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Physics, Chemistry and Math!
Physics: Problem of the day-18
Jun 15th
Topic: Electrostatics
An electric field line emerges from a positive point charge + q1 at an angle α to the straight line connecting it to a negative charge – q2.
Figure-1
At what angle β will the field line enter the charge – q2?
What we see in the diagram here is a cross-sectional diagram of a 3-dimensional situation. Field lines will emanate from +q1 in all directions in space. Although the presence of – q2 will prevent these field lines from being uniformly distributed, we can still assume that in the immediate proximity of +q1, the contribution to the field from –q2 will be relatively negligible, so that we can further assume that the field lines emerge from +q1 and enter –q2 in spatially homogeneous bundles. Thus, the “number of lines” emerging from +q1 and confined to the spatially conical region of vertical angle 2α will be the same as those entering –q2 and confined to the spatially conical region of vertical angle 2β.
Figure-2
The ratio number of lines emanating from +q1 in the depicted region, to the total number of lines emanating from +q1 is the ratio of the corresponding solid angles:
Therefore, the number of lines emanating from +q1 in the depicted spatial region is proportional to 1 – cos α. This immediately leads to:
Proofs by contradiction
Jun 11th
One of the most powerful of methods of finding mathematical proofs is “proof by contradiction”. Say you have to prove some statement S. You start by assuming that the exact opposite of S is true. Proceeding further, you should reach some logical contradiction, which tells you that assuming that S isn’t true was not correct – so S must be true. On the face of it, the method is pretty simple, but it leads to some pretty useful and non-trivial results, of which two examples are discussed here.
(1) The square root of 2 is irrational
We first suppose that √2 is rational, which means that there must exist integers p and q such that p/q = √2. We can also assume that p and q have no common factors, since if they did have a common factor, it can be canceled out.
Now, a series of simple steps leads us to a logical contradiction:
The contradiction is that we’ve shown that both p and q are even, while we had assumed that p and q have no common factor. Therefore, our assumption that √2 is rational was wrong; it must be irrational.
Is it possible to prove that √3 is irrational using the same method?
(2) There exist infinitely many primes
This was first proved by Euclid. He started by assuming that there are finitely many primes. Therefore, there must exist a largest prime; call it p. Now, consider the number N given by N = 1 × 2 × 3 × …. × (p + 1) = p! + 1
Note that if you divide N by any number from 1 to p, you’ll always be left with a remainder of 1. Therefore, N has no factors between 1 and p, which implies two possibilities:
• Either N has a prime factor larger than p, which shows the existence of a prime number larger than p – a contradiction.
• N has no prime factor other than N itself; which means N is itself prime (and obviously N is larger than p) – a contradiction again
Therefore our original assumption that there are finitely many primes is wrong – the number of primes is infinite.
(3) Seven real numbers are given in the interval (1,13). Prove that at least three of them are the lengths of triangle’s sides.
Let the numbers be a,b,c,d,e,f,g. We can assume that a≤ b ≤c
Suppose by way of contradiction that no three of them are the lengths of a triangle’s sides. We have a + c > b and b + c > a. Then a + b ≤c, because if not, then there would exist a triangle with side lengths a,b,c.
We deduce c≥2. Similarly, we have d ≥ c + b ≥3, e ≥ d + c ≥5, f ≥ e + d≥ 8 and, finally, g ≥ f + e ≥ 13, which is a contradiction.
Cauchy-Schwarz inequality
Jun 10th
In this article, we’ll discuss a very powerful inequality called the Cauchy-Schwarz inequality:
Here, we’ll see how we can justify this inequality, and in another article, we’ll discuss its applications.
Proof – 1
The form of the inequality suggests that if we can somehow construct a quadratic equation where some condition can be imposed on its discriminant, we may be able to generate the required inequality.
Consider the function f (x) given by f (x) = (a1x – b1)2 + (a2x – b2)2 + ….. + (anx – bn)2
Obviously,
Note that f (x) is a quadratic expression, and if we write it in the standard quadratic form, we have
By virtue of f(x) always being non-negative, the discriminant D of f (x) must be non-positive. From this condition, the Cauchy-Schwarz inequality is immediately evident. Also, it should be obvious that the equality will hold if D = 0, which implies that f (x) has a real root, call it x0. Therefore, f (x0) = 0=(a1x0 – b1)2 + (a2x0 – b2)2 + ….. + (anx0 – bn)2
This is only possible if each of the terms on the RHS is zero:
Proof – 2
This proof involves the use of a simple result.
Prove this! This result can be easily generalized:
When does equality hold in this?
Now, we prove the Cauchy-Schwarz inequality using this result:
Its up to you to determine when the equality will hold in this case.
Proof – 3
We’ll only outline this proof briefly. If your general mathematical aptitude is good probably the first thing that should come to your mind when you look at the CS inequality is: Vectors!
Take two vectors:
Now, the dot product of two vectors will always be less than the product of their moduli (lengths):
with equality holding if the two vectors are in the same direction! Thus,
Can this result be generalized to n terms?
Physics: Problem of the day-17
Jun 7th
Topic: Mechanics
Threads of lengths h1, h2 and h3 are fastened to the vertices of a homogeneous triangular plate of weight W. The other ends of the threads are fastened to a common point, as shown in the figure below.
Figure-1
What is the tension in each thread, expressed in terms of the lengths of the threads and the weight of the plate?
Clearly, the centre of mass S of the triangle has to be below the point of suspension. Denote the vectors pointing from the centre of mass S to the vertices of the triangle by r1, r2 and r3, and that to the suspension point by m (see figure).
Figure-2
The forces F1, F2 and F3 exerted on the plate by the threads can now be expressed in terms of the vectors defining the thread directions:
Fi = λi (m – ri), i = 1, 2, 3
Since the plate is in equilibrium, the vector sum of the forces acting on it is zero, i.e.
F1 + F2 + F3 + W = 0
Making use of the fact that the vector pointing to the center of mass of the triangle (the origin of the vector reference frame) is the arithmetic mean of the vectors pointing to its vertices, we have
r1 + r2 + r3 = 0
We note that W and m are parallel and, therefore, W = –km. Eliminating r3 from the above equations gives
(λ3 – λ1) r1 + (λ3 – λ2) r2 + (λ1 + λ2 + λ3 – k) m = 0
Since r1, r2 and m, are not in the same plane, a linear combination of them can only be zero if the coefficient of each is zero.
Thus, λ1 = λ2 = λ3 which implies that the tensions in the threads are proportional to their lengths. This deduction would become invalid if one of the threads were slack, since the plane of the plate would then become vertical and m would lie in it.
Probability: The Monty Hall Paradox
May 21st
<Problem taken from LOCUS Math study material>
The problem we are going to discuss is publicly one of the most well known problems in probability. This version of the problem and its solution have been taken from Wikipedia.
Suppose you are in a game show, and you’re given the choice of three doors. Behind one door is a new car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you’ve chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you choose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door 2?” Is it to your advantage to change your choice?
Figure-1
Before going through the solution, give the problem a great deal of thought. Will switching change the probability of wining? Or will it not matter because it is equally likely that the car may be behind Door 1 or Door 2 ?
O.K., now the solution. We are assuming that the player initially picked Door 1. (However, the player may initially choose any of the three doors). We’ll now calculate the probability of winning by switching, by listing out all cases explicitly.
Case 1: The car is behind Door 1. In this case, the game host Monty Hall must open one of the two remaining doors randomly
Case 2: The car is behind Door 2. Then the host must open Door 3
Case 3: The car is behind Door 3. Then the host must open Door 2
If the player chooses to switch, he’ll win only if the car is behind either of the two unchosen doors rather than the one that was originally picked.
The following diagram visually depicts the various cases and what happens with switching.
Figure-2
Let us explain how we arrived at the various probabilities using a probability tree:
Figure-3
Thus, switching leads to winning with probability 2/3 , that is, the player should switch for a higher probability of winning!
This result may seem very counter-intuitive to you. After all, you may say, the remaining two doors must each have a probability 1/2 of containing the car. However, this intuition is wrong.
In fact, you are not alone in this intuition . When first presented with this problem, an overwhelming majority of people assume that each door has an equal probability and conclude that switching does
not matter. Even Nobel laureates have been known to have given the wrong answer and to have insisted upon it! For more interesting facts about this problem, you’ll find a lot of resources online. Here’s a different version of the same problem call “The Prisoner’s Dilemma”
There are three condemned prisoners, a random one of whom has been secretly chosen to be pardoned. One of the prisoners begs the warden to tell him of one of the others who will be executed, arguing that this reveals no information about his own fate, but increases his chances of being pardoned from 1/3 to 1/2. The warden obliges, secretly flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned. Does the warden’s answer change the prisoner’s chances of being pardoned?