- #1
etotheipi
I was helping (or at least attempting to help...) my sister with this question but ended up getting confused about one part myself! The idea is that you have a mass ##m## at a height ##h## which you drop, and on contact with the ground it is converted into a photon of equivalent energy (frequency ##f##) that is shot back up to the original height before its new frequency ##f'## is measured.
The first way I did it was to take the mass as its own system, then its initial energy is just ##mc^2##. On falling to the ground ##mgh## of work is done by gravity so its energy increases to ##mc^2 + mgh##, and this then becomes the ##hf## of the photon (i.e. ##mc^2 + mgh = hf##). If the energy of the photon at the original height is then ##hf'##, we see that this must equal the initial energy so ##mc^2 = hf'## and we finally obtain ##mgh = h(f-f')##. This is then sufficient to calculate ##f'## if we know ##h## and ##f##, and agrees with the answer on the mark scheme.
However, I tried then attempting it by considering the whole system (including the Earth). The initial energy of the configuration is ##mc^2 + M_E c^2 + mgh##, which is of course also the energy of the system just when the mass hits the ground. Once the photon is created, I reasoned that the total energy is now expressed as ##hf + M_E c^2## (because we're not converting any of the Earth's mass), and then the final energy once the photon reaches the top again is ##hf' + M_E c^2##.
This second approach becomes very problematic to me at least. If we are considering the system as a whole, it's total energy in all four snapshots should be equal. The only resolution is that ##f = f'##, which is evidently wrong. I am sure there is an oversight in the second method on my part, but I can't see it. I have the feeling that my treatment of the rest energy of the Earth is questionable. I wondered if anyone could clarify? Thank you!
The first way I did it was to take the mass as its own system, then its initial energy is just ##mc^2##. On falling to the ground ##mgh## of work is done by gravity so its energy increases to ##mc^2 + mgh##, and this then becomes the ##hf## of the photon (i.e. ##mc^2 + mgh = hf##). If the energy of the photon at the original height is then ##hf'##, we see that this must equal the initial energy so ##mc^2 = hf'## and we finally obtain ##mgh = h(f-f')##. This is then sufficient to calculate ##f'## if we know ##h## and ##f##, and agrees with the answer on the mark scheme.
However, I tried then attempting it by considering the whole system (including the Earth). The initial energy of the configuration is ##mc^2 + M_E c^2 + mgh##, which is of course also the energy of the system just when the mass hits the ground. Once the photon is created, I reasoned that the total energy is now expressed as ##hf + M_E c^2## (because we're not converting any of the Earth's mass), and then the final energy once the photon reaches the top again is ##hf' + M_E c^2##.
This second approach becomes very problematic to me at least. If we are considering the system as a whole, it's total energy in all four snapshots should be equal. The only resolution is that ##f = f'##, which is evidently wrong. I am sure there is an oversight in the second method on my part, but I can't see it. I have the feeling that my treatment of the rest energy of the Earth is questionable. I wondered if anyone could clarify? Thank you!