Set 6: Relativity The Metric • The metric defines a measure on a space, distance between points, length of vectors. • 3D Cartesian coordinates: separation vector between 2 points dxi (upper or contravariant indices)
3 X ds2 = dxidxi i=1 • Generalize to curvilinear coordinates, e.g. for spherical coordinates (r, θ, φ) the distances along an orthonormal set of vectors
eˆr : dr
eˆθ : rdθ
eˆφ : r sin θdφ The Metric
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• Length in spherical coordinates
ds2 = dr2 + r2dθ2 + r2 sin2 θdφ2 dφ X i j = gijdx dx ij er rsinθdφ eθ defines the metric eφ rdθ 1 0 0 dr 2 gij = 0 r 0 0 0 r2 sin2 θ The Metric • This would look like an ordinary dot product if we introduce the dual or “covariant” components of the vector
j dxi ≡ gijdx
2 X i i ds = dxidx ≡ dxidx i where the Einstein summation convention is that repeated pairs of upper and lower indices are summed • Similarly the dot product between two vectors is given by
i i j i V Xi = gijV X = ViX Special Relativity
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• In space time, the coordinates run from 0 − 3 with µ = 0 as the x=ct
temporal coordinate, and O [unprimed frame] 2 emission at ds represents the space-time separation t=t'=0 x=x'=0 µ 1 2 3 dx = (cdt, dx , dx , dx ) x'=ct' v • Metric is defined by the requirement that O' [primed frame] two observers will see light propagating at the speed of light. • Spherical pulse travels for time dt at the speed of light c
2 2 i 2 2 i 2 02 0 0i c dt = dxidx → −c dt + dxidx = 0 = −c dt + dx idx Special Relativity • The space time separation for light is null and invariant • So as an invariant measure on the space time, the temporal coordinate has the opposite sign: in Cartesian coordinates −1 0 0 0 0 1 0 0 [gµν =]ηµν = 0 0 1 0 0 0 0 1
2 µ µ ν 2 2 i ds = dxµdx = ηµνdx dx = −c dt + dxidx Lorentz Transformation • Set of all linear coordinate transformations that leave ds2, and hence the speed of light, invariant • 3D example: rotations leave the length of vectors invariant, generalization of a 4D rotation is a Lorentz transformation • Begin with a general linear transformation (excluding a constant term) ∂x0µ x0µ = Λµ xν → Λµ = ν ν ∂xν 02 0α 0β ds = ηαβdx dx α µ β ν = ηαβΛ µdx Λ νdx 2 µ ν = ds = ηµνdx dx α β ηµν = ηαβΛ µΛ ν Lorentz Transformation
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• Suppose a particle at rest in frame O is viewed in frame O0 O [rest frame] moving with a velocity v -v 0µ µ ν dx = Λ νdx ν v dx = (cdt, 0, 0, 0) O' [primed frame] • Take µ = 0, µ = i
0 0 0 0 i 0 cdt = Λ 0dx + Λ idx = Λ 0cdt 0i i dx = Λ 0cdt Lorentz Transformation • Combine
0i i i dx v Λ 0 0 ≡ − = 0 cdt c Λ 0 vi Λi = − Λ0 0 c 0 • Recall invariance of ds2 implies
α β ηµν = ηαβΛ µΛ ν • Evaluate 00 component
α β η00 = −1 = ηαβΛ 0Λ 0 0 2 i 2 −1 = −(Λ 0) + (Λ 0) Lorentz Transformation • Plug in relationship between Λ’s v2 −1 = −(Λ0 )2 + (Λ0 )2 0 c2 0 v2 1 = (Λ0 )2(1 − ) 0 c2
0 • Solve for Λ 0 1 Λ0 = ≡ γ 0 p1 − v2/c2 i i i Λ 0 = −γv /c = −γβ
α β • Spatial components determined from ηµν = ηαβΛ µΛ ν, excluding 0 1 rotation before boost in eˆ1 direction: Λ 1 = −βΛ 1, 0 2 1 2 1 = −(Λ 1) + (Λ 1) Lorentz Transformation
• Lorentz transformation for boost in eˆ1 direction γ −βγ 0 0 −βγ γ 0 0 µ Λ ν = 0 0 1 0 0 0 0 1
0µ µ ν or with dx = Λ νdx β ct0 = γct − βγx = γc(t − x) c v t0 = γ(t − x) c2 x0 = −βγct + γx x0 = γ(x − vt) Lorentz Transformation • Relativity paradoxes by holding various things fixed: simultaneity in one frame not same as another • Lorentz contraction: (primed: rest frame) length measured at fixed t
0 0 ∆x |t = γ∆x → ∆x = ∆x /γ|t • Time dilation measured at x0 = 0 v ∆t0 = γ(∆t − x| ) c2 x=vt v2 1 = γ(∆t − ∆t) = ∆t c2 γ 1 ∆t0 = ∆t γ Lorentz Transformation • Boost of a covariant vector ∂x0 0 ˜ ν ˜ ν µ xµ = Λµ xν → Λµ ≡ ∂xν α 0 0µ ˜ α µ β xαx = xµx = Λµ xαΛ βx ˜ α µ α ˜ ˜ −1 Λµ Λ β = δ β → ΛΛ = I → Λ = Λ ∂xν Λ˜ ν ≡ µ ∂xµ0 γ βγ 0 0 βγ γ 0 0 ˜ ν Λµ = 0 0 1 0 0 0 0 1 Lorentz Transformation • Tensors: multi-index objects that transform under a Lorentz transformation as, e.g.
0µ µ ˜ τ σ T ν = Λ σΛν T τ • Special relativity: laws of physics invariant under Lorentz transformation = laws of physics can be written as relationships between scalars, 4 vectors and tensors • Like ds2 the contraction of a set of 4 vectors or tensors is a Lorentz invariant • What about laws involving derivatives? Derivative Operator • Derivative operator on a scalar transforms as a covariant vector
∂ ∂xβ ∂ ∂ = = Λ˜ β ∂x0α ∂x0α ∂xβ α ∂xβ 0 ˜ β ∇α = Λα ∇β • Derivative operator on a vector transforms as a tensor
∂ T β = ∇ V β = V β α α ∂xα ! ∂ ∂ ∂x0β T 0 β = ∇0 V 0β = V 0β = V µ α α ∂x0α ∂x0α ∂xµ ∂x0β ∂ ∂ ∂x0β = V µ + V µ ∂xµ ∂x0α ∂x0α ∂xµ Derivative Operator • For a Lorentz transformation, the coordinates are linearly related
∂ ∂x0β ∂ = Λβ = 0 ∂x0α ∂xµ ∂x0α µ
∂x0β ∂xν ∂ T β = V µ = Λβ Λ˜ νT α α ∂xµ ∂x0α ∂xν µ α ν so that the derivative of a vector transforms as a tensor as long as the coordinate transformation is “special”, i.e. linear. In general relativity this condition is lifted by promoting the ordinary derivative to a covariant derivative through the connection coefficients 4 Velocity
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• Four velocity: Orest [rest frame] µ 2 2 2 β dx is a vector dτ = −ds /c is a scalar β u [=- uc] u O [lab frame] µ dx β ≡ U µ u' dτ O' [boosted frame] (c, 0, 0, 0)restframe in a boosted frame (particle velocity is opposite to
boost u = −βuc)
µ µ ν U = Λ νUrest = (γuc, −βuγuc, 0, 0) = γu(c, u) 4 Velocity • Boost again by β to show how velocity transforms
00 0 1 1 U = γ(U − βU ) = γγu(c − βu ) ≡ γu0 c 01 0 1 1 01 U = γ(−βU + U ) = γγu(−βc + u ) ≡ γu0 u 02 2 2 02 U = U = γuu = γu0 u 03 3 3 03 U = U = γuu = γu0 u which imply
1 2 γu0 = γγu(1 − vu /c ) 01 1 γu0 u = γγu(u − v) 02,3 1 2 02,3 2,3 γu0 u = γγu(1 − vu /c )u = γuu 4 Velocity • Divide two relations u1 − v u01 = 1 − vu1/c2
02 γu 2 1 2 u = u = 1 2 u γu0 γ(1 − vu /c ) 1 u03 = u3 γ(1 − vu1/c2)
µ 2 2 2 • Note that U Uµ = −(γc) + (γu) = −c also dot product is a way of evaluating rest frame time component of a vector Aµ: µ 0 0 i i 0 U Aµ = −U A + U A = −cA . 4 Acceleration, Momentum, Force • Acceleration dU µ = aµ dτ • Momentum (finite rest mass)
P µ = mU µ = γm(c, u)
• Force dP µ F µ = dτ E&M • Wavevector: phase of wave is a Lorentz scalar
i φ = kix − ωt µ φ = kµx Kµ = (ω/c, k) • Momentum q =h ¯k
P µ = (E/c, q) =hK ¯ µ
• Doppler shift cos θ = uˆ · kˆ
0µ µ ν K = Λ νK u ω0 = γ(ω − k ui) = γω(1 − cos θ) i c γ factor purely relativistic E&M • Charge conservation and 4 current ∂ρ + ∇ ji = 0 ∂t i µ µ ∇µJ = 0 J = (ρc, j) • 4 potential: Aµ = (φ, A) obeys the wave equation 1 ∂2 4π ∇ ∇αAµ = [∇2 − ]Aµ = − J µ α c2 ∂t2 c • Lorentz gauge condition 1 ∂φ ∇ Ai + = ∇ Aµ = 0 i c ∂t µ • Fields are related to derivatives of potential: field strength tensor
Fµν = ∇µAν − ∇νAµ E&M • E and B field 0 −Ex −Ey −Ez E 0 B −B x z y Fµν = Ey −Bz 0 Bx Ez By −Bx 0 • Maxwell Equations 4π ∇µF = J , ∇ F + ∇ F + ∇ F = 0 µν c ν σ µν ν σµ µ νσ • Lorentz scalars (an electromagnetic field cannot be transformed away)
µν 2 2 2 FµνF = 2(B − E ) , detF = (E · B) E&M • Lorentz force e F µ = F µ U ν c ν • Phase space occupation
d3x = γ−1d3x0 , d3q = γd3q0 so d3xd3q is a Lorentz scalar and photon number is conserved so f is a Lorentz scalar (note that the energy spread is negligible in the rest frame) • Field transformation
0 ˜ α ˜ β Fµν = Λµ Λν Fαβ 0 0 Ek = Ek ,Bk = Bk 0 0 E⊥ = γ(E⊥ + β × B) , B⊥ = γ(B⊥ − β × E) Coulomb Field Transformation • Take Coulomb field in the particle rest frame E0 = (q/r02)ˆr0 and boost in the x direction qx0 qγ(x − vt) E = = ,B = 0 x r03 r03 x qγy0 qγy0 qγβz0 qγβz0 E = = ,B = − = − y r03 r03 y r03 r03 qγz0 qγz qγβy0 qγβy E = = ,B = = z r03 r03 z r03 r03 r02 = γ2(x − vt)2 + y2 + z2
This may be rewritten as the velocity term in the Lienard-Wiechart fields Power • Power is a Lorentz scalar (4 momentum transformation with zero spatial momentum from symmetry)
dW dW 0 dW = γdW 0 , dt = γdt0 ,P = = = P 0 dt dt0 • In particular in the instananeous rest frame the power can be calculated using the Larmor formula
2q2 2q2 P 0 = |a0|2 = aµa 3c2 3c3 µ µ 2 µ 2 0 3 0 2 • given a = d x /dτ one can show ak = γ ak and a⊥ = γ a⊥
2q2 P = γ4(a2 + γ2a2) 3c3 ⊥ k so that a parallel acceleration causes a much larger power radiation Relativistic Beaming
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• Isotropic emission also becomes beamed: u' by the addition of θ' u⊥' u||' velocities, the angle changes with a boost O' [primed frame] u0 + v u0 k ⊥ u θ ⊥ uk = u⊥ = u 0 2 0 2 || 1 + vuk/c γ(1 + vuk/c ) u 0 0 0 O [lab frame] v u⊥ u⊥ 1 u sin θ tan θ = = 0 = 0 0 uk uk + v γ γ(u cos θ + v) for light u0 = c and the aberration formula is c sin θ0 tan θ = γ c cos θ0 + v Relativistic Beaming • For θ0 = π/2 then tan θ = c/γv and for γ 1, v → c and so tan θ ≈ θ ≈ 1/γ - beamed a tight half angle • Explains the differential power transformation: Larmor in [primed] rest frame • Solid angle transformation: again apply addition for light to get
u u u0 cos θ = k = k , cos θ0 = k q 2 2 c c u⊥ + uk cos θ0 + v/c cos θ = 1 + v/c cos θ0 • So dΩ = d cos θdφ and 1 dΩ = dΩ0 γ2(1 + β cos θ0)2 Differential power • Energy and arrival time as (µ = cos θ)
0 0 0 0 dW = γ(dW + vdPx) = γ(1 + βµ )dW 0 dtA = γ(1 − βµ)dt • Identity 1 γ(1 − βµ) = γ(1 + βµ0) • Transformation of differential power
dP dW 1 dP 0 = = 4 4 0 dΩ dΩdtA γ (1 − βµ) dΩ 1 q2a02 = sin2 Θ0 γ4(1 − βµ)4 4πc3 Differential power
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• Acceleration parallel to γ velocity: dipole pattern gets perpendicular 1/ lobes bent toward the velocity direction a • Acceleration v γ perpendicular to velocity: forward 1/ dipole enhanced, second lobe distorted a O' [rest frame] O [lab frame]