# Proof of vector calculus identities

Here is the all identities : http://en.wikipedia.org/wiki/Vector_calculus_identities

I need help concerning vector functions and indexing notations.

Let $\overrightarrow{a}$ be a (smooth) vector field and $\varphi$ be a (smooth) scalar function. Show $$\overrightarrow {\nabla }\cdot \left( \varphi\,\overrightarrow {a}\right) = \varphi \left( \overrightarrow {\nabla }\cdot \overrightarrow {a}\right) +\overrightarrow {a} \cdot \overrightarrow {\nabla }\varphi.$$

I have to use this notation to prove this, but how?

I don't really understand.

My second identity is ; $$\overrightarrow {\nabla }\times \left( \phi \cdot \overrightarrow {a}\right)$$

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Is that thing on the left-hand side meant to be $\phi$? If so, $\phi \cdot a$ doesn't make sense. $\phi$ is a scalar field, $a$ is a vector (is it also a field)? Usually, the dot product of a scalar and a vector is not defined. – Muphrid Jan 14 '13 at 17:30
I edit the problem statement so that it makes sense. Please roll back/edit if I misinterpreted what was intended. – JohnD Jan 14 '13 at 17:52
John,thanks for edit.Thanks for comment,I'm looking here phys.ufl.edu/courses/phy3063/spring12/Lecture2-CovariantNot.pdf for basics.Then I will turn here again to see the answers.Becuase now,it's not make any sense on me.I don't understand,what does it mean? – Erbil Jan 14 '13 at 18:07
@Erbil: unfortunately, what's happened is that ordinary vector calculus is simply inadequate for some things, particularly when you get outside of 3d (for instance, in relativity, as that reference describes). So, what you're doing is converting dot and cross products into expressions with indices and learning how to work with those indexed expressions. Index notation is one way to do multivariable calculus outside of 3d in a way that makes sense. – Muphrid Jan 14 '13 at 18:23
@Muphrid : Get outside from 3d? Hmm.It's hard to imagine.But it is not important for now.It's enought to know how to express these multiplications.I will try to do what you did.And then I wil try for second identity.Thanks! Note : I apologies for my English. – Erbil Jan 14 '13 at 18:44

Here's what's happening in $\mathbb{R}^3$ with rectangular coordinates. You can tweak as needed.

Let $g(x,y,z)$ be a smooth scalar function and $\mathbf{F}(x,y,z)=(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z))$ be a smooth vector field. Then \begin{align} \nabla \cdot (g\,\mathbf{F})&=\nabla\cdot((gF_1,gF_2,gF_3))\\ &=(gF_1)_x+(gF_2)_y+(gF_3)_z\\ &=g_xF_1+g(F_1)_x+g_yF_2+g(F_2)_y+g_zF_3+g(F_3)_z, \end{align} while \begin{align} \nabla g\cdot \mathbf{F}&=g_x F_1+g_y F_2+g_z F_3,\\ g\,(\nabla\cdot \mathbf{F})&=g\,((F_1)_x+(F_2)_y+(F_3)_z)=g(F_1)_x+g(F_2)_y+g(F_3)_z. \end{align} Adding these last two yields the first.

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I noticed now! This is a great answer. – Erbil Jan 15 '13 at 16:07

If JohnD has interpreted the problem correctly, then here's how you would work it using index notation. Here, $i$ is an index running from 1 to 3 ($a^1$ might be the x-component of $a$, $a^2$ the y-component, and so on).

$$\nabla \cdot (\varphi a) = \nabla_i (\varphi a^i)$$

Since these are all components (not vectors), you can attack this with the product rule.

$$\nabla_i (\varphi a^i) = (\nabla_i \varphi) a^i + \varphi (\nabla_i a^i)$$

The first term is $a \cdot \nabla \varphi$ and the latter is $\varphi \nabla \cdot a$.

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You probably know the product rule $(uv)'=u'v +uv'$.

I have been taught to use such formula (which follows from derivative-like nature of $\nabla$):

$\nabla(\underline{uv}) = \nabla(\underline{u}v) + \nabla(u\underline{v})$

where derivatives work on the underlined part of the brackets. In your case you can factor $\varphi$ and $\vec{a}$ when they are not under effect of nabla, but in other cases (like $\nabla (\vec{a}\cdot\vec{b})$) it can't be done that easily, but you can derive a formula for dot product starting from formula with underlines.

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Putting a smooth vector field U in the UNIT VECTORS, and a scalar variable ϕ $$\boldsymbol{U}=u_1\boldsymbol{i}+u_2\boldsymbol{j}+u_3\boldsymbol{k}$$ Now showing that

# $$\boldsymbol{\nabla}\times(\boldsymbol{U}\phi)=\phi\boldsymbol{(\nabla\times U)}+\boldsymbol{\nabla}\phi\times\boldsymbol{U}$$

$$\boldsymbol{\nabla}\times(\boldsymbol{U}\phi)=\begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ u_1\phi&u_2\phi&u_3\phi \end{vmatrix}\\ =\left(\boldsymbol{i}\frac{\partial(u_3\phi)}{\partial y}+\boldsymbol{j}\frac{\partial(u_1\phi)}{\partial z}+\boldsymbol{k}\frac{\partial(u_2\phi)}{\partial x}\right)-\left(\boldsymbol{i}\frac{\partial(u_2\phi)}{\partial z}+\boldsymbol{j}\frac{\partial(u_3\phi)}{\partial x}+\boldsymbol{k}\frac{\partial(u_1\phi)}{\partial y}\right)\\ =\phi\left(\boldsymbol{i}\left(\frac{\partial u_3}{\partial y}-\frac{\partial u_2}{\partial z}\right)+\boldsymbol{j}\left(\frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x}\right)+\boldsymbol{k}\left(\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\right)\right)+\left(\boldsymbol{i}\left(u_3\frac{\partial\phi}{\partial y}-u_2\frac{\partial\phi}{\partial z}\right)+\boldsymbol{j}\left(u_1\frac{\partial\phi}{\partial z}-u_3\frac{\partial\phi}{\partial x}\right)+\boldsymbol{k}\left(u_2\frac{\partial\phi}{\partial x}-u_1\frac{\partial\phi}{\partial y}\right)\right)\\ =\phi\begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ u_1&u_2&u_3 \end{vmatrix}+\begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\ \frac{\partial\phi}{\partial x}&\frac{\partial\phi}{\partial y}&\frac{\partial\phi}{\partial z}\\ u_1&u_2&u_3 \end{vmatrix}=\phi\boldsymbol{(\nabla\times U)}+\boldsymbol{\nabla}\phi\times\boldsymbol{U}$$ Having funs with the vector calaulus!! O:DDD

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Consider editing your answer and use Latex. See here for a basic tutorial. – Stefan Hansen Apr 29 '13 at 7:09