I thought I should pass this on to the list.
Maxima 5.18.1 http://maxima.sourceforge.net
Using Lisp GNU Common Lisp (GCL) GCL 2.6.8 (aka GCL)
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) display2d:false;
(out1) false
(%i2) elliptic_kc(-1)*4/3 = 2*integrate(sqrt(1-z^4),z,0,1);
(out2) 2^(3/2)*gamma(1/4)^2/(12*sqrt(%pi)) = beta(1/4,3/2)/2
(%i3) float(%);
(out3) 1.74803836952808 = 1.748038369528082
(%i4)
It would be nice if Maxima could tell if a function is even and be able to do integrate(f(x),x,-a,a) = 2*integrate(f(x),x,0,a) since
Maxima can do the latter integral already in this case without help.
Rich
----- Original Message -----
From: "Richard Hennessy" <rich.hennessy at verizon.net>
To: "Barton Willis" <willisb at unk.edu>
Sent: Tuesday, August 04, 2009 4:09 AM
Subject: Re: [Maxima] Quantum number and energy of electron
Oh, I didn't notice that Maxima can do integrate(sqrt(1-z^4),z,0,1) case.
The binomial theorem can get the rest, so my number is not so interesting, it's 4/3. But since I used mnewton() to get it I didn't
realize this.
---- Original Message -----
From: "Richard Hennessy" <rich.hennessy at verizon.net>
To: "Barton Willis" <willisb at unk.edu>
Sent: Monday, August 03, 2009 9:33 PM
Subject: Re: [Maxima] Quantum number and energy of electron
My result E[n] is asymptotic to yours for very large n. It looks better for smaller n too. I can't derive the Born Sommerfeld
formula anyway, I just did the integration differently and then substituted C = elliptic_kc(-1)*4/3 = integrate(sqrt(1-z^4),z,-1,1)
but I can't remember why. . .
Thanks for the help.
Rich
----- Original Message -----
From: "Richard Hennessy" <rich.hennessy at verizon.net>
To: "Maxima List" <maxima at math.utexas.edu>; "Barton Willis" <willisb at unk.edu>
Sent: Monday, August 03, 2009 8:29 PM
Subject: Re: [Maxima] Quantum number and energy of electron
I have done this calculation too. I had to check my old wxm files to be sure but I did and I found that this is the Born
Sommerfield quantification formula.
I did it for the mu = 1b15 case and got
(%i17) solve([ans], [E[n]])[3];
`rat' replaced -1.864029228766461B-15 by -1/536472274451275 = -1.864029228766462B-15
(%o17) E[n]=(%pi^2*(2*n-1)^(4/3))/(1072944548902550*1072944548902550^(1/3)*gamma(1/4)^(8/3)).
or in the general case
E[n]=(3*3^(1/3)*%pi^2*hbar^(4/3)*mu^(1/3)*(2*n-1)^(4/3))/(2*2^(1/3)*gamma(1/4)^(8/3)*m^(2/3))
This is not the same thing, I have to check my work to see if I made a mistake.
My posted formula works better for small n (< 70) but the constant exponent I cannot derive yet.
Rich
----- Original Message -----
From: "Barton Willis" <willisb at unk.edu>
To: "Richard Hennessy" <rich.hennessy at verizon.net>
Sent: Monday, August 03, 2009 7:49 PM
Subject: Re: [Maxima] Quantum number and energy of electron
A bit nicer:
(%i10) assume(e > 0, mu > 0, %h_bar > 0, n > 0, m > 0)$
(%i11) 2* integrate(sqrt(e - mu * x^4),x,0, (e/mu)^(1/4)) = %pi * %h_bar *
(n+ 1/2)/sqrt(2*m);
(%o11) (beta(1/4,3/2)*e^(3/4))/(2*mu^(1/4))=(%h_bar*%pi*(n+1/2))/(sqrt
(2)*sqrt(m))
(%i12) solve(%,e);
(%o12) [e=((sqrt(3)*%h_bar^(4/3)*%pi^(4/3)*%i-%h_bar^(4/3)*%pi^(4/3))*mu^
(1/3)*(2*n+1)^(4/3))/(2*2^(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3)),e=-((sqrt
(3)*%h_bar^(4/3)*%pi^(4/3)*%i+%h_bar^(4/3)*%pi^(4/3))*mu^(1/3)*(2*n+1)^
(4/3))/(2*2^(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3)),e=(%h_bar^(4/3)*%pi^
(4/3)*mu^(1/3)*(2*n+1)^(4/3))/(2^(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3))]
(%i13) ans : last(%);
(%o13) e=(%h_bar^(4/3)*%pi^(4/3)*mu^(1/3)*(2*n+1)^(4/3))/(2^(2/3)*beta
(1/4,3/2)^(4/3)*m^(2/3))
(%i14) subst([e=e[0], n = 0],ans);
(%o14) e[0]=(%h_bar^(4/3)*%pi^(4/3)*mu^(1/3))/(2^(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3))
(%i15) solve(%,mu);
Is e[0] positive, negative, or zero?pos;
(%o15) [mu=(4*e[0]^3*beta(1/4,3/2)^4*m^2)/(%h_bar^4*%pi^4)]
(%i16) subst(%,ans);
(%o16) e=(4^(1/3)*e[0]*(2*n+1)^(4/3))/2^(2/3)
(%i17) factor(radcan(%));
(%o17) e=e[0]*(2*n+1)^(4/3)
This is a standard QM textbook problem, so I knew the answer before I
did it. Mostly, I did the calculation to see if Maxima could do the
integral.
Barton
-----"Richard Hennessy" <rich.hennessy at verizon.net> wrote: -----
To: "Barton Willis" <willisb at unk.edu>
From: "Richard Hennessy" <rich.hennessy at verizon.net>
Date: 08/03/2009 06:31PM
cc: <Maxima at math.utexas.edu>
Subject: Re: [Maxima] Quantum number and energy of electron
Thanks for this. My formula is tailored to work for the first 70 quantum
numbers and just for one value of mu (but I have rederived
it for other values of mu, hbar and m and the exponent does not depend on
mu, hbar or m as long as you stick to the first 70 qn's).
This is food for thought, perhaps for both of us (definitely for me),
thanks again. This is quite a surprise.
Rich
----- Original Message -----
From: "Barton Willis" <willisb at unk.edu>
To: "Richard Hennessy" <richhen2008 at gmail.com>
Cc: <Maxima at math.utexas.edu>
Sent: Monday, August 03, 2009 9:56 AM
Subject: Re: [Maxima] Quantum number and energy of electron
Here is a Maxima calculation of the WKB energy of the mu * x^4 oscillator.
One hitch: Maxima handles integrate(sqrt(e - ,mu * x^4),x,0, (e/mu)^(1/4))
OK, but not integrate(sqrt(e - mu * x^4),x,-(e/mu)^(1/4), (e/mu)^(1/4)).
Similarly, Maxima can compute the next two terms in the WKB approximation.
(%i1) assume(e > 0, mu > 0)$
(%i2) 2* integrate(sqrt(e - mu * x^4),x,0, (e/mu)^(1/4)) = %pi * hbar * (n
+ 1/2)/sqrt(2*m);
(%o2) (beta(1/4,3/2)*e^(3/4))/(2*mu^(1/4))=(%pi*hbar*(n+1/2))/(sqrt(2)*sqrt
(m))
(%i3) solve(%,e);
Is hbar*m*(2*n+1) positive, negative, or zero?pos;
(%o3) [e=((sqrt(3)*%pi^(4/3)*%i-%pi^(4/3))*hbar^(4/3)*mu^(1/3)*(2*n+1)^
(4/3))/(2*2^(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3)),e=-((sqrt(3)*%pi^(4/3)*%i
+%pi^(4/3))*hbar^(4/3)*mu^(1/3)*(2*n+1)^(4/3))/(2*2^(2/3)*beta(1/4,3/2)^
(4/3)*m^(2/3)),e=(%pi^(4/3)*hbar^(4/3)*mu^(1/3)*(2*n+1)^(4/3))/(2^
(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3))]
Barton
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