Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 50 / 2

In geometrized units, a mass is converted to a length by multiplying its SI value by $G/c^2$, or to a time by multiplying by $G/c^3$. Using the supplied constants, the solar mass becomesThus the characteristic scales are kilometres and a few microseconds. With potential zero at infinity, the Newtonian gravitational potential at the surface, in units of $c^2$, isIts very small magnitude is the relevant weak-field measure. In SI potential units the same number corresponds to about $-1.91\times 10^{11}{\mathrm{m}}^2{\mathrm{s}}^{-2}$.

Power is energy per time. In geometrized units, energy has the dimension of length and time is converted to length using $c$, so the dimensionless geometrized luminosity isEquivalently, $c^5/G$ is the corresponding SI power unit. The lifetime for emitting the entire rest energy at the given constant luminosity isThe same result follows from $M_{\odot }^{(\mathrm{time})}/L_{\odot ,\mathrm{geom}}$. The stipulated full-mass radiation time is of order $10^{13}$ years; this is the constant-luminosity energy budget asked for, not a stellar-evolution calculation.

Here $I_{ij}$ denotes the unreduced second mass moment tensor, distinct from its trace-free mass quadrupole moment. In coordinates referred to the chosen origin,This expression uses $c=1$. In SI units the leading nonrelativistic mass density is $T^{00}/c^2$. For slowly moving point masses, $T^{00}\simeq {\sum }_A{m}_{(A)}{\delta }^{(3)}(\mathbf{y}-{\mathbf{y}}_{(A)}(t))$, soKinetic corrections to the energy density are higher order in the velocity. The second mass moment tensor is not the mechanical moment of inertia tensor, which instead has components ${\delta }_{ij}{I}_{kk}-I_{ij}$.

In the center of mass frame, put the two positions at $(0,0,z(t))$ and $(0,0,-z(t))$, with $z>0$ during infall. Their separation is $2z$, so Newtonian gravity givesMultiplying by $\dot{z}$ and integrating givesThe negative square root is essential for the falling branch. All second mass moment tensor components except $I_{zz}=2m{z}^2$ vanish. Differentiating three times,Substitution yields ${\stackrel{...}{I}}_{zz}=-m^2\dot{z}/z^2$, hencewith every other component of ${\stackrel{...}{I}}_{ij}$ zero. The distance entering each acceleration is $2z$, which accounts for the factor one quarter in the equation of motion.

The trace-free mass quadrupole moment is diagonal:The same constant factors multiply its third derivatives, giving ${\stackrel{...}{Q}}_{ij}{\stackrel{...}{Q}}_{ij}=(2/3)({\stackrel{...}{I}}_{zz}{)}^2$. The quadrupole formula and the preceding infall result therefore giveFor infall from infinity, $P=m^5/(15z^5)$ and $|\dot{z}|=\sqrt{m/(2z)}$. To find the emitted energy, integrate power over time, using $dt=|dz|/|\dot{z}|$:The total initial mass is $2m$, so the radiated fraction is $1/840\simeq 1.19\times 10^{-3}$. The Sun would emit that fraction of its rest energy inunder the stipulated constant luminosity. This is the head-on quadrupole radiation from equal masses prediction with the prescribed stopping rule. At the endpoint $|\dot{z}|=1/2$ and $m/z=1/2$, so extending the slow-motion weak-field formula that far is an extrapolation, not a controlled strong-field prediction.