Questions: Relativistic Stars and the Tolman-Oppenheimer-Volkoff Equation
4 questions to test your understanding
Score: 0 / 4
Question 1 Multiple Choice
The TOV equation is dP/dr = −G(ρ + P/c²)(m + 4πr³P/c²) / [r²(1 − 2Gm/(rc²))]. What does it reduce to when P ≪ ρc² and Gm/(rc²) ≪ 1?
AdP/dr = 0
BdP/dr = −Gmρ/r²
CdP/dr = −GmP/(r²c²)
DdP/dr = −4πGρr
Dropping the P/c² terms and the factor 1 − 2Gm/(rc²) leaves the Newtonian equation of hydrostatic equilibrium, dP/dr = −Gm(r)ρ/r².
Question 2 True / False
A static perfect-fluid star whose density does not increase outward can have radius R = 1.1 × (2GM/c²).
TTrue
FFalse
Answer: False
Buchdahl's theorem requires 2GM/(Rc²) < 8/9, that is, R > (9/8) × 2GM/c² = 1.125 × 2GM/c². A radius of 1.1 times the Schwarzschild radius is too compact.
Question 3 Multiple Choice
Why do the relativistic terms in the TOV equation make collapse easier than in Newtonian gravity?
AThey remove the pressure from the equation
BPressure contributes to the gravitating mass-energy, and the factor 1 − 2Gm/(rc²) in the denominator strengthens the gradient
CThey make the star rotate
DThey reduce the density at the centre
The terms P/c² in two places and the denominator 1 − 2Gm/(rc²) all increase the required pressure gradient. Gravity grows faster than the pressure that resists it.
Question 4 Short Answer
Using GM_sun/c² ≈ 1.48 km, estimate the compactness 2GM/(Rc²) of a 1.4 solar-mass neutron star with radius 12 km.
Think about your answer, then reveal below.
Model answer: About 0.35
2GM/c² = 2 × 1.4 × 1.48 km ≈ 4.1 km. Dividing by 12 km gives about 0.35, so relativistic effects are strong.