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Topic 4: Special Relativity and Thomas Precession

Topic 4: Special Relativity and Thomas Precession

Lecture Series: The of the Matter, 4250, Fall 2010 1

Topic 4: and Thomas

Dr. Bill Pezzaglia CSUEB Physics Updated Nov 2, 2010

2

If you can't explain it simply, you don't understand it well enough.

Albert Einstein (1879-1955)

1905 Special Relativity

1 Index: Rough Draft 3

A. Lorentz Transformations

B. Addition

C. &

D. References

4 A. Lorentz Transformations

1. Special Relativity Effects

2.

3. form of LT

2 A1. Lorentz-FitzGerald Contraction 5

1889 FitzGerald, 1892 Lorentz

•Propose a moving meter stick will appear to shrink in length

L’ = L [1-v2/c 2]1/2

•1905 Einstein deduces this from his postulates of relativity.

A2. Dilation 6

Proposed 1897 by Larmor 1904 by Lorentz

• A moving clock will appear to run slower t t ′ = v 2 1 − c 2

Paradoxically, two observers each moving with a clock, sees the OTHER clock running slowly

3 7 A.3 Special Relativity

1905 Einstein postulates:

• is relative

• of is same for all observers

From these postulates he recovers Lorentz’s transformations

Lorentz Transformation (1904) 8

• 1904 Lorentz, 1905 Poincare & Einstein • Motion is equivalent to a hyperbolic in • Angle βββ is the “”: tanh βββ=v/c

ct ' cosh β sinh β 0 ct         x'  =  sinh β cosh β 0  x         y'   0 0 1  y 

1 cosh β = γ = 1 − v 2 / c 2

4 Clifford Algebra Form of Rotation 9

• Rotation of any quantity “Q” in “xy ” by angle φ can be expressed in exponential “half angle” form: Q'= R Q R−1

ˆ ˆ φ φ φ = e1e2 2/ = + R e cos 2 e1eˆˆ 2 sin 2

Note “spacelike” square negative: 2 = − (e1eˆˆ 2 ) 1

Clifford Algebra Form of LT 10

• Hyperbolic Rotation in “TX ” plane by rapidity angle β has same exponential “half angle” form:

Q'= L Q L−1

ˆ ˆ β β β = e4e1 2/ = + L e cosh 2 eˆ4eˆ1 sinh 2

Note “Timelike” bivectors square positive: (eˆ eˆ )2 = +1 because 4 1 2 = − (eˆ4 ) 1 2 = + (eˆ1) 1

5 Review 11

• Spacetime is Hyperbolic, not trigonometric

β = 1 ( β − −β ) sinh 2 e e β = 1 ()β + −β cosh 2 e e sinh tanh = cosh cosh 2 −sinh 2 = −1

B. Velocity Addition Formula 12

≠ + • don’t add: V V1 V2

β = β + β • DO add: 1 2

6 B.1 Addition of Velocities 13 • add: V=

• Adding anything to gives speed of light

Parallel Velocity Addition (Einstein 1905) 14

• Rapidity angles add, not velocities

tanh β + tanh β tanh ()β + β = 1 2 1 2 + β β 1 tanh 1 tanh 2 V tanh β = c V +V V = 1 2 VV 1+ 1 2 c2

7 Velocity Composition 15

• The general addition of velocities in two different directions is neither commutative or associative:

1  γ  ⊕ = + + 1 ()×()× u v u v 2 u u v  u •v c + λ 1+  1  c2 γ = 1 1− u2 c2

Thomas- 1939 16

• 1775 Euler: The composition of two is a single rotation.

• Wigner shows that in general the composition of two Lorentz transformations is a Lorentz transformation plus a rotation

• WHY isn’t it just a Lorentz transformation?

Spacelike bivectors are Timelike bivectors are not closed under multiplication closed under multiplication, they yield rotations ( )( ) = (e e )(e e ) = e e e1e2 e2e3 e1e3 4 1 4 2 1 2 ( )( ) = (e e )(e e ) = e e e2e3 e3e1 e2e1 4 2 4 3 2 3 ( )( ) = (e e )(e e ) = e e e3e1 e1e2 e3e2 4 3 4 1 3 1

8 Derivation using Half Angles 17

• I have not seen this done anywhere, but possibly Abraham A. Ungar has it in his book somewhere. αˆα 2/ βˆβ 2/ ˆγγ 2/ θˆθ 2/ • Want to show: e e = e e

• where { αβγ } are timelike bivectors for Lorentz αˆ 2 = βˆ 2 = γˆ2 = +1 • θ is spacelike of rotation θˆ2 = −1

• Expand:

α +α α β + βˆ β = γ +γ γ θ +θˆ θ (cosh 2 ˆ sinh 2 )(cosh 2 sinh 2 ) (cosh 2 ˆsinh 2 )(cos 2 sin 2 )

Solution of Wigner Rotation 18

• Solution is analogous to the Rodrigues formula!

α β • Net Velocity γ αˆ tanh + βˆ tanh γˆ tanh = 2 2 is symmetric! 2 + α • βˆ α β 1 ˆ tanh 2 tanh 2

αˆ ⊗ βˆ tanh α tanh β • Rotation θˆ tan θ = 2 2 2 + α • βˆ α β 1 ˆ tanh 2 tanh 2

9 19 C. Thomas Precession

Llewellyn Hilleth Thomas (1903-1992)

• 1926 proposes “Thomas Precession ” to explain the difference between predictions made by spin- orbit coupling theory and experimental observations.

Thomas Precession in Brief 20

• Acceleration perpendicular r r to velocity causes an extra ω ≈ 1 a ×v rotation of the reference th 2c2 frame, with :

• The change in any vector r r dV dV r r in lab frame is related to = +ω ×V dτ dτ th change in rotating (due to lab body accelerating) body frame by:

10 Thomas Precession in Orbits 21

• Atomic: After completes one closed orbit (in period “T”), there will be an “anomalous” net rotation of the (spin axis of the) electron due to the acceleration being perpendicular to the velocity.

v2 ∆θ ≈ ω T = 2π th c2

• This is also true for any circular orbit, for example a , which precesses at rate of 222 ° a day at Hayward, would have an additional Thomas precession due to rotation of earth on the order of: ∆θ ≈ 5.1 ×10 −11 rads / day = 3×10 −6 arc sec/ day

1. Fermi-Walker Transport 22

• vectors ek fixed to an e& = 1 (u ∧ a)• e accelerating NON-rotating k c2 k rigid body will change = 1 ()a u − u a according to: c2 k k – 4 velocity “u” – 4 acceleration: a = u&

• The change in any vector r r dV dV r in lab frame is related to = + 1 ()u∧a •V change in (accelerating) dτ dτ c2 body frame by: lab body

11 Clifford Algebra derivation 23

• Lorentz Transformation of e (t) = L e )0( L−1 basis vector (at time “t”) k j

• Operator for boost in βˆ β direction of velocity = 2 described by rapidity β L e

• I think you can show that the Fermi-Walker transport bivector is given: 1 u ∧ a ≡ 2L&R−1 = −2LL&−1 c2

Incomplete derivation 24

• Differentiating exponential operator I get:

− dβˆ dβˆ 2L&L 1 = β& βˆ + sinh β + ()1− cosh β βˆ ⊗ dτ dτ • The last term is the Thomas Precession

• I haven’t yet shown this gives exactly 1 u ∧ a ≡ 2L&R−1 = −2LL&−1 c2

12 Thomas Term from Fermi-Walker 25

• Expanding the 4-vectors one gets r r r ∧ ≈ γ 3 ∧ + γ 3 k u a v a ce 4ek a • The last term is simply the linear acceleration (timelike bivector)

• The first term is a spacelike bivector,r i.e. a rotationalr r angular velocity ω = 1 γ 3v × a th c2 • Ouch, not quite right. Should be: r γ 2 r r ω = 1 a ×v th c2 γ +1

Larmor Precession in Atomic Orbits 26

• An electron, with µµµ r r r which is moving with constant & = µ ×( − 1 × ) velocity in electromagnetic will S B 2 v E have a on it: c • In rest frame of , there is no r r , only . = This is proportional to the eE ma acceleration of the electron. Substitute this for “E” in above r e r • The magnetic moment is µ ≈ S proportional to the spin. Substitute this for µµµ m r r • We get precession rate: × ω = + v a Larmor c2 & = ω × S Larmor S

13 Larmor plus Thomas 27

r r • Note the is v × a ω = + = −2ω just double the Thomas, and Larmor c2 th with opposite sign.

• The total precession is hence the algebraic sum r r v × a ω = ω +ω = 1 net Larmor th 2 c2 • Thus, observed precession is half the expected Larmor precession (“The Thomas half”)

References 28

• Baylis, Electrodyanmics, has a coherent description of Thomas Precession, using 3D Clifford algebra (i.e. Pauli Algebra). The Wigner rotation is derived in equation (4.29). Thomas Precession in section 4.6. • L. H. Thomas, "Motion of the spinning electron", Nature 117, 514, (1926) • E. P. Wigner, "On unitary representations of the inhomogeneous Lorentz ", Ann. Math. 40, 149–204 (1939). • Tomonaga, The Story of Spin, Lecture 2 and 11.

14 References 29

• Foucault http://en.wikipedia.org/wiki/Foucault_pendulum • Thomas: http://www.columbia.edu/acis/history/thomas.html • Abraham A. Ungar, Beyond the Einstein Addition Law and Its Gyroscopic Thomas Precession

30 Hmk Ideas

• Calculate the Thomas precession rate for the moon. How many years would it take to have the side of the moon facing us change by 180 degrees? • Calculate the Thomas precession rate for a point on the equator of the earth • Does the Kimbal experiment measure Thomas precession due to earth’s rotation? (this is separate from the Foucault precession)

15 31 Things to do

• Obviously need to clean up the derivation of Thomas precession term.

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