Reciprocal Vectors: Representing $\mathbf{u}_i$ in Indicial Notation

In summary, the conversation discusses how to represent the u's in indicial notation and introduces the concept of a reciprocal basis. The conversation then shows how to calculate the reciprocal basis for specific base vectors, and goes on to explain how to verify the conditions for the reciprocal basis. It also includes a discussion about vector division and the dot product between basis vectors. Ultimately, the conversation concludes that the dot product between two perpendicular vectors is always zero, satisfying the conditions for the reciprocal basis.
  • #1
Dustinsfl
2,281
5
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*}
 
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  • #2
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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  • #3
ILikeSerena said:
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.
$$
\mathbf{u}^1 = \frac{\mathbf{u}_2\times\mathbf{u}_3}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^2 = \frac{\mathbf{u}_3\times\mathbf{u}_1}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]};\quad
\mathbf{u}^3 = \frac{\mathbf{u}_1\times\mathbf{u}_2}{[\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3]}
$$
 
  • #4
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}
 
  • #5
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.
 
  • #6
ILikeSerena said:
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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  • #7
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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FAQ: Reciprocal Vectors: Representing $\mathbf{u}_i$ in Indicial Notation

What are reciprocal vectors?

Reciprocal vectors are a mathematical concept used to represent a vector in a different coordinate system. They are typically used in physics and engineering to simplify calculations and equations.

How are reciprocal vectors represented?

Reciprocal vectors are typically represented using indicial notation, where a vector is represented by a set of numbers or indices. These indices correspond to the coefficients of the unit vectors in a specific coordinate system.

What is the relationship between a vector and its reciprocal vector?

The reciprocal vector of a given vector is perpendicular to the original vector and has a magnitude equal to the inverse of the magnitude of the original vector. This means that the dot product of a vector and its reciprocal vector is always equal to 1.

How do you find the components of a reciprocal vector?

To find the components of a reciprocal vector, you can use the formula:
ui = (1/|u|) * ei
where ui is the component of the reciprocal vector, |u| is the magnitude of the original vector, and ei is the unit vector in the direction of the i-th axis.

What is the purpose of using reciprocal vectors?

Reciprocal vectors are used to simplify calculations and equations in physics and engineering. They allow for easier representation of vectors in different coordinate systems and can simplify vector operations such as dot and cross products.

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