I Quadrupole moment tensor calculation for ellipsoid

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The discussion focuses on calculating the element Q_{11} of the quadrupole moment tensor for a homogeneously charged rotationally symmetric ellipsoid. The formula for the quadrupole tensor is provided, and the integration process is outlined, including the transformation of coordinates. A key point of clarification is the introduction of the Heaviside step function, θ(1 - ρ²/a² - z²/c²), which restricts the integral to the volume of the ellipsoid. This function ensures that the integration only considers points within the defined ellipsoidal shape. The conversation highlights the importance of correctly applying the Heaviside function in such calculations.
LeoJakob
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Determine the element ##Q_{11}## of the quadrupole tensor for a homogeneously charged rotationally symmetric ellipsoid,
$$\rho=\rho_{0}=\text { const. for } \frac{x_{1}^{2}}{a^{2}}+\frac{x_{2}^{2}}{a^{2}}+\frac{x_{3}^{2}}{c^{2}} \leq 1 $$

The formula is $$Q_{i j}=\int \rho(\mathbf{r})\left(3 x_{i} x_{j}-\|\mathbf{x}\|^{2} \delta_{i j}\right) d^{3} \mathbf{r}$$

I would calculate: $$ Q_{11}=\rho_{0} \int d z \int \rho d \rho \int \limits_{0}^{2 \pi} d \phi \left(3 \rho^{2} \cos ^{2} \phi-\left(\rho^{2}+z^{2}\right)\right) $$

With ##x_1=\rho \cos \phi,\quad x_2=\rho \sin \phi, \quad x_3= z##, but in the solution they calculate:

$$ Q_{11}=\rho_{0} \int d z \int \rho d \rho \int \limits_{0}^{2 \pi} d \phi \theta\left(1-\frac{\rho^{2}}{a^{2}}-\frac{z^{2}}{c^{2}}\right)\left(3 \rho^{2} \cos ^{2} \phi-\left(\rho^{2}+z^{2}\right)\right) $$

Where does the term ##\theta\left(1-\frac{\rho^{2}}{a^{2}}-\frac{z^{2}}{c^{2}}\right)## come from?
 
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That is the restriction of the integral to the ellipsoid.
 
Orodruin said:
That is the restriction of the integral to the ellipsoid.
Ahhh, thank you very much! So it is a Heaviside step function , please correct me if I misunderstood you.
 
Yes, it is the Heaviside function.
 
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