Expressing this vector integral as a tensor involving the quadrupole

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SUMMARY

The discussion focuses on expressing the vector integral \(\vec{I}\) as a tensor involving the quadrupole moment tensor \(Q\). The integral is defined as \(\vec{I} = \int (\hat{r} \cdot \vec{r'}) \vec{r'} \rho(\vec{r'}) \, d^3r'\), with components \(I_x\), \(I_y\), and \(I_z\) derived from the density function \(\rho(\vec{r'})\). The conclusion reached is that the vector can be compactly represented as \(\vec{I} = Q \hat{r}\), where \(Q\) is the quadrupole moment tensor, simplifying the expression significantly.

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PhDeezNutz
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Homework Statement
In the process of trying to derive an expression for the vector potential of a radiating magnetic dipole(/electric quadrupole)

I came up with this term

$$\vec{I} = \int \left(\hat{r}\cdot\vec{r'}\right) \vec{r'} \rho\left( \vec{r'} \right)\, d^3r'$$

and I want to massage it into a form that looks like a tensor product of sorts.
Relevant Equations
The non-traceless quadrupole is defined as

$$Q_{ij} = \int r_i r_j \rho\left( \vec{r'}\right) \, d^3r'$$
Before writing out each component I'm going to simplify ##\vec{I}## to the best of my abilities

$$\vec{I} = \int \left(\hat{r}\cdot\vec{r'}\right) \vec{r'} \rho\left( \vec{r'} \right)\, d^3r'$$

$$\vec{I} = \hat{r} \cdot \int \vec{r'} \left( x' , y', z' \right) \rho\left( \vec{r'} \right)\, d^3r'$$

Now I'm going to concentrate on each component

$$I_x =\hat{r} \cdot \int x' \vec{r}' \rho \left( \vec{r'} \right) \, d^3r'$$

$$I_y =\hat{r} \cdot \int y' \vec{r}' \rho \left( \vec{r'} \right) \, d^3r'$$

$$I_z=\hat{r} \cdot \int z' \vec{r}' \rho \left( \vec{r'} \right) \, d^3r'$$

Written out completely that is

$$I_x = \left( \hat{r} \cdot \int x'^2 \rho \left( \vec{r'} \right)\, d^3r' \right) \hat{x} + \left(\hat{r} \cdot \int x'y' \rho \left( \vec{r'} \right)\, d^3r' \right) \hat{y} + \left(\hat{r} \cdot \int x'z' \rho \left( \vec{r'} \right)\, d^3r'\right) \hat{z}$$

$$I_y = \left( \hat{r} \cdot \int y'x' \rho \left( \vec{r'} \right)\, d^3r' \right) \hat{x} + \left(\hat{r} \cdot \int y'^2 \rho \left( \vec{r'} \right)\, d^3r' \right) \hat{y} + \left(\hat{r} \cdot \int y'z' \rho \left( \vec{r'} \right)\, d^3r'\right) \hat{z}$$

$$I_z = \left( \hat{r} \cdot \int z'x' \rho \left( \vec{r'} \right)\, d^3r' \right) \hat{x} + \left(\hat{r} \cdot \int z'y' \rho \left( \vec{r'} \right)\, d^3r' \right) \hat{y} + \left(\hat{r} \cdot \int z'^2 \rho \left( \vec{r'} \right)\, d^3r'\right) \hat{z}$$

I'm trying to get this in a compact form that involves the quadrupole tensor but I can't seem to do it.

As always any help is appreciated in advanced.
 
Last edited:
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Would it be

$$\vec{I} = \sum_{i=1}^{3} \sum_{j=1}^3 Q_{ij} \left( \hat{r} \cdot \hat{x_j} \right) \hat{x_i}$$

?

Where ##\hat{r}## is the field point unit vector and the ##\hat{x_j}##'s are the actual basis unit vectors.
 
Last edited:
Wow...I made it too complicated

The integral in question (which is ultimately a vector quantity) is simply

$$\vec{I} = Q \hat{r}$$

where of course Q is the quadrupole moment tensor. I wasn't very successful in manipulating index notation in order to come to this conclusion so I wrote out each component and dot product and it seems to work.
 
Last edited:

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