Planck Units - an algebraic exercise in unit conversion

I'm reading Leonard Susskind's "The Black Hole War". It's a really entertaining and fascinating accounting of his fruitful years long argument with Stephen Hawking over what happens to information when it falls into a black hole.

In Chapter 5 "Planck Invents a Better Yardstick" he describes how Planck came up with a much more canonical system of units than the metric system. In this system the 3 big constants all have value 1. These are: c the speed of light, ℏ Planck's reduced constant, and G Newton's gravitational constant.

It is an interesting exercise both algebraically and in unit conversion to figure out what the basic units of length, time and mass should be in order to make this happen. Start with the following definitions of the basic constants:

• c = 299 792 458 m / s
• G = 6.67300 × 10^-11 m³ kg^-1 s^-2
• ℏ = 1.05457148 × 10^-34 m²

Then your goal is to convert units so that these three constants all end up equaling 1. You'll get 3 non-linear equations in three conversion factors. A very capable 9th grader should be able to do it.

I won't show you how to do it, but I'll give you the answer so you can check your work:

Letting l, t, and m are respectively the Planck units of length, time and mass and 1 meter = L l, 1 second = T t, and 1 kg = M m, and  c, h, and G are respectively the speed of light, Planck's constant and Newton's gravitational constants as expressed in SI, you'll get the following results:

• L =  √ ( c³ / Gℏ )  = 6.19 × 10³⁴ Planck lengths per meter
• T =  √ (c⁵ / G )  = 1.86 × 10⁴³ Planck time intervals per second
• M =  √ ( G / ℏ c ) = 4.59 × 10⁷ Planck mass units per kilogram

NOD to the QROD - a mnemonic for converting improper fractions

I was trying to think of how to teach my kids about converting improper fractions to mixed form. The whole process is kind of confusing and easy to forget. I came up with the following mnemonic to help remember how to do it:

NOD to QROD

What does that mean? It sounds like a mantra that fans might chant at a baseball game as their favorite baseball star QROD approaches the plate. "NOD to QROD, NOD to QROD, NOD to QROD"... Then QROD goes up to the plate, here's the pitch... it's a long drive deep to center field, it's going, it's going, it's gone. QROD hits a home run! Everybody in the park explodes with excitement as QROD circles the bases.

So with NOD to QROD, you can hit a home run every time you want to convert an improper fraction. So again, what does that mean?

Numerator Over Denominator to Quotient Remainder Over Denominator

You always start with an improper fraction of the form Numerator Over Denominator. Your goal is to convert the fraction into mixed form. What that really means is that you should express the fraction as Quotient Remainder Over Denominator. For example suppose you wanted to convert the improper fraction 37/5 into mixed form. Reading the expression out loud we have "37 Over 5". The Numerator 37 is Over the Denominator 5. As I'll explain in the next paragraph, you'll need to do some long division and get a Quotient 7 and Remainder 2. Using the mnemonic, we need to write the answer as QROD. So Quotient 7 Remainder 2 Over Denominator 5 or 7 2/5. Summarizing:

N O D to Q R O D

37/ 5 = 7 2/5

So that tells you how to write the final answer, but it doesn't tell you how to get Q the quotient, or R the remainder. For that you'll need to know how to do long division, something that you will need to learn to do first. But once you know how to do long division, the Quotient is what you get on top, and the Remainder is what you get at the very end, when you've finished dividing the numerator as much as possible into the denominator. For example, the 37/5 example, you would do something like

      7    ---------Quotient

   _______

 5 | 3 7  

     3 5    

     ---

       2    --------Remainder 

Or in general:

                    Q   Quotient

                _______

 Denominator D |   N    Numerator    

                   .

                     .

                       .  

                         R  Remainder

so visually the NOD to QROD means:

                ___Q___

N/D     =    D |   N         =   Q  R/D     

                    ... R

The Importance of Getting Your Units Right According to Spinal Tap

One of my favorite movie skits of all time. It's very applicable to physics!

Warning: It's borderline R-rated due to foul language.

Spinal Tap - Stonehenge

A Stable Physics Demo

I was wondering about the following idea for a physics demo.

You drop a metal ball from the ceiling along a vertical ramp which guides the ball and launches it at the bottom at some angle. Then you set up a small basket to catch the metal ball a certain distance away. When you do the math you get the following formula for the distance d that the basket should be away from the launch point:

d = 2 h sin 2θ

where d is the distance between the launch point and the basket, h is the original height of the ball, and θ is the launch angle from the horizontal.

In particular, if you set up the launch angle as 45 degrees, you get the nifty formula:

d = 2 h

meaning that the horizontal distance is exactly twice the height!

The last formula makes it extremely easy to set up (or to try to verify as a lab experiment). The question I had was whether this will actually work? In particular, when you set up the demo, you're bound to make small mistakes in the set up. You'll lose the "WOW" factor if your metal ball doesn't fall into the basket on your first attempt. The most brittle part of the set up is the launch angle. So it's worthwhile to compute the error introduced into d from an error in θ. I.e., how far will the ball fly off the mark if you set up the launch angle as 44 degrees instead of 45.

It turns out, that since sin 2θ has a maximum at 45 degrees, the error is actually quite small. If you do the calculus of variations you'll get

Δd / d ≈ -4 (Δθ)²

so that the proportional error in distance varies quadratically with the angular error. For an error of 1 degree and a distance of 5 m, this is less than 1 cm error -so we can expect a bullseye! For an error of 5 degrees and a distance of 5m, this is around 15 cm, so that there is a high probability of getting into a small trash can.

Even though the most brittle part of the set up is the launch angle, there are some other issues I could see interfering with perfect results:

• The angle of the launcher on the horizontal plane. You could get the right total distance but still be off the mark because you end up on left or right of target.
• Measuring h and d accurately.
• Friction on the way down or during launch that will reduce the ball's kinetic energy and hence reduce d.
• To a lesser extent, air resistance.

I'll try this experiment and report back!

Josephus Problem in Binary

One of my favorite discrete math problems is Josephus's problem. I love to introduce the problem to the class telling the story of how the clever Josephus Flavius was able to survive the mass suicide, and then act out the story with the entire class using some implement of violence - such as a whiffle ball bat. Of course, I rig the game so that I win. The class is typically is quite amazed and claps enthusiastically when they realize that I'm the last man standing.

So how do I do pull it off? I don't simulate the entire game in my head! Instead, I use a handy little formula for the solution of the Josephus problem:

J(n) = left_rot( n )

where left_rot is the operation on n obtained by expressing n as a binary number, then rotating all the digits one bit to the left (so in particular bringing the left-most bit all the way to the right)

That means, that if there are 52 students present add me to make 53, then consider the binary expression [53]₂ = 110101, then rotate one bit to the left obtaining 101011 which represents 43. I just need to position myself so that I'm in the 43rd position and I'll be the last man standing!

Proof of formula.

For the proof we start with the partial recursive formula for J(n) which you can obtain by simply considering what happens to an even or odd number of people after the sword goes around the circle once:

J(n) =

• 2*J( (n-1)/2 ) + 1, in the case when n is odd
• 2*J( n/2 ) - 1, in the case when n is even

Use induction on the length L of [n]₂.

The base cases are length L = 1 and 2 in which case one can verify directly the three cases:

• J(1) = 1 is the left_rot of itself
• J(2) = 1 which is the left_rot of 10 = [2]₂
• J(3) = 3 which is the left_rot of 11 = [3]₂

Suppose we know the theorem to be true up to length L. Consider J(n) where n has L+1 bits base 2. Express [n]₂ a the string 1ml where l is the least significant bit, and m is the string consisting of the middle L-1 bits.

Case 1 of the inductive step [n is odd]. So l=1. By the recursive formula for J(n) we have

J(n) = 2*J((n-1)/2)+1

= 2*J( ( number(1m1) - 1) / 2 )+1

= 2*J( number(1m0) / 2 ) + 1

= 2*J( number(1m) ) + 1

= 2*number(m1) + 1

Where in the last step we applied the inductive hypothesis. Continuing on we have

2*number(m1) + 1 = number(m10) + 1

= number(m11)

= rot_left( number(1m1 )

= rot_left( n )

and finishes up the first case.

Case 2 of the inductive step [n is even]. So l=0. By the recursive formula for J(n) we have

J(n) = 2*J(n/2)-1

= 2*J( number(1m0) / 2 ) - 1

= 2*J( number(1m) ) - 1

= 2*number(m1) - 1

Where in the last step we applied the inductive hypothesis. Continuing on we have

2*number(m1) - 1 = number(m10) - 1

= number(m01)

= rot_left( number(1m0 )

= rot_left( n )

and finishes up the second case and the proof in its entirety.

When an altitude, bisector, and median make equal angles

Now that the year is over, I'd like to describe a solution to a challenging problem discussed in class.

Given: a triangle ABC, the altitude AD, the angle bisector AE, and the median AF divide the angle ∠BAC into four equal parts.

Prove: Prove that ∠BAC is a right angle.

Sketch of proof: There are several ways to solve this problem, but regardless of the method, it is a good idea to first start by drawing a figure involving all the given points and line segments:

We might be interested in investigating whether the sides satisfy Pythogoras's theorem, so it is a good idea to label lengths of line segments. So let the length AB = c, and the length AC = b. Next, one should label lengths of line segments as defined by the given. We start by setting the length BD = x. Since we are given that ∠BAD = ∠DAE, and that AD⊥BE, we can use ASA to ascertain that the triangles ADB and ADE are congruent. Hence the length DE = x. Furthermore, setting EF = y and using the fact that AE is the median, we get the length FC = 2x + y. Thus we arrive at the following figure:

Now consider the triangle AEC. AF is the bisector at A. Furthermore, recall the following.

Bisector proportionality theorem: An angle bisector in a triangle divides the opposite side in the ratio of the two sides.

Apply this theorem to our figure:

Thus conclude that ( y + 2x ) / b = y / c .

Consider the triangle ABC with bisector AE and apply the same theorem.

Conclude therefore that (2y + 2x) / b = 2x / c . We can get rid of the factor 2 on both sides and reduce to (y + x) / b =x / c .

Now add these two equations:

[ ( y + 2x ) / b = y / c ] + [ (y + x) / b = x / c ]

⇒

( y + 2x ) / b + (y + x) / b = y / c + x / c

⇒

( 2y + 3x ) / b = ( y + x ) / c

This equation will be useful below.

Next construct the point G which is the intersection of AC and the altitude of the triangle AEF from the vertex E. By an argument similar to the above using ASA surmise that AG is congruent to AE. Thus when considering the figure obtained by constructing the line EG and FG we get:

In particular, notice that in terms of lengths of line segments, the equation resolves to:

DC / AC = DF / AG.

That is, the lines AD and FG cut the transversals DC and AC proportionally. Thus using the converse of the transversal proportionality theorem, we can surmise that the lines AD and FG are parallel.

We are ready for the final part of the proof. Let the measured angle ∠BAD =∠DAE=∠EAF=∠FAC = θ. We are trying to prove that 4θ = 90°. Since ∠DFA is the complementary angle to ∠FAD in the right triangle ADF we have ∠DFA = 90 - 2θ. Furthermore, since the angles at H are right angles and EH = HG we can use SAS to surmise that the triangles EFH and GFH are congruent and therefore ∠EFA =∠GFA. Finally, since AD and FG are parallel ∠EFG is a right angle. Putting this altogether we have

∠EFA +∠GFA = ( 90 - 2θ )+ ( 90 - 2θ ) = 90

which becomes

180 - 4θ = 90

therefore

90 = 4θ

QED

Mme. de Saint Juery Method for Long Division

Fellow Buckley faculty member Jacqueline de Saint Juery has been known to baffle our best math whizzes with a very efficient method for doing long division. Here is an example using her method (or actually a slightly modified version that flips the positions of dividend and quotient). Can you figure out how the method works?

For example here is what 1078 / 12 would look like using the de Saint Juery method:

          8 9.8 3 3 ...
   ___________
1 2|  1 0 7 8
        1 1 8
          1 0 0
              4 0
                4 0
                  .. etc ..

Pool Hall Pondering: Why does a cue ball make a 90 degree angle with the ball it hits?

I'm teaching my 9th graders how to analyze billiard ball collisions. Specifically, we are looking at the case when the cue ball hits one other lone ball (say the eight ball). I teach the following principles:

1. momentum should be conserved
2. the cue ball should go off at 90 degrees relative the motion of the ball it hits. Call this the 90 degree rule.

As for #1 -this is well established. But where did #2 -the 90 degree rule- come from? I remember this fact from my college days when I played a lot of pool and it was crucial in assessing the likelihood of a scratch. It was important to make sure the cue ball didn't land in a pocket, so I needed to know the direction it headed after impact. Thus the 90 degree rule was very handy.

I thought about why this rule should be true and came up with the following vector theoretic proof. It's a little beyond your typical 9th grade physics class, but not terribly advanced:

GIVEN: Vectors p1,i , p2,i , p1,f , p2,f respectively denoting the cue ball's and eight ball's momenta before (i) and after (f) the impact. By assumption, the eight ball is initially at rest so that p2,i = 0. We assume conservation of momentum, as well as an elastic collision (i.e. conservation of kinetic energy). We also assume that all the balls have the same mass, denoted by m.

PROVE: The cue ball and eight ball shoot of perpendicularly to each other after impact. In other words, p1,f , p2,f are orthogonal.

PROOF: Since p2,i = 0, the initial total momentum is just p1,i . Conservation of momentum implies that the initial total momentum equals the final total momentum. Therefore

p1,f + p2,f = p1,i .

Furthermore, with a little vector algebra, one can express a particle's kinetic energy in terms of the magnitude of its momentum and its mass as follows:

E = || p || ² / ( 2 m )

Elasticity -i.e. conservation of energy- and the fact that p2,i = 0 therefore imply that

|| p1,i || ² / ( 2 m ) = || p1,f || ² / ( 2 m ) + || p2,f || ² / ( 2 m )

We can get rid of the common factor and re-write as:

|| p1,f || ² + || p2,f || ² = || p1,i || ²

Thinking about it a little we see that we have a case of two vectors whose norms add up to the norm of their sum. In other words, let a = p1,f and b = p2,f. We have a + b = p1,i and ||a|| ² + ||b|| ² = || p1,i || ². This can simply be restated as:

||a|| ² + ||b|| ² = || a + b || ².

Draw this as a vector addition diagram. If a makes one side and b makes the second side starting from the head of a, then the third side -the vector sum a + b - has length that satisfies the Pythagorean theorem! But the Pythagorean theorem has a converse, which I'll show next, and which will guarantee that a is orthogonal to b thus completing the proof.

The final part is:

VECTOR VERSION OF PYTHAGORAS'S THEOREM:

||a||² + ||b||² = || a + b ||² if and only if a is orthogonal to b.

Proof: By definition the norm of the sum of the two vectors is:

|| a + b ||² = ( a + b ) º (a + b ).

Simplify the right hand side by FOIL'ing to a º a + 2 a º b + b º b . Using the definition of norm in terms of the dot product again we get:

|| a + b ||² = || a ||² + 2 a º b + || b ||².

Setting this equal to ||a||² + ||b||² we get

||a||² + ||b||² = || a ||² + 2 a º b + || b ||².

Subtracting ||a||² + ||b||² we get

0 = 2 a º b .

In other words, the original equation holds exactly when a º b = 0, so exactly when a is orthogonal to b.

QED

Remembering Matt Adler

I will always be grateful to Matt for welcoming me as his friend when no one else did. My formative years up to the age of nine were spent in Israel. Then my mother Sareva decided to return the U.S. and be closer to most of her family. When I started going to Solomon Schechter Day School, my English was halting and my social skills were lacking. While others teased me for my body shape or my unusual background, Matt was accepting. We became friends and he was my only friend at school during fourth and fifth grades. We remained close friends throughout elementary school and later at Akiba Hebrew Academy. Matt’s gentle spirit and acceptance of differences is a credo I try to live my life by and teach my children to strive for. I was glad to learn that he maintained his gentle approach even while achieving tremendous success in the harsh world of international corporate law. I am sure he touched others as he did me.

I mourn his loss and give my deepest condolences to Jenn, Jake and Zoe and to Phyllis, Gary, and Marc.

Google Voice - Leave me a message

If you want to leave me a message just click on the following widget and give your phone number. Google Voice will call you and connect you to my voice mail. It's kind of weird at first --getting a call that makes you leave a voicemail for someone else. But it works, and you can contact me this way even if you don't know my cell!

Factoring by Grouping

I always like to understand what I'm teaching. Friday, David Elgart showed me an elegant approach to factoring rational quadratics called "Factoring by Grouping." Below I explain the technique and why it always works.

First here's a summary of the method:

Suppose you wanted to factor the trinomial ax² + bx + c into a product of two rational binomials or prove that it cannot be done.

1. Pull out the common term, possibly negative -throw it back after the last step. So WLOG (a, b, c) are relatively prime and a is positive
2. Find the factor pairs of |ac|
3. Determine what the sign of each term in the factor pair should be
4. Decide which pair to use by requiring m+n = b (in addition to mn = ab). Or if this cannot be done, stop and declare that the trinomial is irreducible.
5. (TRICKY for 8th graders) Use #3 to break up bx into mx + nx
6. (TRICKIER) Group the original trinomial using #4 as (ax² + mx) + (nx + c)
7. (SUPER TRICKY) Extract common factors and reverse the distributive law to get the final answer (aka the banana trick)

Let's illustrate the method by example. E.g., consider the trinomial 4 x² - 12 x - 27:

1. You should always pull out any common factor from a, b, and c and make sure that a is positive

   In the case of 4 x² - 12 x - 27 there was nothing to do as (4, -12, -27) are relatively prime and 4 is positive.

   
   
2. Find the factor pairs of |ac|:

   Factors of 4 * 27 = 108:
   

   • 1*108
     
   • 2*54
     
   • 3*36
   • 4*27
   • 6*18
   • 9*12

3. Determine what the sign of each term in the factor pair should be

   In this case since c is negative, one of m, n must be negative and the other positive. Since b is negative, the larger of m, n must be negative. I.e. we have the following possibilities:
   

   • 1 * -108
     
   • 2 * -54
     
   • 3 * -36
   • 4 * -27
   • 6 * -18
   • 9 * -12

   
   

4. Find numbers m, n satisfying
   
   • mn = ac
   • m+n = b
   In particular |m|, |n| will be one of the factor pairs found in part 2. Assign signs for mn based on the signs of b and c that make this possible (leaving this part vague)

   In our case b and c are both negative so that we need the smaller factor m to be positive and the bigger factor n to be negative. So our list of possibilities from #2 is:
   

   • 1,-108
     
   • 2,-54
     
   • 3,-36
   • 4,-27
   • 6,-18
   • 9,-12

   Look at the list to see which of this satisfy m + n = -12. The answer is 6, -18
   

5. If no such m,n were found in #3, the quadratic is irreducible over the integers. Otherwise we will be able to factor by grouping as follows ... - 12 x = 6x - 18 x
6. Group the original trinomial using #4 as (ax² + mx) + (nx + c) ... 4 x² - 12 x - 27 = ( 4 x² + 6x ) + ( -10 x - 27 )
7. Extract common factors and reverse the distributive law to get the final answer (aka the banana trick ) ... ( 4 x² + 6x ) + ( -18 x - 27 ) = 2 x ( 2 x + 3 ) + - 9 ( 2 x + 3 ) ... here ( 2 x + 3 ) is the "banana" (the banana trick simply says that a*banana + b*banana = (a+b)*banana .... ( 2 x - 9 ) ( 2 x + 3 ) this is the solution.

And here's the algebraic reason that this always works:

• ax² + bx + c
• ac = mn
• b = m+n

Implies

ax² + bx + c

=

ax² + mx + nx + mn/a

=

x(ax + m) + n/a(ax + m)

=

(x + n/a )(ax + m)

=

(1/a)(ax + n)(ax + m)