Saturday, May 31, 2014

Linear Algebra: #1 Introduction


  • Linear Algebra: #1 Introduction


The mathematical idea of a vector plays an important role in many areas of physics.

  • Thinking about a particle traveling through space, we imagine that its speed and direction of travel can be represented by a vector v in 3-dimensional Euclidean space ℜ3. Its path in time t might be given by a continuously varying line — perhaps with self-intersections — at each point of which we have the velocity vector v(t).

  • A static structure such as a bridge has loads which must be calculated at various points. These are also vectors, giving the direction and magnitude of the force at those isolated points.

  • In the theory of electromagnetism, Maxwell’s equations deal with vector fields in 3-dimensional space which can change with time. Thus at each point of space and time, two vectors are specified, giving the electrical and the magnetic fields at that point. 

  • Given two different frames of reference in the theory of relativity, the transformation of the distances and times from one to the other is given by a linear mapping of vector spaces. 

  • In quantum mechanics, a given experiment is characterized by an abstract space of complex functions. Each function is thought of as being itself a kind of vector. So we have a vector space of functions, and the methods of linear algebra are used to analyze the experiment. 


Looking at these five examples where linear algebra comes up in physics, we see that for the first three, involving “classical physics”, we have vectors placed at different points in space and time. On the other hand, the fifth example is a vector space where the vectors are not to be thought of as being simple arrows in the normal, classical space of everyday life. In any case, it is clear that the theory of linear algebra is very basic to any study of physics.

 But rather than thinking in terms of vectors as representing physical processes, it is best to begin these lectures by looking at things in a more mathematical, abstract way. Once we have gotten a feeling for the techniques involved, then we can apply them to the simple picture of vectors as being arrows located at different points of the classical 3-dimensional space.

Friday, May 30, 2014

Complex Analysis: #31 Functions of Finite Order with Zeros

  • Complex Analysis: #31 Functions of Finite Order with Zeros

Theorem 53
Let f be an entire function of strict order α. Then there exists a C > 0 such that, for all R > 0 and DR = {z ∈ ℂ : |z| ≤ r}, we have

vf(R) ≤ CRα

where vf(R) is the number of zeros of f in DR.

Proof
Assume first that f(0) ≠ 0. According to Jensen’s inequality (leaving out the part where we integrate from 0, and remembering that |f(z)| ≤ Ke|z|α , for some K > 0), we have, for R > 1

Complex Analysis: #31 Functions of Finite Order with Zeros equation pic 1

Finally, if we do have f(0) = 0, then take g(z) = f(z)/zm, where m is the order of the zero at 0. Then the theorem will apply to g, and since m is then fixed, to f as well.


Theorem 54
Again, f is an entire function of strict order α with zeros {zn}, listed (with multiplicity) in order of increasing absolute value. We assume that f(0) ≠ 0. Then for every δ > 0 we have the series

∑ |zn|−α−δ

converging.

Proof
Using partial summation and the previous theorem, where we assume that R ∈ ℕ, we have

Complex Analysis: #31 Functions of Finite Order with Zeros equation pic 2

and this last sum is convergent.

This, combined with the discussion concerning the genus of a canonical product of discrete elements of ℂ (see section 24), leads to:

Corollary (Hadamard)
Let f be an entire function of finite order α with zeros {zn}n∈ℕ. Then

Complex Analysis: #31 Functions of Finite Order with Zeros equation pic 3







IMPORTANT NOTE:
This series on Complex Analysis has been taken from the lecture notes prepared by Geoffrey Hemion.  The document can be found at his homepage.

Thursday, May 29, 2014

Complex Analysis: #30 Jensen`s Formula

  • Complex Analysis: #30 Jensen`s Formula

But before doing that, we look at the easier Jensen’s inequality.


Theorem 51 (Jensen’s Inequality)
Let R > 0 be given and let the (non-constant) analytic function f be defined in a region containing the closed disc DR = {z ∈ ℂ : |z| ≤ R}. Assume f(0) ≠ 0 and also f(z) ≠ 0 for all z with |z| = R. Let the zeros of f in DR be z1, . . ., zn . (Here a zero is listed m times if it is a zero of order m.) We assume the zeros are ordered according to their increasing absolute value. Let ||f||R = max{|f(z)| : |z| = R}. Then we have

Complex Analysis: #30 Jensen`s Formula equation pic 1

More generally, thinking about various values of R, let vf(R) = n be the number of zeros of f in DR, where R, thus n, is allowed to vary. Then we have Jensen’ inequality:

Complex Analysis: #30 Jensen`s Formula equation pic 2

It is obviously analytic in DR. Furthermore, we have |g(z)| = |f(z)| when |z| = R. This implies that |g(w)| ≤ ||f||R for all w ∈ DR. [If we had a point w in the interior of DR with |g(w)| > ||f||R, then we can assume that it is maximal with respect to this property. However, that would contradict theorem 19.] Therefore

Complex Analysis: #30 Jensen`s Formula equation pic 3


Theorem 52 (Jensen’s Formula)
The same assumptions as in the previous theorem. Then we have

Complex Analysis: #30 Jensen`s Formula equation pic 4

Proof
In this proof, we will first look at two very simple cases:
1. We first prove Jensen’s formula in the simple case that there are no zeros of f in DR. Then again, as in exercise 12.1, we have an analytic function g, defined in a neighborhood of DR, with f = eg. Or put another way, g = log f. (To be definite, we could specify that log f(0) should be in the principle branch of the logarithm.) Then Cauchy’s formula is simply

Complex Analysis: #30 Jensen`s Formula equation pic 5

which establishes the theorem in the first case.

2. The second case is even simpler. Namely, let ζ be a complex number with |ζ| < R. That is, ζ is some point in the interior of DR. This second case is that the function f is simply f(z) = z − ζ. We then define a new function, namely

Complex Analysis: #30 Jensen`s Formula equation pic 6

Wednesday, May 28, 2014

Complex Analysis: #29 Order of Entire Function

  • Complex Analysis: #29 The Order of an Entire Function

Theorem 49
Let f : ℂ → ℂ be an entire function. Assume that there exist real constants C > 0, λ > 0, such that Re(f(z)) ≤ C(1 + |z|λ), for all z ∈ ℂ. Then f is a polynomial, at most of degree [λ].

Proof
To begin with, we know that, for k, l ∈ Z, we have

Complex Analysis: #29 The Order of an Entire Function equation pic 1

Therefore, taking r → ∞, we see that if k > λ then ak = 0. An analogous argument shows also that bk = 0.

Definition 17
An entire function f is said to have finite order if there exists some real number ρ > 0 and a constant C > 0, such that

|f(z)| ≤ Ce|z|ρ

for all z ∈ ℂ. The infimum over all such ρ is the order of f. That is to say, α is the order of f if |f(z)| ≤ Ce|z|α+∈ for all ∈ > 0 and z ∈ ℂ. If |f(z)| ≤ Ce|z|α for all z ∈ ℂ then α is the strict order of f.


Theorem 50
Let f be an entire function of finite order with no zeros. Then f = eg, where g is a polynomial whose degree is the order of f.

Proof
According to exercise 12.1, given f, then there exists an entire function g with f = eg. But then we must have

Complex Analysis: #29 The Order of an Entire Function equation pic 2

So Re(g(z)) ≤ |z|α if α is the order of f, and therefore the result follows from theorem 49.

Going beyond this, we would like to think about entire functions of finite order, but with zeros This leads us to Hadamard’s theorem. But before we arrive there, let us think about Jensen’s formula.

Monday, May 26, 2014

9702 November 2013 Paper 23 Worked Solutions | A-Level Physics

  • 9702 November 2013 Paper 23 Worked Solutions | A-Level Physics



Question 1
Diameter of cylindrical disc = 28mm
Thickness of cylindrical disc = 12mm
Density of material of disc = 6.8 x 103 kgm-3

Density = mass, m /volume

Volume = π(14 x 10-3) 2 x 12 x 10-3 = 7.389 x 10-6 m3   
Mass, m = density x volume = 6.8 x 103 x 7.389 x 10-6 = 0.0502kg
Weight = mg = 0.0502 x 9.81 = 0.49N




Question 2
(a)
SI units for T : s, R : m and M : kg
K = T2M/R3     units : s2kgm-3

(b)
Uncertainty in T : 0.5%, R : 1% and M : 2%
K = T2M/R3    
{(ΔK / K) x 100% = [2(ΔT / T) + (ΔM / M) + 3(ΔR / R)] x 100%} 
% uncertainty in K = 2(0.5)% {for T} + 2% {for M} +3(1)% {for R} = 6%
K = [864002 x 6 x 1024] / (4.23 x 107) = 5.918 x 1011
{(ΔK / K) x 100% = 6%. So, ΔK = (6/100)  x K} 
(Uncertainty in K =) 6 % of K = 0.355 x 1011
K = (5.9 ± 0.4) x 1011 s2kgm-3




Question 3
(a)
(i)
Velocity is the rate of change of displacement OR displacement change per unit time taken.

(ii)
Acceleration is the rate of change of velocity OR velocity change per unit time taken

(b)
(i)
Initially , the car moves with constant velocity (gradient constant). Then the velocity decreases in the middle section (gradient decreases). Finally, in the last section, the car moves with a constant velocity, which is smaller than the initial constant velocity (gradient is constant but smaller).

(ii)
Velocity =  45/1.5 = 30ms-1

(iii)
At t = 6s, x = 122m and at t = 4s, x = 98m
Velocity at 4s = (122-98) / 2 = 12ms-1

at t = 1.5s, velocity = 30ms-1 and at t = 4s, velocity = 12ms-1
Acceleration = (12-30) / (4.0-1.5) = (12-30) / 2.5 = -7.2ms-2

(iv)
F = ma = 1500 x (-7.2) = -10800N ≈  -11000N




Question 4
{Detailed explanations for this question is available as Solution 390 at Physics 9702 Doubts | Help Page 72 - http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html}



Question 5
{Detailed explanations for this question is available as Solution 660 at Physics 9702 Doubts | Help Page 132 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-132.html}





Question 6
(a)
e.m.f is the total energy available per unit charge. The potential difference across the battery is not equal to the e.m.f of the battery because some of the available energy is used / lost / wasted / given out in the internal resistance of the battery. Hence the potential difference available is less than the e.m.f.

(b)
(i)
V =IR
Current in circuit, I = V/R = 6.9 / 5.0 = 1.38A ≈ 1.4A

(ii)
r = lost volts / current = (9 - 6.9) / 1.38 = 1.5(2)Ω

(c)
(i)
P = EI = 9 x 1.38 = 12 (12.4) W

(ii)
Efficiency of battery = output power / total power = VI /EI
                 = 6.9 / 9      or (9.52) / (12.4)  = 0.767 (76.7%)




Question 7
(a)
(i)
The 6 vertical lines from one plate to the other should be equally spaced with the arrow downwards 

(ii)
Electric field strength E = V /d = 1200 / 40 x 10-3 = 3.0 x 104Vm-1  (allow 1 sf)

(b)
(i)
Force acting on A, F = Ee = 3.0 x 104 x 1.6 x 10-19 = 4.8 x 10-15 N

(ii)
Torque of the couple acting on the rod = F x separation of charges
             =   4.8 x 10-15 x 15 x 10-3 = 7.2 x 10-17 Nm

(iii)
The rod will be vertically aligned, parallel to the electric field, with A will be at the top, close to the +ve plate and B will be at the bottom, close the –ve plate.

The forces are equal and opposite in the same line. So, there is no resultant force and no resultant torque. 

Currently Viewing: Physics Reference | May 2014