- #1
LeoJakob
- 24
- 2
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?
$$\rho=\rho_{0}=\text { const. for } \frac{x_{1}^{2}}{a^{2}}+\frac{x_{2}^{2}}{a^{2}}+\frac{x_{3}^{2}}{c^{2}} \leq 1 $$
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?