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The wikipedia formula for the gradient of a dot product is given as

$$\nabla(a\cdot b) = (a\cdot\nabla)b +(b\cdot \nabla)a + a\times(\nabla\times b)+ b\times (\nabla \times a)$$

However, I also found the formula $$\nabla(a\cdot b) = (\nabla a)\cdot b + (\nabla b)\cdot a $$

So... what is going on here? The second formula seems much easier. Are these equivalent?

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    $\begingroup$ are they from the same article? Which is to say, where did you get these? $\endgroup$ – Will Jagy Sep 17 '13 at 4:21
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    $\begingroup$ It bears mentioning that the second formula works for any combination of dimensions, while the first works only when $a,b$ are $3$-vectors and you are taking the gradient with respect to a $3$-vector. $\endgroup$ – user7530 Aug 27 '14 at 16:35
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    $\begingroup$ Gradient is a vector and the second formula is scalar. It can not be right. $\endgroup$ – Herman Jaramillo Mar 16 '17 at 1:44
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    $\begingroup$ @HermanJaramillo, Gradient is a vector, and the second formula IS a vector, since $\nabla a$ is a dyadic. $\endgroup$ – Vladimir Vargas Nov 20 '17 at 23:28
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    $\begingroup$ One may have a look at the original Wikipedia article $\endgroup$ – Harry49 Aug 16 '18 at 12:55
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They are basically the same. For the first identity, you could refer to my proof using Levi-Civita notation here. And for the second, you should know that $\nabla a=\left(\frac{\partial a_j}{\partial x_i}\right)=\left(\frac{\partial a_i}{\partial x_j}\right)^T$ is a matrix and dot product is exactly matrix multiplication. So the proof is $$(\nabla a)\cdot b+(\nabla b)\cdot a=\left(\frac{\partial a_j}{\partial x_i}b_j+\frac{\partial b_j}{\partial x_i}a_j\right)e_i=\frac{\partial(a_jb_j)}{\partial x_i}e_i=\nabla(a\cdot b)$$

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  • $\begingroup$ Wouldn't the last expression become $(\nabla(a\cdot b))_i$? (Otherwise you're loosing the $i$) $\endgroup$ – HelloGoodbye Aug 27 '14 at 16:24
  • $\begingroup$ @HelloGoodbye Strictly speaking, I missed $e_i$ in the middle expressions. $\endgroup$ – Shuchang Aug 27 '14 at 16:25
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    $\begingroup$ @Shuchang: In fact, they are vastly different since in the first formula $a$ and $b$ are vector fields (and the dot means "scalar product"), while in the second formula $a$ and $b$ are functions and the dot is the usual product of functions. $\endgroup$ – Alex M. Nov 16 '17 at 15:04
  • $\begingroup$ @AlexM. Am I right in thinking this is a problem with syntax. That they evaluate to the same thing but are not syntactically equivalent? $\endgroup$ – samerivertwice Dec 5 '17 at 17:34
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    $\begingroup$ @Alex M. You've guessed what the poster meant in assuming that $a$ doesn't have the same meaning in both statements. And of course the gradient is defined for functions, which $a\cdot b$ is regardless of interpretation-I have no idea what you're getting at with that comment. $\endgroup$ – Kevin Carlson Feb 27 '18 at 14:26
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Since there are not many signs that one may easily use in mathematical notations, many of these symbols are overloaded. In particular, the dot "$\cdot$" is used in the first formula to denote the scalar product of two vector fields in $\mathbb R^3$ called $a$ and $b$, while in the second formula it denotes the usual product of the functions $a$ and $b$. This means that both formulae are valid, but each one is so only in its proper context.

(It is scary to see that the answers and comments that were wrong collected the most votes!)

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  • $\begingroup$ If they were scalar functions, the second identity would say simply $\nabla(ab)=b\nabla a+a\nabla b$ ; no need for dots or parentheses. The Wikipedia link does show a direct "$=$" between the two formulas (with slightly different notation), showing that $a$ and $b$ are still vectors. $\endgroup$ – mr_e_man Oct 23 '18 at 3:26
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Let us use the following index and shorthand notation. $u_{,i}=\displaystyle{\frac{\partial u}{\partial x_i}}$ . $x_1=x ,x_2=y, x_3=z$. Einstein notation. Repeated index means summation over it, and $[.]_i$ the i-th compnent of whatever is inside the square brackets $[]$.

Then \begin{eqnarray} [\nabla (\mathbf{a} \cdot \mathbf{b})]_i = (a_j b_j)_{,i} = a_{j,i}b_j + a_j b_{j,i}. \end{eqnarray}

That is all, I do not see anything more complicated than this. The two sums are matrix vector multiplications. Note that $a_{j,i} b_j$ means the matrix $\partial a_j/\partial x_i$ times the vector $b_j$. You can write this in two different forms \begin{eqnarray} (\nabla \mathbf{a}) \cdot \mathbf{b}= (\mathbf{b} \cdot \nabla) \mathbf{a} = \left ( \begin{array}{c} \displaystyle{b_1 \frac{\partial a_1}{\partial x} + b_2 \frac{\partial a_1}{\partial y} + b_ 3 \frac{\partial a_1}{\partial z}} \\ \\ \displaystyle{b_1 \frac{\partial a_2}{\partial x} + b_2 \frac{\partial a_2}{\partial y} + b_3 \frac{\partial a_2}{\partial z}} \\ \\ \displaystyle{b_1 \frac{\partial a_3}{\partial x} + b_2 \frac{\partial a_3}{\partial y} + b_3 \frac{\partial a_3}{\partial z}} \end{array} \right ) \end{eqnarray} Where the symbol $\nabla \mathbf{a}$ means a matrix. The matrix whose rows are gradients of the components $a_1,a_2,a_3$ respectively. To be more precise the vector $\mathbf{b}$ on the left side is a column vector and that on the center is a row vector, so we can call the vector on the center instead $\mathbf{b}^T$ or transposed of the column vector $\mathbf{b}$, the whole expression in the center should be transposed as well...but this is a minor detail. I do not see any difference between these two things. So we can say \begin{eqnarray} \nabla (\mathbf{a} \cdot \mathbf{b}) = (\nabla \mathbf{a}) \cdot \mathbf{b} + \mathbf{a} \cdot \nabla \mathbf{b} = (\mathbf{a} \cdot \nabla) \mathbf{b} + (\mathbf{b} \cdot \nabla) \mathbf{a}. \end{eqnarray}

I do not see where the curl $(\nabla \times)$ enter in this analysis. Can someone point out an example where if we do not add the curl terms we get different values on the left and on the right?

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  • $\begingroup$ The problem is with the transpose. $\nabla(a\cdot b)_i = a_{j,i}b_j+a_jb_{j,i}$ has a fixed index (no sum) for differentiation, like $\frac{\partial a_1}{\partial x}b_1+\frac{\partial a_2}{\partial x}b_2+\frac{\partial a_3}{\partial x}b_3$ . This is the first component of the matrix product $J_a^Tb$ (using the Jacobian). The directional derivative $(b\cdot\nabla)a=J_ab$ has first component $\frac{\partial a_1}{\partial x}b_1+\frac{\partial a_1}{\partial y}b_2+\frac{\partial a_1}{\partial z}b_3$. This is a sum over the differentiating index, $a_{i,j}b_j\neq a_{j,i}b_j$. $\endgroup$ – mr_e_man Oct 23 '18 at 3:59
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The second equation presented by you, $\boldsymbol{\nabla} \bigl( \boldsymbol{a} \cdot \boldsymbol{b} \bigr) = \: \bigl( \boldsymbol{\nabla} \boldsymbol{a} \bigr) \! \cdot \boldsymbol{b} \, + \: \bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr) \! \cdot \boldsymbol{a}$, is the primary one and it is pretty easy to derive (*

*) here I use the same notation as I did in my previous answers divergence of dyadic product using index notation and Gradient of cross product of two vectors (where first is constant)

$$\boldsymbol{\nabla} \bigl( \boldsymbol{a} \cdot \boldsymbol{b} \bigr) \! \, = \, \boldsymbol{r}^i \partial_i \bigl( \boldsymbol{a} \cdot \boldsymbol{b} \bigr) \! \, = \, \boldsymbol{r}^i \bigl( \partial_i \boldsymbol{a} \bigr) \! \cdot \boldsymbol{b} \, + \, \boldsymbol{r}^i \boldsymbol{a} \cdot \bigl( \partial_i \boldsymbol{b} \bigr) \, = \: \bigl( \boldsymbol{r}^i \partial_i \boldsymbol{a} \bigr) \! \cdot \boldsymbol{b} \, + \, \boldsymbol{r}^i \bigl( \partial_i \boldsymbol{b} \bigr) \! \cdot \boldsymbol{a} \, =$$ $$= \: \bigl( \boldsymbol{\nabla} \boldsymbol{a} \bigr) \! \cdot \boldsymbol{b} \, + \, \bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr) \! \cdot \boldsymbol{a}$$

Again, I use the expansion of nabla as linear combination of cobasis vectors with coordinate derivatives ${\boldsymbol{\nabla} \! = \boldsymbol{r}^i \partial_i}$ (as always ${\partial_i \equiv \frac{\partial}{\partial q^i}}$), the product rule for $\partial_i$ and the commutativity of dot product of any two vectors (for sure, coordinate derivative of some vector $\boldsymbol{w}$, $\partial_i \boldsymbol{w} \equiv \frac{\partial}{\partial q^i} \boldsymbol{w} \equiv \frac{\partial \boldsymbol{w}}{\partial q^i}$, is a vector and not some more complex tensor) – here ${\boldsymbol{a} \cdot \bigl( \partial_i \boldsymbol{b} \bigr) = \bigl( \partial_i \boldsymbol{b} \bigr) \cdot \boldsymbol{a}}$. Again, I swap multipliers to get full nabla $\boldsymbol{\nabla}$ at the second term

For your first equation, the one with cross products, I need to mention the completely asymmetric isotropic Levi-Civita (pseudo)tensor of third complexity, ${^3\!\boldsymbol{\epsilon}}$

$${^3\!\boldsymbol{\epsilon}} = \boldsymbol{r}_i \times \boldsymbol{r}_j \cdot \boldsymbol{r}_k \; \boldsymbol{r}^i \boldsymbol{r}^j \boldsymbol{r}^k = \boldsymbol{r}^i \times \boldsymbol{r}^j \cdot \boldsymbol{r}^k \; \boldsymbol{r}_i \boldsymbol{r}_j \boldsymbol{r}_k$$

or in orthonormal basis with mutually perpendicular unit vectors $\boldsymbol{e}_i$

$${^3\!\boldsymbol{\epsilon}} = \boldsymbol{e}_i \times \boldsymbol{e}_j \cdot \boldsymbol{e}_k \; \boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k = \;\in_{ijk}\! \boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k$$

(some more details about this (pseudo)tensor can be found at Question about cross product and tensor notation)

Any cross product, including “curl” (a cross product with nabla), can be represented via dot products with the Levi-Civita (pseudo)tensor (**

**) it is pseudotensor because of $\pm$, being usually assumed “$+$” for “left-hand” triplet of basis vectors (where ${\boldsymbol{e}_1 \times \boldsymbol{e}_2 \cdot \boldsymbol{e}_3 \equiv \;\in_{123} \: = -1}$) and “$-$” for “right-hand” triplet (where ${\in_{123} \: = +1}$)

$$\pm \, \boldsymbol{\nabla} \times \boldsymbol{b} = \boldsymbol{\nabla} \cdot \, {^3\!\boldsymbol{\epsilon}} \cdot \boldsymbol{b} = {^3\!\boldsymbol{\epsilon}} \cdot \! \cdot \, \boldsymbol{\nabla} \boldsymbol{b}$$

For the pair of cross products that “pseudo” is compensated. As the very relevant example

$$\boldsymbol{a} \times \bigl( \boldsymbol{\nabla} \! \times \boldsymbol{b} \bigr) = \boldsymbol{a} \cdot \, {^3\!\boldsymbol{\epsilon}} \cdot \, \bigl( \boldsymbol{\nabla} \cdot \, {^3\!\boldsymbol{\epsilon}} \cdot \boldsymbol{b} \bigr) = \boldsymbol{a} \cdot \, {^3\!\boldsymbol{\epsilon}} \cdot \, \bigl( {^3\!\boldsymbol{\epsilon}} \cdot \! \cdot \, \boldsymbol{\nabla} \boldsymbol{b} \bigr) = \boldsymbol{a} \cdot \, {^3\!\boldsymbol{\epsilon}} \cdot {^3\!\boldsymbol{\epsilon}} \cdot \! \cdot \, \boldsymbol{\nabla} \boldsymbol{b}$$

Now I’m going to dive into components, and I do it by measuring tensors using some orthonormal basis (${\boldsymbol{a} = a_a \boldsymbol{e}_a}$, ${\boldsymbol{b} = b_b \boldsymbol{e}_b}$, ${\boldsymbol{\nabla} \! = \boldsymbol{e}_n \partial_n}$, ...)

$$\boldsymbol{a} \cdot \, {^3\!\boldsymbol{\epsilon}} \cdot {^3\!\boldsymbol{\epsilon}} \cdot \! \cdot \, \boldsymbol{\nabla} \boldsymbol{b} = a_a \boldsymbol{e}_a \; \cdot \in_{ijk}\! \boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k \; \cdot \in_{pqr}\! \boldsymbol{e}_p \boldsymbol{e}_q \boldsymbol{e}_r \cdot \! \cdot \, \boldsymbol{e}_n \left( \partial_n b_b \right) \boldsymbol{e}_b = a_a \! \in_{ajk}\! \boldsymbol{e}_j \!\in_{kbn}\! \left( \partial_n b_b \right)$$

There’s a relation (too boring to derive it one more time) for contraction of two Levi-Civita tensors, saying

$$\in_{ajk} \in_{kbn} \: = \: \bigl( \delta_{ab} \delta_{jn} \! - \delta_{an} \delta_{jb} \bigr)$$

Thence

$$a_a \! \in_{ajk} \in_{kbn}\! \left( \partial_n b_b \right) \boldsymbol{e}_j = \, a_a \bigl( \delta_{ab} \delta_{jn} \! - \delta_{an} \delta_{jb} \bigr) \! \left( \partial_n b_b \right) \boldsymbol{e}_j = \, a_a \delta_{ab} \delta_{jn} \! \left( \partial_n b_b \right) \boldsymbol{e}_j - a_a \delta_{an} \delta_{jb} \! \left( \partial_n b_b \right) \boldsymbol{e}_j =$$ $$= \, a_b \! \left( \partial_n b_b \right) \boldsymbol{e}_n - \, a_n \! \left( \partial_n b_b \right) \boldsymbol{e}_b = \left( \boldsymbol{e}_n \partial_n b_b \right) a_b - \, a_n \! \left( \partial_n b_b \boldsymbol{e}_b \right) = \left( \boldsymbol{e}_n \partial_n b_b \boldsymbol{e}_b \right) \cdot a_a \boldsymbol{e}_a - \, a_a \boldsymbol{e}_a \! \cdot \left( \boldsymbol{e}_n \partial_n b_b \boldsymbol{e}_b \right)$$

Back to the direct invariant tensor notation

$$\left( \boldsymbol{e}_n \partial_n b_b \boldsymbol{e}_b \right) \cdot a_a \boldsymbol{e}_a = \: \bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr) \! \cdot \boldsymbol{a}$$

$$a_a \boldsymbol{e}_a \! \cdot \left( \boldsymbol{e}_n \partial_n b_b \boldsymbol{e}_b \right) = \boldsymbol{a} \cdot \bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr)$$

Sure, the latter one can also be written as

$$\boldsymbol{a} \cdot \bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr) \! \, = \, a_a \boldsymbol{e}_a \! \cdot \left( \boldsymbol{e}_n \partial_n b_b \boldsymbol{e}_b \right) \, = \, \left( a_a \boldsymbol{e}_a\! \cdot \boldsymbol{e}_n \partial_n \right) b_b \boldsymbol{e}_b \, = \: \bigl( \boldsymbol{a} \cdot \boldsymbol{\nabla} \bigr) \boldsymbol{b}$$

And finally (***

***) it looks like meanwhile I also answered to Formula of the gradient of vector dot product

$$\boldsymbol{a} \times \bigl( \boldsymbol{\nabla} \! \times \boldsymbol{b} \bigr) = \: \bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr) \! \cdot \boldsymbol{a} \: - \: \boldsymbol{a} \cdot \bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr)$$

or

$$\boldsymbol{a} \times \bigl( \boldsymbol{\nabla} \! \times \boldsymbol{b} \bigr) = \: \bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr) \! \cdot \boldsymbol{a} \: - \: \bigl( \boldsymbol{a} \cdot \boldsymbol{\nabla} \bigr) \boldsymbol{b}$$

or

$$\bigl( \boldsymbol{\nabla} \boldsymbol{b} \bigr) \! \cdot \boldsymbol{a} = \: \bigl( \boldsymbol{a} \cdot \boldsymbol{\nabla} \bigr) \boldsymbol{b} \: + \: \boldsymbol{a} \times \bigl( \boldsymbol{\nabla} \! \times \boldsymbol{b} \bigr)$$

I hope now it’s easy enough for everyone to get similar relations for $\bigl( \boldsymbol{\nabla} \boldsymbol{a} \bigr) \! \cdot \boldsymbol{b}$ and “yes” for question Are these equivalent?

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protected by Alex M. Nov 16 '17 at 15:01

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