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[SOLVED] Indicial notation

dwsmith

Well-known member
Feb 1, 2012
1,673
Using indicial notation, I am trying to show that $\mathbf{v}\cdot\mathbf{v} = a^2b^2\sin^2\theta$ where $ \mathbf{v} = \mathbf{a}\times\mathbf{b}$ and $\mathbf{v}_i\hat{\mathbf{e}}_i = a_j \hat{\mathbf{e}}_j\times b_j\hat{\mathbf{e}}_k = \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i$.

So
\begin{alignat}{3}
\mathbf{v}\cdot\mathbf{v} & = & \varepsilon_{ijk} a_jb_k\hat{\mathbf{e}}_i\cdot\varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\\
& = &
\end{alignat}
We have to have a kronecker delta since the only surviving terms are when the unit vectors that are dotted with themselves but that is all I have.
 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
Using indicial notation, I am trying to show that $\mathbf{v}\cdot\mathbf{v} = a^2b^2\sin^2\theta$ where $ \mathbf{v} = \mathbf{a}\times\mathbf{b}$ and $\mathbf{v}_i\hat{\mathbf{e}}_i = a_j \hat{\mathbf{e}}_j\times b_j\hat{\mathbf{e}}_k = \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i$.

So
\begin{alignat}{3}
\mathbf{v}\cdot\mathbf{v} & = & \varepsilon_{ijk} a_jb_k\hat{\mathbf{e}}_i\cdot\varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\\
& = &
\end{alignat}
We have to have a kronecker delta since the only surviving terms are when the unit vectors that are dotted with themselves but that is all I have.
Your expression for [tex]\overrightarrow{v} \cdot \overrightarrow{v}[/tex] isn't correct. Your cross products in the individual terms are different in general (since we can always rotate those.) But the i components have to match. So we get:
[tex]\overrightarrow{v} \cdot \overrightarrow{v} = \epsilon _{ijk} a_i b_j \overrightarrow{e_i} \cdot \epsilon _{imn} a_m b_n \overrightarrow{e_i}[/tex]

[tex]\overrightarrow{v} \cdot \overrightarrow{v} = \epsilon _{ijk} \epsilon _{imn} a_i b_j a_m b_n [/tex]

Now we have the relation:
[tex]\epsilon _{ijk} \epsilon _{imn} = \delta _{jm} \delta _{kn} - \delta _{jn}\delta _{km}[/tex]

So
[tex]\overrightarrow{v} \cdot \overrightarrow{v} = \left ( \delta _{jm} \delta _{kn} - \delta _{jn}\delta _{km} \right ) a_i b_j a_m b_n [/tex]
etc.

-Dan
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Now we have the relation:
[tex]\epsilon _{ijk} \epsilon _{imn} = \delta _{jm} \delta _{kn} - \delta _{jn}\delta _{km}[/tex]
Where does this relation come from?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Where does this relation come from?
That is a theorem you have to prove. But it is a fact, and quite helpful when using indices to prove vector identities.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
Where does this relation come from?
It's listed somewhere in my Mathematical Methods text and my Electrodynamics text as well. But frankly I was lazy and looked it up on Wikipedia. You can prove it by writing all the terms for all the indicies... If you've got about a year's worth of your time. :p Obviously there are better ways to prove it.

-Dan
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
Now I have
$$
a_2^2b_3^2+a_3^2b_2^2 - 2a_2b_3a_3b_2 + a_1^2b_3^2+a_3^2b_1^2 - 2a_1b_3a_3b_1 + a_1^2b_2^2+a_2^2b_1^2 - 2a_1b_2a_2b_1
$$
I cant seem to extract a $a^2b^2$ out of this though.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
Now I have
$$
a_2^2b_3^2+a_3^2b_2^2 - 2a_2b_3a_3b_2 + a_1^2b_3^2+a_3^2b_1^2 - 2a_1b_3a_3b_1 + a_1^2b_2^2+a_2^2b_1^2 - 2a_1b_2a_2b_1
$$
I cant seem to extract a $a^2b^2$ out of this though.
The expression [tex]\delta _{jm} \delta _{kn} a_i b_j a_m b_n = a_i b_j a_j b_k [/tex] for example. When the deltas are multiplied you get only one term. You use both on form a_ b_ a_ b_ . I used the first delta on the last a term and the second delta on the last b. So your expansion will only have two terms.

By the way there are other ways to arranged this product but they all come out to be the same in the end when you rearrange the ab products. For example [tex]b_j a_j = a \cdot b[/tex].

-Dan
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The expression [tex]\delta _{jm} \delta _{kn} a_i b_j a_m b_n = a_i b_j a_j b_k [/tex] for example. When the deltas are multiplied you get only one term. You use both on form a_ b_ a_ b_ . I used the first delta on the last a term and the second delta on the last b. So your expansion will only have two terms.

By the way there are other ways to arranged this product but they all come out to be the same in the end when you rearrange the ab products. For example [tex]b_j a_j = a \cdot b[/tex].

-Dan
The symmetric group of 1,2,3 is
$$
(123),(132),(213),(231),(312),(321)
$$
So ijk and imn run through these. My first option is ijk = 123 then I have imn = 123 or 132. By doing this, I get all these terms. I dont see how there is fewer.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
Using indicial notation, I am trying to show that $\mathbf{v}\cdot\mathbf{v} = a^2b^2\sin^2\theta$ where $ \mathbf{v} = \mathbf{a}\times\mathbf{b}$ and $\mathbf{v}_i\hat{\mathbf{e}}_i = a_j \hat{\mathbf{e}}_j\times b_j\hat{\mathbf{e}}_k = \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i$.

So
\begin{alignat}{3}
\mathbf{v}\cdot\mathbf{v} & = & \varepsilon_{ijk} a_jb_k\hat{\mathbf{e}}_i\cdot\varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\\
& = &
\end{alignat}
We have to have a kronecker delta since the only surviving terms are when the unit vectors that are dotted with themselves but that is all I have.
I see now. What we did with all that came before wasn't wrong, just inefficient. That sine function was killing me. However I finally found what was holding me up: [tex]\theta[/tex] is obviously the angle between the vectors a and b. And that lead me to another approach.

[tex]\overrightarrow{v} \cdot \overrightarrow{v} = (\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{a} \times \overrightarrow{b}) [/tex]

As it happens we can evaluate cross products by the following relation:
[tex](\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{c} \times \overrightarrow{d}) [/tex]

Applying the same trick with the Levi-Civita and Kronicker deltas we can easily derive:
[tex](\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{c} \times \overrightarrow{d}) = ( \overrightarrow{a} \cdot \overrightarrow{c}) ( \overrightarrow{b} \cdot \overrightarrow{d}) - ( \overrightarrow{a} \cdot \overrightarrow{d}) ( \overrightarrow{b} \cdot \overrightarrow{c})
[/tex]

Our problem is a special case of this where [tex]\overrightarrow{c} = \overrightarrow{a}[/tex] and [tex]\overrightarrow{d} = \overrightarrow{b}[/tex]

Recalling [tex](\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{a} \times \overrightarrow{b}) = ( \overrightarrow{a} \cdot \overrightarrow{a}) ( \overrightarrow{b} \cdot \overrightarrow{b}) - ( \overrightarrow{a} \cdot \overrightarrow{b}) ( \overrightarrow{b} \cdot \overrightarrow{a})[/tex]

and recalling the dot product between vectors a and b:

[tex] = a^2 b^2 - (ab~cos(\theta)) (ab~cos(\theta)) = (ab)^2(1 - cos^2(\theta))[/tex]

[tex]= a^2 b^2~sin^2(\theta)[/tex]

I've never seen this problem (I think.) What do a and b represent or is this purely a Mathematical exercise?

-Dan
 

dwsmith

Well-known member
Feb 1, 2012
1,673
It was mathematical but there may be meaning to it. I just don't know yet. It is in the book Continuum Mechanics for Engineers 3rd ed by Mase
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I had refrained from posting here since it clearly said indicial notation, but after your approach topsquark I would like to mention that another attempt at a solution is to remember that $\vec{v} \cdot \vec{v} = |\vec{v}|^2$, but if $\vec{v} = \vec{a} \times \vec{b}$ then $|\vec{v}| = | \vec{a} \times \vec{b} | = ab \sin \theta$, where $\theta$ measures the smallest angle between vectors $\vec{a}$ and $\vec{b}$. Thus, $\vec{v} \cdot \vec{v} = a^2 b^2 \sin^2 \theta$.

Cheers! (Yes)

Fantini