The Pythagorean Theorem and Special Relativity (Tuklas Vol. 13, No. 6 - January 21, 2012)

THE PYTHAGOREAN THEOREM AND SPECIAL RELATIVITY

One of the greatest hallmarks of 20th century physics was the development of a branch of Modern Physics called "Special Relativity." Briefly, it is a theory of space, time and light-– all of them bridging the gap and asking the most fundamental questions in physics. The Theory of Special Relativity was proposed in 1905 by Albert Einstein, then an unknown clerk in a Swiss patent office. In that moment he made three important breakthroughs in Physics: a consolidation of various ideas individually contributed by scientific giants including Lorentz, Poincare, and Maxwell, among others. Of course with great unexpected success also comes some controversies and issues that didn't sit well with fellow scientists and philosophers. However, various experiments have been conducted to this day and have affirmed the three very important contributions of Albert Einstein. The year 1905 was then known as Annus Mirabilis, or "Year of Wonders."
The propositions of this theory are far-reaching and almost impossible to understand. Many of the theory's consequences are counter-intuitive, but the basic foundations and proofs of some propositions are surprisingly elementary. This article aims to connect the astoundingly simple mathematics that gave rise to a superstar. 
We consider two rectangular and parallel plates in space that emit light pulses. The light pulse emitted from one plate is caught and immediately reflected back to the original plate. The time it takes for the light to travel from plate 1 to 2 and back to 1 is \(\Delta t_0\). Since the light pulse makes a round trip, it travels a distance \(2d\). If the speed of light is \(c\), then \(\Delta t_0 = \frac{2d}{c}\). Figure 1 shows the current setup of our thought experiment. 

What then if this setup is placed in a rocket that flies horizontally with a sufficiently fast velocity \(v\)? In order to understand this better, we will welcome the help of two laboratory assistants Albert and Hendrik. Albert stays on the ground and watches the rocket fly past him, while Hendrik wants a joyride and rides the rocket. The experiment of the light pulses is activated while both gentlemen watch, and the rocket is flying horizontally. (And oh yes, Albert must see Hendrik for the entire duration of the trip for the purposes of our thought experiment!) While Hendrik is perhaps enjoying inflight service or eating junk food, he will observe that both he and the plates aren't moving any farther from each other. Therefore what he will see is exactly the same setup as Figure 1. 
Albert's observations, on the other hand, will be totally different from Hendrik's. Since Albert is sitting down on Earth and perhaps doing Facebook or combing his funky hair, he will see that the light pulses will travel in a way illustrated in Figure 2. 

\(v\Delta t\) here is the distance travelled by the rocket. \(\Delta t_0\) and \(\Delta t\) are two different things!
Hendrik Lorentz first suggested this proposition, and by using an incredibly simple Pythagorean Theorem, he was able to arrive at the following relationships: \[ L= \sqrt{d^2+\left (\frac{v\Delta t}{2} \right )^2}.\]
From the equation used in Figure 1, the following equation is derived: \[ \Delta t = \frac{\Delta t_0}{\sqrt{1-v^2/c^2}}. \]
The consequences of this equation are difficult to stomach. It tells us that if \(v = c\), then the equation becomes undefined! This brings us a natural speed limit equal to the speed of light, \(c\). For more human speeds, \(v \ll c\), the effects of this equation is not immediately felt. 
Let us say that I have a rocket travelling at 99% the speed of light, \(v= 0.99c\). Inside the rocket is an observer telling me it takes 1 second for light to travel roundtrip between plates. For an observer on the ground, how long will it take for the light to make a roundtrip? Using the same equation, it will take 2.29 seconds! This equation is called the Time Dilation Equation and has been proven to be accurate in many experiments around the world. 
Still from Pythagorean Theorem and the Time Dilation Equation, a natural consequence would be the following phenomenon: \[ L= L_0 \sqrt{1-\frac{v^2}{c^2}}. \]
For those who are craving for an optical illusion, this is known as the Length Contraction Equation! 
Maybe the only time we will be able to see these phenomena is when our cars or airplanes are flying at at least around 70% the speed of light -- that is around 210,000,000 m/s! Bon Voyage!

ABOUT THE AUTHOR:
Clark Kendrick Go is an Instructor at the Ateneo de Manila University. He is currently taking his M.S. in Physics at the same university.

OLYMPIAD CORNER

from the Asian Pacific Mathematics Olympiad, 2005

Problem: Prove that there exists a triangle which can be cut into 2005 congruent triangles.

Solution: 

Suppose that one side of a triangle has length \(n\). Then it can be cut into \(n^{2}\) congruent triangles which are similar to the original triangle, the ratio of the corresponding sides given by \(n:1\). 
Since \(2005\) is a product of two primes (\(5\) and \(401\)) both of which are of the form \(4k+1\), then it can be represented as a sum of two integer squares \[ \begin{align*} 2005 &= 5\times 401 \\ &= \left( 2^{2}+1\right) \left( 20^{2}+1\right)  \\ &= 40^{2}+20^{2}+2^{2}+1 \\ &= \left( 40-1\right) ^{2}+2\times 40+20^{2}+2^{2} \\ &= 39^{2}+22^{2}. \end{align*} \] Let \(ABC\) be a right-angled triangle with legs \(AB\) and \(BC\) having lengths \(39\) and \(22\), respectively. We draw the altitude \(BK\), which divides \(ABC\) into two similar triangles. Now we divide \(ABK\) into \(39^{2}\) congruent triangles as described above and \(BCK\) into \(22^{2}\) congruent triangles.
Since \(ABK\) is similar to \(BKC\), then all \(39^{2}+22^{2}= 2005\) triangles will be congruent.

PROBLEMS

16. Let \(T\) be a triangle of perimeter 2, and \(a\), \(b\), \(c\) be the lengths of its three sides. Prove that \[ abc + \frac{28}{27} \geq ab + bc + ca \text{ and } abc + 1 \leq ab +bc +ca. \]
    
17. If \(a\), \(b\) are two different positive integers, and the two quadratic equations \[ (a-1)x^2 - (a^2+2)x +(a^2+2a) = 0, (b-1)x^2 - (b^2+2)x +(b^2+2b) = 0 \] have one common root, find the value of \[ \frac{a^b + b^a}{a^{-b} + b^{-a}}. \]
18. In the convex quadrilateral \(ABCD\), the midpoints of \(BC\) and \(AD\) are \(E\) and \(F\) respectively. Prove that \([EDA] + [FBC] = [ABCD]\), where [] denotes the area of the region enclosed by the specified points.

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is January 28, 2012.