Einstein and Gravitation — Question 4

Question 4

Upward light is redshifted when compared with the receiving clock

A stationary source emits light of local frequency $f_e=6\times10^{14}\,\mathrm{Hz}$. It is received at a higher stationary location with $\Delta\Phi/c^2=2\times10^{-10}$. Use the weak-field relation $f_r/f_e\simeq1-\Delta\Phi/c^2$.

1. Find the received frequency change and its sign.

2. Explain why this redshift is consistent with higher stationary clocks running faster.

Question 4 - Solution

Upward light is redshifted when compared with the receiving clock

(a) Derivation and calculation

Let $\eta=\Delta\Phi/c^2=2\times10^{-10}$. The supplied relation gives $$\Delta f=f_r-f_e\simeq-\eta f_e =-(2\times10^{-10})(6\times10^{14}).$$ $$\boxed{\Delta f=-1.2\times10^5\ \mathrm{Hz}},\qquad f_r\simeq5.9999999988\times10^{14}\ \mathrm{Hz}.$$ The negative frequency change is an upward redshift.

(b) Result and interpretation

For stationary signal exchange, successive cycles correspond to source and receiver proper-time intervals. Their first-order ratio is $$\frac{d\tau_r}{d\tau_e}\simeq1+\eta.$$ Since frequency is cycles per local second, $$\frac{f_r}{f_e}=\frac{d\tau_e}{d\tau_r} \simeq\frac1{1+\eta}\simeq1-\eta.$$ The faster upper clock therefore measures fewer arriving cycles per local second. Local light speed remains $c$.

Physical check

Reversing the height difference reverses the first-order frequency shift.