Question

Consider the earth moon system quantum mechanically! Treat the earth and moon as point masses.

1. Write down the potential energy function for the earth/moon. Compare it to the potential energy function for the hydrogen atom.
2. What is the “Bohr” radius for the earth/moon system?
3. Estimate the principle quantum number (n) of the earth/moon system.

Answer 1

Solution

Two-body sys­tems, like the earth-moon sys­tem of ce­les­tial me­chan­ics or the pro­ton-elec­tron hy­dro­gen atom of quan­tum me­chan­ics, can be an­a­lyzed more sim­ply us­ing re­duced mass. In this note both a clas­si­cal and a quan­tum de­riva­tion will be given. The quan­tum de­riva­tion will need to an­tic­i­pate some re­sults on multi-par­ti­cle sys­tems.

In two-body sys­tems the two bod­ies move around their com­bined cen­ter of grav­ity. How­ever, in ex­am­ples such as the ones men­tioned, one body is much more mas­sive than the other. In that case the cen­ter of grav­ity al­most co­in­cides with the heavy body, (earth or pro­ton). There­fore, in a naive first ap­prox­i­ma­tion it may be as­sumed that the heavy body is at rest and that the lighter one moves around it. It turns out that this naive ap­prox­i­ma­tion can be made ex­act by re­plac­ing the mass of the lighter body by an re­duced mass. That sim­pli­fies the math­e­mat­ics greatly by re­duc­ing the two-body prob­lem to that of a sin­gle one. Also, it now pro­duces the ex­act an­swer re­gard­less of the ra­tio of masses in­volved.

The clas­si­cal de­riva­tion is first. Let and be the mass and po­si­tion of the mas­sive body (earth or pro­ton), and and those of the lighter one (moon or elec­tron). Clas­si­cally the force be­tween the masses will be a func­tion of the dif­fer­ence

in their po­si­tions. In the naive ap­proach the heavy mass is as­sumed to be at rest at the ori­gin. Then , and so the naive equa­tion of mo­tion for the lighter mass is, ac­cord­ing to New­ton’s sec­ond law,

Now con­sider the true mo­tion. The cen­ter of grav­ity is de­fined as a mass-weighted av­er­age of the po­si­tions of the two masses:

It is shown in ba­sic physics that the net ex­ter­nal force on the sys­tem equals the to­tal mass times the ac­cel­er­a­tion of the cen­ter of grav­ity. Since in this case it will be as­sumed that there are no ex­ter­nal forces, the cen­ter of grav­ity moves at a con­stant ve­loc­ity. There­fore, the cen­ter of grav­ity can be taken as the ori­gin of an in­er­tial co­or­di­nate sys­tem. In that co­or­di­nate sys­tem, the po­si­tions of the two masses are given by

  

be­cause the po­si­tion of the cen­ter of grav­ity must be zero in this sys­tem, and the dif­fer­ence must be .(Note that the sum of the two weight fac­tors is one.) Solve these two equa­tions for and and you get the re­sult above.

The true equa­tion of mo­tion for the lighter body is , or plug­ging in the above ex­pres­sion for in the cen­ter of grav­ity sys­tem,

That is ex­actly the naive equa­tion of mo­tion if you re­place in it by the re­duced mass , i.e. by

The re­duced mass is al­most the same as the lighter mass if the dif­fer­ence be­tween the masses is large, like it is in the cited ex­am­ples, be­cause then can be ig­nored com­pared to in the de­nom­i­na­tor.

The bot­tom line is that the mo­tion of the two-body sys­tem con­sists of the mo­tion of its cen­ter of grav­ity plus mo­tion around its cen­ter of grav­ity. The mo­tion around the cen­ter of grav­ity can be de­scribed in terms of a sin­gle re­duced mass mov­ing around a fixed cen­ter.

The next ques­tion is if this re­duced mass idea is still valid in quan­tum me­chan­ics. Quan­tum me­chan­ics is in terms of a wave func­tion that for a two-par­ti­cle sys­tem is a func­tion of both and . Also, quan­tum me­chan­ics uses the po­ten­tialin­stead of the force. The Hamil­ton­ian eigen­value prob­lem for the two par­ti­cles is: