Physics 7C Spring 2015 Discussion Section Notes

Physics 7C Spring 2015 Discussion Section Notes Kevin T. Grosvenora;b aBerkeley Center for Theoretical Physics and Department of Physics University of California, Berkeley, CA, 94720-7300, USA bTheoretical Physics Group, Lawrence Berkeley National Laboratory Berkeley, CA 94720-8162, USA Abstract: Some discussion section notes for Physics 7C. Contents 1. Vectors 1 2. The Wave Equation5 3. Solving the Wave Equation7 3.1. Electromagnetic Plane Waves 10 4. Poynting Vector and Flux 11 4.1. Red Laser Pointer 12 5. Ray Tracing Diagrams for Mirrors 13 6. Ray Tracing Diagrams for Lenses 14 7. Compound Optical Systems 16 7.1. Two-Lens Problem 16 7.2. Two-Lens Demonstration 17 8. Midterm 1 Quiz 22 9. Interference 25 9.1. Laser Wavelength Measurement via Metal Ruler 25 10.Thin-Film Interference 27 11.Relativity 28 11.1. How to Measure the Length of a Moving Object 28 11.2. Relativistic Train 30 11.3. Passing Trains 32 12.Midterm 2 Quiz 36 13.Energy and Momentum 40 13.1. 4-Vectors 40 13.2. Colliding Photons 42 14.Quantum Mechanics 45 14.1. The Wacky World of the Double Slit 45 14.2. Blackbody Radiation and the Ultraviolet Catastrophe 47 14.3. Stephan-Boltzmann Law 48 14.4. Bohr Model 49 14.5. Time-Evolution in 1D Infinite Square Well 53 { i { 15.Final Review 57 15.1. Human Eye Optics 57 15.2. Optical Fiber 58 15.3. Modified Michelson Interferometer 59 15.4. Diffraction Grating 61 15.5. Optical Spectroscopy 62 15.6. Relativity and Current-Carrying Wires 64 15.7. Pi Decay 66 15.8. Relativistic Doppler Effect 69 15.9. Quantum Tunneling and Frustrated Total Internal Reflection 70 15.10.Wavefunction Shapes 73 16.Final Exam Solutions 75 16.1. The Pole Vaulter Paradox 75 16.2. Pion Decay 77 1. Vectors For our purposes, a vector will be something that has several components (usually three; or four in relativity). We must be able to add two vectors (component by component) and we must be able to multiply a vector by a real number. There are a slew of other requirements, but they are usually trivially satisfied, at least for the main vector space we will care about: R3, or Rn for general dimension. The symbol Rn means the set of all n-component expressions, A = (A1;A2;:::;An), such that each component is a real number (i.e. Ai 2 R for i = 1; : : : ; n.) In three-dimensional space, let us replace x; y; z with x1; x2; x3 in order to make it easier to generalize to any dimension. We often denote the unit vector in the direction of th xi by xî, whose components are all zero except for the i one, which is a one. Then, A may be written, in a general dimension, n, n X A = Aixî: (1.1) i=1 So that we don't have to keep writing summation signs everywhere, we will usually fol- low Einstein's convention that repeated indices are summed over, unless stated otherwise. Then, (1.1) becomes neater: A = Aixî: (1.2) The dimensionality is nowhere to be found now, so you must make sure you know what it is from context. { 1 { Next, we introduce the Kronecker delta symbol: ( 1 if i = j; δij = (1.3) 0 if i 6= j: Rn has what's called an inner product structure. This is a map (·; ·): Rn ×Rn ! R. That is, you take two vectors, put one in the first slot and the other in the second slot of (·; ·), and you will get a real number, called their inner product. It is also often called their dot product, especially in three dimensions, and we will denote it by A · B. It is defined by A · B = δijAiBj = AiBi; (1.4) where you need to keep in mind that repeated indices are summed over. In addition, R3 has a special structure called a cross product. This is given by a map (· × ·): Rn × Rn ! Rn, so it takes two vectors and spits out another vector.1 In order to talk about cross products, we must introduce the Levi-Civita symbol, and to do that, we must understand cyclic indices. There is a mnemonic for this. Think of a clock that goes from 1 through 3 instead of 1 through 12. Starting at any of the numbers, if you traverse the clock in a clockwise fashion, then the order is declared to be cyclic. If you traverse the clock in the counter-clockwise direction, then the order is anti-cyclic. So, (123), (231) and (312) are cyclic, whereas (132), (213), (321) are anti-cyclic. By the way, and for example, (231) is the permutation that sends 1 (the first slot) to 2 (the first number appearing), 2 to 3 and 3 to 1. The Levi-Civita symbol is defined to be 8 > 1 if (ijk) is cyclic; <> ijk = −1 if (ijk) is anti-cyclic; (1.5) > :> 0 if any index is repeated. Roughly speaking, the Levi-Civita symbol is to the cross product what the Kronecker delta is to the dot product: A × B = ijkxîAjBk: (1.6) Let us take a moment to ensure that this definition of the cross-product agrees with the definition of the cross-product that we are likely to have learned before, namely A × B = (AyBz − AzBy)x^ + (AzBx − AxBz)y^ + (AxBy − AyBx)z^: (1.7) To aid in comparison, let us first rewrite (1.7) in terms of our new notation where x^ is replaced with x^1, and y^ with x^2 and so on. Also, the x-component is the 1-component, the y-component is the 2-component and so on. Then, A × B = (A2B3 − A3B2)x^1 + (A3B1 − A1B3)x^2 + (A1B2 − A2B1)x^3: (1.8) 1Actually, the result of a cross product is what's called a pseudovector, but no matter. { 2 { Now, let us expand out the right hand side of (1.6) to see that it is in fact the same as the right hand side of (1.8): ijkxîAjBk = 123x^1A2B3 + 132x^1A3B2 + 231x^2A3B1 + 213x^2A1B3 + 312x^3A1B2 + 321x^3A2B1 = x^1A2B3 − x^1A3B2 + x^2A3B1 − x^2A1B3 + x^3A1B2 − x^3A2B1: (1.9) Here, we used 123 = 231 = 312 = 1 and 132 = 213 = 312 = −1. Now, it is quite easy to see that (1.8) is the same as (1.9) just organized slightly more neatly. Okay, so far all we have done is introduce a bunch of notation in order to express the dot product and the cross product more compactly. However, this notation is actually useful once we have to deal with multiple products (like multiple cross-products). For such purposes, the following identity is very useful: ijki`m = δj`δkm − δjmδk`: (1.10) This identity is actually reasonably easy to understand. Remember that i, j and k have to take on different values or else the Levi-Civita symbol would be zero anyway (also i, ` and m have to take on different values). Well, since they can only take on three different values, namely 1, 2 or 3, either j is equal to ` and k is equal to m or ` is equal to m and k is equal to `. There just aren't any other possibilities! That's what the right hand side of the equation says, except for the signs, which you can figure out by plugging in some specific set of values for the indices, say i = 1, j = ` = 2 and k = m = 3, and then try i = 1, j = m = 2 and k = ` = 3. In fact, this can be extended to any dimension. In n + 1 dimensions, we write δ δ ··· δ i1j1 i1j2 i1jn δ δ ··· δ i2j1 i2j2 i2jn ii1···in ij1···jn = . ; (1.11) . .. δinj1 δinj2 ··· δinjn where the vertical lines surrounding the matrix means \take the determinant". The identity (1.7) will allow you to deal with situations involving multiple cross products. A number of very important vector identities can be proven using these formulas. For example, let us prove the \BAC-CAB" rule: A × (B × C) = B(A · C) − C(A · B): (1.12) Of course, you could prove this identity component by component by literally expanding both sides out completely. It will take some time and is tedious, but should be pretty straightforward. On the other hand, it is quite easy to prove in our new notation. Define D = B × C = ijkxîBjCk: (1.13) Of course, as in (1.2), we can write D in terms of its components: D = Dixî: (1.14) { 3 { Comparing (1.13) with (1.14) gives us the components of D in terms of the components of B and C: Di = ijkBjCk: (1.15) Getting back to the left hand side of (1.12), we have A × (B × C) = A × D = `mix^ÀmDi: (1.16) Notice that I have used `; m; i instead of the customary i; j; k. These are just dummy indices, so it does not matter what you call them. The reason why I have written i last is because that index is the same as the index on D and I have already written Di in (1.15), so might as well keep that index as i. The reason why I have used ` and m instead of j and k is because j and k already appear in (1.15), so I don't want to use them again or else I will get confused as to which pair of i's are supposed to be summed over and, similarly, which pairs of j's are supposed to be summed over.

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