makhoma
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vector identity??
The text that I'm reading has a line that reads
\left(\mathbf{b}\mathbf{k}\cdot-\mathbf{b}\cdot\mathbf{k}\right)\mathbf{v}=\omega\mathbf{B}
and I'm not sure what it means by \mathbf{b}\mathbf{k}; it's clearly not the dot product nor the cross product. A line or two below it gives a matrix of the equation:
\left(\begin{array}{ccc}-k_{||}b & 0 & 0 \\ 0 & -k_{||}b &0 \\ k_\perp b & 0 & 0 \end{array}\right)\left(\begin{array}{c} v_x \\ v_y \\ v_z\end{array}\right)=\omega\left(\begin{array}{c}B_x \\ B_y \\ B_z\end{array}\right)
for \mathbf{b}=(0,0,b) and \mathbf{k}=(k_\perp,0,k_{||}) which looks like maybe \mathbf{b}\mathbf{k}\cdot\mathbf{v}=0 for x and y directions but not for z??Any suggestions?
unknown
see above
Homework Statement
The text that I'm reading has a line that reads
\left(\mathbf{b}\mathbf{k}\cdot-\mathbf{b}\cdot\mathbf{k}\right)\mathbf{v}=\omega\mathbf{B}
and I'm not sure what it means by \mathbf{b}\mathbf{k}; it's clearly not the dot product nor the cross product. A line or two below it gives a matrix of the equation:
\left(\begin{array}{ccc}-k_{||}b & 0 & 0 \\ 0 & -k_{||}b &0 \\ k_\perp b & 0 & 0 \end{array}\right)\left(\begin{array}{c} v_x \\ v_y \\ v_z\end{array}\right)=\omega\left(\begin{array}{c}B_x \\ B_y \\ B_z\end{array}\right)
for \mathbf{b}=(0,0,b) and \mathbf{k}=(k_\perp,0,k_{||}) which looks like maybe \mathbf{b}\mathbf{k}\cdot\mathbf{v}=0 for x and y directions but not for z??Any suggestions?
Homework Equations
unknown
The Attempt at a Solution
see above
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