Wednesday, May 23, 2007

Finding a Church

Moving to a new city means finding a new church, and even though my wife and I met in church, our religious backgrounds are slightly different. In this case, come last Sunday morning, my wife recommended that we attend here: I wish I had brought my camera to take a color photograph here. Note that the sanctuary has not one, not two, but three video screens at the front. During the service I attended, the round screen in the middle carried kaleidoscopic images of . . . I'm not sure what, but it looked like something from Windows Media Player. The two on the side carried the words to the praise songs set against a background of clouds, trees, oceans and whatnot.

That evening, I picked the church:

We'll probably look some more when we move here, try to find a compromise.

We Bought a House . . .

. . . in the historic district of the city, no doubt surrounded by wealthy communists. The house was built in 1927. Its detached garage is accessed through the alley behind the house--first house I've had like that. Supposedly it has the best public schools in the nation (i.e. the whitest children), a priority for us even though we homeschool (sorry, Dizzy). A relatively long commute for me, but the wife is pleased as punch, so it's worth it.

A big investment for a three year stay. I hope to live there again someday, and in the mean time get slumlord income.

Saturday, May 19, 2007

I've scored my last final exam . . .

. . . posted the grades, written the feedback forms, and driven half-way across the country. Our family mission for the next week is to buy or rent a house near the university where, starting this autumn, I will work on a PhD in engineering.

So, having spent the last three years as, contextually*, the uber alpha, I will spend the next three as just another shit-eating grad student, and at the bottom of that totem pole as well, considering that I barely got in.

*The context being the classes I teach, of course. Back in the dorms, no doubt I am "Professor so-and-so: what a dweeb. I bet his daughters are adopted . . ."

Tuesday, May 15, 2007

Infidelity Around the World

Interesting post from Gene Expression. An excerpt:
There is another issue in Africa which I think needs be brought up: Christianity. I've just finished reading some material on the period in Europe between 500 and 1000, and one point to note is that it was rather difficult for Christian clergy of a Greco-Roman orientation to stamp out polygyny amongst elite males in "barbarian" societies. That is, the nobility of Ireland and Francia were commonly polygynous, even on high up to the Merovingian dynasty. Sometimes this tension between Christian priests and the rulers upon whose patronage they depended played out centuries after the introduction of Christianity. In Africa Christianity is generally less than a century old, with much of the conversion occurring within the last two generations. While the churches preach monogamy, in keeping with Christian models ultimately derived from the Greco-Roman precedent, elite males still tend to enter into operationally polygynous relationships. Because these males are often Christian (Christianity often correlates with high socioeconomic status, and so ability to support extra wives and mistresses) they do not solemnize their relationships with their "secondary" wives. So by definition, if not operation, these are extramarital relationships.

From the File of Rejected Final Exam Questions

Problem 4 [20 pts] It is the year 2017. You are a Captain assigned to the postwar reconstruction of Iran. Your task is to design a distribution system to supply the town of Sirjan from a power generating station in the city of Shiraz. . . .

Problem 10 [20 pts]: It is the year 2027. You are now a Lieutenant Colonel leading counter-insurgency operations in support of the newly independent Republica de Mexifornia. Intelligence reports indicate the breakaway province of Orange County has received covert military supplies from the Greater Chinese Co-Prosperity League (GCCPL). You must interdict these supplies at their storage depot in the Anaheim Angel Stadium.

The rules of engagement forbid you to cross into Mexifornian airspace. You must launch your attack using GPS-guided AGM-86D air-launched cruise missiles from the border town of Bullhead, AZ, approximately 200 miles away from the target.

You suspect that the Orange County insurgents are receiving radar information from a GCCPL KJ-2000 airborne radar flying at 20,000 ft MSL, just outside of Mexifornian territorial waters, and approximately 300 miles from Bullhead . . . .

Monday, May 14, 2007

Waveguide Math

And now for a totally non-relationship oriented post (let's see how many comments I get). Taking the curl of Faraday's Law says: Applying Ampere’s Law and a vector identity give: Assuming no sources and losses inside the waveguide requires that: So: None of this is new; we worked out all this math for plane waves. But for plane waves, we said that for a wave propagating in the z-direction, there was no x and y dependence. In other words, E = Ae-γz, where A was a complex constant. Now we are instead going to assume that E does have an x and y dependence: E = X(x)Y(y)e-γz, where X(x) and Y(y) are functions of x and y. It follows that the derivatives with respect to x and y contained in the vector Laplacian have non-zero values: Consider only the vectors in the z direction. Rearranging terms gives: Likewise: Now, given the form of our expected solution, Ez = X(x)Y(y)e-γz, it follows that Dividing through by X(x)Y(y) e-γz gives: Note that –(γ2 + ω2με) must be a constant. Because X”/X and Y”/Y are each only dependent on x and y, respectively, it follows that both X”/X and Y”/Y must also be constants. Therefore let’s assign the constants –M2 = X”/X and –N2 = Y”/Y so that: M2– N2 = –(γ2 + ω2με) and γ = √(M2 + N2 ω2με). Note: this is the first equation we have developed for γ in this context. You may recall from plane waves that, for a dielectric, γ = = √(με). But now have some additional factors to consider. Note that a propagating wave requires that γ have an imaginary component; if γ is real, we simply have an attenuating field. It follows that M2 + N2 < ω2με. Now let’s assume that X(x) takes the form Aeλx. It follows that X” = λ2Aeλx = –M2 Aeλx. So λ = ±jM and X(x) = AejMx+ Be-jMx = X1sin(Mx) + X2cos(Mx), where X­1 and X2 are complex. Likewise, Y(y) = Y1sin(Ny) + Y2cos(Ny). Therefore: Similarly, for different functions X(x) and Y(y). We could give them subscripts, but we never use them at the same time, so why bother. Armed with these new equations, let’s return to Faraday’s Law and expand the curl of E as described on the back cover of the text: Remembering our z dependency, and separating the vector components, we can write: Likewise, from Ampere’s Law Substituting equation (5) into equation (1) yields: Similarly: Point: If we know the Ez and Hz components, we can find the transverse components! Question: Is there such a thing as a TEM mode (i.e. Ez = Hz = 0) in a waveguide?

Answer: Let Hz = 0; i.e. Gauss’ Law requires that i.e. there are no magnetic sources. Law requires that It follows that if the transverse components of H are non-zero, the z component of E must also be non-zero!

Likewise, by Faraday’s Law, if Ez = 0 and the transverse components of E are non-zero, then Hz ≠ 0.

Therefore, a waveguide cannot support TEM mode.

Boundary Conditions

Remember that Faraday’s Law requires that the E fields tangent to a surface boundary be equal on both sizes of the boundary, i.e. Ez, hollow part = Ez, metal part. But inside the metal, σ → ∞, so σE = J = a finite value requires Ez, inside metal = 0. Therefore Ez, hollow part = 0 at the inside surface of the waveguide. Furthermore, for a wave guide of cross-sectional dimensions a and b, Ex = 0 when y = {0, b} and Ey = 0 when x = {0, a}.

TM Mode: Hz = 0 everywhere inside the waveguide.

Let’s use our equation for Ez we developed earlier: For the boundary x = 0, For the boundary y = 0, So: For the boundary x = a: For the boundary y = b: Plugging in these values for N and M into the equation we developed for Remember that in order to propagate , so the effective cutoff frequency occurs when where
vp = propagation velocity in the dielectric. Note that, because Ez uses the sine function, and sin(0) = 0, both m and n must be greater than zero; otherwise, Ez = 0 and we have no fields for the reason explained earlier.

TE Mode: Ez = 0 everywhere inside the waveguide.

Note: without knowing the surface currents, we cannot directly make any conclusions about Hx and Hy at the boundary. However, we can use the equations we developed earlier, and to say that at boundaries x = {0, a}, At the boundaries y = {0, b}: Given the format of: we can say that: At x = 0, At y = 0, So: At x = a, At y = b, So Notice: because Hz in the TE mode uses the cosine function instead of the sine function, and cos(0) = 1, we can get a wave if either m or n is equal to zero, but not if both are equal to zero. This means that our formula for the cutoff frequency yields as the lowest possible propagation frequency for both the TM and TE cases when a is the larger of the two dimensions, m = 1 and n = 0. We therefore refer to TE10 as the dominant mode.

Spungen writes: "Nice try. There you go with another attempt to reduce propagation, magnetism and dominance to an equation."

Average Age of First Marriage

Year

Men

Women

2003

27.1

25.3

2002

26.9

25.3

2001

26.9

25.1

2000

26.8

25.1

1999

26.9

25.1

1998

26.7

25.0

1997

26.8

25.0

1996

27.1

24.8

1995

26.9

24.5

1994

26.7

24.5

1993

26.5

24.5

1992

26.5

24.4

1991

26.3

24.1

1990

26.1

23.9

1989

26.2

23.8

1988

25.9

23.6

1987

25.8

23.6

1986

25.7

23.1

1985

25.5

23.3

1984

25.4

23.0

1983

25.4

22.8

1982

25.2

22.5

1981

24.8

22.3

1980

24.7

22.0

1979

24.4

22.1

1978

24.2

21.8

1977

24.0

21.6

1976

23.8

21.3

1975

23.5

21.1

1974

23.1

21.1

1973

23.2

21.0

1972

23.3

20.9

1971

23.1

20.9

1970

23.2

20.8

1969

23.2

20.8

1968

23.1

20.8

1967

23.1

20.6

1966

22.8

20.5

1965

22.8

20.6

1964

23.1

20.5

1963 22.8 20.5
hat tip: Bobvis

Friday, May 11, 2007

Illegitimacy Rate by Year

Year Number #/1000 unwed women % of Births
2002 1,358,768 43.6 33.8
2001 1,349,249 43.8 33.5
2000 1,347,043 44.0 33.2
1990 1,165,384 43.8 28.0
1980 665,747 29.4 18.4
1970 398,700 26.4 10.7
1960 224,300 21.6 5.3
1950 141,600 14.1 3.9

Saturday, May 05, 2007