## Archive for July, 2012

### Transitive Evidence

July 31, 2012

A snippet from a conversation, paraphrased:

A: I’m worried about my posture. People will think I’m not attractive because I slouch.

B: Don’t worry, you can improve your posture because you’re intelligent.

A: What? How does that follow?

B: I notice that rich people tend to be able to improve their posture. Meanwhile, it is usually easy for intelligent people to become rich. Therefore, intelligent people can usually improve their posture.

Regardless of the somewhat-questionable factualness of these assertions, is the statement logically sound? If A is evidence for B and B is evidence for C, is A evidence for C? Mathematically, it is quite easy to see this is not the case. Check out this probability distribution, for example:   A few moments of staring will show you that it’s a counterexample (A is evidence for B, B is evidence for C, but A is not evidence for C). Good thing, too! Imagine if it were true:

• Being a Native Hawaiian is evidence for being in Hawaii. Being in Hawaii is evidence for being a tourist. Therefore, being a Native Hawaiian is evidence for being a tourist.
• If an object is an insect, that’s evidence that it can fly. If an object can fly, that’s evidence that it’s an airplane. Therefore, being an insect is evidence that an object is an airplane.
• Having sex is evidence that you are breathing hard. Breathing hard is evidence that you’re jogging. Therefore, having sex is evidence that you’re jogging.
• If it’s raining, that’s evidence that there are umbrellas around. If there are umbrellas around, that’s evidence that you’re in an umbrella factory. Therefore, rain is evidence that you’re in an umbrella factory.

July 30, 2012

Let’s do a quick bit of math related to Dropping a Slinky. Last time, I estimated that it takes about 0.3 seconds for the slinky to collapse. To get a more precise answer, note that however the slinky falls, its center of mass must accelerate downwards at gravitational acceleration.

Where is the slinky’s center of mass? When it’s just hanging, the slinky is in equilibrium, so the derivative of the tension is proportional to the density. Also, if we assume an ideal spring with zero rest length, the tension is inversely proportional to the density (why?). Therefore, we write

$\frac{\mathrm{d}T}{\mathrm{d}x} = g \rho$

$T = \frac{\alpha}{\rho}$

This can be solved to show that the density follows

$\rho \propto \frac{1}{\sqrt{x}}$

Integrating, we find that the center of mass is one third the way up the slinky. The time for the slinky to collapse is the same as the time for the center of mass to fall to the bottom, or

$t = \sqrt{\frac{2 (1/3 l)}{g}}$

This is the same answer, but modified by a factor of 0.81. Notice that this only depends on the “slinkiness” – the zero rest length ideal spring. We expect thick and thin slinkies of different stiffnesses to act in essentially the same way.

July 30, 2012

A short video of what it looks like to drop a slinky. It’s surprising and elegant.

So what’s the speed that waves propagate in a slinky? A slinky is a bit tricky, because as you pull on it, it stretches out so that the density goes down. Meanwhile, the tension goes up. Both these effects increase the speed of wave propagation, so waves travel much more quickly at the top of a hanging slinky than at the bottom.

Since the only material properties around are the linear density $\lambda$ and the tension $T$, we must put these together to get a velocity, which we do by

$v = \sqrt{\frac{T}{\lambda}}$

As the slinky hangs, it should be in equilibrium, so the gradient of the tension at any point is equal to gravity times the density there. This yields the result that the tension is a square root of how far up you go from the bottom of the spring. As a rough estimate, though, the tension should on average be about half the spring’s weight, while the density on average is the spring’s weight divided by its length. Thus

$v = \sqrt{\frac{weight}{mass/length}} = \sqrt{g l}$

where $g$ is gravitational acceleration and $l$ is the slinky’s length. The characteristic time of such a slinky is

$t = \frac{l}{v} = \sqrt{\frac{l}{g}}$

For a one meter slinky we get a time of .3 seconds (and a speed of only a few meters per second), meaning it’s an effect we can see quite well even without high-speed photography!

The same basic mechanism is there in dropping anything else, but typical sound speeds are on the order of thousands of meters per second, so usually it’s much too fast.