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[SOLVED] Indicial notation 2

dwsmith

Well-known member
Feb 1, 2012
1,673
Trying to show that $\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} = 0$.
\begin{alignat}{3}
\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} & = & \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot(a_1\hat{\mathbf{e}}_i + a_2\hat{\mathbf{e}}_j+a_3\hat{\mathbf{e}}_k)\\
& = & \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot a_1\hat{\mathbf{e}}_i + \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot a_2\hat{\mathbf{e}}_j + \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot a_3\hat{\mathbf{e}}_k\\
& = & \varepsilon_{ijk}a_jb_ka_1(\hat{\mathbf{e}}_i\cdot\hat{\mathbf{e}}_i) + \varepsilon_{ijk}a_jb_ka_2(\hat{\mathbf{e}}_i\cdot \hat{\mathbf{e}}_j) + \varepsilon_{ijk}a_jb_ka_3(\hat{\mathbf{e}}_i\cdot \hat{\mathbf{e}}_k)\\
& = & \varepsilon_{ijk}a_jb_ka_1
\end{alignat}
Why is this last term 0?
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,143
Trying to show that $\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} = 0$.
\begin{alignat}{3}
\mathbf{a}\times\mathbf{b}\cdot\mathbf{a} & = & \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot(a_1\hat{\mathbf{e}}_i + a_2\hat{\mathbf{e}}_j+a_3\hat{\mathbf{e}}_k)\\
& = & \varepsilon_{ijk}a_jb_k\hat{\mathbf{e}}_i\cdot a_i\hat{\mathbf{e}}_i
Again, a small error.

[tex]\overrightarrow{a} \times \overrightarrow{b} \cdot \overrightarrow{a} = \epsilon_{ijk}a_jb_k \hat{\mathbf{e}}_i \cdot a_i \hat{\mathbf{e}}_i [/tex]

[tex]\overrightarrow{a} \times \overrightarrow{b} \cdot \overrightarrow{a} = \epsilon _{ijk}a_i a_j b_k[/tex]

Now, [tex]a_i a_j[/tex] is symmetric in i and j, but we are taking an antisymmetric product (from the epsilon) and summing it over a symmetric expression. This will always be zero. So

[tex]\overrightarrow{a} \times \overrightarrow{b} \cdot \overrightarrow{a} = 0[/tex]

-Dan
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Now, [tex]a_i a_j[/tex] is symmetric in i and j, but we are taking an antisymmetric product (from the epsilon) and summing it over a symmetric expression. This will always be zero.
I see.
 
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