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Einstein Summation Convention 
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#1
Feb2810, 12:14 PM

P: 28

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 (X_{1}^2 + x_{2}^2 + x_{3}^2)^n.. Just not sure what to do now.. thanks 


#2
Feb2810, 12:54 PM

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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
Feb2810, 01:21 PM

P: 28

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


#4
Feb2810, 01:25 PM

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Einstein Summation Convention



#5
Feb2810, 01:47 PM

P: 28




#6
Feb2810, 02:29 PM

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Start by writing [itex]\textbf{r}[/itex] in unit vector notation...



#7
Feb2810, 02:47 PM

P: 28




#8
Feb2810, 02:48 PM

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#9
Feb2810, 02:56 PM

P: 28




#10
Feb2810, 03:06 PM

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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\text bf{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
Feb2810, 03:15 PM

P: 28

so del is e_{i}d_{i}?? 


#12
Feb2810, 03:18 PM

P: 28

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



#13
Feb2810, 03:19 PM

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


#14
Feb2810, 04:13 PM

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#15
Feb2810, 04:19 PM

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


#16
Feb2810, 04:34 PM

P: 28

Sorry I imagine that this is painful for you..sorry :( so i think it is e_{i}d_{i}(x_{i}x_{i})^n/2 But i don't know how to simplify this? ahhhhhh :S Is there some key concept im missing.. 


#17
Feb2810, 04:45 PM

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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
Feb2810, 05:16 PM

P: 28

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 :( :( 


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