1. Lecture 21 Revision session http://www.hep.shef.ac.uk/Phil/PHY226.htm Remember Phils Problems and your notes = everything Remember I’m available for questions all through Christmas

2. Revision for the exam http://www.shef.ac.uk/physics/exampapers/2007-08/phy226-07-08.pdf Above is a sample exam paper for this course There are 5 questions. You have to answer Q1 but then choose any 2 others Previous years maths question papers are up on Phils Problems very soon Q1: Basic questions to test elementary concepts. Looking at previous years you can expect complex number manipulation, integration, solving ODEs, applying boundary conditions, plotting functions, showing ‘x’ is solution of PDE. Easy stuff. Q2-5: More detailed questions usually centred about specific topics: InhomoODE, damped SHM equation, Fourier series, Half range Fourier series, Fourier transforms, convolution, partial differential equation solving (including applying an initial condition to general solution for a specific case), Cartesian 3D systems, Spherical polar 3D systems, Spherical harmonics The notes are the source of examinable material – NOT the lecture presentations I wont be asking specific questions about Quantum mechanics outside of the notes

3. Revision for the exam The notes are the source of examinable material – NOT the lecture presentations Things to do now Read through the notes using the lecture presentations to help where required. At the end of each section in the notes try Phils problem questions, then try the tutorial questions, then look at your problem and homework questions. If you can do these questions (they’re fun) then you’re in excellent shape for getting over 80% in the exam. Any problems – see me in my office or email me Same applies over holidays. I’ll be in the department most days but email a question or tell me you want to meet up and I’ll make sure I’m in. Look at the past exam papers for the style of questions and the depth to which you need to know stuff. You’ll have the standard maths formulae and physical constants sheets (I’ll put a copy of it up on Phils Problems so you are sure what’s on it). You don’t need to know any equations e.g. Fourier series or transforms, wave equation, polars.

4. Concerned about what you need to know? Look through previous exam questions. 2008/2009 exam will be of very similar style. You don’t need to remember any proofs or solutions (e.g. Parseval, Fourier series, Complex Fourier series) apart from damped SHM which you should be able to do. You don’t need to remember any equations or trial solutions, eg. Fourier and InhomoODE particular solutions. APART FROM TRIAL FOR COMPLEMENTARY 2 ND ORDER EQUATION IS You don’t need to remember solutions to any PDE or for example the Fourier transform of a Gaussian and its key widths, etc. However you should understand how to solve any PDE from start to finish and how to generate a Fourier transform. Things you need to be able to do: Everything with complex numbers; solve ODEs and InhomoODEs, apply boundary conditions; integrate and differentiate general stuff; know even and odd functions; understand damped SHM, how to derive its solutions depending on damping coefficient and how to draw them; how to represent an infinitely repeating pattern as a Fourier series, how to represent a pulse as a sine or cosine half range Fourier series; how to calculate a Fourier transform; how to (de)convolve two functions; the steps needed to solve any PDE and apply boundary conditions and initial conditions (usually using Fourier series); how to integrate and manipulate equations in 3D cartesian coordinates; how to do the same in spherical polar coordinates; how to prove an expression is a solution of a spherical polar equation.

5. Let’s take a quick look through the course and then do the exam from last year

6. Binomial and Taylor expansions

7. Integrals Try these integrals using the hints provided

8. More integrals Summary Previous page Remember odd x even function Previous page

9. Even and odd functions So even x even = even even x odd = odd odd x odd = even An even function is f(x)=f(-x) and an odd function is f(x)= -f(-x),

10. Complex numbers Real Imaginary Argand diagram a b Cartesian a + ib  r Polar so where

11. Working with complex numbers Add / subtract Multiply / divide Powers

12. Working with complex numbers Roots Example : Step 1: write down z in polars with the 2πp bit added on to the argument. Step 2: say how many values of p you’ll need (as many as n) and write out the rooted expression ….. Step 3: Work it out for each value of p…. here n = 2 so I’ll need 2 values of p; p = 0 and p = 1. If what is z ½ ? p = 0 p = 1

13. 1 st order homogeneous ODE 1 st method: Separation of variables e.g. radioactive decay gives 2 nd method: Trial solution Guess that trial solution looks like Substitute the trial solution into the ODE Comparison shows thatso write

14. Step 3: General solution is or if m 1 =m 2 For complex roots solution is which is same as or Step 1: Let the trial solution be Now substitute this back into the ODE remembering that and This is now called the auxiliary equation 2 nd order homogeneous ODE Solving Step 2: Solve the auxiliary equation for and Step 4: Particular solution is found now by applying boundary conditions

15. 2 nd order homogeneous ODE Step 1: Let the trial solution be So and Example 3: Linear harmonic oscillator with damping Step 2: The auxiliary is then with roots Step 3: General solution is then……. HANG ON!!!!! In the last lecture we determined the relationship between x and t when meaning that will always be real What if or ???????????????????

16. (i) Over-damped gives two real roots 2 nd order homogeneous ODE Example 3: Damped harmonic oscillator Auxiliary is roots are BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!! Both m 1 and m 2 are negative so x(t) is the sum of two exponential decay terms and so tends pretty quickly, to zero. The effect of the spring has been made of secondary importance to the huge damping, e.g. aircraft suspension

17. (ii) Critically damped gives a single root 2 nd order homogeneous ODE Example 3: Damped harmonic oscillator Auxiliary is roots are BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!! Here the damping has been reduced a little so the spring can act to change the displacement quicker. However the damping is still high enough that the displacement does not pass through the equilibrium position, e.g. car suspension.

18. 2 nd order homogeneous ODE Example 3: Damped harmonic oscillator Auxiliary is roots are BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!! (iii) Under-damped This will yield complex solutions due to presence of square root of a negative number. Let so thus The solution is the product of a sinusoidal term and an exponential decay term – so represents sinusoidal oscillations of decreasing amplitude. E.g. bed springs. As before general solution with complex roots can be written as We do this so that  is real

19. 2 nd order homogeneous ODE Example 3: Damped harmonic oscillator Auxiliary is roots are BE CAREFUL – THERE ARE THREE DIFFERENT CASES!!!!!

20. Inhomogeneous ordinary differential equations Step 4: Apply boundary conditions to find the values of the constants Step 1: Find the general solution to the related homogeneous equation and call it the complementary solution. Step 2: Find the particular solution of the equation by substituting an appropriate trial solution into the full original inhomogeneous ODE. e.g. If f(t) = t 2 try x p (t) = at 2 + bt + c If f(t) = 5e 3t tryx p (t) = ae 3t If f(t) = 5e iωt tryx p (t) =ae iωt If f(t) = sin2t try x p (t) = a(cos2t) + b(sin2t) If f(t) = cos  ttryx p (t) =Re[ae iωt ] see later for explanation If f(t) = sin  ttryx p (t) =Im[ae iωt ] If your trial solution has the correct form, substituting it into the differential equation will yield the values of the constants a, b, c, etc. Step 3: The complete general solution is then.

21. Extra example of inhomo ODE Solve Step 1: With trial solution find auxiliary is Step 2: So treating it as a homoODE Step 3: Complementary solution is Step 4: Use the trial solution and substitute it in FULL expression. so cancelling Comparing sides gives…. Solving gives Step 5: General solution is

22. Finding partial solution to inhomogeneous ODE using complex form Sometimes it’s easier to use complex numbers rather than messy algebra Since we can write then we can also say that and where Re and Im refer to the real and imaginary coefficients of the complex function. Let’s look again at example 4 of lecture 4 notes Let’s solve the DIFFERENT inhomo ODE If we solve for X(t) and take only the real coefficient then this will be a solution for x(t) Sustituting so Therefore since take real part

23. Finding coefficients of the Fourier Series… Summary The Fourier series can be written with period L as The Fourier series coefficients can be found by:- Let’s go through example 1 from notes…

24. Finding coefficients of the Fourier Series Find Fourier series to represent this repeat pattern. 0  x 1 Steps to calculate coefficients of Fourier series 1. Write down the function f(x) in terms of x. What is period? 2. Use equation to find a 0 ? Period is 2  3. Use equation to find a n ? 4. Use equation to find b n ?

25. Finding coefficients of the Fourier Series 5. Write out values of b n for n = 1, 2, 3, 4, 5, …. 4. Use equation to find b n ? 6. Write out Fourier series with period L, a n, b n in the generic form replaced with values for our example

26. Fourier Series applied to pulses If the only condition is that the pretend function be periodic, and since we know that even functions contain only cosine terms and odd functions only sine terms, why don’t we draw it either like this or this? Odd function (only sine terms) Even function (only cosine terms) What is period of repeating pattern now?

27. Fourier Series applied to pulses Half-range sine series We saw earlier that for a function with period L the Fourier series is:- where In this case we have a function of period 2d which is odd and so contains only sine terms, so the formulae become:- where Remember, this is all to simplify the Fourier series. We’re still only allowed to look at the function between x = 0 and x = d I’m looking at top diagram

28. Fourier Series applied to pulses Half-range cosine series Again, for a function with period L the Fourier series is:- where Again we have a function of period 2d but this time it is even and so contains only cosine terms, so the formulae become:- where Remember, this is all to simplify the Fourier series. We’re still only allowed to look at the function between x = 0 and x = d I’m looking at top diagram

29. Fourier Series applied to pulses Summary of half-range sine and cosine series The Fourier series for a pulse such as can be written as either a half range sine or cosine series. However the series is only valid between 0 and d Half range sine series Half range cosine series where

30. Fourier Transforms where The functions f(x) and F(k) (similarly f(t) and F(w)) are called a pair of Fourier transforms k is the wavenumber, (compare with ).

31. Fourier Transforms Example 1: A rectangular (‘top hat’) function Find the Fourier transform of the function given that This function occurs so often it has a name: it is called a sinc function.

32. Can you plot exponential functions? For any real number a the absolute value or modulus of a is denoted by | a | and is defined as The ‘one-sided exponential’ function What does this function look like? The ‘****’ function What does this function look like?

33. Fourier Transforms Example 4: The ‘one-sided exponential’ function Show that the function has Fourier transform

34. Complexity, Symmetry and the Cosine Transform We know that the Fourier transform from x space into k space can be written as:- We also know that we can write So we can say:- What is the symmetry of an odd function x an even function ? If f(x) is real and even what can we say about the second integral above ? Will F(k) be real or complex ? If f(x) is real and odd what can we say about the first integral above ? Will F(k) be real or complex ? Odd 2 nd integral is odd (disappears) and F(k) is real 1 st integral is odd (disappears), F(k) is complex What will happen when f(x) is neither odd nor even ? Neither 1 st nor 2 nd integral disappears, and F(k) is usually complex

35. Complexity, Symmetry and the Cosine Transform f(x) is real and even Since we say As before the 2 nd integral is odd, disappears, and F(k) is real let’s see if we can shorten the amount of maths required for a specific case … so But remember that So Now since F(k) is real and even it must be true that were we to then find the Fourier transform of F(K), this can be written:- LET’S GO BACKWARDS

36. Complexity, Symmetry and the Cosine Transform Fourier cosine transform Here is the pair of Fourier transforms which may be used when f(x) is real and even only Example 5: Repeat Example 1 using Fourier cosine transform formula above. Find F(k) for this function

37. Dirac Delta Function The delta function  (x) has the value zero everywhere except at x = 0 where its value is infinitely large in such a way that its total integral is 1. The Dirac delta function  (x) is very useful in many areas of physics. It is not an ordinary function, in fact properly speaking it can only live inside an integral.  (x – x 0 ) is a spike centred at x = x 0  (x) is a spike centred at x = 0

38. Dirac Delta Function The product of the delta function  (x – x 0 ) with any function f(x) is zero except where x ~ x 0. Formally, for any function f(x) Example: What is ?

39. Dirac Delta Function The product of the delta function  (x – x 0 ) with any function f(x) is zero except where x ~ x 0. Formally, for any function f(x) Examples (a) find (b) find (c) find the FT of

40. Convolutions If the true signal is itself a broad line then what we detect will be a convolution of the signal with the resolution function: True signal Convolved signal Resolution function We see that the convolution is broader then either of the starting functions. Convolutions are involved in almost all measurements. If the resolution function g(t) is similar to the true signal f(t), the output function c(t) can effectively mask the true signal. http://www.jhu.edu/~signals/convolve/index.html

41. Deconvolutions We have a problem! We can measure the resolution function (by studying what we believe to be a point source or a sharp line. We can measure the convolution. What we want to know is the true signal! This happens so often that there is a word for it – we want to ‘deconvolve’ our signal. There is however an important result called the ‘Convolution Theorem’ which allows us to gain an insight into the convolution process. The convolution theorem states that:- i.e. the FT of a convolution is the product of the FTs of the original functions. We therefore find the FT of the observed signal, c(x), and of the resolution function, g(x), and use the result that in order to find f(x). If then taking the inverse transform,

42. Deconvolutions Of course the Convolution theorem is valid for any other pair of Fourier transforms so not only does ….. and therefore allowing f(x) to be determined from the FT but also and therefore allowing f(t) to be determined from the FT

43. Example of convolution I have a true signal between 0 < x < ∞ which I detect using a device with a Gaussian resolution function given by What is the frequency distribution of the detected signal function C(ω) given that ? Let’s find F(ω) first for the true signal … Let’s find G(ω) now for the resolution signal …

44. So if then ….. Example of convolution What is the frequency distribution of the detected signal function C(ω) given that ? Let’s find G(ω) now for the resolution signal … so We solved this in lecture 10 so let’s go straight to the answer and …..

45. Poisson’s equation Introduction to PDEs In many physical situations we encounter quantities which depend on two or more variables, for example the displacement of a string varies with space and time: y(x, t). Handing such functions mathematically involves partial differentiation and partial differential equations (PDEs). As Laplace but in regions containing mass, charge, sources of heat, etc. Electromagnetism, gravitation, hydrodynamics, heat flow. Laplace’s equation Heat flow, chemical diffusion, etc. Diffusion equation Quantum mechanics Schrödinger’s equation Elastic waves, sound waves, electromagnetic waves, etc. Wave equation

46. For such equations there is a fundamental theorem called the superposition principle, which states that if and are solutions of the equation then is also a solution, for any constants c 1, c 2. The principle of superposition The wave equation (and all PDEs which we will consider) is a linear equation, meaning that the dependent variable only appears to the 1st power. i.e. In x never appears as x 2 or x 3 etc. Waves and Quanta: The net amplitude caused by two or more waves traversing the same space (constructive or destructive interference), is the sum of the amplitudes which would have been produced by the individual waves separately. All are solutions to the wave equation. Can you think when you used this principle last year?? Electricity and Magnetism: Net voltage within a circuit is the sum of all smaller voltages, and both independently and combined they obey V=IR.

47. Find the solution to the wave equation to predict the displacement of the guitar string at any later time t The One-Dimensional Wave Equation A guitarist plucks a string of length L such that it is displaced from the equilibrium position as shown at t = 0 and then released. Let’s go thorugh the steps to solve the PDE for our specific case …..

48. Step 1: Separation of the Variables Since Y(x,t) is a function of both x and t, and x and t are independent of each other then the solutions will be of the form where the big X and T are functions of x and t respectively. Substituting this into the wave equation gives … Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t Step 2: Rearrange equation Rearrange the equation so all the terms in x are on one side and all the terms in t are on the other: The One-Dimensional Wave Equation

49. (i) (ii) Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 3: Equate to a constant Since we know that X(x) and T(t) are independent of each other, the only way this can be satisfied for all x and t is if both sides are equal to a constant: Suppose we call the constant N. Then we have which rearrange to … and (i) (ii)

50. Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 4: Decide based on situation if N is positive or negative We have ordinary differential equations for X(x) and T(t) which we can solve but the polarity of N affects the solution ….. Linear harmonic oscillator Unstable equilibrium Which case we have depends on whether our constant N is positive or negative. We need to make an appropriate choice for N by considering the physical situation, particularly the boundary conditions. If N is positive If N is negative Decide now whether we expect solutions of X(x) and T(t) to be exponential or trigonometric …..

51. Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 4: Continued ….. Remember that if N is negative, solutions will pass through zero displacement many times, whilst if N is positive solutions only tend to zero once. From this we deduce N must be negative. Let’s write So (i) becomes From lecture 3, this has general solution and in the same way and From before

52. Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 5: Solve for the boundary conditions for X(x) In our case the boundary conditions are Y(0, t) = Y(L, t) = 0. This means X(0) = X(L) = 0, i.e. X(x) is equal to zero at two different points. (This was crucial in determining the sign of N.) Now we apply the boundary conditions: X(0) = 0 gives A = 0. Saying B ≠ 0 then X(L) = 0 requires sin kL = 0, i.e. kL = n . So k can only take certain values where n is an integer for n = 1, 2, 3, …. So we have

53. From previous page Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 6: Solve for the boundary conditions for T(t) By standard trigonometric manipulation we can rewrite this as

54. Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 7: Write down the special solution for Y(x,t) Hence we have special solutions: We see that each Y n represents harmonic motion with a different wavelength (different frequency). In the diagram below of course time is fixed constant (as it’s a photo not a movie!!): (Mistake in notes – please correct harmonic numbers in diagram below)

55. This is the most general answer to the problem. For example, if a skipping rope was oscillated at both its fundamental frequency and its 3 rd harmonic, then the rope would look like the dashed line at some specific point in time and its displacement could be described just by the equation :- (mistake in notes at top of page 4, 3 rd not 2 nd harmonic) Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 8: Constructing the general solution for Y(x,t) We have special solutions: Bearing in mind the superposition principle, the general solution of our equation is the sum of all special solutions: NB. The Fourier series is a further example of the superposition principle.

56. Since Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation At t = 0 the string is at rest, i.e., if we differentiate we find For this to be true for all n and x, and this is only true if So the general solution becomes Step 8 continued: Constructing the general solution for Y(x,t)

57. The guitarist plucked the string of length L such that it was displaced from the equilibrium position as shown and then released at t = 0. This shape can therefore be represented by the half range sine (or cosine) series. Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 9: Use Fourier analysis to find values of B n Half-range sine series where It can be shown (see Phils Problems 5.10) that this shape can be represented by

58. and we see above that the coefficients B n are the coefficients of the Fourier series for the given initial configuration at t = 0. Therefore we can write the general solution at t = 0 as ….. Since Fourier series at t = 0 is Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 9 continued: Use Fourier series to find values of B n for and the general solution is then at t = 0 the general solution is

59. Hence, by trusting the superposition principle treating each harmonic as a separate oscillating sinusoidal waveform which is then summed together like a Fourier series to get the resulting shape, we deduce that at later times the configuration of the string will be:- The solution at t = 0 is Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation Step 10: Finding the full solution for all times But we also know that the general solution at all times is

60. Find the solution to the wave equation to predict the displacement of a guitar string of length L at any time t The One-Dimensional Wave Equation What does this all mean ???? This means that if you know the initial conditions and the PDE that defines the relationships between all variables, the full solution can be found which describes the shape at any later time. Want to know how heat passes down a rod, how light waves attenuate and interfere through a prism, how to define time dependent Schrodinger eigenfunctions, or how anything else that is a linear function with multiple variables interacts ???? Then this is what you should use.

61. SUMMARY of the procedure used to solve PDEs 9. The Fourier series can be used to find the full solution at all times. 1. We have an equation with supplied boundary conditions 2. We look for a solution of the form 3. We find that the variables ‘separate’ 4. We use the boundary conditions to deduce the polarity of N. e.g. 5. We use the boundary conditions further to find allowed values of k and hence X(x). 6. We find the corresponding solution of the equation for T(t). 7. We hence write down the special solutions. 8. By the principle of superposition, the general solution is the sum of all special solutions.. so

62. 3D Coordinate Systems 2. Integrals in 3D Cartesian Coordinates We have dV = dx dy dz, and must perform a triple integral over x, y and z. Normally we will only work in Cartesians if the region over which we are to integrate is cuboid. Example 1 : Find the 3D Fourier transform, The integral is just the product of three 1D integrals, and is thus easily evaluated: This is therefore a product of three sinc functions, i.e. if Just integrating over x gives and Mistake in notes

63. Polar Coordinate Systems 1. Spherical Polar Coordinates Spherical polars are the coordinate system of choice in almost all 3D problems. This is because most 3D objects are shaped more like spheres than cubes, e.g. atoms, nuclei, planets, etc. And many potentials (Coulomb, gravitational, etc.) depend on radius. Physicists define r,  as shown in the figure. They are related to Cartesian coordinates by:. 2. 3D Integrals in Spherical Polars The volume element is (given on data sheet). To cover over all space, we take Example 1 Show that a sphere of radius R has volume 4  R 3 /3. So

64. Polar Coordinate Systems 3.  2 in Spherical Polars: Spherical Solutions As given on the data sheet, (Spherically symmetric’ means that V is a function of r but not of  or .) Example 3 Find spherically symmetric solutions of Laplace’s Equation  2 V(r) = 0. Therefore we can say Really useful bit!!!! If (as in the homework) we were given an expression for V(r) and had to prove that it was a solution to the Laplace equation, then we’d just stick it here and start working outwards until we found the LHS was zero. If on the other hand we have to find V(r) then we have to integrate out the expression.

65. Polar Coordinate Systems Multiplying both sides by r 2 gives Integrating both sides gives where A is a constant. This rearranges to and so …. Integrating we get the general solution:. We’ve just done Q3(i) of the homework backwards!!! (see earlier note in red)

66. Extra tips for the exam When we write and say We mean that y = 5 when t = 0 When we say prove that is a solution of then to answer the question STICK IT IN BOTH SIDES When you solve the complementary solution of a 2 nd order differential equation, you need to know that the trial solution is ALWAYS You also need to know the different forms of the complementary solutions