhhhh :( :(

 

[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]

 

[joe:)]

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]

Thanks..

Is that 2(d_ix_j)x_j?

What now:S?!

 

[gabbagabbahey]

Thanks..

Is that 2(d_ix_j)x_j?

What now:S?!

Yup

Now, calculate [tex]\partial_ix_j=\frac{\partial x_j}{\partial x_i}[/itex]...what happens if $i\neq j$? What about if $i= j$? (If you aren't sure, remember how $x_1$, $x_2$ and $x_3$ are defined)

 

[joe:)]

Yup

Now, calculate [tex]\partial_ix_j=\frac{\partial x_j}{\partial x_i}[/itex]...what happens if $i\neq j$? What about if $i= j$? (If you aren't sure, remember how $x_1$, $x_2$ and $x_3$ are defined)

i=j then it equals 1 otherwise 0?

So is it (x_kx_k)^n/2 -1 times 6x_j??

ahh i need to finish to go to bed :(

 

[gabbagabbahey]

i=j then it equals 1 otherwise 0?

Yup.

This is the exact property that defines the Kronecker delta function (get used to seeing this function),

$$\delta_{ij}=\left\{\begin{array}{llr}1 & , & i=j \\ 0 & , & i\neq j\end{array}\right.$$

So is it (x_kx_k)^n/2 -1 times 6x_j??

ahh i need to finish to go to bed :(

Not quite,

$$\partial_i(x_jx_j)=2(\partial_i x_j)x_j=2\delta_{ij}x_j$$

Now, $\delta_{ij}$ is zero for $i\neq j$, and equals one for $i=j$. So, $\delta_{ij}x_j=x_i$ (You are summing over all values of $j$, and the only non-zero term is the $j=i$ term)

 

[joe:)]

Yup.

This is the exact property that defines the Kronecker delta function (get used to seeing this function),

$$\delta_{ij}=\left\{\begin{array}{llr}1 & , & i=j \\ 0 & , & i\neq j\end{array}\right.$$



Not quite,

$$\partial_i(x_jx_j)=2(\partial_i x_j)x_j=2\delta_{ij}x_j$$

Now, $\delta_{ij}$ is zero for $i\neq j$, and equals one for $i=j$. So, $\delta_{ij}x_j=x_i$ (You are summing over all values of $j$, and the only non-zero term is the $j=i$ term)

Ahh yes thank you..

So how do i simplify the whole thing..is it (x_kx_k)^n/2-1 times 2deltaij xj..but what then is the final simplified answer?

Thanks.. :) I think we are nearly there? :P

 

[gabbagabbahey]

Ahh yes thank you..

So how do i simplify the whole thing..is it (x_kx_k)^n/2-1 times 2deltaij xj..but what then is the final simplified answer?

You tell me, $\delta_{ij}x_j$ simplifies to $x_i$...what about $(x_k x_k)^{n/2-1}$?...that looks like a power of $r$ to me

What does that give you for $\mathbf{\nabla}r^n$?

 

[joe:)]

You tell me, $\delta_{ij}x_j$ simplifies to $x_i$...what about $(x_k x_k)^{n/2-1}$?...that looks like a power of $r$ to me

What does that give you for $\mathbf{\nabla}r^n$?

2x_ie_ir^n-1?

 

[gabbagabbahey]

2x_ie_ir^n-1?

Why $r^{n-1}$? Remember, $r^2=(x_kx_k)$...what exponent do you have to raise each side of this equation to in order for the LHS to be $(x_kx_k)^{n/2-1}$?

Also, what is $x_i\textbf{e}_i$?