- #1

etotheipi

The method that was implied involved calculating the GPE's of the photon (assuming the photon to have a mass h/[c lambda]) at the start and end points in order to calculate the overall decrease in its GPE, with M

_{ph}, M

_{e}and M

_{s}being the photon, Earth and Sun masses respectively and r

_{s}, r

_{e}and r

_{se}being the Sun's radius, Earth's radius and Earth-Sun distance respectively. This turns out to be: $$\Delta U = GM_{ph}[-\frac{M_{e}}{r_{e}}-\frac{M_{s}}{r_{se}}+\frac{M_{s}}{r_{s}}+\frac{M_{e}}{r_{se}}]$$ Since the term M

_{s}/R

_{s}is substantially larger than the others, we omit all of the other terms and find the following approximate expression for change in GPE: $$\Delta U = GM_{ph}\frac{M_{s}}{r_{s}}$$The last step is to equate this change in GPE to the change in energy of the photon from which we can approximate the change in wavelength of that photon:$$\Delta E_{ph} \approx hc \frac{\Delta \lambda}{\lambda^{2}} = \frac{GM_{s}}{r_{s}}\frac{h}{\lambda c}$$ This yields the result $$\frac{\Delta \lambda}{\lambda} = 2.1\cdot10^{-6}$$Whilst I can understand the mathematical steps, I have trouble understanding the intuition for this last part. Why can we equate the increase in the photon's GPE to the decrease in the energy associated with its wavelength? Thanks a bunch.