EQUATIONS THAT SHOOK THE WORLD, PART II: \(t' = t/\sqrt{1-v^2/c^2}\)
Another equation that shook the world, particularly our beliefs about time, is part of a system defined in Einstein's famous "Principle of Special Relativity." It reconciled a particular quirk about time that 19th century mathematicians and physicists wondered about. Namely, if time were really an absolute quantity, how come light does not seem to obey it?
Consider two cars A and B whose speeds are actually the same (as indicated by their respective speedometers). Suppose you are riding in car A, and you see car B approaching from the other end of the road. How fast does car B seem? Consider another situation. Suppose car B is running exactly by your side and toward the same direction. How fast does car B seem now?
In the first situation, it would appear that car B ran faster although the speedometer inside car B registered the same speed as car A. In the second situation, and with the same agreed speeds, car B does not seem to be moving from your point of view. Yet nothing has changed with the number indicated in the speedometers. The two situations seem different because of what you perceive is the relative difference in speeds. In the first, \(v_A = v\) and \(v_B = -v\) so that \(\left|v_B - v_A\right| = |-v - v| = 2v\) (car B apparently moved two times faster than it should be to the opposite direction). In the second situation, \(v_A = v_B = v\); hence, \(\left|v_B - v_A\right| = |v - v| = 0\) (car B does not seem to be moving from your point of view).
Now replace car B with a light beam. Suppose light speed is \(\left|v_B\right| = c\). You would be surprised that your measurements in both situations would actually be the same: \(\left|v_B - v_A\right| = c\). Wait! That could not be right. Let us check the math. In the first situation, light runs against you so we should perceive its speed as: \(\left|v_B - v_A\right| = |-c - v| = c+v\). But in the second situation, it should have been: \(\left|v_B - v_A\right| = |c - v| = c-v\). Does it mean, mathematically, that: \(c = c+v = c-v\)? We would come to the absurd conclusion: \(c = 0\) and \(v=0\), which are obviously inconsistent with the premise that light and car are moving.
The solution to the puzzle came from a group of physicists, most notable of which are George Fitzgerald and Hendrik Lorentz. They were trying to come up with an explanation of a baffling result from an experiment on light quite similar to the hypothetical situation above described.
The solution shook the beliefs of the entire world until that period, and involved a certain transformation of coordinates. Let \(t'\) be the time it took light to travel a distance \(x'\) for a non-moving observer. Its speed should be the ratio between distance and time traveled; hence, \(x'/t' = c\) The transformation addresses what you (in car A) are supposed to observe; namely, a generally different light speed, \(x/t = v\). Consider \(\tilde{x} = x-ct\) and \(\tilde{t} = t-\dfrac{vx}{c^2}\). Then, \[\begin{eqnarray*}
\frac{x'}{t'} = c &=& \frac{x-vt}{t-vx/c^2} \\
&=& \frac{c^2x-c^2vt}{c^2t-vx} \\
&=& \left(\frac{cx - cvt}{c^2t - vx}\right)c
\end{eqnarray*} \]In order for the equation to be true, then we must impose \(cx - cvt = c^2t - vx\). Consequently, \[ \begin{eqnarray*}
cx + vx &=& c^2t + cvt\\
(c+v)x &=& (c+v)ct\\
x &=& ct\\
\frac{x}{t} &=& c = v
\end{eqnarray*} \]which leads to the conclusion that the observer must be seeing the same speed regardless of its motion relative to the light beam. But the coordinate transformation (known today as the Lorentz transformation) implies that we must give up the concept of absolute time. Now we have to accept that time depends on the relative motion of the observer: \(t' = t/\sqrt{1 - v^2/c^2}\). If they are moving fast enough, then they would see everything around them slow down, i.e., the inequality \(t' > t\) becomes perceptible. Fortunately, in most typical speeds we experience in our travels, \(v\ll c\) so that \(1-v^2/c^2 \approx 1\) and \(t' \approx t\). Thus, the illusion that time is absolute under normal circumstances.
One particular hypothetical consequence of the Lorentz transformation is the following. When you only have \(5\) minutes left to go to class, try running very fast so that everyone around seems slower. Your \(5\) minutes will seem less than \(5\) minutes to your teacher. But isn't running fast what we usually do when we only have \(5\) minutes left to be somewhere on time?
ABOUT THE AUTHOR:
Dranreb Earl Juanico is an Assistant Professor at the Ateneo de Manila University. He obtained his B.S., M.S., and Ph.D. in Physics at the University of the Philippines in 2002, 2004 and 2007, respectively.
OLYMPIAD CORNER
from the Asian Pacific Mathematics Olympiad, 2013
Problem: For \(2k\) real numbers \(a_{1},a_{2},....,a_{k},b_{1},b_{2},...,b_{k},\) define the sequence of numbers \(X_{n}\) by \[ X_{n}=\sum_{i=1}^{k}\,\left\lfloor a_{i}n+b_{i}\right\rfloor, n=1,2,\ldots . \]If the sequence \(X_{n}\) forms an arithemetic progression, show that \(\sum_{i=1}^{k}a_{i}\) must be an integer.
Note that for all \(i\), \[ a_{i}\left( n+1\right) +b_{i}-1<\left\lfloor a_{i}\left( n+1\right) +b_{i}\right\rfloor \,\ \leq a_{i}\left( n+1\right) +b_{i}. \]Hence \[ \begin{eqnarray*} \sum_{i=1}^{k}\left(a_{i}\left(n+1\right)+b_{i}-1\right) & < & \sum_{i=1}^{k}\left\lfloor a_{i}\left(n+1\right) + b_{i}\right\rfloor \leq \sum_{i=1}^{k}\left(a_{i}\left (n+1\right) + b_{i}\right) \\ A\left(n+1\right)+B-k & < & X_{n+1} \leq A\left(n+1\right) + B \\ An + \left(A+B-X_{1}\right)-k & < & nd \leq An+\left(A+B-X_{1}\right) \\ \end{eqnarray*} \]However, note that \(A+B-k<X_{1}\leq A+B\). Therefore \(A+B-X_{1}\geq 0\), which implies that \[ An-k\leq An+\left( A+B-X_{1}\right) -k \]Furthermore, \(A+B-X_{1}<k\), and therefore \[ An+\left( A+B-X_{1}\right) <An+k. \]Combining these inequalities, we have \[ \begin{eqnarray*} An-k &<&nd<An+k \\ -k &<&-An+nd<k \\ \left| \left( -A+d\right) n\right| &<&k \\ \left| A-d\right| &<&\frac{k}{n}. \end{eqnarray*} \]But since \(\left| A-d\right| <\frac{k}{n}\) holds true for all positive integers \(n\), it follows that \(A=d\). Since \(\{X_{n}\}\) is a sequence of integers, then \(d\) is also an integer. Hence \(A=\sum_{i=1}^{k}a_{i}\) must be an integer.
Solution:
Let \(A=\sum_{i=1}^{k}a_{i}\) and \(B=\sum_{i=1}^{k}b_{i}\). Moreover, suppose that \(\{X_{n}\}\) is an arithmetic progression with the common difference \(d\); that is, \(X_{n+1}=X_{1}+nd\).Note that for all \(i\), \[ a_{i}\left( n+1\right) +b_{i}-1<\left\lfloor a_{i}\left( n+1\right) +b_{i}\right\rfloor \,\ \leq a_{i}\left( n+1\right) +b_{i}. \]Hence \[ \begin{eqnarray*} \sum_{i=1}^{k}\left(a_{i}\left(n+1\right)+b_{i}-1\right) & < & \sum_{i=1}^{k}\left\lfloor a_{i}\left(n+1\right) + b_{i}\right\rfloor \leq \sum_{i=1}^{k}\left(a_{i}\left (n+1\right) + b_{i}\right) \\ A\left(n+1\right)+B-k & < & X_{n+1} \leq A\left(n+1\right) + B \\ An + \left(A+B-X_{1}\right)-k & < & nd \leq An+\left(A+B-X_{1}\right) \\ \end{eqnarray*} \]However, note that \(A+B-k<X_{1}\leq A+B\). Therefore \(A+B-X_{1}\geq 0\), which implies that \[ An-k\leq An+\left( A+B-X_{1}\right) -k \]Furthermore, \(A+B-X_{1}<k\), and therefore \[ An+\left( A+B-X_{1}\right) <An+k. \]Combining these inequalities, we have \[ \begin{eqnarray*} An-k &<&nd<An+k \\ -k &<&-An+nd<k \\ \left| \left( -A+d\right) n\right| &<&k \\ \left| A-d\right| &<&\frac{k}{n}. \end{eqnarray*} \]But since \(\left| A-d\right| <\frac{k}{n}\) holds true for all positive integers \(n\), it follows that \(A=d\). Since \(\{X_{n}\}\) is a sequence of integers, then \(d\) is also an integer. Hence \(A=\sum_{i=1}^{k}a_{i}\) must be an integer.
PROBLEMS
- Suppose \(n \geq 3\). Show that an even number of the fractions\[ \frac{1}{n},\frac{2}{n},\frac{3}{n},\ldots,\frac{n-1}{n}, \]are in lowest terms.
- For every positive integer \( n > 1\), prove that there exists \(n\) factorials, each \(> 1\), whose product is also a factorial, i.e. \(x_{1}!x_{2}! \ldots x_{n}! = y!\).
- Find all prime numbers \(a\) and \(b\) such that \(a^{b}+b^{a}\) is also a prime and show that there are no other prime pairs \(\left(a,b\right)\) satisfying this property other than the pairs that you have.
- Suppose that \(A\) is a finite set of points in the plane with the property that evey line determined by two points of \(A\) contains a third point of \(A\). Prove that \(A\) is a set of collinear points.
- Find the real roots of the equation \[ x^{3}+2ax+\frac{1}{16}=-a+\sqrt{a^{2}+x-\frac{1}{16}}\quad\left(0 < a < \frac{1}{4}\right). \]
- Let \(ABCD\) be a square, and let \(k\) be the circle with center \(B\) passing through \(A\), and let \(l\) be the semicircle inside the square with diameter \(AB\). Let \(E\) be a point on \(l\) and let the extension of \(BE\) meet circle \(k\) at \(F\). Prove that \(\angle DAF\cong\angle EAF\).
We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions may be submitted to the PEM facilitators on the deadline date or online via vantonio1992@gmail.com. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:30 PM January 11, 2014.
SOLUTIONS
(for November 23, 2013)
- Two circles have exactly two points \(A\) and \(B\) in common. Find a straight line \(L\) through \(A\) such that the circles cut chords of equal lengths out of \(L\). How many solutions can the problem have? (Taken from Mathematics as Problem Solving by Alexander Soifer)(solved by Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jan Kendrick Ong [Chang Kai Shek College])SOLUTION:Trivially, \(AB\) is a solution.
We denote the first circle by \(C_{1}\) and the second by \(C_{2}\).
Suppose the line \(L\) does not pass through \(B\). Based from the conditions, \(\left|AC\right|=\left|AD\right|\), where \(C\) is the points where \(L\) intersects with \(C_{2}\).
Construct a symmetric image of \(C_{1}'\) of \(C_{1}\), as in the figure. Now, since \(\left|AC\right| = \left|AD\right|\), then \(C\) must be on \(C_{1}'\).
Now, note that there must exist \(X\), a point in \(C_{1}\), which is inside \(C_{2}\). This is true because they intersect at exactly two points. Combining this with \(A\) in \(C_{2}\), then we can say that the circle \(C_{1}'\) will have a point \(X'\) outside of \(C_{2}\). This point is actually the symmetric image of \(X\) with respect to \(A\). Therefore, \(C_{1}'\) and \(C_{2}\) have exactly one more point of intersection in addition to \(A\). Thus, \(L\) is another solution to the problem. So there are two solutions. - For nonnegative real numbers \(a\), \(b\), \(c\) with \(a+b+c=1\), prove that \[ \sqrt{a+\frac{\left(b-c\right)^{2}}{4}}+\sqrt{b}+\sqrt{c}\leq3. \](solved by Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jan Kendrick Ong [Chang Kai Shek College])SOLUTION:Let \[ x =\frac{\left(\sqrt{b}+\sqrt{c}\right)^{2}}{2},\quad y=\frac{\left(\sqrt{b}-\sqrt{c}\right)^{2}}{2}. \]From the given expression for \(x\), we have \[\begin{align}
2x & = \left(\sqrt{b}+\sqrt{c}\right)^{2}\\
\sqrt{2x} & = \sqrt{b}+\sqrt{c},
\end{align}\]since \(\sqrt{b}\) and \(\sqrt{c}\) are nonnegative. In addition, \[\begin{align}
x+y & = \frac{\left(\sqrt{b}+\sqrt{c}\right)^{2}}{2}+\frac{\left(\sqrt{b}-\sqrt{c}\right)^{2}}{2}\\
& = \frac{\left(b+2\sqrt{bc}+c\right)+\left(b-2\sqrt{bc}+c\right)}{2}\\
& = b+c.
\end{align}\]This means that \(a+b+c=a+x+y=1\).
Lastly, \[\begin{align}
xy & = \frac{\left(\sqrt{b}+\sqrt{c}\right)^{2}}{2}\cdot\frac{\left(\sqrt{b}-\sqrt{c}\right)^{2}}{2}\\
& = \frac{\left[\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)\right]^{2}}{4}\\
& = \frac{\left(b-c\right)^{2}}{4}.
\end{align}\]So the left-hand side of the inequality reads \[\begin{align}
\sqrt{a+\frac{\left(b-c\right)^{2}}{4}}+\sqrt{b}+\sqrt{c} & = \sqrt{a+xy}+\sqrt{2x}\\
& = \sqrt{a\left(1\right)+xy}+\sqrt{2x}\\
& = \sqrt{a\left(a+x+y\right)+xy}+\sqrt{2x}\\
& = \sqrt{\left(a+x\right)\left(a+y\right)}+\sqrt{2x}\\
& = \sqrt{3}\left[\sqrt{\frac{\left(a+x\right)}{3}\left(a+y\right)}+\sqrt{\frac{2}{3}x}\right].
\end{align}\]By AM-GM Inequality, \[ \sqrt{\frac{\left(a+x\right)}{3}\left(a+y\right)}\leq\frac{1}{2}\left(\frac{a+x}{3}+a+y\right) \]and\[ \sqrt{\frac{2}{3}x}\leq\frac{1}{2}\left(\frac{2}{3}+x\right). \]So \[\begin{align}
\sqrt{a+\frac{\left(b-c\right)^{2}}{4}}+\sqrt{b}+\sqrt{c} & \leq \sqrt{3}\left(\frac{1}{2}\right)\left(\frac{a+x}{3}+a+y+\frac{2}{3}+x\right)\\
& = \sqrt{3}\left(\frac{1}{2}\right)\left(\frac{a+x}{3}+1+\frac{2}{3}\right)\\
& = \sqrt{3}\left(\frac{1}{2}\right)\left(1+\frac{a+x+2}{3}\right)\\
& = \sqrt{3}\left(\frac{1}{2}\right)\left(1+\frac{a+x-1+3}{3}\right)\\
& = \sqrt{3}\left(\frac{1}{2}\right)\left(2-\frac{y}{3}\right)\\
& = \sqrt{3}\left(1-\frac{y}{6}\right)\\
& \leq \sqrt{3}.
\end{align}\] - If \(2n-1\) is a prime number, then for any group of distinct positive integers \(a_{1}\), \(a_{2}\), \(\dots, a_{n}\) there exist \(i,j\in\left\{ 1,2,\dots,n\right\}\) such that \[ {\displaystyle \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}}\geq2n-1. \](solved by Farrell Eldrian Wu [MGC New Life Academy]; partial credit for Jan Kendrick Ong [Chang Kai Shek College])SOLUTION:Let \(2n-1\) be a prime \(p\). Without loss of generality, we can assume \(\left(a_{1},a_{2,}\dots,a_{n}\right) = 1\). Otherwise, \(d=\left(a_{1},a_{2,}\dots,a_{n}\right)\) can be divided from all terms, leaving \({\displaystyle \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}}\) unchanged.
Let us first consider the case where \(p|a_{i}\) for some \(i\) from \(1\) to \(n\). If this is the case, then there must exist \(j\) such that \(p\nmid a_{j}\). This means that \(p\nmid\left(a_{i},a_{j}\right)\). So \[ \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}\geq\frac{a_{i}}{\left(a_{i},a_{j}\right)}\geq p=2n-1. \]
Now suppose \(\left(a_{i},p\right)=1\) for all \(i\) from \(1\) to \(n\). Then \(p\nmid\left(a_{i},a_{j}\right)\). Note that there are \(n\) numbers from \(a_{i},\dots,a_{n}\).
Suppose there are no two numbers in this set such that \(a_{i}+a_{j}\equiv0(\mod p)\).
By Pigeonhole Principle, there are two numbers from \(\left\{ a_{1},\dots,a_{n}\right\}\) such that their difference is divisible by \(p\).
This means that there are two numbers, suppose \(a_{i}\) and \(a_{j}\), such that their sum or difference is divisible by \(p\).
If \(a_{i}-a_{j} \equiv 0(\mod p)\), then \(a_{i}\equiv a_{j}(\mod p)\) and \[ \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}\geq\frac{a_{i}-a_{j}}{\left(a_{i},a_{j}\right)}\geq p = 2n-1. \]
If \(a_{i}+a_{j}\equiv0(\mod p)\), then \[ \frac{a_{i}+a_{j}}{\left(a_{i},a_{j}\right)}\geq p=2n-1. \]
ERRATA
Problem-solvers for Week 3 were incorrectly attributed.
