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:
+q
+
+
+
+
vI
B
Lab Frame
+
+
+
+
+
+q
In Frame ofTest Charge
Lorentzexpanded
Lorentzcontracted
F(pure magnetic)
F(pure electrostatic)
Electricity & Magnetismare identically the sameforce, just viewed from different reference frames
UNIFICATION !!(thanks to Lorentz invariance)
Symmetry:The effect of a force looks the same whenviewed from reference frames boosted in the perpendicular direction
+q
+
+
+
+
vI
B
Lab Frame
+
+
+
+
+
+q
In Frame ofTest Charge
Lorentzexpanded
Lorentzcontracted
F(pure magnetic)
F(pure electrostatic)
Relativity !!
Symmetry ofMaxwell’s Equations
m1
m2
x1 x
2| |
F
If the force does not change when translated to a different point in space, then
¨ force felt at x2
: F = m2 x
2
Translational Invariance Conservation of Linear Momentum
Space-Time Symmetries
¨ recoil felt at x1 : F = m
1x
1
subtracting: m2 x
2 + m
1x
1 = 0 ¨¨
m2 v
2 + m
1 v
1 = constant
˙ ˙ d/dt [ m2 x
2 + m
1 x
1 ] = 0
E = m x2 + V12
dVdx
= 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 basicdescription also holds at other times
dE dV dxdt dx dt
= m x x +
Consider a system with total energy
Gauge Invariance in Electromagnetism:
A A + ∇(x,t) (x,t)t
E = ∇ A t
∇[x,t A + ∇(x,t)t
t
= ∇ A = Et
B = ∇ A ∇ [A + ∇(x,t)]
= ∇ A = B
Gauge Invariance Conservation of Charge
''local"symmetry
PoP !
So a charge could be created here by inputing energy E
x1
E = q(x1)
q
Thus we will have created an overall energy E E = q { (x2) (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 ofsome energy E
E = q(x2)
x2
Take the gauge transformation of a wavefunction to be eiq where is an arbitrary ''phase-shift" as a function of space and time
∇2 i 2m∂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: = Aei(px-Et)
Aei(pxEt+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 )
here’s the problem!
Special Relativity: Invariance with respect to reference framesmoving at constant velocity global symmetry
Generalize to allow velocity to vary arbitrarily at different pointsin space and time (i.e. acceleration) local gauge symmetry
Require an interaction to make this work
Another example...
GRAVITY!
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
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 !
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 mp = 938.3 MeV m
n = 939.6 MeV
So there appears to an ''approximate" symmetry here
IsospinBoth are found in the nucleus, apparently held to each other by pions
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 nI
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
Impose isospin conservation
I3(+) = +1
I3() = 0
I3() = -1
Example: What are the possible values of the isotopic spinand it’s z-component for the following systemsof particles:
a) + + p b) + p
a) + : I = 1, I3 = +1p : I = 1/2, I
3 = +1/2
+
p
n
total I3 = 1 + 1/2 = 3/2
thus, the only value of total Isospin we can have is also I = 3/2
b) : I = 1, I3 = 1p : I = 1/2, I
3 = +1/2
total I3 = 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
P F(x) = F(-x)
discreet symmetry (no conserved ''currents")
y
xm
1m
2
x
y
m1
m2
P x = x
P dx/dt = dx/dt
x
y
m1
m2
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
Also note that parity does not reverse the direction of spin!+z
z
z
+zflip z
z
+zflip ''velocity" direction
parity
+z
z(stand on your head)
+z
z
z
+zflip z
z
+zflip ''velocity" direction
parity
z
+zflip x & ypositions
+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!!
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(kxt)]
Thus, the intrinsic parity of the photon is 1 (or = 1 )
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!!
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)
L12
L3
1
L12
L
3
but final state must have zero total angularmomentum since the initial state has spin 0
Ltot
= L12
+ L3 = 0 L
12 = L
3
= ()3 (1)
L3{ }
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 + + 0and
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
But the observed magnetic moment is consistent with a superposition of only S and D waves (l=0, 2). This can be reconciled if
Parity is a different animal from other symmetries in many respects...
Weak interaction violates parity !!
(so, ''left" and ''right" really matter... weird!)
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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