What is the formula for the norm of a vector cross product?

[Lambda96]

Homework Statement

Use the identity $\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kj}$ to proof $||\vec{a} \times \vec{b}||=||\vec{a}|| \cdot ||\vec{b}|| \sin{\alpha}$

Relevant Equations

$\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kj}$

Hi everyone,

I'm having problems with task c

In the task, the norm has already been defined, i.e. $||\vec{c}||=\sqrt{\langle \vec{c}, \vec{c} \rangle }$ I therefore first wanted to calculate the scalar product of the cross product, i.e. $\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle$ first

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = \epsilon_{ijk}a^{i}b^{j} \vec{e}_k \cdot \epsilon_{ijk}a^{i}b^{j} \vec{e}_k$$
$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = \epsilon_{ijk}a^{i}b^{j} \cdot \epsilon_{ijk}a^{i}b^{j}$$
$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = \epsilon_{ijk} \epsilon_{ijk} ||a^{i}||^2 ||b^{j}||^2 $$

If I look at my calculation now, I've definitely made a mistake, but I don't know how else to arrive at the desired result.
I know I didn't use the identity $\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kj}$, but if I take the scalar product with itself, the epsilon would be $\epsilon_{ijk} \epsilon_{ijk}$ and not $\epsilon_{ijk} \epsilon_{ilm}$, or not?

 

[Orodruin]

I strongly suggest you to read this: https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

In particular about the 5th commandment that is extra emphazised as it is a very common and usually quite a grave mistake.

 

[fresh_42]

How about starting with what is given?
\begin{align*}
\langle \mathrm{a}\times \mathrm{b}\ ,\ \mathrm{a}\times \mathrm{b} \rangle&=\epsilon_{ijk}\,a^i\,b^j\, \epsilon_{lmn}\,a^l\,b^m \,\langle \mathrm{e}_k\ ,\ \mathrm{e}_n \rangle\\
&=\epsilon_{ijk}\,a^i\,b^j\, \epsilon_{lmk}\,a^l\,b^m= \epsilon_{kij}\,\epsilon_{klm}\,a^i\,b^j\,a^l\,b^m\\
&=(\delta_{il}\delta_{jm} - \delta_{jk} \delta_{km})\,a^i\,b^j\,a^l\,b^m \\
&=a^i\,b^j\,a^i\,b^j \,-\, a^i\,b^k\,a^l\,b^k
&=\ldots
\end{align*}
... if I made no mistakes. I think your mistake was in the formula under "relevant equations". It should have been $$\epsilon_{ijk}\epsilon_{ilm}=\delta_{jl} \delta_{km}- \delta_{jm} \delta_{kl} $$ like in $\delta_{22} \delta_{33}- \delta_{23} \delta_{32}.$

 

[Orodruin]

How about starting with what is given?
\begin{align*}
\langle \mathrm{a}\times \mathrm{b}\ ,\ \mathrm{a}\times \mathrm{b} \rangle&=\epsilon_{ijk}\,a^i\,b^j\, \epsilon_{lmn}\,a^l\,b^m \,\langle \mathrm{e}_k\ ,\ \mathrm{e}_n \rangle\\
&=\epsilon_{ijk}\,a^i\,b^j\, \epsilon_{lmk}\,a^l\,b^m= \epsilon_{kij}\,\epsilon_{klm}\,a^i\,b^j\,a^l\,b^m\\
&=(\delta_{il}\delta_{jm} - \delta_{jk} \delta_{km})\,a^i\,b^j\,a^l\,b^m \\
&=a^i\,b^j\,a^i\,b^j \,-\, a^i\,b^k\,a^l\,b^k
&=\ldots
\end{align*}
... if I made no mistakes.

Typo, but the final expression violates the third commandment.

I think your mistake was in the formula under "relevant equations".

That is a typo, yes, but there are worse mistakes as well.

 

[Lambda96]

Thank you Orodruin and fresh_42 for your help

Thanks also Orodruin for the link, it helped me a lot

I have now proceeded as follows:

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = \bigl( \vec{a} \times \vec{b} \bigr)_i \cdot \bigl( \vec{a} \times \vec{b} \bigr)_i$$

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = \epsilon_{ijk} a^{j}b^{k} \epsilon_{ilm} a^{l} b^{m}$$

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = \epsilon_{ijk} \epsilon_{ilm} a^{j}b^{k} a^{l} b^{m}$$$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = \bigl( \delta_{jl} \delta_{km} -\delta_{jm} \delta_{kl} \bigr) a^{j}b^{k} a^{l} b^{m}$$

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = \delta_{jl} \delta_{km} a^{j}b^{k} a^{l} b^{m} -\delta_{jm} \delta_{kl} a^{j}b^{k} a^{l} b^{m} $$

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = a^{l}b^{m} a^{l} b^{m} - a^{m}b^{l} a^{l} b^{m}$$

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle =\bigl( \vec{a} \cdot \vec{a} \bigr) \bigl( \vec{b} \cdot \vec{b} \bigr) - \bigl( \vec{a} \cdot \vec{b} \bigr) \cdot \bigl( \vec{a} \cdot \vec{b} \bigr) $$$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = ||\vec{a}||^2 \cdot ||\vec{b}||^2 - ||\vec{a}||^2 \cdot ||\vec{b}||^2 \cos^2{\alpha}$$

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = ||\vec{a}||^2 \cdot ||\vec{b}||^2 \bigl( 1 - \cos^2{\alpha} \bigr) $$

$$\langle \vec{a} \times \vec{b} , \vec{a} \times \vec{b} \rangle = ||\vec{a}||^2 \cdot ||\vec{b}||^2 \sin^2{\alpha}$$

$$|| \vec{a} \times \vec{b}|| = ||\vec{a}|| \cdot ||\vec{b}|| \sin{\alpha}$$

 

[vanhees71]

It's a bit simpler to use the index-free calculus in this case and use the additional rule
$$\vec{u} \cdot (\vec{v} \times \vec{w})=(\vec{u} \times \vec{v}) \cdot \vec{w}.$$
Setting $\vec{u}=\vec{a} \times \vec{b}$ and $\vec{v}=\vec{a}$ and $\vec{w}=\vec{b}$ this gives
$$(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) =[ (\vec{a} \times \vec{b})\times \vec{a}] \cdot \vec{b}.$$
Now you use the formula for the triple vector product (which is equivalent to the given formula for the Levi-Civita symbols),
$$(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) = [\vec{b} (\vec{a} \cdot \vec{a}) - \vec{a} (\vec{a} \cdot \vec{b}] \cdot \vec{b} = \|vec{a} \|^2 \|\vec{b}|^2 [1-\cos^2 \angle (\vec{a},\vec{b})]=\|\vec{a}|^2 \|\vec{b} \|^2 \sin^2 \angle (\vec{a},\vec{b}).$$
Since by definite $\angle(\vec{a},\vec{b}) \in [0,\pi]$ the sine is $\geq 0$ and thus
$$\|\vec{a} \times \vec{b} \| = \|\vec{a} \| \|\vec{b} \| \sin \angle(\vec{a},\vec{b}).$$