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Comment by stcredzero | Hacker News Reader

I had an ultimate fallback for high school physics. Start putting stuff in likely equations where the units work out to the desired answer. That always worked, save for two or three occasions. Unfortunately, once I figured that out, that let me be lazy and kept me from learning as well as I should have.

When Wolfram Alpha starts solving word problems, time to start worrying. Not about machines rebelling, but about cheating in grade school math classes.

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> Start putting stuff in likely equations where the units work out to the desired answer.

Working physicists call this 'dimensional analysis' and it's pretty much the foundational skill of theoretical physics.

How true is this? Isn't dimensional analysis basically one kind of bookkeeping like Haskell's type checking?

What makes it particularly important in theoretical physics? Is it "foundational" because equations become unintuitive?

With known equations, then yes, dimensional analysis is essentially just a form of bookkeeping that helps you verify your math as you go.

But once you get into using quantum field theory to study new systems, the standard approach is to concoct a conservation of energy equation by summing terms, where each term is a combination of the system's variables with units of energy.

From there, you can discretize the coordinates and predict the existence of various particles (or pseudo-particles, depending on what kind of system you're describing) and their dynamics.

Thanks, but...

my guess is nobody will really get what you mean until they actually try to learn QFT, huh. I'll revisit your comment in a few years.

"When Wolfram Alpha starts solving word problems..."

http://www.cs.utexas.edu/users/novak/cgi/physdemo.cgi

It's no Wolfram|Alpha, but it solves word problems fairly well.

Unfortunately both this and Alpha are trivial to confuse:

  What is the area of a circle with radius "2x"

yields 3.1415926535897931 * 2x^2 which is obviously incorrect. Alpha on the other hand doesn't give a result unless simplified to

  area of a circle with radius (2x)

where it gives a nonsensical display alongside the fact that 4pi is about 12.5664.

It may give you the wrong answer, but it makes it hard to be led astray:

  Evaluated area = pi * radius^2  giving AREA = 3.1415926535897931 * 2x^2

That description makes it obvious that a pair of parentheses are the only thing lacking.

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