Monday, February 9, 2009

What I don't like about physics

One of the first things my wife ever asked me was the question, "Why do you like physics?" I don't think I ever answered that question for her- or at least not directly. After all, I don't think I really had a good answer back then.

The reason I chose to study physics was pretty simple: I felt that I was good at it. I chose to study physics while taking an advanced placement physics course in high school with my older brother. The classroom was filled with some of the smartest kids in school, many of whom were struggling pathetically- asking stupid questions with obviously no clue as to how to do any of the problems. I don't mean to brag, but I slept through class every other day and aced pretty much every test. Yes, I suppose I must be a genius, and that's just me being modest.

But, the question my wife asked was "Why do you like physics?", which requires a different answer entirely. I think I have a much better idea today than I did way back then, but it is still a question that I am hesitant to answer.

After all, when was the last time you heard someone say the phrase "I love physics" without having to wonder how long it has been since he last brushed his teeth?....or at least wonder how many hours he spends leveling up his elven rogue. I also feel that people who talk about physics a lot do so under the impression that people are impressed by it. Take the following conversation:

"So, what's your major?"
"Oh Really?!!!...........You must be really smart!"

Yeah, so you may think that the person in the conversation above is impressed with my intellectual mastery, but if you could see the facial expressions that accompany the dialogue, you would probably have a different idea. Lets revisit the above conversation, but instead here's what the people involved are actually thinking:

"So, what's your major?"
"Oh Really?!!!...........Why would you do that to yourself?"
"I suppose our conversation is over, huh?"
"Yeah.....So what level is your elven rogue?"

In college I learned to hate that conversation. Nevertheless, I'm still going to try to answer that question that my wife asked so long ago. Just keep in mind that physicists can still be cool. Richard Feynman is proof.

I've been typing for a while, and its late so maybe I'll do it later.


  1. Answer will come as soon as I figure out how to write equations in html. If only we could just use LaTeX...

  2. Haha. You're geeking out soooo hard. Let's go on a WoW run later and pwn some n00bs.

  3. Hey, Player, your spelling is really getting awful. You spell it "pawn some boobs."

  4. Matt, here is a serious question. My understanding is that Einstein's equations for General Relativity use a complex number line for time space. Is that true?

  5. From my understanding of special relativity, the time space can be expressed mathematically as a complex number. However, I think this is more of a mathematical trick, and doesn't really explain what is going on. I'll try to explain.

    Here's the basic idea: So, imagine a vector in a 2-dimensional euclidean space. The vector has an x-component and a y-component. Now imagine that the vector itself stays stationary, while the space around it rotates to yield a new coordinate system. The vector now has to be expressed in terms of new components, say x' and y', which are completely different from the original x and y. However, the vector did not change, so there must be SOMETHING about the vector that is mathematically constant within each coordinate system. The answer of course, is the length of the vector. In other words, x^2+y^2=x'^2+y'^2.

    Thus, we claim that the length of a vector is "rotation transformation invariant" (the official definition of euclidean space is literally any space in which length is invariant under rotation transformations).

    In special relativity, though, we know that the time axis is not euclidean in nature, so we must use the Lorentz transformation to solve for coordinates in alternate inertial spaces. So, like in the rotation transformation analogue, we would like to find an "invariant", which does not change when we apply a Lorentz transformation.

    The "Lorentz invariant" that we're looking for is the quantity x^2+y^2+z^2-(ct)^2. We should note that the only difference between the the Lorentz invariant and Euclidean invariant is the minus sign before the time component. It is for reasons relating to this that it is beneficial to use complex numbers when dealing with four vectors, but I think it is more accurate to understand that the time component of a four-vector is simply not euclidean in nature.

    Someone else might disagree with me, though.