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Mathematical Methods for Physicists
Henrik Jeldtoft Jensen Department of Mathematics Imperial College London
(Dated: June 28, 2010)
Contents
I. Introduction 2
II. Complex functions 3
A. Differentiation of Complex Functions 3
B. Residue theory 5
III. Fourier transformation 6
A. The power of the Fourier transform 7
- Differential equations 7
- Response functions 8
- Correlation functions 8
IV. Linear vector spaces 10
V. Calculus of variation 13
VI. Suffix notation 16
VII. Transformation Under Rotation 19
VIII. Numerical Analysis 19
IX. Some Mathematical Notation 19
I. INTRODUCTION
The following is a set of notes to be read in conjunction with the lectures delivered to the 2nd, 3rd and 4th year Students of Physics at Imperial College London.
Let us make it clear from the beginning; mathematics is the language needed to be able to formulate the laws of Nature and it is the language needed in order to be able to think about the subtleties of Nature. As with any other language it is difficult to separate the content of a message from the notation and the syntax. Try to formulate the content of Newton’s 2nd law without using any mathematical notation.
Physicists need mathematics in order to be able to talk and to think. The basic mathematics covered in this course is not only for the theoretical physicists. It is a prerequisite for anyone who wants to be able to read about physics in books or scientific papers or to be able to follow a physics seminar. I have met students who thought they were better physicists because they didn’t waist their time or energy on mathematics. This is a misunderstanding. Physics, as any other reasonably developed science, can only be appreciated by the use of mathematics. Not only is mathematics needed to be able to carry out simple analytic analysis of problems. More importantly, one is only able to understand the concepts of physics, if one possesses some degree of mathematical flexibility.
It is as impossible to learn mathematics solely by watching others perform, as it is to learn to play a musical instrument exclusively by listening to other people play. This is not a problem. The mind expanding experience of digesting mathematical concepts through the contemplation of the lecture notes and the exercises will easily capture the open minded person and lead to a sound engagement with one of humankind’s oldest and most profound activities. The musing of mathematical ideas and constructions can be done anywhere. All what is needed is to activate this most marvellous of all instruments: our brain. And sometimes pencil and paper comes in handy too. In fact it is hardly possible to understand a mathematical text by reading it. It is much better to copy every single detail and to think carefully about what it all means while coping.
The Cambridge book Riley, Hobson and Bence: Mathematical Methods for Physics and Engineering, together with the Students Solution Manual for Mathematical Methods for Physics and Engineering by Riley and Hobson is magnificent. If one gets hold of this pair of books and read and work with the problems, one will become greatly accomplished in the math used by physics.
But there exist many excellent text books on mathematics written for physicists and other scientist. I list a few which I know and have found readable.
Introductory level:
I1 G. Arfken, ”Mathematical Methods for Physicists”, Academic Press.
I2 H. Jeffreys and B.S. Jeffreys, Methods of Mathematical Physics, Cambridge University Press.
I3 D. A. McQuarrie, Mathematical Methods for Scientists and Engineers”, University Science Books.
I4 J. Mathews and R.L Walker, ”Mathematical Methods of Physics”, Benjamin/Cummings Publishing Company.
Comprehensive more advanced level:
A1 Courant and Hilbert, ”Methods of Mathematical Physics”.
A2 Morse and Feshbach, ” Methods of Theoretical Physics”.
Here follows a few books that might help one to appreciate the nature of mathematics.
E1 J. Dieudonn´e, ”Mathematics - The music of reason”, Springer.
x
f(x)
x x+h
f(x)
f(x+h)
FIG. 1 The line approaches the tangent as h → 0. The slope of the tangent is defined as the derivative.
A. Differentiation of Complex Functions
Recall how one differentiates a real function. The derivative of the function f (x) at the point x 0 can be interpreted as the slope of the tangent at this point. The reason for this is clear. Consider the slope of the line from the point (x, f (x)) to the point (x + h, f (x + h)). See Fig. 1. The derivative is obtained as the value of the slope of this line in the limit h → 0. Now note the following important point. We only consider f (x) to be differentiable if the same value for the slope is obtained irrespectively of how the limit h → 0 is assumed. We can take the limit while demanding h > 0, or we can take the limit while insisting that h remains non-positive. The function is only differentiable at x 0 if this makes no difference. 1
The essential point is that the limit
limδx→ 0
f (x 0 + δx) − f (x) δx
df dx
|x 0 (6)
is independent of how the limit is taken. The derivative of a complex function f (z) at the point z 0 is defined in exactly the same way
df dz
|z 0 = limδz→ 0
f (z 0 + δz) − f (z) δz
with the same insistence that the limit must not depend on the manner in which the limit is taken, This has remarkably restrictive consequences in the case of complex functions. From the requirement that the same value is obtained for df /dz when we let δz vary parallel to the x-axis as when we let δz vary parallel to the y-axis it follows, (as will be discussed in the lectures) that the real part and the imaginary part of f (z) are linked together in the following intriguing way.
∂U ∂x
∂V
∂y ∂V ∂x
∂U
∂y
We used the following notation z = x + iy together with
f (z) = Ref (z) + iImf (z) = U (x, y) + iV (x, y). (9)
The relations in Eq. (8) are called the Cauchy-Riemann conditions. Not only does a differentiable function satisfy these conditions, we will find that if a complex function satisfies Eq. (8) then the limit in Eq. (7) will be independent of how δz is taken to zero, and, hence, a function that satisfies Eq. (8) will also be differentiable.
Thus, to check if a complex function is differentiable one simply check if its real and imaginary parts fulfil the Cauchy-Riemann condition.
(^1) One can define the derivative from the left or from the right and they might be different if the function has a cusp at x 0 , but in that case the function is not differentiable, only differentiable from the left or the right respectively.
S
E
W
N
FIG. 2 Many ways around the mountain, but only two different classes
Z_N! 1
Z_ Z_ Z A
Z_B
FIG. 3 Integration along a path in the complex plane.
B. Residue theory
Complex differentiable functions posses a spectacular sensitivity to theirs global environment. Think of it this way. You are going to cross a big plain from South to North, see Fig. (2). In the centre there is a high mountain. It is early morning and the sun is in the East. You can either pass to the East of the mountain or to the West. Whether you pass two kilometres or three kilometres to the East of the mountain won’t make much difference to you. But there is a big difference between passing the mountain to East or to the West. If you decide on the Western route you’ll find yourself in the shadow of the mountain for a while, whereas if you stay on the Eastern route you’ll be warmed by the sun during the entire trip.
It is the same with integrals in the complex plane of complex functions. The integral is define much in the same way as the Riemann sum used to define the integral of a real function. Assume that the path P leads from point zA to point zB in the complex plane. Chop the path up in N − 1 segments from zA to z 1 , and from z 1 to z 2 etc. see Fig. 3. We introduce dz 1 = z 1 − za, dz 2 = z 2 − z 1 ,..,dzN = zB − zN − 1. The following sum
IN =
∑^ N
i=
f (zi)dzi, (10)
is used to define the integral from zA to zB along the path P by assuming the limit N → ∞, i.e. we define ∫
P
f (z)dz ≡ lim N →∞
IN. (11)
Consider now the integral along different paths leading from point zA to point zB in the complex plane. If the complex function f (z) is differentiable everywhere in the plane then, as we will see in the lectures,
The inverse transform can be considered equivalent to the determination of the coefficients in Eq. (12). In the case of vectors we determine the coordinates, say along the 2nd basis direction e 2 from the scalar product. If e 2 has length 1 and ei are mutually orthogonal, we have y = r · e 2. The integral between f (x) and e−ikx^ in Eq. (15) is like a scalar product between f (x) and e−ikx.
A. The power of the Fourier transform
Fourier transformation is a very powerful tool when analysing physical problems. We often have to solve both ordinary as well as partial differential equations. In condensed matter physics and statistical mechanics theoreticians as well as experimentalists use response functions and correlation functions to characterise the behaviour. The mathematical analysis relies on some rather impressive properties of the Fourier transform. E.g. that differentiation of a function corresponds to a multiplication of its Fourier transform. That certain types of integrals of functions correspond to multiplication of Fourier transforms. We list a few of the details below.
- Differential equations
Consider the ordinary differential equation
df 2 dx^2
We know that the solution is obtained from the homogeneous Eq.
d^2 f dx^2
by adding a particular solution to the inhomogeneous Eq. (16). We also know how to find the solution to the homogeneous Eq. by solving the auxiliary eq. λ^2 + 1 = 0. But how do we obtain the solution to the inhomogeneous Eq. (16 directly. In introductory courses all what is normally said is that one tries to guess. Instead one may use Eq. (15). First we notice that
df dx
d dx
−∞
dk 2 π
fˆ (k)eikx
−∞
dk 2 π
fˆ (k) d dx
eikx
−∞
dk 2 π
fˆ (k)[ik]eikx. (17)
From this we conclude that the Fourier transform of the derivative of a function is equal to ik times the Fourier transform of the function:
dfˆ dx
(k) = ik fˆ (k). (18)
Thus, if we substitute the left hand side of Eq. (15) for f (x) and the equivalent for g(x) we obtain from Eq. (16)
(ik)^2 fˆ (k) + fˆ (k) = ˆg(k). (19)
Or equivalently
fˆ (k) = ˆg(k) 1 − k^2
It is now just a matter of performing the inverse Fourier transformation to obtain the sought of function
f (x) =
−∞
dk 2 π
g ˆ(k) 1 − k^2
eikx. (21)
- Response functions
Let us consider the gravitational potential φ(r) at position r produced by a set of N masses m(ri) located at the positions ri with i = 1, 2 , ..., N ,
φ(r) =
∑^ N
i=
−G 0
|r − ri|
m(ri). (22)
Here G 0 is the gravitational constant. If we instead consider a continuous mass distribution the equation will become an integral equation:
φ(r) =
V
−G 0
|r − r′|
ρ(r′)dr′. (23)
Where ρ(r′) is the mass density at position r′. The structure of Eq. (23) is frequently encountered in physics and is of the form
φ(r) =
V
G(r − r′)q(r′)dr′. (24)
This type of Eq. describes how the effect of a “charge” at position r′ is “transmitted” through space by the response function G(r − r′) to the position R where we “monitor” the response φ(r), accumulated from all the charges present. Note that the response function depends only on the relative position vector R − r′. In Eq. (23) the gravitational response function is obviously given by G(r − r′) = −G 0 /|r − r′| and the gravitational charge is simply the mass density q(r′) = ρ(r′). The response Eq. (24) is particularly simple when Fourier transformed. Assume one dimension for simplicity. Substitute the expression in Eq. (15) into Eq. (24) ∫ (^) ∞
−∞
dk 2 π
φˆ(k)eikr^ =
V
−∞
dk 2 π
Gˆ(k)eik(r−r′)
−∞
dk′ 2 π
ˆq(k′)eik′r′dr′
−∞
dk 2 π
Gˆ(k)eikr
−∞
dk′ 2 π
qˆ(k′)
V
ei(k′−k)r′dr′. (25)
In the lectures we introduce Dirac’s delta function and discuss that ∫ dr′ei(k′−k)r′^ = 2πδ(k′ − k). (26)
We shall also see that the Dirac delta function when integrated simply zeros its argument: ∫ (^) ∞
−∞
dxf (x)δ(x − x 0 ) = f (x 0 ). (27)
We can make use of this property to finish the calculation in Eq. (25) ∫ (^) ∞
−∞
dk 2 π
φˆ(k)eikr^ =
−∞
dk 2 π
Gˆ(k)ˆq(k)eikr^ (28)
from which we conclude that
φˆ(k) = Gˆ(k)ˆq(k). (29)
The conclusion is that if we know the Fourier transform of the response function and of the charges, all we need to do is to multiply these together in order to obtain (the Fourier transform) of the resulting physical effect.
- Correlation functions
In the study of natural phenomena we are often concerned with how much one part of a system influences another. Or we want to know how much of what happens at a specific moment in time, is influenced by what
To obtain the correlation coefficient between A and B empirically we would have to perform N mea- surements of A and N measurements of B and finally N measurements of A and B simultaneously. As an example think of trying to settle whether a person’s height and distance between the ears are related to each other. We would then first determine the average height, then the average distance between ears and finally the average of the product of height times distance between ears.
The empirical estimate of the correlation coefficient obtained by this procedure is given by
CAB =
N
∑^ N
i=
aibi − [
N
∑^ N
i=
ai][
N
∑^ N
i=
bi]
N
∑^ N
i=
[ai − 〈A〉][bi − 〈B〉]. (35)
In the second equality we made use of Eq. (34) for both A and B.
We now return to the temporal signal f (t). We measure the signal again and again at times separated by T time units. We think of the measurements of f at times t = 1, 2 , 3 , ... as ai and the measurements of F at the time t + T as bi. We will now use Eq. (35) to obtain an estimate of the correlation between the two measurements for the signal separated by T time units. To make life simple we will neglect the 1/N factors in Eq. (35). A justification for this is that we are interested in the functional dependence of the correlations on the time interval T and not so much interested in the actual specific value of the correlation coefficient. Since the correlation coefficient will depend on T we talk about the correlation function. Moreover, since we are correlating the signal with itself we talk about the autocorrelation function given by:
C(T ) =
t
[f (t) − 〈f (t)〉][f (t + T ) − 〈f (t + T )〉]
dt[f (t) − 〈f (t)〉][f (t + T ) − 〈f (t + T )〉]
dt[f (t) − 〈f (t)〉][f (t + T ) − 〈f (t)〉]. (36)
In the last equality we made use of the fact that the average of value f (t) and f (t + T ) are identical.
It is hopefully by now clear why the autocorrelation function is an important object for the study of memory effects or causality effects in a signal. Much time is spend experimentally and theoretically on the determination of correlation functions or the equivalent power spectrum whichever is most easy to determine. The power spectrum of the signal f (t) is defined as
Sf (ω) = | fˆ (ω)|^2. (37)
That is the square of absolute value of the Fourier transform of the signal. The reason it doesn’t matter which of the two C(T ) of Sf (ω) one gets hold of is that, as we show in the lectures, they are related in the following simple way
Sf (ω) = Cˆ(ω). (38)
In words: the power spectrum is equal to the Fourier transform of the correlation function.
IV. LINEAR VECTOR SPACES
The prototype space is IR^3 , that is
RI^3 = {(x 1 , x 2 , x 3 )|xi ∈ RI, i = 1, 2 , 3 }. (39)
A basis in IR^3 is
e 1 = (1, 0 , 0), e 2 = (0, 1 , 0), e 3 = (0, 0 , 1). (40)
since an arbitrary vector v = (x 1 , x 2 , x 3 ) in IR^3 can be written as a linear combination
v = x 1 e 1 + x 2 e 2 + x 3 e 3. (41)
The two vectors e 1 and e 2 span a subspace of IR^3 :
span(e 1 , e 2 ) = {(x 1 , x 2 , 0) | x 1 , x 2 ∈ RI}. (42)
This subspace is equivalent to the two dimensional plane.
From the theory of Fourier series we know that the set of functions
e 0 (x) = 1, e 1 (x) = cos x, e 2 (x) = cos 2x, ... (43)
together with the functions
˜e 1 (x) = sin x, ˜e 2 (x) = sin 2x, ˜e 3 (x) = sin 3x, ... (44)
constitute a basis in C^0 ([−π, π], R) since any continuous functionI f : [−π, π] → R can be written asI
f (x) = a 0 e 0 (x) +
∑^ ∞
n=
[anen(x) + bnen(x)]. (45)
Consider the scalar product
〈f, g〉 =
∫ (^) π
−π
f (x)g(x)dx. (46)
By use of this scalar product one can show that the basis
e 0 (x), e 1 (x), ..., ˜e 1 (x), ˜e 2 (x), ... (47)
consists of orthogonal (but not normalised) vectors. For example
〈en, em〉 =
∫ (^) π
−π
cos(nx) cos(mx)dx ∝ δn,m. (48)
Normed vector spaces We are used to being able to measure the distance between two points X = (x 1 , x 2 , x 3 ) and Y = (y 1 , y 2 , y 3 ) in IR^3 by use of Pythagoras’ theorem, namely as [(y 1 − x 1 )^2 + (y 2 − x 2 )^2 + (y 3 − x 3 )^2 )]^1 /^2. This is of course
the same as the length of the vector XY~ = (y 1 − x 1 , y 2 − x 2 , y 3 − x 3 ) or in other words the distance between
point X and point Y is equal to the norm of the vector XY~ :
‖ XY~ ‖ = [(y 1 − x 1 )^2 + (y 2 − x 2 )^2 + (y 3 − x 3 )^2 )]^1 /^2. (49)
We introduce generalised norms. Consider the following example. The subspace C odd^0 of C^0 ([−π, π], R)I consisting of odd functions. We know from the theory of Fourier series that these functions can be written as
f (x) =
∑^ ∞
n=
bn sin(nx), (50)
and we can use
‖f ‖ =
∑^ ∞
n=
|bn|^2 (51)
a_p2! (^) a_p2+
a_p3+
a_p1! a_p1+
a_p3!
!!
! (^)!
!!
FIG. 5 The limit is confined to a smaller and smaller interval
procedure, see Fig. 5 forever and construct a sequence of intervals within intervals of shorter and shorter length:
[0, 1] ⊃ I 1 ⊃ I 2 ⊃ I 3 ⊃ · · ·. (59)
Since the length of the interval Ik decreases to zero and an ∈ Ik for all n > pk, the sequence an must converge to a number a larger than 0 and smaller than 1, since [0, 1] is an interval we know that a ∈ [0, 1] and hence that the sequence is strongly convergent to a limit within the set [0, 1]. This completes the proof.
Scalar products and projections For ordinary vectors in IR^2 or in IR^3 we use the scalar product to find the projection b‖ of a vector b along a vector a. We have
b‖ = (b · a)
a |a|
Consider quantum mechanics. Assume a particle is in a quantum state described by the normalised wave function ψ. In order to calculate the probability that a measurement of the energy will yield the value E = En, for some particular energy level En, we need to know what the component of ψ is along the normalised eigenstate φn corresponding to the eigenvalue En. I.e., we must calculate the scalar product 〈ψ, φn〉 to obtain the probability |〈ψ, φn〉|^2 for measuring En.
and δ expressions As a way to get used to the − δ formalism we list here a few examples:
Continuity A function f (x) is said to be continuous at a point x 0 if and only if
∀ > 0 ∃δ > 0 | |x − x 0 | < δ ⇒ |f (x) − f (x 0 )| < . (61)
Convergence A sequence of numbers xn is said to converge to the point x if and only if
∀ > 0 ∃n 0 | n > n 0 ⇒ |x − xn| < . (62)
Example: The sequence an = 1 + 1/n converges to 1. Proof: given > 0 we can choose n 0 as the smallest integer larger than 1/ to obtain that
n > n 0 ⇒ | 1 − xn| < . (63)
V. CALCULUS OF VARIATION
How do we find the most efficient way of doing something. We first need to decide exactly what type of efficiency we are interested in. We might want to calculate the fastest way of getting from one point to
t
r_i
r_f
r
r*(t)
t i t_f
r(t)=r*(t)+! r(t)
FIG. 6 Variations about a trajectory
another. Or we might be interested in which shape of a string suspended at its endpoints minimises its total potential energy.
A very important example is to find the physical path followed by a particle as it travels from an initial position ri to a final position rf through a potential field U (r). Sir William Rowan Hamilton (1805-1865) realised that the particle travels in such a way as to minimise the difference between the kinetic energy and the potential energy, i.e. a path that minimises the time integral of the Lagrangian. At every time instance you monitor your kinetic energy, T [˙r(t)], as well as your potential energy, U [r(t)]). You calculate the difference L(t) = T (t) − U (t) and add it all up to obtain what is called the action
S =
∫ (^) tf
ti
L[r, r˙]dt. (64)
In order to minimise S one should spend long time in regions where the potential is large and short times where the potential assumes small values. Hence a small action is obtained if we can design our trajectory r(t) and velocity v(t) = dr/dt such that we move with low velocity through regions of large potential energy. The question is how do we determine r(t) from U (r) in order to obtain the smallest possible value of S?
We need to establish an equation for r(t) from the requirement that S is minimal for that particular function r(t). For ease of notation we confine ourselves to one dimension^4 , and specify the position by the scalar r(t). Imagine that we somehow had found the trajectory which minimises S[r(t)] and denote this specific functional relation between position and time by r∗(t). We will then have that for all other choices r(t) of ways to get from ri to rf the action of r∗(t) is smaller than the action of r(t), or
S[r∗(t)] < S[r(t)]. (65)
Imagine we choose another trajectory r(t) which only slightly differs from r∗(t). This is easiest illustrated by plotting the position as function of time as in Fig. 6. The difference between r∗(t) and r(t) we denote by δr(t), i.e,
r(t) = r∗(t) + δr(t). (66)
The difference between S[r(t)] and S[r∗(t)] will be small if δr(t) is small. In fact we can imagine to write the difference between S[r(t)] and S = [r∗(t)] as an expansion in deviation δr(t) between the two paths. We will the have something that schematically would look like
δS = S[r(t)] − S[r∗(t)] = a 1 δr(t) + a 2 [δr(t)]^2 + · · ·. (67)
(^4) This is really not an essential restriction, but we avoid to have to think about the three dimensional vector r
We assume, as in the Figure, that δr(ti) = δr(tf ) = 0 and arrive at the following expression
δS =
∫ (^) tf
ti
∂L
∂r
∗
d dt
∂L
∂ r˙
∗
δr(t)dt. (76)
Since δS must be equal to zero for all perturbations δr(t) if r∗(t) is an extremal path we conclude that the integrand in Eq. (76) must vanish:
( ∂L ∂r
∗
d dt
∂L
∂ r˙
∗
This equation is called the Euler-Lagrange equation for the action in Eq. (64). In general whenever one wants to minimise or maximise a certain cost function, the starting point is to determine the cost function and then derive the Euler-Lagrange equation. All the fundamental equations of physics can be obtained from the action, i.e. the time integral of the Lagrangian. The Maxwell equations for electrodynamics can be derived in this way, the Einstein equations for gravitation can be obtained this way and so can the equations for QCD and for the electro-weak forces. The Euler-Lagrange Eq. (77) is in fact equivalent to Newton’s 2nd law when the Lagrangian in Eq. (64) is the usual mechanical Lagrangian:
L[r(t), r˙(t)] =
m r˙^2 − U (r), (78)
where m denotes the mass of the particle and U (r) the potential energy. We immediately obtain Newton’s law from Eq. (77) by noticing that for L[r, r˙] of the form in Eq. (78) we have
∂L ∂r
∂U (r) ∂r ∂L ∂ r˙
= m r,˙ (79)
which upon substitution into Eq. (77) leads to
∂U (r) ∂r
d dt
m r˙ = 0. (80)
VI. SUFFIX NOTATION
When dealing with vectors and matrices and tensors, calculations easily become very cumbersome if one trys to bulldoze one’s way through in coordinate notation without an efficient notation. Suffix notation is meant to supply this efficient and economical notation. After a bit of practise the machinery will hopefully become a cherished device - especially if the study of electromagnetism or other field theories are undertaken.
As often in physics we start out with a contribution by Albert Einstein, namely the so called repeated index convention. As an example consider the familiar scalar product between the two vectors a = (a 1 , a 2 , a 3 ) and b = (b 1 , b 2 , b 3 ), both of three dimensions:
a · b = a 1 b 1 + a 2 b 2 + a 3 b 3 =
∑^3
i=
aibi ≡ aibi. (81)
Einstein’s contribution consists in ignoring the summation sign and simply as of convention assume sum- mation over any index that is repeated. Einstein may have made more important contributions to our understanding of the univers, but hardly any more straight forward.
There are two other bits of notation that turn out to be very handy when analysing vector identities. The first is the Kronecker delta (after the German mathematician Leopold Kronecker 1823-1891):
δij =
1 if i = j 0 if i 6 = j.
One notice that δi,j can be thought of as a discrete version of the Dirac delta function δ(x − x 0 ) in the sense that also δ(x − x 0 ) = 0 for all values of x and x 0 except when x = x 0. Another way of think about δij is as the elements of the unit matrix
I =
(^) = {δij }. (83)
The permutation symbol is slightly more involded to define
ijk =
0 if any two i, j, k are the same 1 if i, j, k is a cyclic permutation of 1,2,3 (an even permutation) − 1 if i, j, k is an anti-cyclic permutation (an odd permutation)
One often finds great use of ijk in relation to vector cross products and determinants of matrices.
To get some feeling of how these symbols are manipulated we will present the proof of the following identity. The proof should definitely be read with a pencil in hand. The best way of absorbing the idea is to simply copy the proof bit by bit.
ijkpqk = δipδjq − δiq δjp. (85)
Proof: STEP 1. By a bit of perseverance one can show that for any matrix A = {aij }
detA = ijka 1 ia 2 j a 3 k. (86)
This is done in the following way. Proof Step 1: We note that in the sum ijka 1 ia 2 j a 3 k there are 3 × 3 × 3 = 27 terms. However only 6 are non-zero, namely:
ijka 1 ia 2 j a 3 k = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 −a 11 a 23 a 32 − a 12 a 21 a 33 − a 13 a 22 a 31
= a 11
a 22 a 23 a 32 a 33
∣ −^ a^12
a 21 a 23 a 31 a 33
∣ +^ a^13
a 21 a 22 a 31 a 32
a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33
= detA (88)
STEP 2.
It is not very difficult to generalise this result to the more general
ijkapiaqj ark = detApqr. (89)
Proof Step 2: One simply need to distinguish between the three different cases:
i) If pqr is a cyclic permutation of 123, say pqr = 231, then
ijka 2 ia 3 j a 1 k = ijka 1 ka 2 ia 3 j = kij a 1 ka 2 ia 3 j = detA. (90)
δip δiq δik δjp δjq δjk δkp δkq 3
= δipδjq 3 + δiq δjkδkp + δjpδkq δik −δikδjq δkp − δiq δjp 3 − δjkδkq δip = 3(δipδjq − δiq δjp) + δiq δjp + δjpδiq − δipδjq − δipδjq = δipδjq − δiq δjp. (97)
Which completes the proof of Eq. (85).
To become better acquainted with this sections vector notation it may be useful to look through Jefferyes and Jeffreys book Chap. 2 (See p. 2 for reference).
VII. TRANSFORMATION UNDER ROTATION
The very important point to keep in mind is that a vector is a geometrical object: a direction in space. From this simple fact follow all the transformation properties discussed in the lectures. Most of which are likely to be known from A-levels.
A look at Chap. 3 in Jeffreys and Jeffreys book (See p. 2 for reference) would probably help one to be come acquainted with tensors.
VIII. NUMERICAL ANALYSIS
See separate set of notes.
IX. SOME MATHEMATICAL NOTATION
Example: Continuity of a fuction:
∀ > 0 ∃δ > 0 | |x 0 − x| < δ ⇒ |f (x 0 ) − f (x)| <
Translations: ∀ means “for all” ∃ means “there exists” | means “such that”
Example: Set theory: If S 1 = { 1 , 2 , 3 , 4 } and S 2 = { 2 , 3 } then S 2 ⊂ S 1 since 2 ∈ S 2 and 3 ∈ S 2 are also in S 1.
Translations: {a, b} denotes the set consisting of the elements a and b. a ∈ S means that a is an element in the set S.
S 2 ⊂ S 1 means that the elements of the set S 2 are all members in S 1 and that S 1 contains elements that are not found in S 2. If we write S 2 ⊆ S 2 the two sets might be identical.
More set theory: [0, 1] denotes the closed interval 0 ≤ x ≤ 1. ]0, 1[ denotes the open interval 0 < x < 1. [0, 1[ denotes the half open/closed interval 0 ≤ x < 1.
