2. Sect. 7.9: Lagrangian Formulation of Relativity (input from Marion!) We now see, in principal at least, how to generalize Newton’s 2 nd Law Equations of motion to Special Relativity. –But, see Goldstein Sect. 7.8, where Relativistic Angular Momentum is discussed. It’s NONTRIVIAL! Relativistic angular momentum & torque are 2 nd rank tensors! A natural question: How to generalize Lagrangian mechanics to Relativity? Or, what is an appropriate Relativistic Lagrangian? Two ways people have done this: 1. Try to obtain a covariant Hamilton’s Principle & Lagrange’s Equations. Treat space & time on equal footing in a 4d configuration space. 2. Don’t worry about 1. Try to find a Lagrangian which will (in some inertial frame) reproduce the Newton’s 2 nd Law Equations of motion obtained by using the spatial part of the Minkowski 4-force: K = (dp/dτ) = γ u (dp/dt)

3. Here, we focus on method 2. See Goldstein, Sect. 7.10, p. 318-324 for the details of method 1. My understanding is that both methods lead to the same PHYSICS! Consider the case of Conservative Forces.  A Velocity independent potential energy V = V(x i ) exists. For simplicity, work in Cartesian spatial coordinates. Single particle only. Recall: In the Lagrange formalism, for Lagrangian L, the Cartesian momentum components (for 3d velocity u) are: p i  (∂L/∂u i ) (1) Recall: The Relativistic Momentum: Cartesian components of the 3d vector: p = γmu [= m(dr/dτ)] p i = γmu i = mu i [1 - β 2 ] -½ = mu i [1 - β 2 ] -½ (2) Goal: Find a Lagrangian L for which (1) & (2) give the same momentum. (1) & (2) should be the SAME!

4.  p i  (∂L/∂u i )  mu i [1 - β 2 ] -½ (3) β = (u/c) What Lagrangian L gives Equation (3)? Actually, there is no UNIQUE answer to this question! One thing we can say is that (3) involves only the velocity of the particle.  Expect the velocity independent part of L to be unchanged from the non-relativistic case. That is, the velocity independent part of L will still be the negative of the PE, -V. However, it is not clear that the velocity dependent part of L will be the Relativistic KE T! So, write: L  T* - V (4) Choose T* = T*(u i ) so that (3) is satisfied:  (∂T*/∂u i )  mu i [1 - β 2 ] -½ (5)

5. We can easily show that this gives: T* = -mc 2 [1 - β 2 ] ½ = -mc 2 γ -1 (6)  L = - mc 2 [1 - β 2 ] ½ - V (7) V = V(x 1,x 2,x 3 ) β 2 = (u 2 /c 2 ) = c -2 [(u 1 ) 2 + (u 2 ) 2 + (u 3 ) 2 ] Using (7), we get the Lagrange Equations of motion in the usual way: (d/dt)[(∂L/∂u i )] - (∂L/∂x i ) = 0 Note: The Relativistic Lagrangian is NOT L = T - V, where T = Relativistic Kinetic Energy. This is because T = ([1 - β 2 ] -½ -1)mc 2 = (γ - 1) mc 2  T*

6. Simple Examples of Dynamics Example 1. Motion under a constant force –Examples: a. A particle of mass m in free fall under the Earth’s gravitational force. b. A particle of mass m & charge q in a constant electric field E. Take x as the direction of the constant force F. The velocity is in the x direction only. Define a  (F/m). V = - max  L = - mc 2 [1 - β 2 ] ½ + max (∂L/∂u) = cβ[1 - β 2 ] -½, (∂L/∂x) = ma Equation of motion: (d/dt)[(∂L/∂u)] - (∂L/∂x) = 0  (d/dt)(β[1 - β 2 ] -½ ) = (a/c) (1) Integrating (1) gives: β[1 - β 2 ] -½ = (at + α)/c (2) α = constant = v 0 = initial velocity. Solve (2) for β = (u/c) = [at + α][c 2 + (at + α) 2 ] -½ (3)

7. (3)  u = (dx/dt) = c[at + α][c 2 + (at + α) 2 ] -½ (3a) Integrate (3a) (0  t, x 0  x): x - x 0 = c ∫ dt[at + α][c 2 + (at + α) 2 ] -½ x - x 0 = (c/a){[c 2 + (at + α) 2 ] ½ - [c 2 + α 2 ] ½ } Pick the initial conditions: t = 0, x 0 = 0, v 0 = α = 0. Then, simple algebra gives: (x + c 2 /a) 2 - c 2 t 2 = (c 4 /a 2 ) (4) This is a hyperbola in the x-t plane! Contrast this to the Freshman physics result: x = (½)at 2 (4a) A parabola in the x-t plane. Student exercise: Prove that (4a) is obtained from (4) in the non-relativistic limit: u = (dx/dt) << c. Hint: Also, (at + α) << c & v 0 = α << c

8. Example 2. The relativistic 1d Simple Harmonic Oscillator.  L = - mc 2 [1 - β 2 ] ½ - (½)kx 2 We could solve this with L. Instead, note that, even in this relativistic problem, the total energy E is conserved. That is: E = T + V = constant, where T = relativistic KE.  E = T + V = (mc 2 )[1 - β 2 ] -½ + (½)kx 2 = const (1) β = (u/c) = c -1 (dx/dt) Solving (1) gives [V = (½)kx 2 ] β 2 = (u/c) 2 = c -2 (dx/dt) 2 = 1 - m 2 c 4 (E-V) -2 (2) Gives u = u(x)!

9. β 2 = (u/c) 2 = c -2 (dx/dt) 2 = 1 - m 2 c 4 (E - V) -2 (2) Before doing the simple harmonic oscillator, first, briefly look at (2) in the general case where V = V(x) is symmetric about x = 0 & has a minimum there.  The motion is oscillatory between 2 (symmetrically placed) turning points x   b where b is determined by E = V(  b). From (2) we can get t = t(x) in general. However, it’s useful to focus on the period of oscillation. Solve for dt & integrate. (t = 0  τ, x = 0  b): τ = (4/c)∫dx[1 - m 2 c 4 (E - V(x)) -2 ] -½ (3) Now, specialize to V(x) = (½)kx 2. In this case, (3) is an elliptic integral! The exact solution to the 1d relativistic simple harmonic oscillator is only possible in terms of elliptic integrals!

10. β 2 = (u/c) 2 = c -2 (dx/dt) 2 = 1 - m 2 c 4 (E - V) -2 (2) Instead of dealing with elliptic integrals, lets look at the 1d relativistic harmonic oscillator in the limit of small PE. That is, in the limit: V(x) = (½)kx 2 << mc 2 = E 0 = rest energy. Keep only the lowest order relativistic corrections to the usual harmonic oscillator solutions. In general, the energy is: E  γmc 2. With γ = [1 - (u 2 /c 2 )] -½ Note: For small β 2 = (u/c) 2, γ  1 + (½)β 2 = 1 + (½) (u/c) 2 and E  mc 2 [1 + (½)(u/c) 2 ] Change notation & write: E  mc 2 [1 + ε], ε  (u/c)  [E - V(x)](mc 2 ) -1  1 + ε – (½)kx 2 (mc 2 ) -1  1 + ε - κx 2 with κ  (½)k(mc 2 ) -1. In terms of the turning point at x = b, this has the form: (E - V(x))(mc 2 ) -1 = 1 + κ(b 2 - x 2 )

11. (E - V(x))(mc 2 ) -1 = 1 + κ(b 2 - x 2 ) (4) The period was: (x = 0  b): τ = (4/c)∫dx[1 - m 2 c 4 (E - V(x)) -2 ] -½ (3) Put (4) into (3) & expand the integrand to order (κb 2 ) 2 : τ  (4/c)∫dx[2κ(b 2 - x 2 ) 2 ] -½ [1 – (¾)(b 2 - x 2 )] This integrates to give: τ  (2π/c)(2κ) -½ [1 – (⅜)κb 2 ] = 2π(m/k) ½ [1 - (3/16) kb 2 (mc 2 ) -1 ] For the non-relativistic case, the period is τ 0 = 2π(m/k) ½ So, we can write: τ  τ 0 [1 - (3/16) kb 2 (mc 2 ) -1 ]  Define: The lowest order relativistic correction to the period: Δτ  τ - τ 0 = - (3/16) τ 0 kb 2 (mc 2 ) -1 = - (⅜)ετ 0 Similarly, the lowest order relativistic correction to the frequency: Δν  ν - ν 0 = (3/16) ν 0 kb 2 (mc 2 ) -1 = (⅜)εν 0

12. Summary: Relativistic 1d simple harmonic oscillator: Lowest order corrections to non- relativistic frequency & period are: (Δν/ν 0 ) = - (Δτ/τ 0 ) = (3/16) kb 2 (mc 2 ) -1 = (⅜)ε Physics: A simple harmonic oscillator, amplitude b. The period (& frequency) of a relativistic oscillator depends on amplitude, in sharp contrast to the non-relativistic case! The functional form of x(t)? Go back to c -2 (dx/dt) 2 = 1 - m 2 c 4 (E - V) -2 (2) Solve for dt & integrate to get t = t(x). In general, an elliptic integral. In the same (lowest order correction) approximation we just did, it can be done in closed form. Then, invert to get x(t). Student exercise!

13. Example 3. A relativistic charged particle in a constant magnetic field B. We could start with the Lagrangian L with the velocity-dependent potential of the form (Ch. 1, with relativistic correction): L = - mc 2 [1 - β 2 ] ½ - q  + qA  v –Use a gauge where the scalar potential  = 0, & the vector potential A = (½)B  r However, we know that this problem results in the Lorentz Force on a charged particle: F = q(v  B)  The equation of motion (from the relativistic force discussion) is: (dp/dt) = q(v  B) = (q/m)γ -1 (p  B) where p = γmv = Relativistic momentum

14. (dp/dt) = q(v  B) = (q/m)γ -1 (p  B) (1) Solving (1) is almost identical to solving the analogous equation in the non-relativistic limit, except for the presence of γ  1. The result is therefore qualitatively similar to the non- relativistic result: Because F  v = 0, no work is done on the particle by the B field  The energy E is constant. Also, the component of p along B is constant.  The motion is in the plane  B. We can easily show that the vector p precesses around the direction of B with a frequency: Ω = qB(mγ) -1. The only difference between this & the non-relativistic result is the γ  1 in the denominator. Ω  Cyclotron Frequency. The particle is moving in a circle in the plane  B. We can easily get the circle radius in terms of the momentum as: r = p(qB) -1 (the same as in the non-relativistic case!). Emphasize: We MUST use the relativistic momentum in all expressions!