4.5 powers of black holes: a pair of black holes orbiting their common center - of - mass can merge and…

4.5 powers of black holes: a pair of black holes orbiting their common center - of - mass can merge and radiate gravitational waves such as those now observed by the ligo - virgo - kagra collaboration. gravitational wave characteristics such as the energy radiated per time, rate of change of the wave frequency, and the amplitude of the wave depend on the individual black hole masses, (m_1) and (m_2), in the form of the chirp mass (mathcal{m}) (mathcal{m}equivmu^{3/5}(m_1 + m_2)^{2/5}) (4.36) where (mu) is known as the reduced mass (muequiv\frac{m_1m_2}{m_1 + m_2}) (4.37) a) if the black hole masses are given in kg, what is the unit for (mu) and (mathcal{m})? b) simplify (mathcal{m}) for the following values i) (m_1=m_2 = m) ii) (m_1=m_2/2 = m) iii) (m_1=m_2/20 = m) c) the gravitational wave amplitude depends on the chirp mass power (mathcal{m}^{5/3}). what is the ratio (mathcal{m}_a^{5/3}/mathcal{m}_b^{5/3}) for binary black hole systems (a) and (b) where the black hole masses are both (10m) for system (a) and both (m) for system (b). this provides a ratio of gravitational wave amplitudes if systems (a) and (b) are at the same distance from the gravitational wave observer with the same orbital period

4.5 powers of black holes: a pair of black holes orbiting their common center - of - mass can merge and radiate gravitational waves such as those now observed by the ligo - virgo - kagra collaboration. gravitational wave characteristics such as the energy radiated per time, rate of change of the wave frequency, and the amplitude of the wave depend on the individual black hole masses, (m_1) and (m_2), in the form of the chirp mass (mathcal{m}) (mathcal{m}equivmu^{3/5}(m_1 + m_2)^{2/5}) (4.36) where (mu) is known as the reduced mass (muequiv\frac{m_1m_2}{m_1 + m_2}) (4.37) a) if the black hole masses are given in kg, what is the unit for (mu) and (mathcal{m})? b) simplify (mathcal{m}) for the following values i) (m_1=m_2 = m) ii) (m_1=m_2/2 = m) iii) (m_1=m_2/20 = m) c) the gravitational wave amplitude depends on the chirp mass power (mathcal{m}^{5/3}). what is the ratio (mathcal{m}_a^{5/3}/mathcal{m}_b^{5/3}) for binary black hole systems (a) and (b) where the black hole masses are both (10m) for system (a) and both (m) for system (b). this provides a ratio of gravitational wave amplitudes if systems (a) and (b) are at the same distance from the gravitational wave observer with the same orbital period

Answer

Explanation:

Step1: Find the unit of $\mu$

Given $\mu=\frac{m_1m_2}{m_1 + m_2}$, since $m_1$ and $m_2$ have unit kg, substituting the units: $\frac{\text{kg}\times\text{kg}}{\text{kg}+\text{kg}}=\text{kg}$.

Step2: Find the unit of $\mathcal{M}$

Given $\mathcal{M}=\mu^{3/5}(m_1 + m_2)^{2/5}$, with $\mu$ in kg and $m_1 + m_2$ in kg. The unit of $\mathcal{M}$ is $\text{kg}^{3/5}\times\text{kg}^{2/5}=\text{kg}$.

Step3: Simplify $\mathcal{M}$ for $m_1 = m_2=m$

First find $\mu=\frac{m\times m}{m + m}=\frac{m^2}{2m}=\frac{m}{2}$. Then $\mathcal{M}=\left(\frac{m}{2}\right)^{3/5}(m + m)^{2/5}=\left(\frac{m}{2}\right)^{3/5}(2m)^{2/5}=m\left(\frac{1}{2}\right)^{3/5}\times2^{2/5}=m\times2^{-3/5 + 2/5}=m\times2^{-1/5}$.

Step4: Simplify $\mathcal{M}$ for $m_1=\frac{m_2}{2}=m$ (i.e., $m_1 = m$, $m_2 = 2m$)

$\mu=\frac{m\times2m}{m + 2m}=\frac{2m^2}{3m}=\frac{2m}{3}$. Then $\mathcal{M}=\left(\frac{2m}{3}\right)^{3/5}(m + 2m)^{2/5}=\left(\frac{2m}{3}\right)^{3/5}(3m)^{2/5}=m\left(\frac{2}{3}\right)^{3/5}\times3^{2/5}=m\times2^{3/5}\times3^{-3/5+2/5}=m\times2^{3/5}\times3^{-1/5}$.

Step5: Simplify $\mathcal{M}$ for $m_1=\frac{m_2}{20}=m$ (i.e., $m_1 = m$, $m_2 = 20m$)

$\mu=\frac{m\times20m}{m + 20m}=\frac{20m^2}{21m}=\frac{20m}{21}$. Then $\mathcal{M}=\left(\frac{20m}{21}\right)^{3/5}(m + 20m)^{2/5}=\left(\frac{20m}{21}\right)^{3/5}(21m)^{2/5}=m\left(\frac{20}{21}\right)^{3/5}\times21^{2/5}=m\times20^{3/5}\times21^{-3/5 + 2/5}=m\times20^{3/5}\times21^{-1/5}$.

Step6: Find the ratio of $\mathcal{M}_A^{5/3}/\mathcal{M}_B^{5/3}$

For system $A$, $m_{1A}=m_{2A}=10m$, $\mu_A=\frac{10m\times10m}{10m + 10m}=5m$, $\mathcal{M}A=(5m)^{3/5}(10m + 10m)^{2/5}=(5m)^{3/5}(20m)^{2/5}=m\times5^{3/5}\times20^{2/5}$. For system $B$, $m{1B}=m_{2B}=m$, $\mu_B=\frac{m\times m}{m + m}=\frac{m}{2}$, $\mathcal{M}_B=\left(\frac{m}{2}\right)^{3/5}(m + m)^{2/5}=\left(\frac{m}{2}\right)^{3/5}(2m)^{2/5}=m\times2^{-1/5}$. $\frac{\mathcal{M}_A^{5/3}}{\mathcal{M}_B^{5/3}}=\frac{(m\times5^{3/5}\times20^{2/5})^{5/3}}{(m\times2^{-1/5})^{5/3}}=\frac{m^{5/3}\times5\times20^{2/3}}{m^{5/3}\times2^{-1/3}}=\frac{5\times(2^2\times5)^{2/3}}{2^{-1/3}}=\frac{5\times2^{4/3}\times5^{2/3}}{2^{-1/3}}=5^{5/3}\times2^{5/3}=10^{5/3}$.

Answer:

a) The unit of $\mu$ and $\mathcal{M}$ is kg. b) i) $\mathcal{M}=m\times2^{-1/5}$ ii) $\mathcal{M}=m\times2^{3/5}\times3^{-1/5}$ iii) $\mathcal{M}=m\times20^{3/5}\times21^{-1/5}$ c) $\frac{\mathcal{M}_A^{5/3}}{\mathcal{M}_B^{5/3}}=10^{5/3}$