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Lecture 6: Symmetries I Symmetry & Unifying Electricity/Magnetism Space-Time Symmetries Gauge Invariance in Electromagnetism Noether’s Theorem Isospin.

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Presentation on theme: "Lecture 6: Symmetries I Symmetry & Unifying Electricity/Magnetism Space-Time Symmetries Gauge Invariance in Electromagnetism Noether’s Theorem Isospin."— Presentation transcript:

1 Lecture 6: Symmetries I Symmetry & Unifying Electricity/Magnetism Space-Time Symmetries Gauge Invariance in Electromagnetism Noether’s Theorem Isospin Parity Sections 5.3, 6.1, Append C.1, C.2 Useful Sections in Martin & Shaw:

2 Symmetry something else is unchanged When something is changed,

3 +q +  +  +  +  v I B Lab Frame +    + + + + +q In Frame of Test Charge Lorentz expanded Lorentz contracted F (pure magnetic) F (pure electrostatic)  Electricity & Magnetism are identically the same force, just viewed from different reference frames UNIFICATION !! (thanks to Lorentz invariance)  Symmetry & Electromagnetism Symmetry: The effect of a force looks the same when viewed from reference frames boosted in the perpendicular direction

4 +q +  +  +  +  v I B Lab Frame +    + + + + +q In Frame of Test Charge Lorentz expanded Lorentz contracted F (pure magnetic) F (pure electrostatic) Relativity !!  Symmetry & Electromagnetism Symmetry of Maxwell’s Equations

5 m1m1 m2m2 x 1 x 2 | F If the force does not change when translated to a different point in space, then ¨ force felt at x 2 : F = m 2 x 2 Translational Invariance  Conservation of Linear Momentum Translational Invariance Space-Time Symmetries ¨ recoil felt at x 1 : F =  m 1 x 1 subtracting: m 2 x 2 + m 1 x 1 = 0 ¨¨ m 2 v 2 + m 1 v 1 = constant ˙ ˙ d/dt [ m 2 x 2 + m 1 x 1 ] = 0

6 E = m x 2 + V 1212 dV dx = m x x + x but dV/dx =  F =  mx (Newton’s 2nd law) dE dt = m x x  m x x = 0  E = constant  Time Invariance  Conservation of Energy assume this basic description also holds at other times dE dV dx dt dx dt = m x x + Consider a system with total energy Time Invariance

7 Gauge Invariance in Electromagnetism: A  A + ∇  (x,t)      (x,t)  tt E =  ∇   A  tt   ∇ [  x,t   A + ∇  (x,t)   tt  tt =  ∇   A = E  tt B = ∇  A  ∇  [A + ∇  (x,t)] = ∇  A = B Gauge Invariance  Conservation of Charge Gauge Invariance in EM ''local" symmetry

8 PoP ! So a charge could be created here by inputing energy E x1x1 E =  q  (x 1 ) q Thus we will have created an overall energy E  E =  q {  (x 2 )   (x 1 ) } So, to preserve energy conservation, if  is allowed to vary as a function of position, charge must be conserved To see this, assume charge were not conserved (Wigner, 1949) PoP ! And destroyed here, with the output of some energy E E =  q  (x 2 ) x2x2 Necessity of Charge Conservation

9 Continuous Symmetries  Conserved ''Currents" (Emmy Noether, 1917) Noether’s Theorem

10 Take the gauge transformation of a wavefunction to be   e iq   where  is an arbitrary ''phase-shift" as a function of space and time ∇ 2  i 2m ∂t∂t ∂   Say we want the Schrodinger equation to be invariant under such a transformation clearly we’re in trouble ! Consider the time-derivative for a simple plane wave:  = Ae i(px-Et)   Ae i(px  Et+q  )  /  t  = i (  E + q  /  t )  Note that if we now introduce an electric field, the energy level gets shifted by  q  But we can transform     /  t, thus cancelling the offending term! (a similar argument holds for the spatial derivative and the vector potential) Gauge invariance REQUIRES Electromagnetism !! Gauge symmetry from another angle...  /  t  = i (  E +  q  + q  /  t )  Importance of Gauge Invariance here’s the problem!

11 Special Relativity: Invariance with respect to reference frames moving at constant velocity  global symmetry Generalize to allow velocity to vary arbitrarily at different points in space and time (i.e. acceleration)  local gauge symmetry Require an interaction to make this work Another example... GRAVITY! Gauge Invariance and Gravity

12 All known forces in nature are consequences of an underlying gauge symmetry !! Gauge symmetries are found to result from all the known forces in nature !! or perhaps Chicken or egg ?

13 Pragmatism : Symmetries (and asymmetries) in nature are often clear and can thus be useful in leading to dynamical descriptions of fundamental processes True even for ''approximate" symmetries ! Symmetry and pragmatism

14 n p )  n p proton & neutron appear to be swapping identities  ''exchange force" This means, to conserve charge, pions must come in 3 types: q = -1, 0, +1 Note that m p = 938.3 MeV m n = 939.6 MeV So there appears to an ''approximate" symmetry here Isospin Both are found in the nucleus, apparently held to each other by pions

15 Noether’s theorem says something must be conserved... Call this ''Isospin" in analogy with normal spin, so the neutron is just a ''flipped" version of the proton p n I 3 : 1/2 -1/2 I=1/2 System  (just 2 states) Some way pions can be produced p + p  p + n +  +  p + p +    p + p +   +    I=1 for the pions (similar arguments for other particle systems) So we can think of these particle ''states" as the result of a (continuous) ''rotation" in isospin-space Assignment of Isospin Impose isospin conservation I 3 (  + ) = +1 I 3 (   ) = 0 I 3 (   ) = -1

16 Example: What are the possible values of the isotopic spin and it’s z-component for the following systems of particles: a)  + + p b)   + p a)  + : I = 1, I 3 = +1  p : I = 1/2, I 3 = +1/2 ++   pnpn total I 3 = 1 + 1/2 = 3/2 thus, the only value of total Isospin we can have is also I = 3/2 b)   : I = 1, I 3 =  1  p : I = 1/2, I 3 = +1/2 total I 3 =  1 + 1/2 =  1/2 thus, possible values of total Isospin are: I = 1  1/2 = 1/2 or I = 1  1/2 = 3/2 Isospin Example

17 P F(x) = F(-x)  discreet symmetry (no conserved ''currents") y x m1m1 m2m2 x y m1m1 m2m2 P x =  x P dx/dt =  dx/dt x y m1m1 m2m2 consider the scattering probability of the following: Parity So, even though x and dx/dt are each odd under parity, the scattering probability, P S, is even (i.e. P P S = P S )  parity is multiplicative, not additive

18 Also note that parity does not reverse the direction of spin! +z zz zz flip z zz +z flip ''velocity" direction parity +z zz (stand on your head) More on Parity +z zz zz flip z zz +z flip ''velocity" direction parity zz +z flip x & y positions +z zz (stand on your head) not the same ! But, for orbital angular momentum in a system of particles, it depends on the symmetry of the spatial wave function!!

19 Intrinsic parity of the photon from ''first-principles": ∇  E(x,t) =  (x,t)/  0 P  (x,t) =  (  x,t) P ∇ =  ∇ But also E =  ∇    A/  t =   A/  t (in absence of free charges) thus, we must have P E(x,t) =  E(-x,t) for Poisson’s equation to remain invariant and since  /  t doesn’t change the parity  P A(x,t) =  A(  x,t) But A basically corresponds to the photon wave function: A(x,t) = N  (k) exp[i(k  x  t)] Thus, the intrinsic parity of the photon is  1 (or   =  1 ) Intrinsic Parity of the Photon

20 However, the effective parity depends on the angular momentum carried away by the photon from the system which produced it: P  =   (  1) l (i.e. radiation could be s-wave, p-wave etc.) but for an isolated photon, this cannot be disentangled!! Effective Parity of the Photon

21 Example: From this, deduce the intrinsic parity of the  and explain why the decays:    0 +  0     +   and are never seen P  = P  P  P    = (   ) 3 (  1) (  1) L 12 L 3  11 L 12  L3L3 but final state must have zero total angular momentum since the initial state has spin 0 L tot = L 12 + L 3 = 0  L 12 = L 3   = (   ) 3 (  1) L3L3 { } 2 = (   ) 3 = (   ) 3 =  1 However, for 2-pion final states we would have: P = (   ) 2 (  1) L but we must have L=0, so P = (   ) 2 = 1 and is thus forbidden The  (547) meson has spin 0 and is observed to decay via the electromagnetic interaction through the channels:    0 +  0 +  0     +   +  0 and Parity Example

22 1) Polarized protons scattering off a nucleus show no obvious asymmetry towards spin-up vs spin-down directions 2) Ground state of deuteron (np) has total angular momentum J=1 and spin S=1. Thus, the orbital angular momentum could take on values of l = 0 (m=1), l = 1 (m=0) or l = 2 (m =  1) P (p+n) = P (d)  p  n =  p  n (  1) l  so l must be even (in strong/electromagnetic interactions) By convention,  p =  n  +1 & also  e-  +1  and the relative parities of the other particles then follow (anti-fermions have the opposite parity, anti-bosons have the same parity) Evidence for Parity Conservation Parity Conservation & Assignment But the observed magnetic moment is consistent with a superposition of only S and D waves ( l =0, 2). This can be reconciled if

23 Parity is a different animal from other symmetries in many respects... Weak interaction violates parity !! (so, ''left" and ''right" really matter... weird!) Basic Approach to Parity It is often impossible to determine the absolute parity (assigned +1 or -1) of many particles or classes of particles. So we essentially just assume that basic physical processes are invariant with respect to parity and construct theories accordingly, making arbitrary assignments of parity when necessary until we run into trouble.


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