Question

3c) 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 should calculate the gravitational shift due to the difference of the potential energies at the surface and at the height of 2 meters. The difference of potentials will be

$\Delta U = -\dfrac{GM}{R} + \dfrac{GM}{R+h} =- \dfrac{GMh}{R(R+H)} \approx -\dfrac{GMH}{R^2}$ .

So the photon will lose this amount of energy, therefore, the change of energy will be

$\Delta E = h\Delta \nu = -\dfrac{GMH}{R^2}\cdot \dfrac{h\nu}{c^2},$ so $\dfrac{\Delta \nu}{\nu} =- \dfrac{GMH}{c^2R^2} = -\dfrac{6.67\cdot10^{-11}\cdot2\cdot2\cdot10^{30}\cdot 2}{(3\cdot10^8)^2\cdot(1.5\cdot10^4)^2} = -2.6\cdot10^{-5}.$

So $\dfrac{\Delta\lambda}{\lambda} =- \dfrac{\Delta\nu}{\nu}, \; \\ \Delta\lambda = - \dfrac{\Delta\nu}{\nu} \cdot\lambda = 2.6\cdot10^{-5} \cdot 6000 Å = 0.16Å.$