MHB Indicial Notation: $\mathbf{v}\cdot\mathbf{v} = a^2b^2\sin^2\theta$

[Dustinsfl]

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.

 

[topsquark]

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 \overrightarrow{v} \cdot \overrightarrow{v} 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:
\overrightarrow{v} \cdot \overrightarrow{v} = \epsilon _{ijk} a_i b_j \overrightarrow{e_i} \cdot \epsilon _{imn} a_m b_n \overrightarrow{e_i}

\overrightarrow{v} \cdot \overrightarrow{v} = \epsilon _{ijk} \epsilon _{imn} a_i b_j a_m b_n

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

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

-Dan

 

[Dustinsfl]

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

Where does this relation come from?

 

[Ackbach]

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]

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

 

[Dustinsfl]

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 can't seem to extract a $a^2b^2$ out of this though.

 

[topsquark]

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 can't seem to extract a $a^2b^2$ out of this though.

The expression \delta _{jm} \delta _{kn} a_i b_j a_m b_n = a_i b_j a_j b_k 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 b_j a_j = a \cdot b.

-Dan

 

[Dustinsfl]

The expression \delta _{jm} \delta _{kn} a_i b_j a_m b_n = a_i b_j a_j b_k 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 b_j a_j = a \cdot b.

-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 don't see how there is fewer.

 

[topsquark]

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: \theta is obviously the angle between the vectors a and b. And that lead me to another approach.

\overrightarrow{v} \cdot \overrightarrow{v} = (\overrightarrow{a} \times \overrightarrow{b}) \cdot (\overrightarrow{a} \times \overrightarrow{b})

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

Applying the same trick with the Levi-Civita and Kronicker deltas we can easily derive:
(\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})<br />

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

Recalling (\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})

and recalling the dot product between vectors a and b:

= a^2 b^2 - (ab~cos(\theta)) (ab~cos(\theta)) = (ab)^2(1 - cos^2(\theta))

= a^2 b^2~sin^2(\theta)

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

-Dan

 

[Dustinsfl]

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]

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