- #1

- Homework Statement:
- Two point masses M are attached to the ends of a spring of spring constant K. In the quadrupole approximation, what fraction of the energy of oscillation of the spring is radiated away per cycle?

- Relevant Equations:
- ~

I'd just like to check my work. Establish coordinates ##(t, x^i)## with spatial origin at the centre of mass; let the two masses have positions$$x_1(t) = (-a - b \cos{\omega t},0,0), \quad x_2(t) = (a + b \cos{\omega t},0,0)$$The quadrupole moment tensor ##q_{\mu \nu}## is calculated by$$T^{00}(x,t) = M[\delta(x-x_1(t)) + \delta(x-x_2(t))] $$ $$q_{\mu \nu}(t) = 3 \int T^{00} x^{\mu} x^{\nu} d^3 x \implies q_{11}(t) = 6M (a + b \cos{\omega t})^2$$and ##q_{11}## is the only non-zero component, hence the trace is ##q = {q^{\mu}}_{\mu} = q_{11}##. Then define

$$Q_{\mu \nu} = q_{\mu \nu} - \frac{1}{3} \delta_{\mu \nu} q$$from which it follows that all the off-diagonal elements are zero, whilst $$Q_{11} = \frac{2}{3} q_{11}, \quad Q_{22} = Q_{33} = -\frac{1}{3} q_{11}$$Then$$P = \frac{1}{45} \sum_{\mu, \nu = 1}^{3} \dddot{Q}_{\mu \nu}^2 = \frac{8}{45} M\omega^3 \left( ab \sin{\omega t} + 2b^2 \sin{2\omega t} \right)^2$$and after expanding and doing the integration$$\Delta E = \int_0^{2\pi / \omega} P dt = \frac{8}{45} M\omega^3 \left( \frac{\pi b^2}{\omega} (4b^2 + a^2) \right)$$Then divide by ##E##, and re-write ##\omega## in terms of ##K##. Does it look right?

$$Q_{\mu \nu} = q_{\mu \nu} - \frac{1}{3} \delta_{\mu \nu} q$$from which it follows that all the off-diagonal elements are zero, whilst $$Q_{11} = \frac{2}{3} q_{11}, \quad Q_{22} = Q_{33} = -\frac{1}{3} q_{11}$$Then$$P = \frac{1}{45} \sum_{\mu, \nu = 1}^{3} \dddot{Q}_{\mu \nu}^2 = \frac{8}{45} M\omega^3 \left( ab \sin{\omega t} + 2b^2 \sin{2\omega t} \right)^2$$and after expanding and doing the integration$$\Delta E = \int_0^{2\pi / \omega} P dt = \frac{8}{45} M\omega^3 \left( \frac{\pi b^2}{\omega} (4b^2 + a^2) \right)$$Then divide by ##E##, and re-write ##\omega## in terms of ##K##. Does it look right?