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[SOLVED] reciprocal vectors

dwsmith

Well-known member
Feb 1, 2012
1,673
How do I even represent the u's in indicial notation since the u's are labeled 1,2,3? I can't have components 1,2,3 and say u is $u_i$.

With respect to the triad of base vectors $\mathbf{u}_1$, $\mathbf{u}_2$, and $\mathbf{u}_3$ (not necessary unit vectors), the triad $\mathbf{u}^1$, $\mathbf{u}^2$, and $\mathbf{u}^3$ is said to be the reciprocal basis if $\mathbf{u}_i\cdot\mathbf{u}^j = \delta_{ij}$.
Show that to satisfy these conditions
$$
\mathbf{u}^1 = \frac{\mathbf{u}_3\times\mathbf{u}_1}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^2 = \frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^3 = \frac{\mathbf{u}_1\times\mathbf{u}_2}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}
$$
and determine the reciprocal basis for the specific base vectors
\begin{alignat*}{3}
\mathbf{u}_1 & = & 2\hat{\mathbf{e}}_1 + \hat{\mathbf{e}}_2,\\
\mathbf{u}_2 & = & 2\hat{\mathbf{e}}_2 - \hat{\mathbf{e}}_3,\\
\mathbf{u}_3 & = & \hat{\mathbf{e}}_1 + \hat{\mathbf{e}}_2 + \hat{\mathbf{e}}_3
\end{alignat*}
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
You can write $\mathbf u_i$ for instance as $u_{ij}$.
A basis can be seen as just a matrix containing the basis vectors.

You appear to have a couple of typos in your problem.
The numerators of the first two reciprocal vectors have the wrong indices.

To verify, consider that the two vectors in a cross product are both perpendicular to the resulting vector.
That should make it easy to calculate each of the dot products.
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
You can write $\mathbf u_i$ for instance as $u_{ij}$.
A basis can be seen as just a matrix containing the basis vectors.

You appear to have a couple of typos in your problem.
The numerators of the first two reciprocal vectors have the wrong indices.

To verify, consider that the two vectors in a cross product are both perpendicular to the resulting vector.
That should make it easy to calculate each of the dot products.
$$
\mathbf{u}^1 = \frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^2 = \frac{\mathbf{u}_3\times\mathbf{u}_1}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^3 = \frac{\mathbf{u}_1\times\mathbf{u}_2}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
All I have to do is show that $\mathbf{u}_1\cdot \mathbf{u}^j = \delta_{1j}$ where j = 1,2,3.
How do I do vector division?
\begin{alignat}{3}
\frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}\cdot \mathbf{u}_1 & = & \frac{\mathbf{u}_2\times\mathbf{u}_3}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)}\cdot \mathbf{u}_1\\
& = & \frac{\mathbf{u}_1}{\mathbf{u_1}}\\
& = & 1\\
\frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}\cdot \mathbf{u}_2 & = & \frac{\mathbf{u}_2\times\mathbf{u}_3}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)}\cdot \mathbf{u}_2\\
& = & \frac{\mathbf{u}_2}{\mathbf{u_1}}\\
& = & ?
\end{alignat}
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
The cross product in the numerator is a vector.
The dot product in the denominator is a scalar.

You are supposed to do a dot product of $\mathbf u_1$ with the nominator, and afterward divide by the scalar denominator.

Btw, the dot product in the denominator represent the volume of the parallellepipedum between the 3 basis vectors.
It is symmetric in each cyclic combination of the basis vectors.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The cross product in the numerator is a vector.
The dot product in the denominator is a scalar.

You are supposed to do a dot product of $\mathbf u_1$ with the nominator, and afterward divide by the scalar denominator.

Btw, the dot product in the denominator represent the volume of the parallellepipedum between the 3 basis vectors.
It is symmetric in each cyclic combination of the basis vectors.
\begin{alignat}{3}
\frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]} \cdot\mathbf{u}_2 & = & \frac{\mathbf{u}_2\times\mathbf{u}_3}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)} \cdot\mathbf{u}_2\\
& = & \frac{\mathbf{u}_2\cdot(\mathbf{u}_2\times\mathbf{u}_3)}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)}\\
& = & \frac{\varepsilon_{ijk}u_ju_k\hat{\mathbf{e}}_i \cdot \mathbf{u_2}}{\mathbf{u}_1 \cdot(\mathbf{u}_2 \times\mathbf{u}_3)}
\end{alignat}
How do I show the numerator is zero now?

Is it because the two left over are always the reverse?
(123) and (132) = 1 - 1 = 0
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
The cross product of $\mathbf u_2$ and $\mathbf u_3$ is perpendicular to $\mathbf u_2$, therefore their dot product is zero $\square$.

Or if you want to do it with indices (your indices of the basis vectors are not quite right):
$\mathbf u_2 \cdot (\mathbf u_2 \times \mathbf u_3) $

$\qquad = u_{2i}\mathbf{\hat e}_i \cdot \varepsilon_{ijk}u_{2j}u_{3k}\mathbf{\hat e}_i$

$\qquad = \varepsilon_{ijk}u_{2i}u_{2j}u_{3k}$​

Since $\varepsilon_{ijk}$ is anti-symmetric in i and j, the result is zero $\square$.
 
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