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December 20th, 2014, 07:53 PM  #1 
Newbie Joined: Jun 2012 Posts: 10 Thanks: 0  Gradient of the dot product of unit vectors
Hello, I hope this is the right place for this question. I need to find the gradient of a dot product of unit vectors, $\displaystyle \nabla({\bf \hat{r}} \cdot {\bf \hat{s}})$, in terms of the original vectors $\displaystyle {\bf r}$ and $\displaystyle {\bf s}$. I have the answer already, but would like to know the quickest / best way of getting there. The answer is: $\displaystyle \frac{1}{{\bf r}}\left[({\bf \hat{r}} \cdot {\bf \hat{s}})\,{\bf \hat{r}} + {\bf \hat{s}}\right] + \frac{1}{{\bf s}}\left[({\bf \hat{r}} \cdot {\bf \hat{s}})\,{\bf \hat{s}} + {\bf \hat{r}}\right]$, I've been reliably told that the derivation is "trivially easy" using tensor derivatives. However, my knowledge of tensor calculus is apparently so woeful that trying my hand at Christoffel symbols, etc, has ended is only frustration. Can anyone help? Cheers 
December 23rd, 2014, 07:31 PM  #2 
Newbie Joined: Jun 2012 Posts: 10 Thanks: 0  I think I have it...
So, after posting the question, I took myself off to YouTube and watched the first ~20 videos of Pavel Grinfeld's wonderful Tensor Calculus course (link to first video) to try to get a handle on it. This is what I ultimately came up with. Comments on correctness/incorrectness and rigour/lack of rigour welcome. (I use unbolded letters for magnitudes ($\displaystyle r \equiv {\bf r}$.) We essentially want the covariant derivative. $\displaystyle \begin{align} \nabla_k({\bf \hat{r}}\cdot{\bf \hat{s}}) &= \nabla_k\left(\frac{{\bf r}}{r}\cdot\frac{{\bf s}}{s}\right) \\ &= \nabla_k\left(\frac{{\bf r}_i {\bf s}^i}{rs}\right) \\ &= \nabla_k{\bf r}_i\left(\frac{{\bf s}^i}{rs}\right) + \nabla_k{\bf s}^i\left(\frac{{\bf r}_i}{rs}\right) + \nabla_k\frac{1}{r}\left(\frac{{\bf r}_i{\bf s}^i}{s}\right) + \nabla_k\frac{1}{s}\left(\frac{{\bf r}_i{\bf s}^i}{r}\right) \\ \end{align}$ In spherical coordinates, the first term is $\displaystyle \begin{align} \nabla_k{\bf r}_i\left(\frac{{\bf s}^i}{rs}\right) &= \left(\frac{\partial{\bf r}_i}{\partial Z^k}  \Gamma^m_{ki}{\bf r}_m\right)\frac{{\bf s}^i}{rs} \\ &= (1  0)\frac{{\bf s}^i}{rs} \\ &= \frac{{\bf s}^i}{rs}, \end{align}$ where the Christoffel term vanishes because the Christoffel symbol is zero whenever both $\displaystyle {\bf r}_m$ and $\displaystyle {\bf s}^i$ are not zero (i.e. whenever $\displaystyle m=i=1$), and where the partial derivative term is unity when $\displaystyle k=1$ (i.e. $\displaystyle Z^k$ is the radial coordinate), and zero otherwise. So, in fact, we could have just said that in spherical coordinates, $\displaystyle \begin{align} \nabla_k{\bf r}_i &= 1, \end{align}$ Similarly, $\displaystyle \begin{align} \nabla_k{\bf s}^i &= 1, \end{align}$ For the other terms, we have $\displaystyle \begin{align} \nabla_k\frac{1}{r} &= \frac{1}{r^2}\nabla_k{r} \\ &= \frac{{\bf \hat{r}}}{r^2} \end{align}$ and $\displaystyle \begin{align} \nabla_k\frac{1}{s} &= \frac{1}{s^2}\nabla_k{s} \\ &= \frac{{\bf \hat{s}}}{s^2}. \end{align}$ Altogether, this makes $\displaystyle \begin{align} \nabla_k({\bf \hat{r}}\cdot{\bf \hat{s}}) &= \frac{{\bf s}}{rs} + \frac{{\bf r}}{rs}  \frac{{\bf \hat{r}}}{r^2}\left(\frac{{\bf r}\cdot{\bf s}}{s}\right)  \frac{{\bf \hat{s}}}{s^2}\left(\frac{{\bf r}\cdot{\bf s}}{r}\right) \\ &= \frac{1}{r}\left[({\bf \hat{r}}\cdot{\bf \hat{s}})\,{\bf \hat{r}} + {\bf \hat{s}}\right] + \frac{1}{s}\left[({\bf \hat{r}}\cdot{\bf \hat{s}})\,{\bf \hat{s}} + {\bf \hat{r}}\right]. \end{align}$ 

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