scratch1
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A short scratch document, apparently Phil's own working note from an E&M junk bin folder, with the author calling the argument a hokey proof. It writes a general vector function as three scalar angular functions, expands each in Y_lm, and uses the gradient and r-cross-gradient relations for each partial wave. It then concludes the resulting coefficients still depend on angles, a big so what, and the text cuts off. Greek symbols and arguments are lost in extraction.
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Scratch stuff
I will try to give a hokey proof right here. Certainly for a particular , we know that
rYm(,) = am(,) + bm(,)
I could compute a and b by doing the gradient, but let's not bother. Then we know that
rxYm(,) = - bm(,) + am(,)
We solve to get [ where cm(,) am(,)2 + bm(,)2]
= - (am(,)/cm(,)) rYm(,) + (bm(,)/cm(,)) rxYm(,)
= (bm(,)/cm(,))rYm(,) + (am(,)/cm(,)) rxYm(,)
Rewrite as
= - a'm(,) rYm(,) + b'm(,) rxYm(,)
= b'm(,) rYm(,) + a'm(,) rxYm(,)
Notice that the above is true for each m except when =0 which we don't include in things normally. Now, the most general vector function can be written:
A(,) = e(,) + f(,) + g(,)
where all functions are dimensionless and do not include any r factors. At this point, we could expand each of the three scalar functions onto the Ym like this
e(,) = em Ym(,) em = d Ym*(,) e(,)
Thus,
A(,) = em Ym(,) + fm Ym(,) +gm Ym(,)
Now comes the trick. We can use the expressions shown above for and with each partial wave, sort of custom to that partial wave. Then we have:
A(,) = em Ym(,) + fm Ym(,) { - a'm(,)rYm(,) + b'm(,)rxYm(,) }
+gm Ym(,) { b'm(,)rYm(,) + a'm(,)rxYm(,) }
Now just group the two vector quantities together to get
A(,) = em Ym(,) + Ym(,) { - fm a'm(,) + gm b'm(,) } rYm(,)
Ym(,){ fm b'm(,) + gm a'm(,) } rxYm(,)
Rewrite as,
A(,) = em Ym(,) + Ym(,) { cm(,) } rYm(,)
Ym(,){ dm(,) } rxYm(,)
Now expand each of these c and d coefficients onto the Ym(,),
A(,) = em Ym(,) + Ym(,) { C'm' Y'm' (,)} rYm(,)
Ym(,){ D'm' Y'm' (,)} rxYm(,)
And after all this work, we arrive at a big "so what!" The coefficients of the basis functions are now functions of and , but that is not what we want to be claiming.
So we now have shown that
A(,) = em