Vectors in Special Relativity

Transcription

1 Chapter 2 Vectors in Special Relativity 2.1 Four - vectors A four - vector is a quantity with four components which changes like spacetime coordinates under a coordinate transformation. We will write the displacement four - vector as: x O c t, x, y, z) = { α }, α = 0, 1, 2, ) x is a frame independent vector joining near by points in spacetime. O means: has components in frame O. x α are the coordinates themselves [ which are coordinate dependent ]. In frame Ō, the coordinates are { xᾱ} so: x { xᾱ} Ō ᾱ = 0, 1, 2, ) The Lorentz transformations can be written as 3 xᾱ = Λᾱβ x β, 2.3) β=0 where Λᾱβ is the [4 4] Lorentz transformation matrix Λᾱβ = γ γv γv 0 0 c γ 0 0 c ) where γ = 1 v 2 /c 2 ) 1/2 and x β, xᾱ can be regarded as column vectors. 18

2 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY 19 The positioning of the indices is explained later but indicates that we can use the Einstein summation convention: sum over repeated indices, if one index is up and one index down. Thus we can write: xᾱ = Λᾱβ x β. 2.5) β is a dummy index which can be replaced by any other index; ᾱ is a free index, so the above equation is equivalent to four equations. For a general four - vector A O A 0, A 1, A 2, A 3), 2.6) we can write Aᾱ = ΛᾱβA β. 2.7) If A and B are two four - vectors, clearly C = A + B and D = µa are also four - vectors with obvious components C O A 0 + B 0, A 1 + B 1, A 2 + B 2, A 3 + B 3), 2.8) and D O µa 0, µa 1, µa 2, µa 3). 2.9) In any frame O we can define a set of four - basis vectors: e 0 O 1, 0, 0, 0), 2.10) e 1 O 0, 1, 0, 0), 2.11) e 2 O 0, 0, 1, 0), 2.12) and e 3 O 0, 0, 0, 1). 2.13) In general we can write e α ) β = δ β α = 1, α = β = 0, α β, 2.14)

3 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY 20 where α labels the basis vector and β labels the coordinate. Any four - vector can be expressed as a sum of four - vectors parallel to the basis vectors i.e. A = A α e α = Aᾱeᾱ. 2.15) The last equality reflects the fact that four - vectors are frame independent. Writing: Aᾱeᾱ = ΛᾱβA β eᾱ = A β Λᾱβeᾱ = A α Λ β α e β, 2.16) where we have exchanged the dummy indices α and β and ᾱ and β, we see that this equals A α e α for all A if and only if This gives the transformation law for basis vectors: e α = Λ β α e β. 2.17) e 0 = γe 0 γv c e 1, e 1 = γv c e 0 + γe 1, e 2 = e 2, e 3 = e ) Note that the basis transformation law is different from the transformation law for the components since A β α takes one from frame Ō to O. So in summary for vector basis and vector components we have: e α = Λ β α e β, A β = Λ β α A α. 2.19) Since O has a velocity v relative to Ō, we have: e β = Λ α β v)e α e ν = Λ β ν v)e β = Λ β ν v)λ α β v)e α Λ β ν v)λ α β v) = δ α ν. 2.20)

4 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY 21 So Λ α β v) is the inverse of Λ β α v). Likewise A ν = Λ ν β v)a β = Λ ν β v)λ β α v)a α = δ ν αa α. 2.21) It follows that the Lorentz transformations with v gives the components of a four - vector in O from those in Ō. The magnitude of a four - vector is defined as A 2 = A A: A 2 = A 02 + A 12 + A 22 + A 32, 2.22) in analogy with the line element ds 2 = c 2 dt 2 + dx 2 + dy 2 + dz ) The sign on the A 0 will be explained later. This is a frame invariant scalar. A is spacelike if A 2 > 0, timelike if A 2 < 0 and null if A 2 = 0. The scalar product of two four - vectors A and B is: A B = A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B ) Since A + B) A + B) = A 2 + B 2 + 2A B, 2.25) A B is frame independent. A and B are orthogonal if A B = 0; they are not necessarily perpendicular in the spacetime diagram [ for example a null vector is orthogonal to itself ], but must make equal angles with the 45 o line. Basis vectors form an orthonormal tetrad since they are orthogonal: e α e β = 0 if α β, normalized to unit magnitude: e α e α = ±1: e α e β = = η αβ. 2.26) We will see later what the geometrical significance of η αβ is.

5 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY 22 ct U α Figure 2.1: The worldline of a particle with four - velocity U α. 2.2 Four - velocity, momentum and acceleration We must express Newton s laws of motion in terms of four - vectors so that they are frame invariant and consistent with Special Relativity. The four - velocity of a particle is the tangent to its worldline of length c [ see Figure 2.1 ]: x U = dx dτ, U U = c2, 2.27) where τ is the proper time [ the time measured in the particle s own rest frame ]. This is the most natural analogue of the three - velocity. It is clearly a four - vector since both dx and dτ are invariant. In the particle s own rest frame Ō, the four - velocity is It follows that in a general frame O: U = c e 0 = c, 0, 0, 0). 2.28) U α = Λ α β c e 0) β = cλ α 0 = γ c, v, 0, 0), 2.29) or U α = γ c, v), 2.30) where v is the particle s three - velocity. For low velocities v c, U 0 c and the spatial part is nearly the same as v. If an observers own four - velocity is written as W = c, 0, 0, 0) in O we have γ = U W c )

6 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY 23 We can then write the particle s three - velocity as [ EXERCISE 2.2 ]: W W) v = U W U W. 2.32) This is a spacelike vector expressed in coordinate - independent form [ although W singles out a particular observer ]. For photons the four - velocity is not defined since dx dx = 0, i.e. there is no frame in which a photon is at rest. The four - momentum is defined by P = m 0 U, P P = m 2 0c ) where m 0 is the rest mass of the particle i.e. the mass in its own rest frame. The spatial part of P [ the three - momentum ] is m 0 γv so the apparent mass exceeds the rest mass. m = γm 0 = m 0 1 v 2 /c ) The time part of the four - momentum is the energy of the particle E divided by c: P 0 = E c = 1 c P W = m 0γc = mc, 2.35) so we have E = m 0 γc 2 = mc ) For v c E = m 0 c 2 1 v 2 /c 2 ) 1/2 m 0 c 2 + 1m 2 0v ) Since the second term is the kinetic energy, we interpret the first term as the rest - mass energy of the particle. In general P P = m 2 0c 2 implies that E 2 = m 2 0 c4 + c 2 p p, 2.38) where p is the particle s three - momentum. Since m as v c, requiring infinite energy, we infer that the particles with non - zero rest mass, can never reach the speed of light. For photons traveling in the x - direction E = cp 1 P.P = 0 m 0 = 0, 2.39) hence photons have zero rest mass.

7 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY Relativistic Doppler shift Consider a photon moving at an angle α with respect to the x - axis. Its three - velocity is c cos α, c sin α, 0), so its four - momentum is P = hν c, hν ) hν cos α, c c sin α, 0 where h is Planck s constant and ν the frequency., 2.40) In a frame Ō with three - velocity v, 0, 0) relative to O the frequency is ν. Using the Lorentz transformations we get therefore so we get the following result: P 0 = Λ 0 β P β = Λ 0 0 P 0 + Λ 0 1 P 1, 2.41) [ h ν hν c = γ c v ] hν cos α c2 ν ν, 2.42) = 1 v/c cos α 1 v 2 /c ) If α = 0, so that the photon moves in the same direction as Ō, we have ν ν = For low velocities v c this reduces to 1 v/c 1 + v/c. 2.44) ν ν = 1 v c, 2.45) This is the usual Doppler shift ν = v, modified at large v. ν c If α = π/2, so the photon moves perpendicular to O, we have ν ν = γ. 2.46) This is the transverse Doppler shift and is a consequence of time dilation.

8 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY Four - acceleration The four - acceleration of a particle is defined as du dτ = dt d [γ c, v)] dτ dt since we have dγ dt = γ d c, v). 2.47) dt = d dt = γ3 c v dv 2 dt 1 v2 c 2 ) 1/2 = γ3 c 2 v dv dt, 2.48) A = γ 2 0, v) + γ4 v v c, v). 2.49) c2 For speeds much less than the speed of light v c, we obtain ) v v A =, v c, 2.50) therefore spatial part approximates the usual three - acceleration at low velocities. Note that U U = c 2 U du dτ to the four - velocity. = 0, so the four - acceleration is orthogonal Finally Newton s second law requires that we define the four - force as 2.3 Relativistic dynamics For interacting particles we postulate that F = m 0 A. 2.51) P i 2.52) i is conserved [ where i labels the particle ] since this is the natural analogue of Newton s law.

9 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY 26 Before u 1 u 2 After U Figure 2.2: The colliding particles in Example 1. Conservation of the time component corresponds to energy conservation [ with rest mass being included in this ]. Conservation of spatial components corresponds to conservation of three - momentum. Note that i P i is evaluated at a particular time by different observers and they will see different events as simultaneous. However i P i will be conserved in all frames. The center of momentum frame for a particular system of particles is the one in which Etotal P i = i c ), 0, 0, ) Example 2.1 As a simple example, consider two particles with rest masses m 1 and m 2, moving with speeds u 1 and u 2 u 1 > u 2 ), which collide and coalesce [ see Figure 2.2 ]. What is the mass M and speed U of the resulting particles. The initial four - momenta are P i = m i γ i c, u i, 0, 0), 2.54) where i = and The final four - momentum is γ i = 1 u2 i c 2 ) 1/ ) P = Mγ c, U, 0, 0). 2.56) Conservation of four - momentum gives P = P 1 + P 2, 2.57)

10 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY 27 and squaring this relation gives P 2 = P P P 1 P ) Now and P 2 i = m2 i c2, P 2 = M 2 c 2, 2.59) P 1 P 2 = m 1 m 2 γ 1 γ 2 u1 u 2 c 2). 2.60) This gives M = m 21 + m 2 + 2m 1 m 2 γ 1 γ 2 1 u ) 1u ) For u 1 c and u 2 c, M m 1 + m 2 which is the usual Newtonian result. From the x and t components of the four - momentum equation we have Dividing we get MγU = m 1 γ 1 u 1 + m 2 γ 2 u 2, Mγc = m 1 γ 1 c + m 2 γ 2 c. 2.62) U = m 1γ 1 u 1 + m 2 γ 2 u 2 m 1 γ 1 + m 2 γ ) For u 1 c and u 2 c we obtain Example 2.2 c 2 U = m 1u 1 + m 2 u 2 m 1 + m ) Consider a collision between a photon with frequency ν moving in the x direction, with an electron of rest mass m e initially at rest [ see Figure 2.3 ]. Before the collision, the four - momenta of the photon and electron are hν Q 1 = c, hν ) c, 0, 0, After the collision we have Q 2 = P 1 = m e c, 0, 0, 0). 2.65) h ν c, h ν c cos θ, h ν ) c sin θ, 0 P 2 = m e γ c, v sin Ψ, v cos Ψ, 0), 2.66),

11 CHAPTER 2. VECTORS IN SPECIAL RELATIVITY 28 Q 2 Θ Q P 1 1 Before Ψ P 2 After Figure 2.3: Sciamatic diagram illustrating Compton scattering. where ν is the photon s frequency after the collision, v is the speed of the electron and θ, Ψ are shown in Figure 2.3. Conservation of four - momentum gives Rearranging and squaring we get Using P 2 1 = P2 2 = m2 e c2 and Q 2 1 = Q2 2 = 0 we obtain which gives P 1 + Q 1 = P 2 + Q ) P 2 2 = P 1 + Q 1 Q 2 ) ) P 1 Q 1 P 1 Q 2 Q 1 Q 2 = 0, 2.69) m e hν + m e h ν = h2 ν ν c 2 cos θ 1). 2.70) Finally after simplifying this expression we arrive at the famous Compton scattering formula sin 2 θ 2 ) = m ec 2 2h 1 ν 1 ). 2.71) ν