How do we differentiate the following vector product with respect to $\boldsymbol r$. \begin{equation} \frac{d}{d\boldsymbol r}\bigg[(\boldsymbol \omega \times\boldsymbol r)\cdot (\boldsymbol \omega \times\boldsymbol r)\bigg] \end{equation} I know that the answer is $-\boldsymbol \omega\times(\boldsymbol\omega\times\boldsymbol r)$. Here is my "not so great" working that lead me to a resemblance of the Lagrange identity. \begin{equation} \frac{d}{d\boldsymbol r}\bigg[(\boldsymbol \omega\times \boldsymbol r)\cdot (\boldsymbol \omega\times\boldsymbol r)\bigg]=\frac{d}{d\boldsymbol r}(\boldsymbol\omega \times\boldsymbol r)\cdot (\boldsymbol \omega\times\boldsymbol r)+(\boldsymbol \omega \times\boldsymbol r)\cdot \frac{d}{d\boldsymbol r}(\boldsymbol \omega\times\boldsymbol r) \end{equation} \begin{equation} =\bigg[\frac{d}{d\boldsymbol r}\boldsymbol\omega\times\boldsymbol r\bigg]\cdot \bigg[\boldsymbol\omega\times\boldsymbol r\bigg]+\bigg[\boldsymbol\omega\times\boldsymbol r\bigg]\cdot\bigg[\frac{d}{d\boldsymbol r}\boldsymbol\omega\times\boldsymbol r\bigg] \end{equation} \begin{equation} =2\bigg\{\bigg[\frac{d}{d\boldsymbol r}\boldsymbol\omega\cdot\boldsymbol\omega\bigg]\bigg[\boldsymbol r\cdot \boldsymbol r\bigg]-\bigg[\frac{d}{d\boldsymbol r}\boldsymbol\omega\cdot \boldsymbol r\bigg]\bigg[\boldsymbol r\cdot \boldsymbol \omega\bigg]\bigg\} \end{equation} Thank you,
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1What you have to differentiate is the dot product of two vectors, each of which are cross products. Now the dot product is a scalar, so differentiating it should give you a scalar, not a vector as you seem to think. – Macavity Aug 01 '15 at 14:03
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@Macavity Not true. The derivative is with respect to the components of the vector $r$, so the result in this case should be a vector because, in essence, there are 3 derivatives being computed. – wltrup Aug 01 '15 at 16:47
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@wltrup right. Missed that it is wrt vector $r$. – Macavity Aug 01 '15 at 16:59
3 Answers
If you know what the Kronecker delta $\delta_{ij}$ and the Levi-Civita symbol $\epsilon_{ijk}$ are, there's an even simpler method than that presented by Kelechi Nze above/below.
If this is your first time with these symbols then what I'm going to show you below may look complicated but, believe me, it's a simple and very powerful method, well worth learning. It will help you prove complicated vector identities in no time.
The derivation below looks long only because I'm being quite detailed. Once you get the hang of this method, you can do these things in just a couple of lines.
As it happens, the scalar product of two vectors $A$ and $B$ can be written as
$$A \cdot B = \sum_{i\,=\,1}^{3} A_{i}\,B_{i} = \sum_{i\,=\,1}^{3} \sum_{j\,=\,1}^{3} \delta_{ij}\,A_{i}\,B_{j} $$
but we'll write it as
$$A \cdot B = \delta_{ij}\,A_{i}\,B_{j} $$
where a sum over repeated indices in the same side of an equation is implied (the so-called Einstein summation convention). The indices represent components.
Likewise, the vector product $A \times B$, in 3 dimensions, can be written as
$$(A \times B)_{i} = \epsilon_{ijk}\,A_{j}\,B_{k}$$
again with a sum over repeated indices implied. Note that $i$ is not repeated on the LHS nor is it repeated on the RHS, so there's no sum over $i$ implied. (L/RHS = left/right-hand side)
Using the above, we can write the expression you want to compute as such:
$$\frac{d}{dr_{m}}\Big[\, (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) \,\Big]$$
Note that $i,j,k,r,s$ are all repeated but there is a free index $m$ left over. So,
$$\frac{d}{dr_{m}}\Big[\, (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) \,\Big] = (\epsilon_{ijk}\,\omega_{j}\,\frac{dr_{k}}{dr_{m}})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) + (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\frac{dr_{s}}{dr_{m}}) $$
But $\frac{dr_{k}}{dr_{m}} = \delta_{km}$ and $\frac{dr_{s}}{dr_{m}} = \delta_{sm}$ (it's the general form of $dx/dy = dx/dz = 0$ and $dx/dx = 1$) so
$$\frac{d}{dr_{m}}\Big[\, (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) \,\Big] = (\epsilon_{ijk}\,\omega_{j}\,\delta_{km})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) + (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,\delta_{sm}) $$
Since a sum over $k$ and a sum over $s$ are implied and $\delta_{km}$ is 0 when $k \ne m$ but 1 when $k = m$ (likewise for $s$), we have
$$\frac{d}{dr_{m}}\Big[\, (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) \,\Big] = (\epsilon_{ijm}\,\omega_{j})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) + (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irm}\,\omega_{r}) $$
Now, repeated indices are dummy indices so we can rename them at will (carefully, of course) so we can make the second expression on the right more like the first by exchanging $k$ with $s$ and $r$ with $j$ in the second term:
$$\frac{d}{dr_{m}}\Big[\, (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) \,\Big] = (\epsilon_{ijm}\,\omega_{j})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) + (\epsilon_{irs}\,\omega_{r}\,r_{s})\, (\epsilon_{ijm}\,\omega_{j}) $$
We now see that they're identical, so
$$\frac{d}{dr_{m}}\Big[\, (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) \,\Big] = 2\,(\epsilon_{ijm}\,\omega_{j})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) $$
Now we convert the result back to the "normal" vector notation:
$$\frac{d}{dr_{m}}\Big[\, (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) \,\Big] = 2\,(\epsilon_{ijm}\,\omega_{j})\, (\omega \times r)_{i} $$
It turns out you can cyclically rotate the indices of $\epsilon$ without changing its sign, so $\epsilon_{ijm} = \epsilon_{mij}$ and
$$\frac{d}{dr_{m}}\Big[\, (\epsilon_{ijk}\,\omega_{j}\,r_{k})\, (\epsilon_{irs}\,\omega_{r}\,r_{s}) \,\Big] = 2\,(\epsilon_{mij}\,\omega_{j})\, (\omega \times r)_{i} = 2\,\epsilon_{mij}\,(\omega \times r)_{i}\,\omega_{j} = 2\,((\omega \times r) \times \omega)_{m} $$
Thus,
$$\frac{d}{dr}\Big[\, (\omega \times r) \cdot (\omega \times r) \,\Big] = 2\,((\omega \times r) \times \omega) $$
Now, as we all know, the vector product is anti-commutative, so $A \times B = - B \times A$. Hence,
$$\frac{d}{dr}\Big[\, (\omega \times r) \cdot (\omega \times r) \,\Big] = -2\,\omega \times (\omega \times r) $$
(Alternatively, we could have used the fact that $\epsilon_{mij} = -\epsilon_{mji}$ in a previous step)
Like I said, this looks complicated at first but really isn't and is a very powerful method.
One last thing: sometimes you'll need the following result, which is a pain to prove on its own, and worth memorising:
$$\epsilon_{ijk}\epsilon_{irs} = \delta_{jr}\,\delta_{ks} - \delta_{js}\,\delta_{kr}$$
There's a sum only over $i$, and it's the first index in both $\epsilon$ symbols.
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Wow this is a brilliant answer, thank you for introducing this notation to me! I have spent some time working through it. I am not too sure of you last step after you rotate the indices $\epsilon _{ijm}=\epsilon _{mij}$, in particular how it is okay to swap the order of $\omega_j$ and the bracketed cross product? This is the origin of the negative sign. – AngusTheMan Aug 01 '15 at 18:42
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Thanks, but the brilliance of the answer comes from the awesomeness of the method, not from me. :) I'll edit the answer to clarify that last step. By the way, I'd appreciate if you accepted my answer since you seem to like and prefer it. – wltrup Aug 01 '15 at 18:44
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"how it is okay to swap the order of $\omega_j$ and the bracketed cross product? This is the origin of the negative sign" - remember that we're dealing with the components now, not with the vectors, so it's ok to re-order any product. The non-commutativity of cross products is all taken care of by the $\epsilon$ and its indices. That's why this method is so cool and powerful. You don't need to worry about vectors any more while you're in the middle of doing "index gymnastics", as it's sometimes called. – wltrup Aug 01 '15 at 18:52
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Oh, and wait until you have to prove some vector calculus identities like $\nabla \times (\nabla \times A) = \nabla(\nabla\cdot A) - \nabla^2 A$. This method does it in a line or two, instead of a full page. – wltrup Aug 01 '15 at 18:54
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How can the derivative of a scalar be a vector? Or is the abusive notation intended to be the gradient of a scalar? – Mark Viola Aug 01 '15 at 18:57
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@Dr.MV in the OP's question, the derivative is with respect to the components of the vector r, so there are really 3 derivatives to be computed. In my comment just above, $\nabla$ is the gradient operator. I just got lazy and didn't write all the $\vec{}$ or bolded the vectors. So, yes, abuse of notation. Should have been $\vec{\nabla}\times(\vec{\nabla}\times\vec{A}) = \vec{\nabla}(\vec{\nabla}\cdot\vec{A}) - \nabla^2\vec{A}$. – wltrup Aug 01 '15 at 18:59
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@wltrup I can't thank you enough, one thing I have noticed on wikipedia is that the indices are raised. Should I raise the indices on everything bar the permutation symbol? (just to be proper). So the derivative $dr^m$ and $\omega^j$ and $r^j$ etc ? – AngusTheMan Aug 01 '15 at 19:10
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@wltrup I was referring to the original question in which abusive notation was pervassive. That is the derivative with respect to $r$ of the inner product renders a scalar. Perhaps the abuse was only that $r$ is intended to represent a Cartesian coordinate, although as written could more likely represent the magnitude of the position vector. No worry. – Mark Viola Aug 01 '15 at 19:11
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2You're most welcome. Glad this method has had a positive impact on you. It did wonders for me when I worked with General Relativity. No, don't worry about raising the indices for now. There is a very important distinction between upper and lower indices but you don't have to worry about that until you learn about curvilinear coordinate systems. If you're dealing with cartesian coordinates, you're fine with all of the indices down. – wltrup Aug 01 '15 at 19:12
Here's an (ugly) method:
First write $$\boldsymbol{w}=\begin{pmatrix}w_1\\w_2\\w_3 \end{pmatrix}, \boldsymbol{r} = \begin{pmatrix}r_1\\r_2\\r_3 \end{pmatrix}$$
Then compute \begin{equation} (\boldsymbol w \times\boldsymbol r)\cdot (\boldsymbol w \times\boldsymbol r) = w_2^2r_3^3 + w_3^2r_2^2 - 2w_2r_3w_3r_2 \\ + w_1^2r_3^3 + w_3^2r_1^2 - 2w_1r_3w_3r_1 \\ + w_1^2r_2^3 + w_2^2r_1^2 - 2w_1r_2w_2r_1 \end{equation}
Assuming that $\boldsymbol{w}$ is independent of $\boldsymbol{r}$ we can now calculate $\frac{d}{d\boldsymbol{r}}\bigg[(\boldsymbol \omega \times\boldsymbol r)\cdot (\boldsymbol \omega \times\boldsymbol r)\bigg]$ with ease.
$$\frac{d}{d\boldsymbol{r}}\bigg[(\boldsymbol \omega \times\boldsymbol r)\cdot (\boldsymbol \omega \times\boldsymbol r)\bigg] = \frac{d}{d\boldsymbol{r}} \bigg[w_2^2r_3^3 + w_3^2r_2^2 - 2w_2r_3w_3r_2 + w_1^2r_3^3 + w_3^2r_1^2 - 2w_1r_3w_3r_1 + w_1^2r_2^3 + w_2^2r_1^2 - 2w_1r_2w_2r_1\bigg] $$
As $\boldsymbol \omega $ is independent of $\boldsymbol r$ all $\frac{\partial \omega_k}{\partial r_k}$ terms are zero. As each $r_k$ is independent of every other $r_j$ where $j \neq k$ all $\frac{\partial r_k}{\partial r_j}$ terms are zero so we obtain $$\frac{d}{d\boldsymbol{r}} \bigg[w_2^2r_3^3 + w_3^2r_2^2 - 2w_2r_3w_3r_2 + w_1^2r_3^3 + w_3^2r_1^2 - 2w_1r_3w_3r_1 + w_1^2r_2^3 + w_2^2r_1^2 - 2w_1r_2w_2r_1\bigg] = \pmatrix{2(w_3^2 + w_2^2)r_1 - 2(r_3w_3 + r_2w_2)w_1 \\ 2(w_3^2 + w_1^2) - 2(r_3w_3+r_1w_1)w_2 \\ 2(w_2^2+w_1^2) - 2(r_2w_2 + r_3w_1)w_3} = -2 \boldsymbol{w}\times(\boldsymbol{w}\times\boldsymbol{r}) $$
Which is close to your answer bar a factor of 2.
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@Kelechi Nze why do you have $r_3^3$ and also $r_2^3$ in your equation? Can you please give a more detailed demonstration of what you did there? I really don't understand how you made the matrix at the end of demonstration.
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