This is so cool I felt I had to share with someone,
Griffiths, Introduction to Quantum Mechanics, 2nd ed., Pearson Prentice Hall (2005), p. 159, Problem 4.17.
Quote:
Consider the earth-sun system as a gravitational analog to the hydrogen atom.
a.) What is the potential energy function?
This of course is Newton's law of universal gravitation,
V(r) = -G * M * m / r
where m and M are the mass of the Earth and Sun, respectively.
Quote:
b.) What is the "Bohr radius", a_g, for this system? Work out the actual number.
In the electric case, there's a factor of e^2 / (4*pi*e_0) in everything, being the Coulomb potential (the electron and proton both have charge e).
Everywhere this appears, simply replace it with G*M*m.
The hydrogen Bohr radius a_0 = 4*pi*e_0*hbar^2 / (m_e*e^2) has it upside down, so multiply and divide by the new stuff. Also, m_e is the electron
mass, which arises from the momentum operator (assuming the proton mass is so large it's negligible, which is a fair assumption), this gets changed to
Earth's mass (assuming the Sun's mass is very large, which is still a good assumption).
a_g = hbar^2 / (m * G*M*m) = 2.348 x 10^-138 m, pretty damn small (a good reason to ignore quantum mechanics in the solar system!).
Quote:
c.) Write down the gravitational "Bohr formula", and, by equating E_n to the classical energy of a planet in a circular orbit of radius r_o, show that
n = sqrt(r_o / a_g). From this, estimate the quantum number n of the Earth.
Easy way to remember it: all the coefficients stack up to -13.6 eV, which should be familiar to anyone who's had a physics or chemistry class that
covers atoms or atomic spectra.
For our purposes, replace m_e with m, and that e^2 / (4*pi*e_0) thing with G*M*m again.
As for the orbital energy, assuming a round orbit, potential will equal kinetic energy, and the total energy will be the sum, which will also also
equal the Sun's potential, which has the same old G*M*m, divided by the Earth's orbital radius r_o (defined as 1 A.U.). kinetic total potential
= 5.3 x 10^33 J.
Now, the problem states to equate energy, so we've got the classical energy equal to the "Bohr" energy:
E_c = -G*M*m / r_o = E_n = -(m / 2*hbar^2) * (G*M*m)^2 * 1/n^2
n^2 = r_o * (m / 2*hbar^2) * (G*M*m)^2 / (G*M*m)
n = sqrt(r_o * m * G*M*m / (2*hbar^2))
Since a_g = hbar^2 / (m * G*M*m),
n = sqrt(r_o / (2*a_g))
Hmm, maybe I wasn't supposed to take the total, just the kinetic. If that's the case, then the 2 should cancel, leaving the equation given.
In any case, using the numbers above, n ~= 2.52 x 10^74.
Now for the interesting part that I've been waiting for since I first saw this problem:
Quote:
d.) Suppose the Earth made a transition to the next lower level (n - 1). How much energy (in Joules) would be released? What would the wavelength of
the emitted photon (or, more likely, graviton) be? (Express your answer in light years -- is the remarkable answer {Thanks to John Meyer for pointing
this out.} a coincidence?)
The next lower level is 10^-74 times insignificant; neither Windows Calculator nor Google Calculator have enough decimal places to find it brute
force. Therefore, let us apply a little algebra:
E_n = E_0 * 1/n^2 and E_(n-1) = E_0 * 1/(n-1)^2 so
E_delta = E_0 * (1/n^2 - 1/(n-1)^2)
Common denominator,
= E_0 * ((n-1)^2 - n^2) / (n^2)(n-1)^2
Expand,
= E_0 * ((n^2 - 2*n + 1 - n^2) / (n^2)(n-1)^2
n^2 cancels -- that's good, it's a HUGE number!
= E_0 * ((1 - 2*n) / (n^2)(n-1)^2
Since n is so big, I will now ignore the -1 term in the denominator, so it reduces to n^4. Distributing,
= E_0 * (1/n^4 - 2/n^3)
Likewise, 1/n^4 is 10^74 times smaller than 2/n^3, so it goes away.
E_delta ~= E_0 * (-2/n^3)
= -1.688 x 10^182 J * (-2 / (2.52 x 10^74)^3)
= 2.1 x 10^-41 J
The wavelength (assuming a massless particle, like a photon, or as the problem aptly suggests, a graviton) is therefore 9.46 x 10^15 m, which is extraordinarily close to 1.000 light years!
And to finish the problem: is this answer a coincidence? Well, a year is a fairly arbitrary measure, it's just what we call that period of time. But
wait, what was that time based on?
You may recall that General Relativity predicts gravity waves being radiated from accelerated masses. Well, the waves certainly should have a
wavelength of 1 ly -- that's how fast they travel and how often they're given off from the Earth. But how was this just derived, above? Well, with
quantum mechanics. But the Earth is stupid huge, surely QM cannot apply?
The most remarkable part of this problem, I think, is that it proves QM works -- or at least, is consistent with Relativity on a large scale --
despite producing numbers so incredibly large or small as to be absurd -- they work out in the end. Truely amazing!
Tim
[Edited on 12-27-2009 by 12AX7]watson.fawkes - 27-12-2009 at 07:19
That's analysis, not algebra... SCNR...12AX7 - 27-12-2009 at 10:18
So QUOTE is case sensitive, but not URL??? WTF? Sigh...
TimVogelzang - 30-12-2009 at 14:26
If quantum mechanics applies to everything in our universe and everything came from the big bang that created our universe, is everything in our
universe entangled? Does this support astrology?
Griffiths, Introduction to Quantum Mechanics, 2nd ed., Pearson Prentice Hall (2005)
What do you think of that book? I have only briefly skimmed it in a bookstore (too expensive for an impulse buy!) However, I have two other textbooks
by Griffiths, one on electromagnetism, and another on elementary particles, and both are EXCELLENT. I am not happy with any of my basic QM textbooks,
so I kind of want to get this one someday. JohnWW - 30-12-2009 at 16:17
Downloads of Griffiths' Introduction To Quantum Mechanics can be found by searching on Google. I found and downloaded it in January 2009
using these URLs:
SCNR...