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Classical Physics

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General Relativity

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Accelerated Reference Frames - Rindler Coordinates
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Particle Physics

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Quantum Field Theory

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Semiconductor Reliability

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Solid State Electronics

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Special Relativity

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4-vectors
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Statistical Mechanics

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String Theory

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Bosonic Strings
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Introduction to String Theory
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Superconductivity

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Supersymmetry (SUSY) and Grand Unified Theory (GUT)

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Chiral Superfields
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Introduction to Supersymmetry
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The Gauge Hierarchy Problem

test

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test

The Standard Model

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Electroweak Unification (Glashow-Weinberg-Salam)
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Gauge Theories (Yang-Mills)
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Gravitational Force and the Planck Scale
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Introduction to the Standard Model
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Isospin, Hypercharge, Weak Isospin and Weak Hypercharge
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Quantum Flavordynamics and Quantum Chromodynamics
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Special Unitary Groups and the Standard Model - Part 1
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Special Unitary Groups and the Standard Model - Part 2
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Special Unitary Groups and the Standard Model - Part 3
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Standard Model Lagrangian
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The Higgs Mechanism
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The Nature of the Weak Interaction

Topology

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Units, Constants and Useful Formulas

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Constants
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Formulas
Last modified: April 26, 2019

Building Groups From Other Groups --------------------------------- Quotient Groups --------------- One way to build finite groups is to perform some kind operation involving smaller simpler groups. Before we go any further we need to discuss the idea of equivancy relations. Equivalence Relations on Sets ----------------------------- Basic idea: If X is the set of all cars, and ~ is the equivalence relation "has the same color as", then one particular equivalence class consists of all green cars. X/~ could be naturally identified with the set of all car colors. An equivalence relation, ~, from a set, S, to a set, T, is a subset of S x T. An equivalence relation, ~, on a set, S, is a subset of S x S. Where ~ is interpreted as 'equivalent to' and not 'approximately equal to'. ~ has to be reflexive, symmetric and transitive. That is, for si ∈ subset: si ~ si - reflexive si ~ sj -> sj ~ si - symmetric si ~ sj and si ~ sk -> si ~ sk - transitive Equivalency Classes ------------------- The equivalence class of an element a is denoted by [a] and is defined as the set: [a] = {s ∈ S|s ~ a} Non-overlapping (disjoint) equivalence classes form a partition of S where elements in the same partition are related. The set of all equivalence classes in S with respect to an equivalence relation ~ is denoted as S/~ and is called S modulo ~ (or the quotient set of S by ~). Example 1: ℤ is the set of all integers. Let m ~ n i.f.f. m - n is divisble by 3 for any m,n ∈ ℤ. Reflexive because m - m = 0 which is divisible by 3. Symmetric because if m - n is divisible by 3 then n - m = -(m - n) is also divisible by 3. Transitive because m - n = 3a and n - p = 3b (a and b are integers) so m - p = (m - n) + (n -p) = 3(a + b). Thus, m ~ p. The equivalency classes are: [0] = {0,3,6,...} [1] = {1,4,7,...} [2] = {2,5,8,...} [3] = {3,6,9,...} [4] = {4,7,10,...} ℤ/~ = {[0],[1],[2],...} Example 2: {(1,1),(2,2),(3,3),(2,3),(3,2)} Reflexive because (1,1), (2,2), (3,3) are present. Symmetric because (2,3), (3,2) are present. Transitive because (2,3), (3,2) -> (2,2) (3,2),(2,3) -> (3,3) The equivalency classes are: [1] = {1} [2] = {2,3} [3] = {3,2} S/~ = {[1],[2]} = {{1},{2,3}} Equivalence Relations on Groups ------------------------------- Define a relation ~ on G by saying a ~ b if and only if there exists g ∈ G such that: Multiplicative: a = gbg-1 Additive: a = g + b + (-g) Proof that ~ is an equivalence relation on G: Reflexive: a = eae-1 a ~ a Symmetric: a = gbg-1 g-1a = bg-1 g-1ag = b g-1a(g-1)-1 = b x-1 ∈ G Therefore a ~ b implies b ~ a Tramsitive: a = gbg-1 and b = hch-1 a = g(hch-1)g-1 = ghch-1g-1 = ghc(gh)-1 gh ∈ G Therefore a ~ c a ~ b are said to be CONJUGATE and, therefore, conjugacy is an equivalence relation on a group. [a] = {b ~ a|b ∈ G and b = gag-1} [a] is called the CONJUGACY CLASS. Normal Subgroup --------------- The conjugacy class containing e is a normal subgroup, if: multiplicative: gng-1 ∈ N for n ∈ N, g ∈ G or, gn = ng after multiplying g from the right additive: g + n + (-g) ∈ N for n ∈ N, g ∈ G or, g + n = n + g Note that e = 1 in the multiplicative case and e = 0 in the additive case (i.e. 0-1 is undefined). In the multiplicative case if G is abelean then, gg-1n = n ∈ N We can also look at the above in a reverse manner. Let G be a group and N be subgroup of G. To be a subgroup N must contain the identity, e. Then, a ~ b is an equivalence relation if and only if: Multiplicative case: ab-1 = e Additive case: a + (-b) = e Where a, b ∈ G and e ∈ N Proof: Reflexive: aa-1 = e ∈ N a ~ a Symmetric: (ab-1)-1 = e-1 (b-1)-1a-1 = e-1 ba-1 = e-1 b ~ a Tramsitive: ab-1 = e and bc-1 = e2 (ab-1)(bc-1) = e2 ac-1 = e2 a ~ c The equivalence class of e, is a normal subgroup N of G and the other equivalence classes are the cosets of N It is important to note that cosets are not subgroups. In summary, the elements of any group may be partitioned into conjugacy classes. Members of the same conjugacy class share many properties, and study of conjugacy classes of non-abelian groups reveals many important features of their structure. For an abelian group, each conjugacy class is a set containing one element. Congruency Relations on Groups ------------------------------ An equivalence relation is just a relation which is reflexive, transitive, and symmetric. A congruence relation is a little bit more than an equivalence relation in that it respects some structure such as a group. In this case we add the additional 2 axioms. a1 ≅ a2 (mod n), b1 ≅ b2 (mod n) Multiplicative: a1b1 ≅ a2b2 (mod n) Additive: a1 + b1 ≅ a2 + b2 (mod n) For all a1, a2, b1, b2 ∈ G. Example: a1 = 1, a2 = 4, b1 = 2, b2 = 5 1 ≅ 4 (mod 3) and 2 ≅ 5 (mod 3) Therefore, (1 + 2) ≅ (4 + 5) (mod 3) and, (1 x 2) = (4 x 5) (mod 3) Now that we have discussed equivalence relations on groups we are in a position to discuss quotient groups. These best way to do this is to use an example. Example: G = (ℤ,+) ℤ = {0,1,-1,2,-2,3,-3,...} N = 3ℤ = {3z|z ∈ ℤ} ℤ: --o---o---o---o---o---o---o---o---o---o---o---o---o-- -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 3ℤ: --o-----------o-----------o-----------o-----------o-- -6 -3 0 3 6 1 + 3ℤ = {1 + x|x ∈ 3ℤ}: ------o-----------o-----------o-----------o---------- -5 -2 1 4 2 + 3ℤ = {2 + x|x ∈ 3ℤ}: ----------o-----------o-----------o-----------o------ -4 -1 2 5 and so on. We can now pick a representative element from each coset and name the coset after it. [0] = {0,3,6} [1] = {1,4,7} [2] = {2,5,8} [0] contains the identity. Therefore, if we pick g = 2 and n = 1 we get: 2 + 1 + (-2) = 1 which proves that [0] is a normal subgroup. In summary, in a quotient group, the equivalence class of the identity element is always a normal subgroup of the original group, and the other equivalence classes are the cosets of that normal subgroup. Together they form a partition of G written as G/N, where G is the original group and N is the normal subgroup. This is also written as G/~. Direct Product of Groups ------------------------ The direct product is another way to build finite groups from smaller simpler groups. Consider 2 groups G and H. The direct product is defined as: G x H = A In terms of elements this is: (g1,h1) x (g2,h2) = (g1◇g2,h1□h2) Where ◇ is the operation on G and □ is the operation on H. The order of A is: |A| =|G| x |H| G and H do not have to be the same size. This operation is the analogue of the Cartesian product of sets and is one of several important notions of direct product in mathematics. Example 1: G = ℤ mod 2 under + = {0,1} H = ℤ mod 2 under + The elements are given by (not a Cayley table). | 0 | 1 --+----+--- 0 | 00 | 01 --+----+--- 1 | 10 | 11 The Cayley table is: + | 00 | 11 | 10 | 01 ----+----+----+----+---- 00 | 00 | 11 | 10 | 01 ----+----+----+----+---- 11 | 11 | 00 | 01 | 10 ----+----+----+----+---- 10 | 10 | 01 | 00 | 11 ----+----+----+----+---- 01 | 01 | 10 | 11 | 00 It is important to recognize the mod of each group in caculating the table entries. For example, the blue cell is: (1,0) x (1,0) = ((1 + 1),(0 + 0) = (2,0) However, 2 (mod 2) = 0 If we let 1 = 11, 0 = 00, 1 = 10 and 2 = 01 we get: + | 0 | 1 | 2 | 3 ---+---+---+---+---- 0 | 0 | 1 | 2 | 3 ---+---+---+---+---- 1 | 1 | 0 | 3 | 2 ---+---+---+---+---- 2 | 2 | 3 | 0 | 1 ---+---+---+---+---- 3 | 3 | 2 | 1 | 0 This is the Klein 4 group. Therefore Z2 x Z2 is isomorphic to the Klein 4 group. Example 2: G = ℤ2 under + H = ℤ3 under + = {0,1,2} The elements are given by (not a Cayley table). | 0 | 1 | 2 --+----+----+--- 0 | 00 | 01 | 02 --+----+----+--- 1 | 10 | 11 | 12 The Cayley table is: + | 00 | 11 | 02 | 10 | 01 | 12 ----+----+----+----+----+----+---- 00 | 00 | 11 | 02 | 10 | 01 | 12 ----+----+----+----+----+----+---- 11 | 11 | 02 | 10 | 01 | 12 | 00 ----+----+----+----+----+----+---- 02 | 02 | 10 | 01 | 12 | 00 | 11 ----+----+----+----+----+----+---- 10 | 10 | 01 | 12 | 00 | 11 | 02 ----+----+----+----+----+----+---- 01 | 01 | 12 | 00 | 11 | 02 | 10 ----+----+----+----+----+----+---- 12 | 12 | 00 | 11 | 02 | 10 | 01 Again, it is important recognize the mod of each group in caculating the table entries. For example, the blue cell is (1,2) x (1,2) = ((1 + 1),(2 + 2) = (2,4) The first number is subject to mod 2 and the second number is subject to mod 3. Therefore: (2,4) -> (0,1) Let 0 = 00, 1 = 11, 2 = 02, 3 = 10, 4 = 01 and 5 = 12. This leads to: + | 0 | 1 | 2 | 3 | 4 | 5 ---+---+---+---+---+---+--- 0 | 0 | 1 | 2 | 3 | 4 | 5 ---+---+---+---+---+---+--- 1 | 1 | 2 | 3 | 4 | 5 | 0 ---+---+---+---+---+---+--- 2 | 2 | 3 | 4 | 5 | 0 | 1 ---+---+---+---+---+---+--- 3 | 3 | 4 | 5 | 0 | 1 | 2 ---+---+---+---+---+---+--- 4 | 4 | 5 | 0 | 1 | 2 | 3 ---+---+---+---+---+---+--- 5 | 5 | 0 | 1 | 2 | 3 | 4 This is the ℤ mod 6 group. There Z mod 2 x Z mod 3 is isomorphic to the Z mod 6 group. If each group is an abelian group, as is the case here, then we can refer to G x H as G ⊕ H. However, this is simply a matter of notation - the concepts are always the same regardless of whether we use additive or multiplicative notation. Since both G and H in the above case are abelian we can also write: G x H ≡ G ⊕ H Example 3: G = ℤ2 under + H = {1,-1,i,-i) under x The elements are given by (not a Cayley table). | 1 | -1 | i | -i ----+-----+------+-----+------ 0 | 0,1 | 0,-1 | 0,i | 0,-i ----+-----+------+-----+------ 1 | 1,1 | 1,-1 | 1,i | 1,-i The Cayley table is: x | 0,1 | 0,-1 | 0,i | 0,-i | 1,1 | 1,-1 | 1,i | 1,-i ------+------+------+------+------+------+------+------+------ 0,1 | 0,1 | 0,-1 | 0,i | 0,-i | 1,1 | 1,-1 | 1,i | 1,-i ------+------+------+------+------+------+------+------+------ 0,-1 | 0,-1 | 0,1 | 0,-i | 0,i | 1,-1 | 1,1 | 1,-i | 1,i ------+------+------+------+------+------+------+------+------ 0,i | 0,i | 0,-i | 0,-1 | 0,1 | 1,i | 1,-i | 1,-1 | 1,1 ------+------+------+------+------+------+------+------+------ 0,-i | 0,-i | 0,i | 0,1 | 0,-1 | 1,-i | 1,i | 1,1 | 1,-1 ------+------+------+------+------+------+------+------+------ 1,1 | 1,1 | 1,-1 | 1,i | 1,-i | 0,1 | 0,-1 | 0,i | 0,-1 ------+------+------+------+------+------+------+------+------ 0,-i | 0,-i | 0,i | 0,1 | 0,-i | 1,-i | 1,i | 1,1 | 1,-1 ------+------+------+------+------+------+------+------+------ 1,1 | 1,1 | 1,-1 | 1,i | 1,-i | 0,1 | 0,-1 | 0,i | 0,-1 ------+------+------+------+------+------+------+------+------ 1,-1 | 1,-1 | 1,1 | 1,-i | 1,i | 0,-1 | 0,1 | 0,-i | 0,i ------+------+------+------+------+------+------+------+------ 1,i | 1,i | 1,-i | 1,-1 | 1,1 | 0,i | 0,-i | 0,-1 | 0,1 ------+------+------+------+------+------+------+------+------ 1,-i | 1,-i | 1,i | 1,1 | 1,-1 | 0,-i | 0,i | 0,1 | 0,-1 ------+------+------+------+------+------+------+------+------ This time H has no mod so only the first entry is involved.