Physics Friday 61

Let us consider a spinning “top”, spinning about an axis of rotational symmetry with angular velocity ω; the mass of the top is m, and the moment of inertia for the top about this axis is I. The top is supported by a surface below, which it contacts at a single “pivot” point on the rotation axis; the distance between this pivot point and the center of gravity is l.
For all following analysis, we ignore dissipative forces.



The only external forces on the top are gravity and the normal force at the pivot.
If the rotational axis is perfectly vertical, these forces cancel and produce no torques, so that the top continues its simple rotation unchanged.
But what if the top is tilted from the vertical by an angle θ?
We choose as our origin the pivot point. We now have a nonzero torque which has magnitude , and is directed perpendicular to the vertical plane containing the rotation axis.

Now, we note that the top has angular momentum L=Iω directed along the rotation axis. Now, we recall that ; thus the angular momentum will change as a result of the torque. We see that τ is horizontal, and so the vertical component of L, L_z, does not change, and as τ and L are perpendicular, neither does the magnitude of L. Thus, we see that the angular momentum vector L will be rotating uniformly about the vertical axis. Due to the object’s symmetry, the permanent axis must remain along L, and so will rotate about the axis. This is classical torque-driven precession.

If the angular velocity of the precession is Ω, then

19683

According to WordPress, this blog has reached 3⁹=19683 page views.

Monday Math 60

Let ω(n) be the number of distinct prime factors of the integer n, with us defining ω(1)=0 (see, for example, here). What, then, is the value of the series ?
How about ?

Physics Friday 60

Let us consider a hydrogen atom: a single electron “orbiting” a single (much heavier) nucleus containing a single proton. The potential energy due to the attractive Coulomb force between these charged particles is , where r is the distance between the particles. The time-independent Schrödinger equation for the electron (ignoring relativistic effects, particle spins, and magnetic moments) is

(Here, the electron mass m should actually be the reduced mass μ of the electron-proton pair; however, the correction involved is small, and even smaller for more massive atomic nuclei). With the way our potential energy is defined (with zero energy at infinite separation), our bound states for the electron (our states of interest for an atom) will have E<0.

Now, we note that the potential is spherically symmetric, and so our Hamiltonian commutes with the angular momentum operators, and we can perform separation of variables in spherical coordinates, with the angular components being the spherical harmonics (see here and here).
As in here, when we perform the spherical coordinate separation , we obtain radial equation


Making the substitution ,

Now, let us define , which has units of length, and dimensionless variable , so that . Then we have

now, let us make the transform (ρ is dimensionless); then we get


We see that as ρ→∞, the equation is approximated by
, which has general solution ; normalizability requires C₂=0. Also, examining ρ→0, our equation is approximately , which has general solution ; considering the origin tells us C₂=0 again. Combining these, we thus try the substitution , giving radial equation

.
Now, letting , we get

Which is the associated Laguerre differential equation

with and ; so the solution to our transformed equation is
, where is an associated Laguerre function. Now, the resulting wavefunction can be normalized only if is a polynomial; this is true when is a non-negative integer; thus, we have , where n is a positive integer, and 0≤l≤n-1. Looking back through our conversions,
we have
,

,
and
,
where , is the constant needed to normalize the wavefunction; with some work involving the properties of associated Laguerre polynomials (see for example here), we can find the normalized wavefunction to be:

Where n=1,2,3,…, l=0,1,2,…,n-1, and m=-l, -l+1,…,l-1,l (for a given n, we have n² different angular momentum states).
One should note that is the Bohr radius.

Now, recall that we had ; with the normalization condition , we have
, and the ground state energy of hydrogen is approximately -13.6 eV, and so the ionization energy of hydrogen is approximately 13.6 eV.

Monday Math 59

Recall the Riemann zeta function
.

Now, consider the product . We see:
,
and we have the zeta function series with the even terms removed. Now, let us multiply the above by . Then we see:

where the remaining terms are those where k is not divisible by 2 or 3. Multiplication of the above by will eliminate those remaining terms divisible by 5; we may continue this procedure through the primes, and (in analogy with the Sieve of Eratosthenes) will, in the infinite product, eliminate every term, k>1, giving
, where p_n is the nth prime; thus the Riemann zeta function may be expressed as the product
.
Thus we see a connection between the primes and the Riemann zeta function. In fact, through this identity (first proved by Euler), the divergence of at s=1 (divergence of the harmonic series) implies that there are infinitely many primes.

Physics Friday 59

Let us consider the isotropic three-dimensional quantum harmonic oscillator: we have (where the particle mass is now m₀ to prevent confusion later). In cartesian coordinates, this becomes:

where H_x, H_y, and H_z are each the hamiltonian for a one-dimensional harmonic oscillator, in the x, y, and z directions respectively. Thus, the energy eigenstates will be products of eigenstates of these three; we have energy levels
. Thus, we have energies , with degeneracy equal to the number of solutions of , which, via combinatorics, is . (Only the ground state n=0 is non-degenerate.)

Now, suppose instead we consider spherical coordinates. As depends on r only (spherical symmetry), we see that under separation of variables, the angular components will be given by the spherical harmonics (see here and here). We will then have . With this in place, our Schrödinger equation becomes, for the radial component:

Defining , we see that this differential equation simplifies to
.
Now, we rescale the radial coordinate by defining dimensionless . Then the above equation becomes
, where is the rescaled u(r). We can expect that far from the origin, we should have something like a Gaussian. If we attempt the substitution (where k is a constant for which we will find later a value convenient for solving the equation), then this becomes, after eliminating a factor of ,
.
We see that this simplifies greatly if , so that , and then
.
Lastly, defining , and , we get
,
which is the associated Laguerre differential equation

with parameters and .
The solutions are associated Laguerre functions , , and to be physically valid, we require that be a non-negative integer. Thus l≤n, and they are both odd or both even integers. Thus, for even n, l can take values 0,2,4,…,n-2,n; and for odd n, l can take values 1,3,5,…,n-2,n. Adding the fact that m is an integer with possible values -l,-l+1,…,l-1,l; we have 2l+1 different possibilities for m. So, for even n, we have
different possible angular momentum states; and for odd n, there are
angular momentum states; the same formula as for even n. Looking back at the start of this post, we note that this formula matches the result for the degeneracies of the energy states we found using cartesian coordinates.

Another Birthday

And a happy 200th birthday to America’s 16th president, Abraham Lincoln

Darwin Day

Happy Darwin Day! Charles Darwin was born 200 years ago to the day.

Monday Math 58

Recall the product formula for the sine function (here and here):

Thus:

or with z=πx,

Taking the logarithm of both sides:
.
Now, , so taking the derivative of both sides of the above equation:
.
Note that via the geometric series,
.
Thus

Note that we can reverse the order of summation:
,
where is the Riemann zeta function.

Now, we use Euler’s formula: as , , , and
.
Previously, we showed that the function can be given by the series . Thus
,
and so
.
Now, as and , we can see that the first two terms of the series in the above are
.
Thus
.
Now, recall that for n≥2, B_n=0 for odd n; thus the series above only has even n terms, and we can rewrite it, using n=2k, as

or thus as

Comparing this to
, we see via the term-by-term comparison that

or solving for the zeta function:

which gives the Riemann zeta function for even positive integers, as previously noted here.

Physics Friday 58

In a previous Friday post, I demonstrated one method of determining the energy eigenvalues for the one-dimensional quantum harmonic oscillator. In particular, we took the (time-independent) Schrödinger equation , and by defining dimensionless parameters , , and then attempting a solution of the form , we derived the differential equation , where the series solution has recursion relation . Then, the requirement that the wavefunction be normalizable requires that the series solution terminate after finitely many terms, requiring that ⇒, for n a non-negative integer.

Now, let us consider the ground state case: n=0. Then , and the series recursion relation becomes , so that , and the terminating solution is that u(ξ) is a constant (a₀). Then , and so
.
Normalizing this wavefunction,

and
.
So the ground state wavefunction of the 1-dimensional quantum harmonic oscillator is a gaussian.

Now, let us examine the uncertainties of position and momentum. The Heisenberg Uncertainty Principle tells us that (see here). For our wavefunction, we see

and

as in both cases the integrand is odd (see here).
Thus

Using and , we see that
, and so
.
Similarly,
.
Now,
,
so

Taking the product,
,
which is exactly the lower limit allowed by the uncertainty principle. Thus, the non-zero value of the ground-state energy (zero-point energy) can be seen as being a result of the Heisenberg Uncertainty Principle.