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# Akano's Blog

## Equation of the Day #6: The Lagrangian

Posted by Akano , in Math/Physics Apr 22 2013 · 149 views
Classical, Mechanics, Physics and 1 more...
You may have learned once that classical mechanics all stems from Newton's laws of motion, and while that is true, it is not necessarily the best way to solve a given physical problem. Often when we look at a physical system, we take note of certain physical parameters: energy, momentum, and position. However, these can be more generalized to fit the physical situation in question better. This is where Lagrange comes in; he thought of a new way to formulate mechanics. Instead of looking at the total energy of a system, which is the potential energy plus the kinetic energy, he instead investigated the difference in those two quantities,

where T is the kinetic energy and V is the potential energy. Since the kinetic and potential energy, in general, depend on the coordinate position and velocity of the particle in question, as well as time, so too does the Lagrangian. You're probably thinking, "okay, what makes that so great?" Well, if we were to plot the Lagrangian and calculate the area under the curve with respect to time, we get a quantity known as the action of the particle.

where t1 and t2 are the starting and ending times of interest. Usually if the motion is periodic, the difference between these times is one period. Now, it turns out that for classical motion, the action is minimized with respect to a change in the path along which the particle moves for the physical path along which the particle actually moves. This sounds bizarre, but what it means is that there is only one path along which the particle can move while keeping the action minimized. Physicists call this the Principle of Least Action; I like to call it "the universe is inherently lazy" rule. When you do the math out, you can calculate an equation related to the Lagrangian for which the action is minimized. We call these the Euler-Lagrange Equations.

These are the equations of motion a particle with Lagrangian L in generalized coordinates qi with velocity components denoted by qi with a dot above the q (the dot denotes taking a time derivative, and the time derivative of a coordinate is the velocity in that coordinate's direction). This is one of the advantages of the Lagrangian formulation of mechanics; you can pick any coordinate system that is best-suited for the physical situation. If you have a spherically symmetric problem, you can use spherical coordinates (altitude, longitude, colatitude). If your problem works best on a rectangular grid, use Cartesian coordinates. You don't have to worry about sticking only with Cartesian (rectilinear) coordinates and then converting to something that makes more sense; you can just start out in the right coordinate system from the get go! Now, there are a couple of special attributes to point out here. First, the quantity within the time derivative is a familiar physical quantity, known as the conjugate momenta.

Note that these do not have to have units of linear momentum of [Force × time]. For instance, in spherical coordinates, the conjugate momentum of longitude is the angular momentum in the vertical direction, which has units of action, [Energy × time]. The Euler-Lagrange equations tell us to take the total time derivative of these momenta, i.e. figure out how they change in time. This gives us a sort of conjugate force, since Newton's second law reads that the change in momentum over time is force. The other quantity gives special significance when it equals zero,

This is just fancy math language for saying that if one of our generalized coordinates, qi, doesn't appear at all in our Lagrangian, then that quantity's conjugate momentum is conserved, and the coordinate is called "cyclic." In calculating the Kepler problem – the physical situation of two particles orbiting each other (like the Earth around the Sun) – the Lagrangian is

Note that the only coordinate that doesn't appear in the Lagrangian is ϕ, the longitude in spherical coordinates. Thus, the conjugate momentum of ϕ, which is the angular momentum pointing from the North pole vertically upwards, is a conserved quantity. This reveals a symmetry in the problem that would not be seen if we used the Lagrangian for the same problem in Cartesian coordinates:

That just looks ugly. Note that all three coordinates are present, so there are no cyclic coordinates in this system. In spherical coordinates, however, we see that there is a symmetry to the problem; the symmetry is that the situation is rotationally invariant under rotations about an axis perpendicular to the plane of orbit. No matter what angle you rotate the physical situation by about that axis, the physical situation remains unchanged.

## Science Journals

Posted by Akano , in Life, Math/Physics Mar 19 2013 · 143 views

So, in an earlier blog entry I talked about a journal article one of our professors presented at our department's journal club discussing neutrons in a purely gravitational potential well. Well, I decided to read it and am going to present it to the math and science grad students on Friday because I think it's pretty dang awesome.

Related: Airy functions are weird. And cool. Perhaps I'll discuss them later...

Also, tomorrow spring is here! (If it weren't for ponies, I would not say that with so much excitement.)

`

## Special Relativity

Posted by Akano , in Math/Physics Mar 01 2013 · 184 views

A particle moving through space at the speed of light (i.e. a "massless" particle) does not experience time, and particles that are at rest travel through time at the speed of light.

So the next time you feel lazy loafing on your couch or computer chair, just remember that you are traveling at light speed, no matter how fast or slowly you move.

This entry brought to you by SCIENCE!

## A note on the life of a graduate student

Posted by Akano , in My Little Pony, Math/Physics, Life Jan 27 2013 · 143 views

There's a conference that my adviser wants my lab group to go to in June. Understandable. What's not understandable is that the deadline for abstracts to present at said conference is tomorrow (it was originally Friday, but got bumped back over the weekend). This means, of course, that I have to have enough data to write an amazingly short paragraph about what I'd be presenting at the conference by tomorrow. At 5 p.m.

Now, I'm not one for stress. As my brothers and roommate can easily tell you, I take it rather easy and do not like to be rushed. This past week has been the exact opposite for me.

So, I'm slightly burned out, but I have some data that will hopefully be enough to make an abstract with. Huzzah...

On a more pleasant note, yesterday's Spike episode was good. I love Gummy. And Tank.

## Jackson E&M Final Progress

Posted by Akano , in Life, Math/Physics Dec 17 2012 · 138 views

Saturday: Received final. Four questions pulled from Jackson and Griffiths' textbooks. Finished one problem (the Griffiths one).

Sunday: Finished two more problems. I am on fire! Worked on the fourth problem. Way uglier than the other three.

Monday: Woke up at noon. Haven't continued working yet. Hungry.

This has been a look into the life of a grad student during finals week! Tune in next time where we see whether said student has eaten in the last three days!

## IT WORKS!

Posted by Akano , in Math/Physics Dec 10 2012 · 157 views

I actually invented something that works!

Er, rather, my code works! Hooray!

And with two days to spare.

## Finite Difference, Y U NO WORK?!

Posted by Akano , in Math/Physics Dec 09 2012 · 136 views

So, I'm working on a computer project for my Electrodynamics course. I'm using a computer method called the Relaxation or Finite Difference Method. It basically takes a physical scenario, divides the space of interest into a grid, and assigns voltages for each grid intersection. Then, using a computer language of choice (I'm using FORTRAN, like a boss), I make a program that essentially takes a weighted average of all the points whose voltages aren't fixed until the program doesn't change those voltages anymore. This gives a surprisingly good approximation for a physical system.

I'm basically modeling a system with two conducting cylindrical shells of equal radius separated by some height and which are at voltages +V0 and -V0. The problem is that my output graphs do not look physical; the voltage just drops to near zero rapidly for points outside and between the cylinders, whereas I expect that the graph should gradually drop.

Curse you, technology!

## Equation of the Day #5: Cardioid

Posted by Akano , in Math/Physics Dec 02 2012 · 681 views
Math, Geometry, Cardioid, Heart and 1 more...

I made it in GIMP. A cardioid is the envelope formed by a set of circles whose centers lie on a circle and which pass through one common point in space. This image shows the circle on which the centers of the circles in the above image lie. A cardioid is also the path traced by a point on a circle which is rolling along the surface of another circle when both circles have the same radius (here is a cool animation of that).

What is the cardioid's significance? Well, it looks like a heart, which is kind of cool. It's also the (2D) pickup pattern of certain microphones (I have a cardioid microphone). If a sound is produced at a given point in space, the pickup pattern shows an equal intensity curve. So, if I place a microphone at the intersection point of all those circles, the outside boundary is where a speaker producing, say, a 440 Hz tone would have to be to be heard at a given intensity. So, the best place to put it would be on the side where the curve is most round (the bottom in this picture) without being too far away from the microphone.

Another interesting fact about the cardioid is that it is the reflection of a parabola through the unit circle (r = 1. Here's what I mean). In polar coordinates, the equation of the above cardioid is given by

where a is a scaling factor, and theta is the angle relative to the positive x-axis. The origin is at the intersection of the circles. The equation of a parabola opening upwards and whose focus is at the origin in polar coordinates is just

which is an inversion of the cardioid equation through r = 1, or the unit circle.

## Lunar Eclipse!

Posted by Akano , in Math/Physics Nov 28 2012 · 139 views

In, like, ten minutes several hours.

GO LOOK AT IT, WEST COAST BZPers!

## Equation of the Day #4: Measurement Standards

Posted by Akano , in Math/Physics Nov 25 2012 · 146 views

If you're building something and want to tell other people how to build it, it's useful to show the dimensions of said something (how big it is) relative to other things that people are familiar with. However, there are very few things in this world that are exactly the same size as other similar things (e.g. not all apples weigh the same or have the same volume). So, some smart people once upon a time decided to make standards of measurement for various properties of matter (which I think we can all agree was a smart decision). I wanted to talk about one of these today: the meter.

The word meter (or metre for those who live across the pond/in Canada) comes from the word for "measure" in Greek/Latin (e.g. speedometers measure speed, pedometers measure steps, &c.), but the meter I'm talking about is the International System (SI) unit of distance. The original definition of the meter was one ten-millionth of the distance from the Earth's equator to the North Pole at sea level (not through the Earth). The first person to measure the circumference of the Earth was the Greek mathematician/astronomer/geographer Eratosthenes (and he was accurate to within 2% of today's known value) circa 240 B.C., so this value was readily calculable in 1791 when this standard was accepted.

In 1668, an alternative standard for the meter was suggested. The meter was suggested to be the length a pendulum needed to be to have a half-period of one second; in other words, the time it took for the pendulum to sweep its full arc from one side to the other had to be one second. The full period of a pendulum is

So, when L = 1 m and T = 2 sec, we get what the acceleration due to gravity, g, should be in meters per second per second (according to this standard of the meter). It turns out that g = pi2 meters per second per second, which is about 9.8696 m/s2. This is very close to the current value, g = 9.80665 m/s2 which are both fairly close to 10. In fact, for quick approximations, physicists will use a g value of ten to get a close guess as to the order of magnitude of some situation.

So, you may be wondering, why is it different nowadays? Well, among a few other changes in the standard meter including using a platinum-iridium alloy bar, we have a new definition of the meter: the speed of light. Since the speed of light in a vacuum is a universal constant (meaning it is the same no matter where you are in the universe, unlike the acceleration due to gravity at a point in space), they decided to make the distance light travels in one second a set number of meters and adjust the meter accordingly. Since the speed of light is 299,792,458 meters per second exactly, this means that we have defined the meter as the distance light travels in 1/299,792,458th of a second.

This is all nice, but it's not a very intuitive number to work with. After all, we humans like multiples of ten (due to having ten fingers and ten toes), so why not make a length measurement of the distance light travels in one billionth (1/1,000,000,000th) of a second (a.k.a. nanosecond)? That seems a bit more intuitive, don't you think? It turns out that a light-nanosecond is about 11.8 inches, or about 1.6% off of the current definition of a foot. In fact, one physicist, David Mermin, suggests redefining the foot to the "phoot," or one light-nanosecond, since it's based off of a universal constant while the current foot is based off the meter by some odd, nonsensical ratio.

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Name: Akano
Real Name: Forever Shrouded in Mystery
Age: 25
Gender: Male
Likes: Science, Math, LEGO, Bionicle, Ponies, Comics, Yellow, Voice Acting
Notable Facts: One of the few Comic Veterans still around
Has been a LEGO fan since ~1996
Bionicle fan from the beginning
Misses the 90's. A lot.