What is Einstein Notation for Curl and Divergence?

1. Jul 4, 2011

JDoolin

Anybody know Einstein notation for divergence and curl?

What I would like to do is give each of these formulas in three forms, and then ask a fairly simple question; What is the Einstein notation for each of these formulas?

The unit vectors, in matrix notation:

$$\vec{u_x}=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} or \begin{pmatrix} 1& 0& 0 \end{pmatrix}$$
Similarly
$$\vec{u_y}=(0,1,0) , \vec{u_z}=(0,0,1)$$

The "del" operator
$$\nabla = \vec{u_x} \frac{\partial }{\partial x} + \vec{u_y} \frac{\partial }{\partial y}+\vec{u_z} \frac{\partial }{\partial z}$$

The del operator in matrix notation:
$$\nabla = \begin{pmatrix} \frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z} \end{pmatrix}$$ or $$\nabla = \begin{pmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} & \end{pmatrix}$$

The divergence, here expressed in four different notations:

$$\nabla \cdot \vec{V}=\begin{pmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \end{pmatrix}\begin{pmatrix} V_x\\ V_y\\ V_z \end{pmatrix} =\sum_{a=\left \{ x,y,z \right \}}\frac{\partial V_a}{\partial a}=\frac{\partial V_x}{\partial x}+\frac{\partial V_y}{\partial y}+\frac{\partial V_z}{\partial z}$$

The first expression, uses the del-dot operator, or a "nabla-dot" as LaTeX uses. The second expression is matrix multiplication. The third expression is a summation, as you sum over the terms as you let a=x, a=y, and a=z in turn. And the last expression is the fully expanded algebraic expression.

The third expression (summation notation) is the one that is closest to Einstein Notation, but you would replace x, y, z with x_1, x_2, x_3 or something like that, and somehow with the interplay of subscripts and superscripts, you imply summation, without actually bothering to put in the summation sign.

However, I'm not quite clear on the details, and I would benefit by seeing exactly what the Einstein Notation is for this case.

--------------------
The next operator is the "curl" or del-cross operator which is somewhat cumbersome in matrix form:

$$\nabla \times \vec {V} = \begin{pmatrix} \frac{\partial }{\partial y} & 0 & 0\\ 0 & \frac{\partial }{\partial z} & 0\\ 0 & 0 & \frac{\partial }{\partial x} \end{pmatrix} \begin{pmatrix} V_z\\ V_x\\ V_y \end{pmatrix}- \begin{pmatrix} \frac{\partial }{\partial z} & 0 & 0\\ 0 & \frac{\partial }{\partial x} & 0\\ 0 & 0 & \frac{\partial }{\partial y} \end{pmatrix} \begin{pmatrix} V_y\\ V_z\\ V_x \end{pmatrix}$$

(I wonder whether there might be some simpler way to express it.)

This simplifies to

$$\nabla \times \vec {V} = \begin{pmatrix} \frac{\partial V_z}{\partial y}-\frac{\partial V_y}{\partial z}\\ \frac{\partial V_x}{\partial z}-\frac{\partial V_z}{\partial x}\\ \frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y} \end{pmatrix} =\left (\frac{\partial V_z}{\partial y}-\frac{\partial V_y}{\partial z} \right )\vec {u_x}+ \left (\frac{\partial V_x}{\partial z}-\frac{\partial V_z}{\partial x} \right )\vec {u_y}+ \left (\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y} \right )\vec {u_z}$$

the last expression, is the expression given in the referenced e-book.

I have seen this expression given in two other ways:

$$\nabla \times \vec {V} = \det \begin{pmatrix} \vec u_x & \vec u_y & \vec u_z \\ \frac{\partial }{\partial x} &\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ V_x & V_y & V_z \end{pmatrix}$$

which represents the cross-product of The vectors
$$\begin{pmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \end{pmatrix}$$

and $$\begin{pmatrix} V_x & V_y & V_z \end{pmatrix}$$

And in summation notation

$$\sum_{a=\left \{ x,y,z \right \}} \sum_{b=\left \{ x,y,z \right \}} \sum_{c=\left \{ x,y,z \right \}} \delta _{abc}\frac{\partial }{\partial a}V_b \vec u_c$$

where
$$\delta_{abc}\overset{def}{=}\left\{\begin{matrix} 1,& abc=\left \{xyz,zyx,zxy \right \}\\ -1,& abc =\left \{zyx,yxz,xzy \right \}\\ 0,& \mathrm a=b , b=c,c=a \end{matrix}\right.$$

Again, if I'm not mistaken, Einstein notation is most similar to the summation notation, but I'm not exactly sure what it would look like.

Last edited: Jul 4, 2011
2. Jul 4, 2011

Mentz114

The generalised curl of a vector A is
$$\partial_\mu A_\nu-\partial_\nu A_\mu$$
and divergence
$$\partial_\mu A^\mu$$

3. Jul 4, 2011

Hmmmm, fascinating. I think I sort of get it. So does the order $$\mu, \nu [\tex] somehow correspond to the forward ordering of the pairs (y,z) (z,x) (x,y), while the [tex] \nu \mu [\tex] correspond to the backward ordering of the pairs (z,y) (y,x) (x,z)? ... and maybe use of a superscript and subscript just tells you to use the same variables? Maybe I should also ask for the formula for gradient. 4. Jul 4, 2011 Mentz114 The subscripts and super scripts are tensor indices, they run over the dimensions. To make sense of the expressions, expand them over their range. In 4D the divergence is [tex] \partial_\mu A^\mu= \partial_0 A^0 +\partial_1 A^1 +\partial_2 A^2 +\partial_3 A^3$$
the summation is because one index is high the other low, and the result is a scalar.
The gen curl is an antisymmetric rank-2 tensor
$$F_{\mu\nu}= \left[ \begin{array}{cccc} 0 & b & c & d \\\ -b & 0 & f & e\\\ -c & -f & 0 & g\\\ -d & -e & -g & 0\end{array} \right]$$
with $F_{01}=b=\partial_0 A_1-\partial_1 A_0$ and so on.

To write the gradient we need a basis, say $\vec{e}_\mu$. This is four vectors, labelled with the index $\mu$. The gradient of a scalar S is just the usual vector
$$\vec{\nabla}S=\partial_0\ S\ \vec{e}_0 + \partial_1\ S\ \vec{e}_1 + \partial_2\ S\ \vec{e}_2 + \partial_3\ S\ \vec{e}_3$$

5. Jul 4, 2011

JDoolin

So, let me see if I can at least repeat the gist of what you are saying, though I'm not entirely sure what it means: (I am assuming, of course, that the form of the 4X4 curl tensor can be generalized back to the 3X3 curl tensor.)

The tensor associated with a 3X3 curl... or should I say "The general curl in three dimensions is an antisymmetric rank-2 tensor:"

$$F_{uv}=\begin{pmatrix} 0 & \left (\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y} \right ) & \left (\frac{\partial V_z}{\partial x}-\frac{\partial V_x}{\partial z}\right ) \\ \left (\frac{\partial V_x}{\partial y}-\frac{\partial V_y}{\partial x}\right ) & 0 & \left (\frac{\partial V_z}{\partial y}-\frac{\partial V_y}{\partial z}\right )\\ \left (\frac{\partial V_x}{\partial z}-\frac{\partial V_z}{\partial x}\right ) & \left (\frac{\partial V_y}{\partial z}-\frac{\partial V_z}{\partial y}\right ) & 0 \end{pmatrix}$$

Does this translate directly, somehow, into:

$$\nabla \times \vec {V} = \left (\frac{\partial V_z}{\partial y}-\frac{\partial V_y}{\partial z} \right )\vec {u_x}+ \left (\frac{\partial V_x}{\partial z}-\frac{\partial V_z}{\partial x} \right )\vec {u_y}+ \left (\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y} \right )\vec {u_z}$$

Last edited: Jul 4, 2011
6. Jul 4, 2011

JDoolin

On the other hand, if somehow,

$$\partial_\mu V_\nu \overset{def?}{=} \partial_x V_y \vec u_z + \partial_y V_z \vec u_x + \partial_z V_x \vec u_y$$

while

$$\partial_\nu V_\mu \overset{def?}{=} \partial_z V_y \vec u_x + \partial_y V_x \vec u_z + \partial_x V_z \vec u_y$$

Then it would make perfect sense to say

$$\nabla \times \vec V = \partial_\mu V_\nu - \partial_\nu V_\mu$$

P.S. I found yet another matrix multiplication form for defining the curl:

$$\nabla \times \vec V =\begin{pmatrix} \partial_y &-\partial_z \\ \partial_z &-\partial_x \\ \partial_x &-\partial_y \end{pmatrix} \begin{pmatrix} V_z &V_x &V_y \\ V_y &V_z &V_x \end{pmatrix}$$

Last edited: Jul 4, 2011
7. Jul 5, 2011

JDoolin

Another version of Einstein Notation for the curl:

8. Jul 7, 2011

JDoolin

I found some possibly relevant equations in Mathematical Methods in the Physical Sciences 2nd Edition

Boas defines the h's as the "scale factors." Sort of a vague definition, but I suppose it must do. For instance, in a cylindrical coordinate system, a length 1 radial component is the same whether you are near the axis, or far away from the axis. But a lenth one angular component is very small near the axis, but very large far away from the axis.

Eq 10.8.4 $$ds^2 = \begin{pmatrix} dx_1 & dx_2 & dx_3 \end{pmatrix} \begin{pmatrix} g_{11} &g_{12} &g_{13} \\ g_{21} &g_{22} &g_{23} \\ g_{31} &g_{32} &g_{33} \end{pmatrix} \begin{pmatrix} dx_1\\ dx_2\\ dx_3 \end{pmatrix}$$

$$\begin{matrix} Cartesian & h_1=g_{11}=1, h_2=g_{22}=1, h_3=g_{33}=1\\ Cylindrical & g_{11}=1, g_{22}=r, g_{33}=1\\ Spherical & g_{11}=1, g_{22}=r, g_{33}=r sin( \theta ) \end{matrix}$$

Generalized forms for gradient, divergence, and Laplacian are given in eq 10.9.2, 10.9.8, 10.9.10, 10.9.11.

$$\nabla u = \begin{pmatrix} \frac{1}{h_1} \frac{\partial }{\partial x_1}\\ \frac{1}{h_2} \frac{\partial }{\partial x_2}\\ \frac{1}{h_3} \frac{\partial }{\partial x_3} \end{pmatrix}u$$

$$\nabla \cdot \vec{V} = \frac{1}{h_1 h_2 h_3}\begin{pmatrix} \frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} &\frac{\partial }{\partial x_3} \end{pmatrix} \begin{pmatrix} h_2 h_3 V_1\\ h_1 h_3 V_2\\ h_1 h_2 V_3 \end{pmatrix}$$

$$\nabla^2 u = \frac{1}{h_1 h_2 h_3}\begin{pmatrix} \frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} &\frac{\partial }{\partial x_3} \end{pmatrix} \begin{pmatrix} \frac{h_2 h_3}{h_1} \frac{\partial }{\partial x_1}\\ \frac{h_1 h_3}{h_2} \frac{\partial }{\partial x_2}\\ \frac{h_1 h_2}{h_3} \frac{\partial }{\partial x_2} \end{pmatrix} u$$

$$\nabla \times \vec V = \frac{1}{h_1 h_2 h_3}\begin{vmatrix} h_1 \vec e_1 & h_2 \vec e_2 & h_3 \vec e_3 \\ \frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3} \\ h_1 V_1 & h_2 V_2 & h_3 V_3 \end{vmatrix}$$

These forms, if worked out for cartesian, cylindrical, and spherical, should generate the table on the inside back cover of Jackson's "Classical Electrodynamics" 3rd edition.

Last edited: Jul 7, 2011
9. Jul 9, 2011

JDoolin

$$\nabla \overset ? = \frac{\partial x^\mu}{\partial x'^\nu}\cdot \frac{\partial }{\partial x^\mu} \overset ? = \begin{pmatrix} \frac{\partial ||\vec r||}{\partial ||\vec r||} \frac{\partial }{\partial r}\\ \frac{\partial ||\vec r||}{\partial ||\vec \theta||} \frac{\partial }{\partial \theta}\\ \frac{\partial ||\vec r||}{\partial ||\vec \phi||} \frac{\partial }{\partial \phi} \end{pmatrix} =\begin{pmatrix} \frac{\partial }{\partial r}\\ \frac{1}{r} \frac{\partial }{\partial \theta}\\ \frac{1}{r sin\theta} \frac{\partial }{\partial \phi} \end{pmatrix}$$

Edit: I left out a summation, I believe.

Last edited: Jul 9, 2011
10. Jul 9, 2011

JDoolin

Correction:

$$\begin{matrix} Cartesian & h_1=\sqrt{g_{11}}=1, h_2= \sqrt{g_{22}} =1, h_3= \sqrt{g_{33}} =1\\ Cylindrical & g_{11}=1, g_{22}=r^2, g_{33}=1\\ Spherical & g_{11}=1, g_{22}=r^2, g_{33}=(r sin( \theta ))^2 \end{matrix}$$

11. Jul 9, 2011

JDoolin

With the summation:

$$\frac{\partial x^\mu}{\partial x'^\nu}\cdot \frac{\partial }{\partial x^\mu} \overset ? = \begin{pmatrix} \frac{\partial r}{\partial x} \frac{\partial }{\partial r}+ \frac{\partial \theta}{\partial x} \frac{\partial }{\partial \theta}+ \frac{\partial \phi}{\partial x} \frac{\partial }{\partial \phi} \\ \frac{\partial r}{\partial y} \frac{\partial }{\partial r}+ \frac{\partial \theta}{\partial y} \frac{\partial }{\partial \theta}+ \frac{\partial \phi}{\partial y} \frac{\partial }{\partial \phi} \\ \frac{\partial r}{\partial z} \frac{\partial }{\partial r}+ \frac{\partial \theta}{\partial z} \frac{\partial }{\partial \theta}+ \frac{\partial \phi}{\partial z} \frac{\partial }{\partial \phi} \end{pmatrix}$$

or

$$\frac{\partial x^\mu}{\partial x'^\nu}\cdot \frac{\partial }{\partial x^\mu} \overset ? = \begin{pmatrix} \frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} & \frac{\partial \phi}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y} & \frac{\partial \phi}{\partial y} & \\ \frac{\partial r}{\partial z} & \frac{\partial \theta}{\partial z} & \frac{\partial \phi}{\partial z} & \end{pmatrix} \begin{pmatrix} \frac{\partial }{\partial r}\\ \frac{\partial }{\partial \theta}\\ \frac{\partial }{\partial \phi} \\ \end{pmatrix}$$

where $$x^\mu = \left \{ r, \theta, \phi \right \}$$ and $$x^\nu  = \left \{ x, y, z \right \}$$

I'm not sure this particularly resembles the equation for gradient after all.

12. Jul 14, 2011

JDoolin

What, precisely, is the derivative of a vector field? Is it the divergence? Is it the curl? Is it something else entirely?

I am still following along

http://www.mathpages.com/rr/appendix/appendix.htm

and I think I've found where the ∇∙ is hidden

In section 4 of the appendix, A is defined as

$$\vec A = A_i \vec u^i$$

...or in the more lengthy notation

$$\vec A = A_x \vec u^x + A_y \vec u^y +A_z \vec u^z$$

Then the product rule of differentiating yields

$$d \vec A = u^i dA_i + A_i d u^i$$

Mathematically, the author changes his variables to j:

$$d \vec A = u^j dA_i + A_j d u^j$$

and replaces

$$\begin{matrix} d \vec A = \frac{\partial \vec A}{\partial x^i}dx^i, \\d A_j = \frac{\partial A_i}{\partial x^i}dx^i,\\ d u^j = \frac{\partial u^j}{\partial x^i}dx^i \end{matrix}$$

$$d \vec A = \frac{\partial \vec A}{\partial x^i}dx^i = \frac{\partial A_j}{\partial x^i}dx^i + \frac{\partial u^j}{\partial x^i}dx^i$$

Now, what is that thing on the left? It is a derivative of a vector field. Is that the divergence? Is it the curl? Is it a combination of both?

Regardless, let's divide out the repeated term:

$$\frac{\partial \vec A}{\partial x^i} = \frac{\partial A_j}{\partial x^i} + \frac{\partial u^j}{\partial x^i}$$

Allowing j to run through x,y,z and i to run through r, θ, Φ, it expands to three equations

$$\frac{\partial \vec A}{\partial \vec u^r} = \left (\frac{\partial A_x}{\partial \vec u^r} + \frac{\partial \vec u^x}{\partial \vec u^r} \right )+ \left (\frac{\partial A_y}{\partial \vec u^r} + \frac{\partial \vec u^y}{\partial \vec u^r} \right )+ \left (\frac{\partial A_z}{\partial \vec u^r} + \frac{\partial \vec u^z}{\partial \vec u^r} \right )$$

$$\frac{\partial \vec A}{\partial \vec u^\theta} = \left (\frac{\partial A_x}{\partial \vec u^\theta} + \frac{\partial \vec u^x}{\partial \vec u^\theta} \right )+ \left (\frac{\partial A_y}{\partial \vec u^\theta} + \frac{\partial \vec u^y}{\partial \vec u^\theta} \right )+ \left (\frac{\partial A_z}{\partial \vec u^\theta} + \frac{\partial \vec u^z}{\partial \vec u^\theta} \right )$$

$$\frac{\partial \vec A}{\partial \vec u^\phi} = \left (\frac{\partial A_x}{\partial \vec u^\phi} + \frac{\partial \vec u^x}{\partial \vec u^\phi} \right )+ \left (\frac{\partial A_y}{\partial \vec u^\phi} + \frac{\partial \vec u^y}{\partial \vec u^\phi} \right )+ \left (\frac{\partial A_z}{\partial \vec u^\phi} + \frac{\partial \vec u^z}{\partial \vec u^\phi} \right )$$

I am not at all sure if I have this notation correct. It might be, instead

$$\frac{\partial \vec A}{\partial ||\vec u{ }^\theta||} = \left (\frac{\partial A_x}{\partial \theta} + \frac{\Delta ||\vec u{ }^x||}{\Delta ||\vec u{ }^\theta||} \right )+ \left (\frac{\partial A_y}{\partial \theta} + \frac{\Delta ||\vec u{ }^y||}{\Delta ||\vec u{ }^\theta||} \right )+ \left (\frac{\partial A_z}{\partial \theta} + \frac{\Delta ||\vec u{ }^z||}{\Delta ||\vec u{ }^\theta||} \right )$$

... for instance. Might this have something to do with covariant and contravariant derivatives?

Last edited: Jul 14, 2011
13. Jul 14, 2011

Mentz114

At the beginning of section 4 the author introduces a basis ui and a cobasis ui and writes vector A in terms of both, then gets the total differential of A by the chain rule.

dA = ui dAi + Ai dui (1a)

and similarly for the cobasis. I think you've followed it up to here, but I suggest that you read carefully from the paragraph after (1b). It's explained the dA is the differential of A.

Crucially, in curved space, the second term on the right does not disappear, and the author introduces the covariant derivative. You might find it a bit strange that the cobasis has appeared. It isn't important unless the space(time) is curved, in which case the curvature can be described by the relationship between the basis vectors and the cobasis 1-forms.

I think you'd get along a bit faster if you studied the 4-D manifold, and the tangent and cotangent spaces of curves ( worldlines). As in Sean Carroll's lecture notes, for instance.

Last edited: Jul 14, 2011
14. Jul 20, 2011

JDoolin

Is there any mathematical difference between that and

dA = ui dAi + Ai dui (2a)

or is it just a notational difference?

Right. The key there is "If the basis vectors have a constant direction relative to a fixed Cartesian frame" I selected spherical coordinates as my example, because the basis vectors ${\vec u }_r , {\vec u }_\theta , {\vec u }_\phi$ all vary from place to place. However with the dependence on ${\vec u }_r , {\vec u }_\theta , {\vec u }_\phi$ and $\theta =\arctan(\frac{z}{\sqrt{x^2+y^2}})$, it is probably a bit difficult to figure out.

So if I understand correctly, the contravariant derivative can be written either way:
ui dAi or ui dAi based only on changes in the vector field; mathematically identical to the divergence in cartesian coordinates.

and the covariant derivative is, Ai dui or Ai dui which would turn out to be zero in cartesian coordinates, but nonzero in curvilinear coordinates.

Is that correct?

The problem I want to understand is how to get the kappa x y terms at the bottom of http://www.mathpages.com/rr/s6-06/6-06.htm, and precisely what they physically represent, conceptually. I'm guessing I'll get along the fastest if I pore over documents where this equation is discussed.

15. Jul 20, 2011

Mentz114

I don't think I'm helping here. The notes are referring to main sections which I haven't got the time to read. Section 4 starts in E3 where tensor indexes can be low or high and the components don't change in rectilinear coords and then abuses notation and makes it difficult for me to follow.

I wish you well with it.