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General Lorentz Transformation

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General Lorentz Transformation

- Consider a Lorentz Transformation, with arbitrary v, :
ct´= γ(ct -βr) r´= r + β-2(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 = -γβγ 0 0

0 0 1 0

0 0 0 1

- For any Lorentz transformation x´ Lx, there exists an Inverse TransformationL-1:
L-1L = LL-1 = 1, x = L-1x´

- Existence of L-1 puts 4 constraints on diagonal elements of L & 6 constraints on the off-diagonal 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 the

components of velocity u of P with

respect 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) = γ-1(u2´)[1 + (vu1´)/c2]-1

u3 = c(dx3/dx0) = γ-1(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 = γ-1(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 = γ-1(u´sinθ´)(u´cosθ´ + v)-1Also:

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. S2moves with v with

respect to S1. S3moves with

v´ with respect to S2.

GOAL: Find the velocity v´´ of S3with respect to S1.

- Transforming from S1to S3Equivalent to applying 2 successive Lorentz boosts: L1-2, to transform from S1to S2 followed by L2-3, to transform from S2to S3. That is, can write (matrix multiplication): L1-3 L2-3L1-2

- Velocity Addition: Lorentz Transformation (Method 2 Goldstein):
Transforming from S1to S3

L1-3 L2-3L1-2

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 theASSUMPTION (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. S2moves with v = cβ with respect to S1. S3moves with v´= cβ´ with respect to S2.
- No loss of generality to have the axes of S1 so that β || x-axis of S1 & so that β´ is in the x´-y´ plane of S2.

- Transform from S1to S3 Apply 2 Lorentz boosts: L, to transform from S1to S2 followed by L´, to transform from S2to S3. Can write (matrix multiplication): L´´ L´L.

- L´´is not symmetric It is a rotation + a velocity boost.
- Let S3move 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 TransformationBACK from S3to 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-1is very small so that:

sin(ΔΩ) (γ - 1)(βy´´)β-1is 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´´)β-1

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 NOTan inertial system & thus, strictly speaking we cannot connect the 2 frames with a Lorentz Transformation (LT)!
- However, we can circumvent this bythe 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: velocity

ΔΩ = (γ - 1)(βy´´)β-1

- 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 angleΔΩgiven 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)v-2 (2)

Here:a (Δv/Δt) = particle acceleration as seen from the lab system S1

The Thomas Precession frequency: velocity

ω = - (γ - 1)(v a)/v2 (2)

- v is usually small enough thatγ = [1 - β2]-½ 1 + (½)β2
The Thomas Precession frequencybecomes:

ω = (½)(a v)c-2 (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!

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