Physics 214 Examples of four-vectors Winter 2017

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1 Physics 214 Examples of four-vectors Winter The velocity four-vector The velocity four-vector of a particle is defined by: u µ = dxµ dτ = γc; γ v), 1) where dτ = γ 1 is the differential proper time which is a scalar quantity). Note that γ 1 v 2 / ) 1/2, where v is the magnitude of the velocity vector v d x/ that appears in eq. 1). The squared magnitude of the velocity four-vector, u 2 g µν u µ u ν = 2) is a Lorentz invariant. Note that it is most easily evaluated in the rest frame of the particle where v = 0, in which case u µ = c1; 0). Let us now consider the following question. Suppose that the velocity vector is u µ = u 0 ; u) in an inertial frame K. That is u 0 γc and u γ v, where v is the velocity as measured in K. Note that γ depends implicitly on v and is also frame dependent. A second inertial frame K is defined to be moving with relative velocity w with respect to K. Note that I have chosen a different symbol for the relative velocity to avoid confusion with v which is the velocity of the particle in the reference frame K. We wish to relate the four vector u µ which describes the velocity of the particle in K and the corresponding four vector u µ which describes the velocity of the particle in K. This is accomplished by a Lorentz boost: u µ = Λ µ νu ν 3) with the boost matrix Λ µ ν given by 1 Λ = γ w γ w β γ w β δ ij +γ w 1) βi β j β 2, 4) where β w/c and γ w 1 β 2 ) 1/2. Then, eqs. 3) and 4) imply that: u 0 = γ w u 0 β u), 5) u = u+ γ w 1) β 2 β u) β γ w βu 0. 6) 1 For consistency, I should really define β w w/c. However, there should be no confusion in the present discussion, so I will omit the subscript w to reduce the clutter in the typography. 1

2 Dividing these two equations yields: [ ] u u = 1 u 0 u 0 β + γ w 1) u γ w γ w β β u) β βu ) Substituting u 0 = γc, u = γ v, and u /u 0 = v /c in eq. 7), we arrive at: [ v 1 v = + γ w 1) ] v w) w w. 8) v w γ 1 w w 2 γ w This result can be rewritten as: [ v 1 1 = v w 1 γ w ) v w v w 1 w 2 ) ] v w w. 9) w 2 This is the law of addition of velocities. In the simple case where v and w are parallel, it follows that: 2 ) v w v = w. 10) w 2 In this case, eq. 9) simplifies immediately to: v = v w 1 v w/. 11) In the non-relativistic limit, γ w 1 and eqs. 9) and 11) both reduce to the expected form: v = v w. 2. The momentum four-vector The momentum four-vector also called the four-momentum) is related in a simple way to the velocity four-vector: p µ = mu µ = E/c; p), 12) where [using eq. 1)] p = γm v, 13) E = γm. 14) Note that by dividing these two equations, one deduces an expression for the particle velocity: v = pc2 E. 15) 2 One can check the correctness of eq. 10) by taking the dot product of both sides of the equation with w. 2

3 Again, it must be emphasized that v, which appears both explicitly and implicitly in the factors of γ in eqs. 13) 15) corresponds to the velocity of the particle. Thus, in the rest frame of the particle, v = 0 and γ = 1, which implies that p µ = mc1; 0). Furthermore, the mass m is a scalar quantity which is Lorentz invariant); it corresponds to the rest energy of the particle divided by. This also follows from the observation 3 that the Lorentz invariant scalar p µ p µ = m 2. Finally, by noting that g µν p µ p ν = p 0 ) 2 p 2 = m 2, 16) and inserting p 0 = E/c, one obtains an expression for the relativistic energy: E 2 = p 2 +m 2 c 4. 17) Taking the square root, and expanding out resulting expression in the limit of v c yields: E m + p 2 2m, 18) which we recognize as the sum of the rest energy and the non-relativistic kinetic energy. More generally, the relativistic energy can be written as E = m +T, which defines the relativistic kinetic energy as: T = p 2 +m 2 c 4 m. 19) 3. The force and acceleration four-vectors The relation between the three-vector force and the three-vector momentum remains valid in special relativity, F = d p. Using eq. 13), it follows that Using it follows that F = d p = d ) d v dγ γm v = γm +m v. 20) dγ = d 1 v v ) 1/2 = γ3 2 v d v c, [ d v F = γm + γ2 v d v ) ] v which is the relativistic generalization of Newton s second law. 3 Since Lorentz scalars do not depend on the reference frame, I may compute it in any frame. By choosing the rest frame of the particle, the computation is trivial. 3 21)

4 Two special cases are noteworthy: 1. v d v/ linear motion). In this case, d v/ = κ v, for some constant κ. This constant is easily determined by taking the dot product with v/v 2 where v 2 v v is the squared magnitude of the velocity three-vector). It follows that d v = v v 2 Plugging this result into eq. 21) yields F = γm d v ] [1+ γ2 v 2 = γ 3 m d v after using γ 2 = 1 v 2 / ) v d v/ circular motion). v d v ). 22) In this case v d v/ = 0. Plugging this result into eq. 21) yields, F = γm d v, for linear motion. 23), for circular motion. 24) In older introductory books on relativity, the concept of relativistic mass, defined as m R γm was introduced. I suppose that this was motivated by eqs. 13) and 14) which could be written as p = m R v which resembles the non-relativistic expression of momentum) and E = m R. However, as eqs. 21) 24) make clear, this is a very silly thing to do. Indeed, there is nothing special about γm that would single out this choice for some definition of the relativistic mass. In the formalism presented in these notes, m 2 = p 2 /u 2 is the ratio of two Lorentz-invariant scalars, and thus is itself Lorentzinvariant and independent of the choice of reference frame). In the older introductory books on relativity, m was called the rest mass. In the formalism presented in these notes, m is a Lorentz invariant quantity that is an intrinsic property of the particle like electric charge). We still have not yet developed an appropriate expression for a four-vector that can be related to the force. To see how to do this, we first consider the dot product of eq. 20) with v, which yields, This should be compared with F v = γm v d v ) 1+ γ2 v 2 de = d ) γmc 2 = mdγ = γ 3 m v d v. 25) = γ3 m v d v, after using eq. 14). Hence, the relation between the power de/ and F v, de = F v, 4

5 remains valid in special relativity, as long as we define the power to be the time rate of change of the relativistic energy. The above results motivate the introduction of Minkowski four-vector force, K µ dpµ dτ = γ F v ; γ F c ) γ 4 = m v d v c ; γ2d v + γ4 v d v ) ) v, 26) where we have used dτ = γ 1 and then employed eqs. 21) and 25) to obtain the final form above. Likewise, we can define the acceleration four-vector, Using dτ = γ 1, it follows that α µ = α µ duµ dτ = γc dγ ; γ d ) ) γ 4 γ v = v d v c ; γ2d v + γ4 c dγ dτ ; d ) ) γ v. 27) dτ v d v ) ) v. The three-vector acceleration is defined as usual by a d v/. Then, one can rewrite the four-vector acceleration as 4 ) γ α µ 4 = c v a; γ2 a+ γ4 ) v a v = γ4 c v a 1; v ) +γ 2 0; a ). 28) c An important property of the four-vector acceleration is u µ α µ = 0. 29) This can be proved trivially as follows. Noting that u µ u µ = which is a constant, it follows that 0 = d u µ u µ) = 2u µ α µ. dτ One can also derive the same result from eqs. 1) and 28), ) u µ α µ = γ 5 v a γ 3 v a γ5 v 2 c v a = 2 γ3 v a γ 2 1 γ2 v 2 = 0, since γ 2 1 v 2 / ) = 1. Finally, by comparing eqs. 26) and 28), we arrive at the four-vector version of Newton s second law, K µ = mα µ. It then follows from eq. 29) that u µ K µ = 0. 4 Most books simply use the notation a µ for the four-vector acceleration. One disadvantage of this notation is that the space component of this four vector, which would be denoted by a i is not the ith component of the vector d v/. 5

6 The concept of constant acceleration must be reconsidered in special relativity. It clearly cannot mean that the three-vector a = d v/ is constant, since this would be a frame-dependent condition. Instead, constant acceleration means that the square of the four0vector acceleration, α µ α µ is a constant. Clearly, the latter is a Lorentz-invariant condition. It is traditional to define, g 2 α µ α µ. 30) To motivate this definition, consider the case of constant acceleration in the direction of motion; i.e., a v corresponding to linear motion). In this case, eq. 22) yields v 2 a = v a) v. Inserting this result in eq. 28) yields after some simplification, α µ = γ 4 v a c ) ; a, for linear motion. It then follows that ) ) g 2 = γ 8 a 2 v a)2 = γ 8 a 1 2 v2 = γ 6 a 2. Hence, constant linear acceleration in special relativity means that g = γ 3 a is constant. Note that in each instantaneous rest frame 5 of the accelerating particle, constant linear acceleration does correspond to constant a as expected. We can easily evaluate the velocity at time t of a constantly accelerating particle, which starts off at rest at t = 0. Since a = d v/, and g = γ 3 a is constant, dv = γ 3 g = ) 3/2 1 v2 g. where v v. Integrating this equation, with the boundary condition, vt = 0) 0 yields gt vt) = ) 1+ g2 t 2 1/2. Note that the non-relativistic limit, vt) gt, is a good approximation when gt c i.e., when the velocity is non-relativistic). Furthermore, in the limit of t, we have lim vt) = c. t and v < c for all finite times t. Thus, in special relativity, a particle that is constantly accelerating never reaches the speed of light in a finite amount of time! 5 At any time t, the instantaneous rest frame corresponds to the inertial frame traveling at the same velocity as the particle at time t. 6