# Vector Calculus

• August 4th 2007, 10:57 AM
physics4life
Vector Calculus
hi this is about vector calculus subscript notation:

i generally understand v.c.s.n but can't do something like this:
Show : (AXB) . (CXD) = (A.C)(B.D) - (A.D)(B.C)

my work so far:
$(AXB)i = \varepsilon_{ijk}A_jB_k$

$(CXD)i = \varepsilon_{ijk}C_jD_k$

so
(AXB) . (CXD) = $(\varepsilon_{ijk}A_jB_k)_i . (\varepsilon_{ijk}C_jD_k)_i$

= $\varepsilon_{ijk}\varepsilon_{ijk}A_jC_jB_kD_k$

= $( \delta_{ij} \delta_{jk} - \delta_{ik}\delta_{jj})(A.C)_j(B.D)k$

= $[ \delta_{ij}(A.C)_j ][ \delta_{jk}(B.D)_k ] - [ \delta_{ik}(B.D)_k ][ \delta_{jj}(A.C)_j ]$

= $(A.C)_i(B.D)j - (B.D)_i(A.C)_j$

=(A.C)(B.D) - (B.D)(A.C)

the first part of the answer (in red) i got right.. but the 2nd part is wrong as you can see

how am i meant to get -(A.D)(B.C)???
• August 4th 2007, 11:10 AM
topsquark
Quote:

Originally Posted by physics4life
hi this is about vector calculus subscript notation:

i generally understand v.c.s.n but can't do something like this:
Show : (AXB) . (CXD) = (A.C)(B.D) - (A.D)(B.C)

my work so far:
$(AXB)i = \varepsilon_{ijk}A_jB_k$

$(CXD)i = \varepsilon_{ijk}C_jD_k$

so
(AXB) . (CXD) = $(\varepsilon_{ijk}A_jB_k)_i . (\varepsilon_{ijk}C_jD_k)_i$

Here's the problem: you are using jk indices for the second sum. And you have the epsilon-epsilon identity wrong. Here's what you should do:
$(A \times B) \cdot (C \times D)$

$\to (\varepsilon_{ijk}A_jB_k)(\varepsilon_{imn}C_mD_n)$

$= \varepsilon_{ijk} \varepsilon_{imn}A_jB_kC_mD_n$

$= (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km})A_jB_kC_mD_n$

$= \delta_{jm} \delta_{kn}A_jB_kC_mD_n - \delta_{jn} \delta_{km}A_jB_kC_mD_n$

$= A_jB_kC_jD_k - A_jB_kC_kD_j$

$\to (A \cdot C)(B \cdot D) - (A \cdot D)(B \cdot C)$

-Dan
• August 4th 2007, 12:35 PM
physics4life
cheers dan, your method is easy to understand. it's frustrating trying to learn all this calculus on your own when there is just a huge pile of books and you're forever researching. so i appreciate your help.
just a question though, on the 2nd epsilon identity you used: $\epsilon_{imn}$
what i wanted to ask is: do you have to use the imn notation.. or can you use any preferred notation.. such as ixy....
i understand the final result will still be the same as long as the notation is consistent throughout the solution, but i just wanted to confirm whether imn has to be used as a standard for vector calculus. thanks once again .:o
• August 4th 2007, 01:53 PM
topsquark
Quote:

Originally Posted by physics4life
cheers dan, your method is easy to understand. it's frustrating trying to learn all this calculus on your own when there is just a huge pile of books and you're forever researching. so i appreciate your help.
just a question though, on the 2nd epsilon identity you used: $\epsilon_{imn}$
what i wanted to ask is: do you have to use the imn notation.. or can you use any preferred notation.. such as ixy....
i understand the final result will still be the same as long as the notation is consistent throughout the solution, but i just wanted to confirm whether imn has to be used as a standard for vector calculus. thanks once again .:o

You can use any indices you like. (You can even make up your own, but they're much harder to type. :p ) The point is that they are different from the ones you used previously since they have no relation to the indices in the other cross product.

By the way, a nice way to remember that epsilon-epsilon identity:
$\varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{jm} - \delta_{jn} \delta_{km}$
Note that the "i" index doesn't appear anywhere in the deltas. Now look at the first delta of each pair. They both have a "j," the first index after the "i" in the first epsilon. The second index of those first deltas is in sequence, m and n... the second and third indices of the second epsilon.

-Dan
• August 4th 2007, 03:23 PM
physics4life
Quote:

Originally Posted by topsquark
By the way, a nice way to remember that epsilon-epsilon identity:
$\varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{jm} - \delta_{jn} \delta_{km}$
Note that the "i" index doesn't appear anywhere in the deltas. Now look at the first delta of each pair. They both have a "j," the first index after the "i" in the first epsilon. The second index of those first deltas is in sequence, m and n... the second and third indices of the second epsilon.
-Dan

hold on a minute.
if we have a standard epsilon-epsilon identity.. like
$\varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$,
then surely the identity you used would be: $\varepsilon_{ijk} \varepsilon_{imn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}$?
but yours is different. is there something i missed while learning this topic?
• August 4th 2007, 03:32 PM
topsquark
Quote:

Originally Posted by physics4life
hold on a minute.
if we have a standard epsilon-epsilon identity.. like
$\varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$,
then surely the identity you used would be: $\varepsilon_{ijk} \varepsilon_{imn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}$?
but yours is different. is there something i missed while learning this topic?

The two are the same. Note that in my statement the common index in the epsilons is the first. Rearranging your expression by noting that $\varepsilon_{kij} = \varepsilon_{ijk}$ you will find we have the same expressions.

-Dan