Question

The mass and radius of a neutron star is 2M and 15 km, respectively. Calculate the 

value of gravitational red shift for light of wavelength 6000 Å at a distance of 2 m 

from its surface.

Answer 1

We look at the photon emitted from the surface and detected on the height of 2 m over the surface.

The gravitational redshift corresponds to the change of energy of photon due to the change of gravitational potential energy.

At the surface of the star the potential is

$U_0 = -\dfrac{GM}{R},$ at the height of 2 m it is $U_1 = -\dfrac{GM}{R+H}$ . The change of potential is

$U_1 - U_0 = -\dfrac{GMH}{R(R+H)} = \dfrac{h\Delta\nu}{h\nu/c^2} = c^2\dfrac{\Delta\nu}{\nu}$ ,

$z = \dfrac{\Delta\nu}{\nu} = \dfrac{GMH}{c^2 R(R+H)}, \\ z = \dfrac{6.67\cdot10^{-11}\cdot2\cdot2\cdot10^{30}\cdot 2}{ (3\cdot10^8)^2\cdot15\cdot10^3\cdot(15\cdot10^3+2)} = 2.6\cdot10^{-5}.$