Trouble with Notation in Vector Analysis

[ExtravagantDreams]

Difficulty with notation

I've always had trouble with notation, since there are so many ways to write the same thing.

This one is for Vector Analysis

First off, when speaking of unit vectos, such as i, j, k is it ok to use the notation;
$${\hat i},{\hat j},{\hat k}$$?

Such as in
$$f(x,y,z) = (z^3+2y){\hat i}+(3x^2+5y){\hat j}...$$

Or is this ment to notate something else?

Secondly, in the folowing, does the ${\varepsilon _{{\bf i,j,k}} }$ (I'm noting now that this is known as the permutation symbol or Levi-Civita density) represent the same thing as ${{\bf e} _{{\bf i,j,k}} }$ or ${{\bf e} _{1,2,3} }$ (unit vectors)?
$$\sum\limits_{{\bf i,j,k}} {\varepsilon _{{\bf i,j,k}} } A_{\bf i} B_{\bf j} C_{\bf k}$$

Also, I am not quite sure how to expand this. I am trying to prove that
$$\vec A\vec B\vec C= \sum\limits_{{\bf i,j,k}} {\varepsilon _{{\bf i,j,k}} } A_{\bf i} B_{\bf j} C_{\bf k}$$
The left side is easy to expand through a matrix. Maybe once I understand the notation the rest will make sense.

Is this the complete expansion:
$$\varepsilon _{{\bf i,j,k}} A_{\bf i} B_{\bf j} C_{\bf k} + \varepsilon _{{\bf k,i,j}} A_{\bf k} B_{\bf i} C_{\bf j} + \varepsilon _{{\bf j,k,i}} A_{\bf j} B_{\bf k} C_{\bf i} + \varepsilon _{{\bf i,k,j}} A_{\bf i} B_{\bf k} C_{\bf j} + \varepsilon _{{\bf j,i,k}} A_{\bf j} B_{\bf i} C_{\bf k} + \varepsilon _{{\bf k,j,i}} A_{\bf j} B_{\bf i} C_{\bf k}$$

Or is there really also all the zero tems?
$$\varepsilon _{{\bf i,i,k}} A_{\bf i} B_{\bf i} C_{\bf k}...$$
I think there is 12 total.

Then the even permutation is equal to 1 and the odd is equal to -1 so:

$$A_{\bf i} B_{\bf j} C_{\bf k} + A_{\bf k} B_{\bf i} C_{\bf j} + A_{\bf j} B_{\bf k} C_{\bf i} - A_{\bf i} B_{\bf k} C_{\bf j} - A_{\bf j} B_{\bf i} C_{\bf k} - A_{\bf j} B_{\bf i} C_{\bf k}$$

Which is exactly the triple product expansion

 

[Galileo]

For unit vectors, you have as a rule:
$$\vec i, \vec j, \vec k$$
for the unit vectors in the positive x,y and z directions respectively.
Or sometimes the following are used:
$$\vec e_1, \vec e_2, \vec e_3$$
Physicists often use
$$\hat x, \hat y, \hat z$$
for these, or $\hat v$ in general for a unit vector in the direction of a vector $\vec v$.
I`m sure everyone will understand what you mean by:
$$\hat i, \hat j , \hat k$$
so
$$f(x,y,z) = (z^3+2y){\hat i}+(3x^2+5y){\hat j}$$
looks fine. It the equation of some vector field.

The equation:
$$\vec A \vec B \vec C= \sum\limits_{{\bf i,j,k}} {\varepsilon _{{\bf i,j,k}} } A_{\bf i} B_{\bf j} C_{\bf k}$$
is correct, but I would write the vector triple product as:
$$\vec A \cdot (\vec B \times \vec C)$$
since it's easier to understand and more logical.
I think the Levi-Civita symbol is usually written as:
$$\epsilon_{ijk}$$, without the comma's (but who cares )

The following sum
$$\sum\limits_{{\bf i,j,k}} {\varepsilon _{{\bf i,j,k}} } A_{\bf i} B_{\bf j} C_{\bf k}$$
is shorthand notation, but it's often used. You are summing over all applicable values of i,j and k. So it's actually a triple sum.
$$\sum\limits_{{\bf i,j,k}} {\varepsilon _{{\bf i,j,k}} } A_{\bf i} B_{\bf j} C_{\bf k}=\sum_{i=1}^{i=3}\sum_{j=1}^{j=3}\sum_{k=1}^{k=3}\varepsilon _{{\bf i,j,k}} }A_{\bf i} B_{\bf j} C_{\bf k}$$

Cheers.

 

[robphy]

I've always had trouble with notation, since there are so many ways to write the same thing.

This one is for Vector Analysis

First off, when speaking of unit vectos, such as i, j, k is it ok to use the notation;
$${\hat i},{\hat j},{\hat k}$$?

Such as in
$$f(x,y,z) = (z^3+2y){\hat i}+(3x^2+5y){\hat j}...$$

Or is this ment to notate something else?

What you wrote is fine, although you probably should write the left hand side as $\vec f(x,y,z)$.
When writing TeX, it's prettier to write $\hat\imath$ and $\hat\jmath$. (Click on the symbols to see how they were written.)

Secondly, in the folowing, does the ${\varepsilon _{{\bf i,j,k}} }$ (I'm noting now that this is known as the permutation symbol or Levi-Civita density) represent the same thing as ${{\bf e} _{{\bf i,j,k}} }$ or ${{\bf e} _{1,2,3} }$ (unit vectors)?
$$\sum\limits_{{\bf i,j,k}} {\varepsilon _{{\bf i,j,k}} } A_{\bf i} B_{\bf j} C_{\bf k}$$

Usually, these sorts of symbols are written without commas, e.g., ${\varepsilon _{{\bf ijk}} }$. In tensor calculus, the comma is often a symbol for "partial derivative". Notations vary on which of $\epsilon$, $\varepsilon$, $e$ represents the totally antisymmetric quantity taking values from $\{+1, 0, -1\}$ or those values times $\sqrt{g}$.

Also, I am not quite sure how to expand this. I am trying to prove that
$$\vec A\vec B\vec C= \sum\limits_{{\bf i,j,k}} {\varepsilon _{{\bf i,j,k}} } A_{\bf i} B_{\bf j} C_{\bf k}$$
The left side is easy to expand through a matrix. Maybe once I understand the notation the rest will make sense.

Is this the complete expansion:
$$\varepsilon _{{\bf i,j,k}} A_{\bf i} B_{\bf j} C_{\bf k} + \varepsilon _{{\bf k,i,j}} A_{\bf k} B_{\bf i} C_{\bf j} + \varepsilon _{{\bf j,k,i}} A_{\bf j} B_{\bf k} C_{\bf i} + \varepsilon _{{\bf i,k,j}} A_{\bf i} B_{\bf k} C_{\bf j} + \varepsilon _{{\bf j,i,k}} A_{\bf j} B_{\bf i} C_{\bf k} + \varepsilon _{{\bf k,j,i}} A_{\bf j} B_{\bf i} C_{\bf k}$$

Or is there really also all the zero tems?
$$\varepsilon _{{\bf i,i,k}} A_{\bf i} B_{\bf i} C_{\bf k}...$$
I think there is 12 total.

Then the even permutation is equal to 1 and the odd is equal to -1 so:

$$A_{\bf i} B_{\bf j} C_{\bf k} + A_{\bf k} B_{\bf i} C_{\bf j} + A_{\bf j} B_{\bf k} C_{\bf i} - A_{\bf i} B_{\bf k} C_{\bf j} - A_{\bf j} B_{\bf i} C_{\bf k} - A_{\bf j} B_{\bf i} C_{\bf k}$$

Which is exactly the triple product expansion

The left-hand side is $\vec A \cdot( \vec B \times \vec C)$, the scalar triple product. I've never seen the notation $\vec A\vec B\vec C$ for that. Maybe $|\vec A\vec B\vec C|$ is better since
$$\vec A \cdot( \vec B \times \vec C) =det \left(<br /> \begin{array}{ccc}<br /> A_1 & A_2 & A_3 \\<br /> B_1 & B_2 & B_3 \\<br /> C_1 & C_2 & C_3<br /> \end{array} \right)$$

As you wrote, the right-hand side is a triple sum with each index running from 1 to 3. This means that there are 3*3*3=27 terms. Many are zero (e.g., $\epsilon_{111}A_1 B_1 C_1$ and $\epsilon_{121}A_1 B_2 C_1$) because of the total-antisymmetry. Note that your final result is a number. There should be no free indices. (Maybe you are using the symbol j as both a summation index and as a component-label.)

 

[ExtravagantDreams]

Thanks for the help guys. Yes, the commas were a dumb error and I should have noticed that.

So the triple sum was understood to go from one to three? I did not know that
$$\sum_{i,j,k}=\sum_{i=1}^{i=3}\sum_{j=1}^{j=3}\sum_{k=1}^{k=3 }$$
Hence me using the symbols i,j, and k for both a summation index and as a component-label.

 

[Dr Transport]

as a general rule for summations (physicists), latin letters $$i, j, k$$ run from 1 to 3, greek letters, from 1 to 4 or 0 to 3.