etotheipi
I came across a question recently which involved calculating the change in wavelength of a photon between being emitted from the surface of the sun and arriving at the Earth.
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 Mph, Me and Ms being the photon, Earth and Sun masses respectively and rs, re and rse 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 Ms/Rs 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.
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 Mph, Me and Ms being the photon, Earth and Sun masses respectively and rs, re and rse 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 Ms/Rs 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.