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
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.
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).
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...
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?
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 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..
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 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 :( :(
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]