Einstein Summation Convention

In summary, to find the gradient of (mod r)^n and a.r, you can use the Einstein summation convention to reduce the sum of three terms to just one term with an implied summation over an index. You can write r in unit vector notation and rewrite the gradient operator, \mathbf{\nabla}, using the same definitions. The modulus of r in index notation is expressed as r^2=x_ix_i, and hence r^n=(x_ix_i)^{n/2}. To simplify, just use the product and chain rules to calculate the derivatives.
  • #1
joe:)
28
0

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
 
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  • #2
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
gabbagabbahey said:
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?
 
  • #4
joe:) said:
How can i write (mod r)^n in unit vector notation?

You can't, it's a scalar. However, [itex]\textbf{r}[/itex] and [itex]\mathbf{\nabla}(r^n)[/itex] are vectors (Are you comfortable with using boldface type to denote vectors, and normal font for scalars?...If so, you can simply write [itex]r[/itex] to represent the modulus of the position vector [itex]\textbf{r}[/itex] as I have done).
 
  • #5
gabbagabbahey said:
You can't, it's a scalar. However, [itex]\textbf{r}[/itex] and [itex]\mathbf{\nabla}(r^n)[/itex] are vectors (Are you comfortable with using boldface type to denote vectors, and normal font for scalars?...If so, you can simply write [itex]r[/itex] to represent the modulus of the position vector [itex]\textbf{r}[/itex] as I have done).

Sorry..I'm not really very familiar with the eistein notation and still can't see how to do this :(
 
  • #6
Start by writing [itex]\textbf{r}[/itex] in unit vector notation...
 
  • #7
gabbagabbahey said:
Start by writing [itex]\textbf{r}[/itex] in unit vector notation...

so r=r'r where r' is a unit vector
 
  • #8
joe:) said:
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.
 
  • #9
gabbagabbahey said:
I should have been more specific, try writing it in terms of Cartesian unit vectors.

r=(x1i, x2j, x3k)?
 
  • #10
joe:) said:
r=(x1i, x2j, x3k)?

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

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

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


[tex]\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[/tex]

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

Now, try rewriting the gradient operator,

[tex]\mathbf{\nabla}=\frac{\partial}{\partial x}\textbf{i}+\frac{\partial}{\partial y}\textbf{j}+\frac{\partial}{\partial z}\textbf{k}[/tex] using the same definitions...
 
  • #11
gabbagabbahey said:
Why do you have an ordered triplet with unit vector inside?

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

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


[tex]\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[/tex]

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

Now, try rewriting the gradient operator,

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

Ahh this helps. Thanks

so del is eidi??
 
  • #12
how do i apply this on the r^n now?
 
  • #13
joe:) said:
Ahh this helps. Thanks

so del is eidi??

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

And how about [itex]r[/itex], the modulus of [itex]\textbf{r}[/itex]...what do you get for that in index notation?
 
  • #14
gabbagabbahey said:
Sure, if you define [itex]\partial_i\equiv\frac{\partial}{\partial x_i}[/itex], you get [itex]\mathbf{\nabla}=\textbf{e}_i\partial_i[/itex]

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

um xi2??
 
  • #15
joe:) said:
um xi2??

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

So, what is [itex]\mathbf{\nabla}(r^n)[/itex] in index notation?
 
  • #16
gabbagabbahey said:
That's not very good notationally (there is no repeated index in your expression, so no summation is implied); I would say [itex]r^2=x_ix_i[/itex], and hence [itex]r^n=(x_ix_i)^{n/2}[/itex].

So, what is [itex]\mathbf{\nabla}(r^n)[/itex] in index notation?

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 I am missing..
 
  • #17
joe:) said:
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 eidi(xixi)^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 [itex]\mathbf{\nabla}r^n=\textbf{e}_i\partial_i(x_jx_j)^{n/2}[/itex]. To simplify this, just use the product and chain rules to calculate the derivatives involved.
 
  • #18
gabbagabbahey said:
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 [itex]\mathbf{\nabla}r^n=\textbf{e}_i\partial_i(x_jx_j)^{n/2}[/itex]. 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 :( :(
 
  • #19
Start by applying the chain rule:

[tex]\partial_i(x_jx_j)^{n/2}=(x_kx_k)^{n/2-1}\partial_i(x_jx_j)[/tex]

(I switched the indices from [itex]j[/itex] to [itex]k[/itex] 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
gabbagabbahey said:
Start by applying the chain rule:

[tex]\partial_i(x_jx_j)^{n/2}=(x_kx_k)^{n/2-1}\partial_i(x_jx_j)[/tex]

(I switched the indices from [itex]j[/itex] to [itex]k[/itex] 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(dixj)xj?

What now:S?!
 
  • #21
joe:) said:
Thanks..

Is that 2(dixj)xj?

What now:S?!

Yup:approve:

Now, calculate [tex]\partial_ix_j=\frac{\partial x_j}{\partial x_i}[/itex]...what happens if [itex]i\neq j[/itex]? What about if [itex]i= j[/itex]? (If you aren't sure, remember how [itex]x_1[/itex], [itex]x_2[/itex] and [itex]x_3[/itex] are defined)
 
  • #22
gabbagabbahey said:
Yup:approve:

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

i=j then it equals 1 otherwise 0?

So is it (xkxk)^n/2 -1 times 6xj??

ahh i need to finish to go to bed :(
 
  • #23
joe:) said:
i=j then it equals 1 otherwise 0?

Yup.:approve:

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

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

So is it (xkxk)^n/2 -1 times 6xj??

ahh i need to finish to go to bed :(

Not quite,

[tex]\partial_i(x_jx_j)=2(\partial_i x_j)x_j=2\delta_{ij}x_j[/tex]

Now, [itex]\delta_{ij}[/itex] is zero for [itex]i\neq j[/itex], and equals one for [itex]i=j[/itex]. So, [itex]\delta_{ij}x_j=x_i[/itex] (You are summing over all values of [itex]j[/itex], and the only non-zero term is the [itex]j=i[/itex] term)
 
  • #24
gabbagabbahey said:
Yup.:approve:

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

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



Not quite,

[tex]\partial_i(x_jx_j)=2(\partial_i x_j)x_j=2\delta_{ij}x_j[/tex]

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

Ahh yes thank you..

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

Thanks.. :) I think we are nearly there? :P
 
  • #25
joe:) said:
Ahh yes thank you..

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

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

What does that give you for [itex]\mathbf{\nabla}r^n[/itex]?
 
  • #26
gabbagabbahey said:
You tell me, [itex]\delta_{ij}x_j[/itex] simplifies to [itex]x_i[/itex]...what about [itex](x_k x_k)^{n/2-1}[/itex]?...that looks like a power of [itex]r[/itex] to me:wink:

What does that give you for [itex]\mathbf{\nabla}r^n[/itex]?

2xieir^n-1?
 
  • #27
joe:) said:
2xieir^n-1?

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

Also, what is [itex]x_i\textbf{e}_i[/itex]?
 

1. What is the Einstein Summation Convention?

The Einstein Summation Convention, also known as Einstein notation or Einstein's summation convention, is a mathematical convention used to simplify expressions involving summations of multidimensional arrays or tensors. It is named after the famous physicist Albert Einstein, who introduced this notation in his theory of general relativity.

2. How does the Einstein Summation Convention work?

In this convention, repeated indices in a tensor expression are implicitly summed over all possible values. This means that instead of explicitly writing out all the summation terms, the summation is implied by repeating the index. This simplifies the notation and makes it easier to write and understand complicated tensor expressions.

3. What are the advantages of using the Einstein Summation Convention?

The main advantage of using this convention is that it reduces the amount of notation required to write a tensor expression, making it more compact and easier to work with. It also helps to avoid errors that can occur when writing out long summations. Additionally, it allows for a more intuitive understanding of tensor operations and makes it easier to perform calculations.

4. Are there any limitations to the Einstein Summation Convention?

While this convention is useful for simplifying tensor expressions, it does have some limitations. It can only be used for tensors with fixed dimensions and indices that are repeated exactly twice in an expression. It also does not work well for expressing operations involving non-linear or non-polynomial functions.

5. Can the Einstein Summation Convention be used in all branches of science?

Yes, this convention can be used in various fields of science, including physics, mathematics, and engineering. It is particularly useful in fields that deal with vector and tensor operations, such as general relativity, electromagnetism, and fluid mechanics. However, it may not be applicable in fields that require more complex mathematical notation or deal with non-linear systems.

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