General Lorentz Transformation Consider a Lorentz Transformation, with arbitrary v, : ct´ = γ(ct ...

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Transcript of General Lorentz Transformation Consider a Lorentz Transformation, with arbitrary v, : ct´ = γ(ct ...
General Lorentz Transformation• Consider a Lorentz Transformation, with arbitrary v, :
ct´ = γ(ct  βr) r´ = r + β2(βr)(γ 1)β  γctβ • Transformation matrix in 4d spacetime: x´ Lx
• Can write L as: L = RL0 = L0´R´. Here R rotation matrix, as in Ch. 4. L0 Pure “boost” or pure velocity transformation. For example: (v  x): γ βγ 0 0
L0 = γβ γ 0 0 0 0 1 0 0 0 0 1
• For any Lorentz transformation x´ Lx, there exists an Inverse Transformation L1:
L1L = LL1 = 1, x = L1x´ • Existence of L1 puts 4 constraints on diagonal
elements of L & 6 constraints on the offdiagonal elements of L.
10 constraints on 16 elements of L Only 6 independent elements of L
PHYSICS: 3 elements of L correspond to 3 components of relative velocity & 3 correspond to the Euler angles of rotation
Sect. 7.3: Velocity Addition & Thomas Precession
• Some of the following is from the E&M book by Jackson!
• Velocity Addition: Recall the Galilean Transformation:
See figure: Moving point P with velocity u´ in frame S´. S´is moving with velocity v in
positive x1 direction with respect to frame S: Velocity u of point P with respect to frame S is obviously:
u = u´ + v
• Velocity Addition: Lorentz Transformation (Method 1) Figure: Moving point P, velocity u´
in S´. S´ moves with velocity v in the +x1 direction with respect to S: Find thecomponents of velocity u of P withrespect to S. Inverse Lorentz transform (differential form) is: (ILT)
dx0 = γ(dx0´ + βdx1´), dx2 = dx2´
dx1 = γ(dx1´ + βdx0´), dx3 = dx3´ β = (v/c), γ = [1  β2]½
u´ components: ui´ (dxi´/dt´) = c (dxi´/dx0´) (i =1,2,3)
u components: ui (dxi/dt) = c (dxi/dx0) (i =1,2,3)
To find relations between these, use them with above ILT equations. Divide dx1, dx2 & dx3 by dx0 & manipulate.
• Velocity Addition: Lorentz Transformation (Method 1) (ILT):
dx0 = γ(dx0´ + βdx1´), dx2 = dx2´
dx1 = γ(dx1´ + βdx0´), dx3 = dx3´
ui´ (dxi´/dt´) = c (dxi´/dx0´) (i =1,2,3)
ui (dxi/dt) = c (dxi/dx0) (i =1,2,3)
For example: u1 = c(dx1/dx0) =
c[γ(dx1´ + βdx0´)][γ(dx0´ + βdx1´)]1 =
c[(dx1´/dx0´) + β][1 + β(dx1´/dx0´)]1
Or: u1 = [u1´ + v]/[1 + (vu1´)/c2]
Similarly: u2 = c(dx2/dx0) = γ1(u2´)[1 + (vu1´)/c2]1
u3 = c(dx3/dx0) = γ1(u3´)[1 + (vu1´)/c2]1
• Velocity Addition: Lorentz Transformation (Method 1) More generally: u component of u  v
u component of u v
Find: u = [u´ + v][1 + (vu´)/c2]1
u = γ1(u´)[1 + (vu´)/c2]1
Einstein Velocity Addition Formula• For u´ & v parallel, get the Goldstein result:
u = [u´ + v][1 + (vu´)/c2]1 (or βu = [βu´ + β][ 1+ βu´β]1) • Get Galilean result: u = u´ + v for v << c (γ 1, β 0)• Can also get relations between spherical angles in S´ & S (see figure). Can
show (student exercise): = ´ and tanθ = γ1(u´sinθ´)(u´cosθ´ + v)1 Also:u = [(u´)2 + v2 + 2u´vcosθ´  (u´vsinθ´)2/c2]½[1 + (u´v/c2)cosθ´]1
• Velocity Addition: Lorentz Transformation (Method 2 Goldstein): 3 inertial systems:
S1, S2,S3, with aligned x
axes. S2 moves with v with
respect to S1. S3 moves with
v´ with respect to S2.
GOAL: Find the velocity v´´ of S3 with respect to S1. • Transforming from S1 to S3 Equivalent to applying 2
successive Lorentz boosts: L12, to transform from S1 to S2 followed by L23, to transform from S2 to S3. That is, can write (matrix multiplication): L13 L23L12
• Velocity Addition: Lorentz Transformation (Method 2 Goldstein):
Transforming from S1 to S3
L13 L23L12 Or: (β = v/c, γ = [1  β2]½ ,
β´ = v´/c, γ´ = [1  (β´)2]½)
γ´´ β´´γ´´ 0 0 β´´ γ´´ 0 0 0 0 1 0 0 0 0 1
(β´´ = v´´/c, γ ´´= [1(β´´)2]½)
The 2 forms must be the same: β´´ = (β + β´)/(1+ ββ´) Or: v´´ = (v + v´)/(1+ vv´/c2)
Einstein Velocity Addition Formula (parallel velocities)
Sect. 7.3: Thomas Precession
• Product of 2 Lorentz Transformations (LT) is a LT. If L& L´ are LT’s: L´´ L´L is also a LT. NOTE! They do not commute! Order matters!
• If L´ & L are pure boosts, but their velocities are not parallel, L´´ will involve a boost PLUS a rotation. This rotation Thomas Precession.
• Consider this under the ASSUMPTION (Approximation): Speed v´ associated with L´ is small compared to speed v associated with L & also v´ << c
• Consider 3 inertial systems: S1, S2,S. S2 moves with v = cβ with respect to S1. S3 moves with v´ = cβ´ with respect to S2.
– No loss of generality to have the axes of S1 so that β  xaxis of S1 & so that β´ is in the x´y´ plane of S2.
• Transform from S1 to S3 Apply 2 Lorentz boosts: L, to transform from S1 to S2 followed by L´, to transform from S2 to S3. Can write (matrix multiplication): L´´ L´L.
• L´´ is not symmetric It is a rotation + a velocity boost. – Let S3 move with v´´ = cβ´´ with respect to S1.
• Off diagonal elements in L´´ corresponding to z motion are zero The rotation is about an axis xy plane.
• Approximation: v´ = cβ´ is small compared to v = cβ . Also v´ << c β´ << 1, γ´ 1. Keep only terms to 1st order in β´. This implies:
• Using this approximation in the result for L´´ gives:
• This is a combination of a boost & a rotation. Now, to find (separate out) the rotation part: Consider (in the same approximation) the inverse Lorentz Transformation BACK from S3 to S1. Can show (see text) this is:
• Can get rotation matrix (under same assumption) by:
• Compare with the rotation matrices in Ch. 4 (generalized to include the 4th dimension) & get:
1 0 0 0 0 cos(ΔΩ) sin(ΔΩ) 0
R 0 sin(ΔΩ) cos(ΔΩ) 0 0 0 0 1
R corresponds to S3 being rotated with respect to S1 about the z axis through an angle ΔΩ
• When comparing the 2 expressions, recall that the approximations we’ve made mean that the off diagonal element
(γ  1)(βy´´)β 1 is very small so that:
sin(ΔΩ) (γ  1)(βy´´)β 1 is very small and thus,
sin(ΔΩ) ΔΩ = (γ  1)(βy´´)β 1 & cos(ΔΩ) 1
• Summary: Rotation part of L´´ L´L corresponds to S3 rotating with respect to S1 about z axis through (very small, in this approx) angle:
ΔΩ = (γ  1)(βy´´)β1
Thomas Precession Angle Very important & very well verified experimentally in atomic physics!
• Briefly consider a particle moving in the lab frame with velocity v which is NOT CONSTANT. Since the particle rest frame is accelerated with respect to the lab frame, it is NOT an inertial system & thus, strictly speaking we cannot connect the 2 frames with a Lorentz Transformation (LT)!
• However, we can circumvent this by the following device: Imagine an number of inertial systems, each moving with uniform velocity relative to the lab system. As the velocity of the particle changes, for each infinitesimal time change dt, we can match the particle velocity to that of one of these inertial frames. At each instant, we can make a LT from the lab to the particle frame. That is, in some sense, we can describe the transformation by a time dependent LT!
• We mention this now because a rotating frame (as in Thomas Precession) is not an inertial frame!
Thomas Precession:
ΔΩ = (γ  1)(βy´´)β1
• Application: S1 = lab system. S2, S3 = 2 of these instantaneous rest frames for a particle undergoing Thomas Precession just described. A small time Δt apart.
• In Δt, a lab observer will see a change Δv in the particle velocity. Our discussion shows that this will have only a y component:
Δv = cβy´´. Initially, choose velocity v = cβ along the x axis. The vector velocity has rotated through the infinitesimal angle ΔΩ given above. Using all of this, we can write the vector angle as:
ΔΩ =  (γ  1)(v Δv)/v2 (1)Dividing (1) by Δt gives an instantaneous angular velocity of the particle: ω (ΔΩ/Δt) or: ω =  (γ  1)(v a)v2 (2) Here: a (Δv/Δt) = particle acceleration as seen from the lab system S1
The Thomas Precession frequency: ω =  (γ  1)(v a)/v2
(2)• v is usually small enough that γ = [1  β2]½ 1 + (½)β2
The Thomas Precession frequency becomes: ω = (½)(a v)c2 (3)
• This was originally proposed to explain the anomalous magnetic moment of the electron. It led to the hypothesis that electrons have an internal quantum number called spin angular momentum. See the discussion in the Relativity chapter in J.D. Jackson’s E&M book for more details!