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Re: physics/pedagogy of coffee-mixing



Right (memory jogged). The conservation argument is the one he used -- furthermore, very counter intuitive, if I understand it, NO ERROR.

bc

John Denker wrote:

At 12:47 PM 6/25/00 -0400, Ludwik Kowalski wrote:

I never heard this question. So let me show how I would answer.
Model each cup as an urn with 1000 balls, black for the coffee and
red for the tea.

An excellent model. It cuts to the core of the problem, avoiding a number
of nit-picks and distractions.

Suppose one spoonful means 10 balls and that good
mixing occurs after the first transfer. Also that volumes of coffee
and tea balls are identical. Pure water balls are white and can be
ignored in this problem.

To be consistent with the model given above, there are no white
balls. Just black and red.

After the first transfer we have 990 black balls in one cup and a
mixture of 1010 balls in another. What happens after the second
transfer? It is unlikely that the second spoon will contain more
than one or two black balls. On that basis my answer is that
there is more coffee in the tea than tea in the coffee.

Oops, sorry, this is going down the wrong track. But I suppose it
illustrates my assertion that the wrong answers are very attractive.

Here's my analysis of the problem

1) The incisive qualitative analysis: There is a conservation law: In the
final state there are 2000 balls total (red+black) and 1000 balls in each
urn. Each red ball in the black urn displaces one black ball, which must
(by conservation) be in the other urn. Therefore there is _exactly_ the
same amount of coffee in the tea as tea in the coffee.

You don't even need to assume perfect mixing.

2) You can do numerical examples until you're convinced that the
concentrations always wind up the same. This alas loses some insights,
such as the (non)role of mixing.

3) The semi-quantitative constructive algebraic analysis starts by
recognizing that the second spoonful (tea-cup -> coffee-cup) transfers a
liquid that is not 100% tea; it contains _some_ coffee molecules.

The question is, how concentrated is it? Assuming we stirred the tea-cup
after the first-stage transfer of coffee, we have a concentration that is a
first-order quantity (first order in the small number which represents the
size of the spoon relative to the size of the cup).

And what is the effect of this second transfer? To first order, it just
puts some tea into the coffee-cup. To second order, it puts some coffee
back into the coffee-cup; the amount of coffee that makes the round-trip
is a second-order quantity. It is however strictly positive. The typical
conclusion is that there must be less tea in the coffee-cup, because the
round-trip coffee displaced some tea from the second spoonful. This is the
most-common mistake; it is diametrically opposite from the answer Ludwik
gave above.

But now we need to take into account another second-order term, namely the
amount of liquid in the cup when the transferred spoonful arrives. The
second step, adding the mostly-tea spoonful to the coffee cup, makes a
first-order change in the concentration of the coffee. The amount of the
transfer (one spoonful) appears in the numerator. But we need to be
careful about the denominator. The first step transferred coffee to a cup
containing 1000 units of tea, but the second step transferred mostly-tea to
a cup containing only 900 units of coffee. There is a first-order
correction in the denominator! A first-order correction to a first-order
quantity is a second-order quantity. In this case we have a second-order
term which magnifies the effect of the step-2 transfer, just cancelling the
second-order (displacement) correction described in the previous paragraph.

So, to second order everything cancels and there is the same amount of tea
in the coffee as coffee in the tea.

=========================

Now to address the psychological and pedagogical points. Here are some DOs
and DON'Ts.

a) If there is a conservation argument, GO WITH THE CONSERVATION ARGUMENT!

b) People have been trained to make first-order approximations. They are
in the habit of throwing away second-order terms without thinking. This is
just fine if the right answer is a first-order effect. But we need to
teach people to make approximations to a CONSISTENT ORDER OF
APPROXIMATION! If things cancel to first order, you can't just pounce on
the simplest second-order term you stumble across. You have to go back to
square one and keep *all* the second-order terms. Doing things to second
order in your head is hard; it's probably time to get out the pencil and
paper.

c) If numerical simulations suggest a simple pattern, see if there is a
rule that guarantees the pattern. See item (a).

d) Similarly, in general, if a long and complex derivation gives a simple
answer, look for a simpler derivation. As an application of this general
rule, if things cancel to first order and look like they might be going to
cancel to second order, it is probably worth checking to see if there is a
conservation law that guarantees cancellation to all orders. See item (a).

------

Anybody else have thoughts on why this riddle is more challenging than it
looks?