Relativistic Dynamics

MIT Course 8.033, Relativistic dynamics Topics • Formula summary • Momentum & energy • Acceleration & force (optional) • Transformation of force (optional) • Transformation of acceleration (optional) Dynamics toolbox: formula summary • Mass-energy unification: E = mc2 = m0 γc2 • Momentum 4-vector: ⎛ ⎞ px ⎜ py ⎟ ⎟ P ≡ m0 U = ⎜ ⎝ pz ⎠ E/c • Energy formula: E= � (m0 c2 )2 + (cp)2 • Velocity formula: β= cp E • Optional material: – Acceleration 4-vector: A≡ dU = γu2 dτ � a 0 � + γu4 u·a c2 � u c � – Force 4-vector: d F≡ P = γu dτ – Power: � F P/c � = m0 A P = Ė = u · F = m0 γu3 u · a – Force 3-vector: F u·a = a + γu2 2 u = m 0 γu c � γu2 a a (u & a parallel) (u & a perpendicular) – Acceleration 3-vector: ma = F − Pu c2 – Force transformation: Fx� = Fy� = Fz� = P� = Fx − cv2 P , 1 − ucx2v F � y ux v � , γ 1 − c2 F � z ux v � , γ 1 − c2 P − vFx 1 − ucx2v Momentum & energy toolbox: • Relativistic mass: m = γm0 • Mass-energy unification: E = mc2 • Momentum 4-vector (momentum-energy unification): ⎛ ⎞ ⎛ ⎞ ⎛ ux ux ⎜ uy ⎟ ⎜ ⎟ ⎜ dX ⎟ = m ⎜ uy ⎟ = ⎜ P ≡ m 0 U = m0 = m 0 γu ⎜ ⎝ ⎠ ⎝ uz uz ⎠ ⎝ dτ c c ⎞ px py ⎟ ⎟, pz ⎠ E/c (Use upper case X, U and P for the 4-vectors to avoid confusion with the x, u and p 3-vectors.) • Handy velocity formula follows straight from this: β= • Rest energy: cp E E0 = m0 c2 is total energy of particle in the frame where it is at rest • Kinetic enery: 1 K = E − E0 = mc − m0 c = m0 c (γ − 1) = m0 u2 + O 2 2 2 2 � u4 c � • Rest mass invariant: m0 = 1� 1� −Pt ηP = 2 E 2 − c2 p2 , c c giving the handy relations � E = (m0 c2 )2 + (cp)2 , � E2 p ≡ |p| = − (m0 c)2 . c2 • Low-speed limit |β| � 1: 1 E ≈ m0 c2 + m0 u2 , 2 p = m0 γu ≈ m0 u. • High-speed limit |β| ≈ 1 (γ � 1, E � E0 ): E ≈ cp This becomes exact (E = cp) for particles moving with speed of light, like photons and gravitons. • −Pt ηP = (E/c)2 −p2 is invariant also for system of particles, since � � � � � � � Ptot ≡ ΛPi = Λ Pi = Pi = ΛPtot . i i i • We derived p = m0 γu only for 1-dimensional collision. But any collision is 1-dimensional in the frame where the total momentum is zero! Acceleration & force (optional!) • The acceleration 4-vector A and the Force 4-vector F are less useful than their 4-vector cousins X, U, P and K. We’ll use F mainly for deriving the force transformation law, which will in turn give us the transformation law for electromagnetic fields. We’ll use upper case A for the acceleration 4-vector to avoid confusion with the the acceleration 3-vector a, and the annoying symbol F for the force 4-vector to avoid confusion with the the force 3-vector F. • Acceleration 4-vector: A ⎛ ⎞ ux � � ⎜ uy ⎟ dU dU d d u ⎜ ⎟ ≡ = γu = γu γu ⎝ = γu γu uz ⎠ c dt dt dt dτ c � � � � � � u̇ u a + γ̇u u 2 = γu + γu γ̇u = γu 0 c γ̇u c � � � � u·a a u = γu2 + γu4 2 , 0 c c where in the last step, we have used the fact that γ̇u = γu3 u·a . c2 • Force 4-vector: F≡ d d d d P = γu P = γu m0 U = m0 U, dτ dt dt dτ so by definition, we have F = m0 A. (Note that this does not apply the Newtonian result F = ma!) • Interpretation of Force 4-vector: � � � � d ṗ F F = γu P = γu = γu , P/c Ė/c dt where F = ṗ is the familiar force 3-vector and P = Ė is the power, the energy change per unit time (in Watts). • Work-energy theorem: dE = F · dr = F · dr dt = F · udt, dt so the power satisfies P = Ė = u · F. • Force 3-vector explicitly: Dividing the above equation F = m0 A by γu gives F u·a = a + γu2 2 u. m 0 γu c • Special case where u and a are parallel, e.g., for linear motion: � F u2 a � = a + γu2 2 = 1 + γu2 β 2 a = γu2 a. m0 γu c • Special case where u and a are perpendicular, eg, for circular mo­ tion: F =a m 0 γu • Note that in relativity, F and a are generally not parallel, but that they are parallel for these two special cases. • Acceleration 3-vector explicitly: a= F u·F F P − u= − u. m0 γu m0 γu c2 m mc2 The last term (the departure from F = ma) is seen to have the form of a friction term proportional to the power put into the particle. Derivation: the three steps below. γ̇u = d m0 γu c2 d E Ė u·F P = = = = . dt m0 c2 dt m0 c2 m0 c2 m0 c2 m0 c 2 Combining this with the other expression for γ̇u above gives u·a= u·F . γu3 m0 The above equation for F now becomes F P u·a =a+ u = a + γu2 2 u. m0 γu m0 γu c2 c Transformation of force • Let’s compute the transformation law for force by transforming to a frame S � moving with velocity v in the x-direction relative to S: ⎞⎛ ⎛ ⎞ ⎛ Fx� γ 0 0 −γβ Fx ⎜ Fy� ⎟ ⎜ 0 ⎟ ⎜ Fy 1 0 0 � ⎟ ⎜ ⎟⎜ F = γu� ⎜ ⎝ Fz� ⎠ = ΛF = γu ⎝ 0 0 1 0 ⎠ ⎝ Fz −γβ 0 0 γ P/c P � /c ⎛ ⎞ ⎛ ⎞ γ[Fx − βP/c] γ[Fx − βP/c] ⎜ ⎟ ⎜ ⎟ � γ F Fy u y ⎟= ⎜ ⎟. = γu ⎜ u v x ⎝ ⎠ ⎝ ⎠ Fz Fz γ(1 − c2 ) γ[P/c − βFx ] γ[P/c − βFx ] In the last step, we used the relation γu� = γu γ[1 − ux v/c2 ] which we proved earlier when transforming the velocity 4-vector U — it followed from the fact that its normalization is Lorentz invariant, i.e., U�t ηU� = Ut ηU. • The 4 components now give our desired force transformation equa­ tions: Fx� = Fy� = Fz� = P� = Fx − cv2 P , 1 − ucx2v F � y ux v � , γ 1 − c2 F � zu v � , γ 1 − cx2 P − vFx , 1 − ucx2v where P = u · F as usual. • If we take S to be the rest frame of the particle, then u = 0, P = u · F = 0 and this simplifies to Fx� = Fx , Fy� = Fy /γ, Fz� = Fz /γ, so in the frame S � where the particle is moving, the force is unaffected in the parallel direction and suppressed by γ in the transverse directions. ⎞ ⎟ ⎟ ⎠ Transformation of acceleration • We could derive expressions using an approach like for force, but the results are so messy that it’s not particularly useful — it’s better to deal with explicit problems as needed. • Here’s a useful special case that you get to derive on a problem set (probably PS7): For an arbitrary acceleration a in S, the acceler­ ation a� in S � is related to a via ax = ay = a�x γ 3 (1 + vu�x /c2 )3 a�y , γ 2 (1 + vu�x /c2 )2 with the important caveat that the expression for ay is only valid for the case where either u�y = 0 or a�x = 0.