Quadrupole moment tensor calculation for ellipsoid

  • #1
LeoJakob
21
1
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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  • #2
That is the restriction of the integral to the ellipsoid.
 
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  • #3
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.
 
  • #4
Yes, it is the Heaviside function.
 
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