Physics 4410, Spring '99 Homework #8 Hints

Issued Wed, Mar 3 Due Wed, Mar 10

1) Are you as surprised at how big the field turns out to be as I was? The earth's magnetic field is 1 gauss, which (in MKS units) is about 10^-4 Tesla. Even strong magnets in the lab are usually well under a Tesla. So, I guess most of the time we really are in the weak field limit...

2) You can use Griffiths' equations (which he references), or the equivalent ones from my lecture notes, which look slightly different but are all equivalent, whichever you find easier to work with

3) Again, you can use my formulas from lecture notes, or Griffiths (e.g. 6.74 and 6.75 for weak field, 6.78 for strong field) whichever seems easier.

4) There are many masses floating around in the formulas - some explicit, and some implicit (like, the ones buried in the definition of the Bohr radius) The key trick here is to carefully look through the derivation and decide where you need to use the reduced mass, and where you need to use the regular mass. E.g, in Griffiths Eq's 6.84, those are regular masses, but hidden inside the Bohr radius "a" is a reduced mass. For this reason, you should probably start from Griffiths' equation 6.88, NOT from 6.92, because in going from one to the other he got rid of the constant "e^2" in favor of one more power of "a", but in doing so it's no longer clear which of the "mass" factors in the denominator of Eq. 6.92 are really reduced, and which are not.

I find it much easier to find the RATIO of the hyperfine splitting of the new "atom" to the old hydrogen one (so that lots of nasty common constants just drop out), and then multiply by Eq. 6.92 in the end to get the answer desired. That seems to reduce the risk of calculator error a lot.

Also, in part c, Griffiths is using the word "muonium" to refer to a positive muon/negative electron "atom". (I might prefer to reserve that name for a positive muon/negative muon "atom", but never mind - do what Griffiths wants)