Archive for October, 2008

A Touch of Humor for Halloween

October 31, 2008

Why do mathematicians confuse Halloween and Christmas?
Answer:

Physics Friday 44

October 31, 2008

Quantum Mechanics and Momentum
Part 1: The De Broglie Relations

Planck’s relation tells us that the energy of a photon is proportional to the frequency: E=, where h is Planck’s constant. In terms of angular frequency ω, we use ω=2πν to find that , where is called the reduced Planck constant, or the Dirac constant.
Classical electrodynamics of an electromagnetic wave and the relativistic energy-momentum relation for a particle with zero rest mass both give the same result for the momentum of a photon: that , and so the momentum is
, where λ is the wavelength. Solving for wavelength, we have . This wavelength can be found for any particle, not just photons, and is called the de Broglie wavelength (see here for details on the historical context and experimental support for this).

Now, let us consider the corresponding angular wavenumber . Then we have

or .

The two equations, and , are known as the de Broglie relations, and the latter one will be important in the later parts of this series.

Monday Math 43

October 27, 2008

Now, we’ve introduced the digamma function, which allows us to find the derivative of the gamma function. But what about higher order derivatives? Here, we find the polygamma functions.
Just as the digamma function is ,
we can define the trigamma function
.
In terms of Γ and its derivatives,
.
So the second derivative of Γ(z) is
.
From the series expression we found for ψ0,

we find

And so we see , where ζ(s) is, as always, the Riemann zeta function.

Further derivatives give us the polygamma function of order n:

From repeated derivatives of , we see that, for n≥1,
.
Thus we have the special values:
.

And from our reccurence relation , we can take the nth derivative to get
.

And we can extend the reflection formula via repeated differentiation to get

From the duplication formula

we find, for n≥1

.
Again letting z=1/2, we have, for n≥1,

Thus , , and .

 

The polygamma functions can be used to find the values of the Euler-Mascheroni integrals: . We already found here that , the Euler-Mascheroni constant. Similarly, we see , where is the nth derivative of the gamma function. We see that

Which, using what we know of these values, gives

Physics Friday 43

October 24, 2008

The average consumer microwave oven uses microwave radiation with a frequency of 2.45 GHz. If you have such an oven running at 1000 W of power (power of the microwaves emmitted into the oven interior), how many photons are emmitted per second?

This is a very simple quantum mechanics problem. The energy of a photon is E=. Here, we have photons of frequency ν=2.45×109 Hz. Therefore, the photons have energy E=(6.626×10-34 J·s)(2.45×109 Hz)=1.62×10-24 J (=1.01×10-5 eV).
The power is P=1000 W=1000 J/s, so we have (1000 J/s)/(1.62×10-24 J)=6.17×1026 photons per second.

Happy Mole Day

October 23, 2008

It’s 6:02 AM, October 23, so, happy Mole Day!

14641

October 20, 2008

This blog has now reached 14641=114 pageviews.

Monday Math 42

October 20, 2008

What is the value of the integral ? How about the general , for non-negative integer n?

First, recall that . Thus,


Recall from the previous Monday Math post the digamma function: ,
and thus
.
In particular, , with γ the Euler-Mascheroni constant.

Testing the nifty new poll feature

October 17, 2008

So, WordPress has given us a new feature for us to put polls in our posts for our readers. So:

Physics Friday 42

October 17, 2008

Last week, we considered a thin ring of radius a and mass M, and the energy of a mass m on the perpendicular axis at a distance z from the center of the ring. We noted that the equilibrium position of the mass m is the center of the ring, and that, at least with regards to motion perpendicular to the plane of the ring, the mass will be attracted toward this point.
Now, let us consider motion within the plane of the ring; in particular, is the equilibrium point at the ring center stable or unstable?

As before, we calculate the gravitational potential energy. Here, due to symmetry, the only element of the position of mass m relative to the ring that will affect the energy is the distance ρ from the center of the ring. Note that we are considering ρ<a.

Let us choose our origin to be the center of the ring, and let us choose an angular coordinate θ so that the mass is on the ray θ=0. Note that a small segment of the ring of arc would have mass . For such a portion at a coordinate angle θ, the distance between that piece and the mass is given by the law of cosines:
.



Thus the gravitational potential energy due to this portion of the ring is
.

And we integrate over all angles to find the energy:

This, however, is an elliptic integral, and cannot be done analytically.
Remember, however, we are interested in stability of the origin, and thus are considering ρa. In particular, to obtain stability, we need to know , which means we need only consider to second order in .
Our integrand is
.

to second order in x is
.
And so our integrand is, to second order,
.
Integrating over 0≤θ≤2π, the terms independent of theta are simply multiplied by 2π, while the and terms integrate to zero; thus, we have:


.
This gives at the center, as expected, and shows us that the potential energy decreases as the mass moves from the center of the ring, and thus the center is an unstable equilibrum. (The Ringworld is unstable!)

Monday Math 41: The Digamma Function

October 13, 2008

Finding the derivative of the gamma function is not an easy task. However, there are reasons to do so. In particular, mathematicians define the digamma function (which I’ve mentioned here), the logarithmic derivative of the gamma function:
\psi_0(z)\equiv\frac{d}{dz}\ln{\Gamma}(x)=\frac{{\Gamma}'(x)}{{\Gamma}(x)}
This is continuous except for the poles at zero and the negative integers.
We can use the Weierstrass form to find some useful expressions for the digamma function. We have:
\Gamma(z)=e^{-\gamma{z}}\frac{1}{z}\prod_{k=1}^{\infty}\frac{e^{z/k}}{1+\frac{z}{k}},
where γ is the Euler-Mascheroni constant, and thus
\begin{array}{rcl}\ln\Gamma(z)&=&\ln\left[e^{-\gamma{z}}\frac{1}{z}\prod_{k=1}^{\infty}\frac{e^{z/k}}{1+\frac{z}{k}}\right]\\&=&-\gamma{z}-\ln{z}+\sum_{k=1}^{\infty}\ln\left[\frac{e^{z/k}}{1+\frac{z}{k}}\right]\\&=&-\gamma{z}-\ln{z}+\sum_{k=1}^{\infty}\left[\frac{z}{k}-\ln\left(1+\frac{z}{k}\right)\right]\end{array}.
Taking the derivative with respect to z, we get
\begin{array}{rcl}\psi_0(z)&=&-\gamma-\frac{1}{z}+\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{z+k}\right]\\&=&-\gamma+\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{z+k-1}\right]\end{array}

Now, note immediately what happens if z is a positive integer n; the terms -\frac{1}{n+k-1}, k=1,2,3,… are canceled by the terms \frac{1}{k} for k=n,n+1,n+2,…, and thus the sum \sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{z+k-1}\right] collapses for positive integer n to the finite sum
\sum_{k=1}^{n-1}\frac{1}{k}=H_{n-1}, where Hn are the harmonic numbers. Thus for positive integer n,
\psi_0(n)=-\gamma+\sum_{k=1}^{n-1}\frac{1}{k}=-\gamma+H_{n-1}
and so we see immediately that ψ0(1)=-γ and ψ0(2)=1-γ (and thus the only positive zero of ψ0 lies between these two).

Using the series form
\psi_0(z)=-\gamma+\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{z+k-1}\right],
we see
\psi_0(z+1)=-\gamma+\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{z+k}\right],
and so we have
\begin{array}{rcl}\psi_0(z+1)-\psi_0(z)&=&\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{z+k}\right]-\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{z+k-1}\right]\\&=&\sum_{k=1}^{\infty}\left[\frac{1}{z+k-1}-\frac{1}{z+k}\right]\\&=&\frac{1}{z}\\\psi_0(z+1)&=&\psi_0(z)+\frac{1}{z}\end{array}
as a recurrence relation for ψ0.

Now, recall the Euler reflection formula we derived previously:
\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi{x}}
Taking the logarithm of both sides, and then differentiating,
\begin{array}{rcl}\ln\left(\Gamma(x)\Gamma(1-x)\right)&=&\ln\left(\frac{\pi}{\sin\pi{x}}\right)\\\ln{\Gamma}(x)+\ln{\Gamma}(1-x)&=&\ln\pi-\ln\sin\pi{x}\\\psi_0(x)-\psi_0(1-x)&=&-\pi\cot\pi{x}\\\psi_0(1-x)&=&\psi_0(x)+\pi\cot\pi{x}\end{array}.

Applying a similar process to the Legendre duplication formula:
\begin{array}{rcl}{\Gamma}(2z)&=&\frac{2^{2z-1}}{\sqrt{\pi}}{\Gamma}(z+\frac{1}{2}){\Gamma}(z)\\\ln{\Gamma}(2z)&=&\ln\left(\frac{2^{2z-1}}{\sqrt{\pi}}\right)+\ln{\Gamma}(z)+\ln{\Gamma}(z+\frac{1}{2})\\\ln{\Gamma}(2z)&=&(2z-1)\ln2-\frac{1}{2}\ln\pi+\ln{\Gamma}(z)+\ln{\Gamma}(z+\frac{1}{2})\\2\psi_0(2z)&=&2\ln2+\psi_0(z)+\psi_0(z+\frac{1}{2})\\\psi_0(2z)&=&\frac{1}{2}\psi_0(z)+\frac{1}{2}\psi_0(z+\frac{1}{2})+\ln2\end{array}
Plugging in z=1/2, we get
\begin{array}{rcl}\psi_0(1)&=&\frac{1}{2}\psi_0\left(\frac{1}{2}\right)+\frac{1}{2}\psi_0(1)+\ln2\\\frac{1}{2}\psi_0(1)&=&\frac{1}{2}\psi_0\left(\frac{1}{2}\right)+\ln2\\\psi_0(1)&=&\psi_0\left(\frac{1}{2}\right)+2\ln2\\\psi_0\left(\frac{1}{2}\right)&=&\psi_0(1)-2\ln2\\\psi_0\left(\frac{1}{2}\right)&=&-\gamma-2\ln2\end{array}

In fact, Gauss proved a formula (Gauss’s DIgamma Theorem) for the exact value of the digamma function for any rational value between 0 and 1: given 0<p<q,
\psi_0\left(\frac{p}{q}\right)=-\gamma-\frac{\pi}{2}\cot\left(\frac{\pi{p}}{q}\right)-\ln(2q)+\sum_{k=1}^{q-1}\cos\left(\frac{2\pi{kp}}{q}\right)\ln\sin\left(\frac{\pi{k}}{q}\right).

Lastly, recall that \gamma=\lim_{n\to\infty}(H_n-\ln{n}). From this, and that \phi_0(n)=H_{n-1}-\gamma, we expect that \phi_0(x) should approach \ln(x-1), and thus approach \ln{x}, as z goes to infinity. Now, using the Stirling’s approximation for the gamma function,
{\Gamma}(x)\approx\sqrt{2\pi}x^{x-\frac{1}{2}}e^{-x}=\sqrt{\frac{2\pi}{x}}\left(\frac{x}{e}\right)^x
we see
\ln{\Gamma}(x)\approx\left(x-\frac{1}{2}\right)\ln{x}-x+\frac{1}{2}\ln(2\pi)
\phi_0(x)\approx\frac{x-\frac{1}{2}}{x}+\ln{x}-1
\phi_0(x)\approx\ln{x}-\frac{1}{2x},
which actually lies sandwiched between \ln{x} and \ln(x-1).