## Archive for September, 2008

### Monday Math 39

September 29, 2008

Suppose we have a group of n people (n≥2). The name of each person is written on a separate slip of paper. These n slips are put in a box, and each person then draws a slip from the box. What, then, is the probability that nobody draws their own name? What happens to this probability when n becomes large?
Solution

### Physics Friday 39

September 26, 2008

Let us consider two point charges of charge –q and q (q>0) separated by a distance d. Let d be the vector from the negative to the positive charge. Choosing the point halfway between the charges as the origin of our coordinate system, then the electric potential is given by:

and the electric field is
.

First, let us examine the field far from the charges (rd). To simplify things, let θ denote the angle between d and r. Then the law of cosines tells us that
 and
. Thus, for rd, we see

and

So we have
.
Using , we see that for distant points, the field is approximated by
.
Defining , and using the unit vector in the direction of r , we have
, or, in terms of θ,
, so we see the magnitude of the distant field goes as the inverse cube of distance. The vector p is the dipole moment, and if one lets d go to zero while holding p constant, the field approaches the one above at all points, and one has a point dipole.

Now, let us instead consider the potential and field near the point between the charges (rd). In this case, , and

Thus

and so
, where  is the unit vector in the direction of d. Note that this approximation is a constant vector. Thus, if we have  while holding  constant, our electric field appoaches a uniform electric field.

### Monday Math 38

September 22, 2008

Find .

The methods I’ve shown before for definite integrals (differentiating under the integral sign, converting to a double integral and reversing order, expanding in infinite series) will not be of much use on this one. However, we can use complex analysis and contour integration to find this.

First, due to symmetry, we see . Further, symmetry also tells us , as the integrand is odd. Thus:
.

So we will compute the integral of  over an appropriate contour in the complex plane. As , the function has poles at ±i. Expanding in partial fractions, , and so we see the residues of  at z=i and z=-i are  and , respectively.

Let us first choose the (counter-clockwise) contour bounding a semicircle of radius R>1 in the upper half plane; that is, the contour consists of the real axis from –R to R, and then the semicircle of radius R in the upper half plane. We can parametrize the semicircle as , 0≤t≤1.

The contour encloses only the pole at z=i, so therefore the integral of our function over our contour is given by the residue theorem as:


Breaking up our contour into its two segments, though, we see:
, where C1 is our semicircle.
Thus 

Now, we will need to use the following theorem of complex analysis:
Suppose f(z) is continuous on the contour C, and there exists a real constant M such that  on C. Then , where L is the length of the contour C.

So, examining |f(z)| on C1, we see:
.
For a>0, the numerator in the above is less than or equal to one for all t in the range 0≤t≤1, and thus
.
Since the length of our semicircle is πR, our theorem tells us
. Now, as R goes to infinity,  goes to zero. Thus, .

Applying this limit to our equation for the real axis portion,

So, for a>0,
.

Note that if a<0, our bound on the semicircle does not work. So now, let us perform the integral over the contour bounding a semicircle in the lower half-plane; it consists of the real axis from R to -R, and then the semicircle of radius R in the lower half plane, which we can parametrize as , 0≤t≤1.

This contour encloses the pole at z=-i only, so via the residue theorem,


Breaking up our contour into its two segments, though, we see:
, where C2 is our semicircle.
Thus .

As to |f(z)| on C2, we see:
.
Here, the numerator in the above is less than or equal to one for all t in the range 0≤t≤1 when a<0, and thus
.
The length of our semicircle is still πR, and so applying the theorem and taking the limit as R goes to infinity, we get
.
Therefore , and so  for a<0.

Combining these two, we see:
.

(Note that a=0 gives the correct equation ).

### Care Bear Stare

September 19, 2008

I was looking through some older bookmarks (specifically, I was looking for some pages for a multi-part work that’s been percolating in my mind for months, and which I’m now putting to (metaphorical) paper), and I found these two posts that describe the “Care Bear Stare” mode of political thought found in segments of the American political spectrum, both Left and Right. Namely, this is the belief that all problems are surmountable if only we care hard enough (Left), or have enough will (Right).
Care Bear Stare!” by Julian Sanchez
The Care Bear Stare Won’t Fix the World” by Rod Dreher

### Physics Friday 38

September 19, 2008

Electrostatics and Charged Surfaces

In the integral form, Gauss’ Law reads:

or in words, that the flux of an electric field through a closed surface equals the total enclosed electric charge divided by ε0. Now, let us consider some surface with a surface charge density σ(r). Let us pick a point on the surface. Let us call the two sides of the surface side 1 and side 2, and let n be a surface normal at the point, directed from side 1 to side 2. Lastly, let us call the electric field immediately on either side of the surface E1 and E2.

Now, let us make a gaussian surface around our point. Our gaussian “pillbox” will be a small, flat cylinder with ends of area dA and axis parallel to n. Letting the height of the cylinder be small, the flux through the “sides” will vanish, and the net flux will be that through the ends. For the end on side 2, we see the (outward) flux will be, for a small “pillbox”, E2·(dAn)=dAE2·n. Similarly, the flux for the end on side 1 is E1·(-dAn)=-dAE1·n. Thus, the total flux is (E2E1n dA. Now, the total enclosed charge is just σ dA, and so:
.
Thus, a surface charge creates a discontinuity in the field perpendicular to the charged surface.

Now, let us consider a perfect conductor in an electric field (the basis of many electrostatics problems). The requirement for a static situation is that the electric potential be constant on the surface of the conductor, and that the electric field be zero within the conductor (otherwise charges in the conductor will be experiencing net electric forces, and thus no static state). As a result, this generally leads to a distribution of charges on the surface of the sphere (“induced charges;” see this previous problem for an example), as required by the resulting discontinuity in the electric field. Note that since the definition of the electric potential Φ is that E=-Φ, the condition that Φ is constant on the surface of the conductor is equivalent to the requirement that the electric field be normal to the surface.

Now, combining these conditions with our finding about surface charges, we see that at the surface of the conductor, with our conventions as before, and side 1 the interior of the conductor, E1=0, and so σ= ε0E2·n. Writing this in terms of the potential outside the conductor, we have σ=-ε0Φ·n=-ε0nΦ; where nΦ=Φ·n is the directional derivative in the direction of the outward surface normal n. This allows us to find the induced charge on the surface of a conductor from the electric potential outside the conductor.

### Happy Constitution Day

September 17, 2008

The United States Constitution was adopted by the Constitutional Convention in Philadelphia 221 years ago today. [My thanks go to Eugene Volokh for the reminder].

### Monday Math 37

September 15, 2008

When written as a decimal, the fraction 1/89 consists of a repeating decimal; specifically with a 44-digit repeat:
1/89=0.01123595505617977528089887640449438202247191…
Note the first few digits after the decimal point: 0, 1, 1, 2, 3, 5. You should recognise from last week the start of the Fibonacci sequence. Examing this, we see:

Or, using our convention last week, it looks like the series

is equal to 1/89. Does the sum actually hold? What if we try it in bases other than base 10?
(more…)

### 50 Years Today

September 12, 2008

Fifty years ago today, Jack Kilby at Texas Instruments demonstrated the first integrated circuit.

### Physics Friday 37

September 12, 2008

Let us consider a pair of parallel, cylidrical metal rails of radius r, the centers of which are separated by a distance d. A movable metal connector is placed between these, so as to slide freely along the rails.

Now, suppose we run a current I through this by connecting a current source to the rails at one end:

What will be the resulting force, if any, on the connector?
(more…)