Question

The ratio of the Schwarzschild radius to the actual radius of a body is the crucial parameter for assessing the importance of general relativity. Show, that for a main sequence star, like the sun, with a typical interior temperature $T_I$, this ratio is approximately given by $\frac{R_{sch}}{R} \approx \frac{kT_I}{m_Hc^2}$ . Show that for a white dwarf $\frac{R_{sch}}{R} \approx \frac{m_e}{m_H}$ , and for a neutron star $\frac{R_{sch}}{R} \approx 1$ .

Answer 1

The Schwarzschild radius is given as r_sch = 2GM/O^2 (i) For sun gravitational force is balanced by pressure of the material So. For H atom in the star gravitational energy is GMmH/r GMmm/r middot m_H c^2/2r = 3/2 KT r_scn/r = 3 KT/m_H c^2 implies r_scn/r TildeEqual KT/m_Hc^2 (ii) For white dwarf gravitational energy of a hydrogen atom is given to the electron so GMmH/r = m_c c^2 2GM/c^2 middot m_H c^2/2r = m_e c^2 r_sch/r TildeEqual m_e/m_H (iii) For neutron star there is neutron so gravitation energy of neutron is total energy of neutron so GMmn/r = m_n c^2 GM/r = c^2 2GM/c^2 middot c^2/r = c^2 r_sch/r TildeEqual 1