Force

Problem:

A relativistic particle accelerates from rest at the origin at t = 0 under the influence of a constant force F = mgi.
(a)  Calculate its velocity and position as a function of time t.
(b)  Do your expressions for the velocity and position as a function of time yield the non-relativistic results if v/c << 1?

Solution:

• Concepts:
  F = dp/dt
• Reasoning:
  The relativistic momentum is  p = mv/(1 - v²/c²)^½.
• Details of the calculation:
  (a)  F = dp/dt,  p = i mv/(1 - v²/c²)^½, F = mgi.
  This problem involves only motion in one dimension, along the x-axis.
  d(v/(1 - v²/c²)^½)/dt = g,  v/(1 - v²/c²)^½ = gt + constant,   v(0) = 0 --> constant = 0.
  v² = (gt)² - (v²/c²)(gt)²,  v(t) = gt/(1 + (gt/c)²)^½.
  dx/dt = gt/(1 + (gt/c)²)^½, 
  x(t) = ∫₀^tgt'dt'/(1 + (gt'/c)²)^½ = (c²/g)[(1 + (gt/c)²)^½ - 1] + x₀.
  x(0) = 0 --> x₀ = 0,  x(t) = (c²/g)[(1 + (gt/c)²)^½ - 1].
  (b)  In the non-relativistic limit v/(1 - v²/c²)^½ = v = gt, neglecting terms higher than second order in an expansion in powers of v/c.
  x(t) = (c²/g)[(1 + (gt/c)²)^½ - 1] = (c²/g)[(1 + ½(gt/c)²) - 1] = ½gt².
  These are the non-relativistic results.

Problem:

A point mass m, at rest at the origin at t = 0 is acted on by a constant force F = ma i.  Find the time T it takes for the mass to reach a speed v = 0.75c and the position of the mass at time T.

Solution:

• Concepts:
  This is a one-dimensional problem:  F = dp/dt,  E = (m²c⁴ + p²c²)^½,  v/c = pc/E
• Reasoning:
  The equation of motion quickly yields v(t).  We integrate to find x(t).
• Details of the calculation:
  (a)  Equation of motion:  F = dp/dt
  p = Ft.  E = (m²c⁴ + (Ft)²c²)^½,  v/c = Ftc/(m²c⁴ + (Ft)²c²)^½ = at/(c² +a²t²)^½.
  v = dx/dt,  x(t) = ∫₀^t Ftc²dt/(m²c⁴ + (Ft)²c²)^½ = (1/F)∫₀^Ftc ydt/(m²c⁴ + y²)^½
  = (1/F)(m²c⁴ + y²)^½|₀^Ftc = (1/F)(m²c⁴ + (Ftc)²)^½ - mc²/F.
  x(t) = (c⁴/a² + (tc)²)^½ - c²/a.
  When v/c = aT/(c² +a²T²)^½ = 0.75 or a²T² - (9/16)a²T² = (9/16)c², then T² = (9/7)(c²/a²),
  T = (3/ √7)(c/a).
  x(T) = (c⁴/a² + (9/7)c⁴/a²)^½ - c²/a = 0.51 c²/a.

Problem:

At t = 0 in the lab frame a particle of mass 1 kg has velocity (v_x, v_y) = (0.6 c, 0) and acceleration (a_x, a_y) = (2, 3) m/s².  What force is acting on the particle in the lab frame and in an inertial frame moving with velocity v = 0.6 c i at t = 0, i.e. instantaneous inertial rest frame of the particle.  (You can assume that at t = 0 the origins of the two frames coincide and the particle is at the origin.)

Solution:

• Concepts:
  F = dp/dt, 4-vectors, frame transformations
• Reasoning:
  We are given dv/dt in the lab frame.  We must find the relationship between dv/dt and dp/dt to find F.
• Details of the calculation:
  In the lab frame:
  F_x = dp_x/dt,  F_y = dp_y/dt,  p_x = γmv_x, p_y = γmv_y.
  dp_x/dt = γma_x + (dγ/dt)mv_x,  dp_y/dt = γma_y + (dγ/dt)mv_y.
  dγ/dt = d(1 - v_x²/c² - v_y²/c²)^-½/dt = [(v_xdv_x/dt + v_ydv_y/dt)/c²]γ³.
  At t = 0 we have dγ/dt = [v_x(dv_x/dt)/c²]γ³, since v_y = 0.
  Therefore
  dp_x/dt = γma_x + γ³ma_x[v_x²/c²] =  γma_x[1 + γ²v_x²/c²] = γ³ma_x.
  dp_y/dt = γma_y.
  (F_x, F_y) = m(γ³a_x, γa_y) , where γ = [1/(√{1 - v_x²/c² - v_y²/c²})] = 5/4 .
  Thus (F_y, F_x) = ((5/4)³*2, (5/4)*3)N = (125/32, 15/4)N.

  To transform to another inertial frame we use
  (p₀, p) = (E/c, p) is a 4-vector.
  At t = 0 we have p_x' = -γ'β'E/c + γ'p_x,  p_y' = p_y.
  Here β' and γ' are constants whose magnitude depends on the relative velocity of the two inertial frames.
  dp_x'/dt = d(-γ'β'E/c + γ'p_x)/dt,  dp_y'/dt = dp_y/dt.
  The proper time interval is measured in the particles rest frame.
  We have dt = γ'dt' = γ'dτ when transforming to the instantaneous rest frame of the particle.
  F_x' = dp_x'/dt' = γ'd(-γ'β'E/c + γ'p_x)/dt.
  We write -γ'β'E/c + γ'p_x = γ'γm(v_x - v'),
  since E = γmc², β'E/c = γmv', and p_x = γmv_x.
  Here γ = (1 - v²/c²)^-½, where v is the speed of the particle in the lab frame.
  dγ'γm(v_x - v')/dt = mγ'[(dγ/dt)(v_x - v') - γ(dv_x/dt)]
  At t = 0 we have (v_x - v') = 0, and
  dγ'γm(v_x - v')/dt = mγ'γa_x = mγ²a_x,
  F_x' = mγ³a_x = F_x.
  For a frame such as the particle's, moving along with speed 0.6c with respect to the lab we have F_x' = F_x and F_y' = γF_y . Thus the force in the instantaneous inertial rest frame of the particle is (F_x', F_y') = (125/32, 75/16)N = 25/16(5/2, 3)N.

Problem:

A particle with mass m is launched upward at t = 0 from rest at y = 0 with relativistic momentum p = p_y > p₀.  It is subject to a constant force F in the negative y direction.
(a)  Find the particle's position as a function of time.
(b)  Show that for short times and a small initial speed your equation approximates the non-relativistic result.
(c)  Determine the time T at which the particle returns to y = 0.
NOTE: Give all answers for the laboratory frame.

Solution:

• Concepts:
  One-dimensional problem:  F = dp/dt,  E = (m²c⁴ + p²c²)^½,  v/c = pc/E

• Reasoning:
  We are asked to solve the equation of motion for a relativistic particle.

• Details of the calculation:
  One-dimensional problem:  F = dp/dy,  E = (m²c⁴ + p²c²)^½,  v/c = pc/E
  (a)  Equation of motion:  F = dp/dt
  p = p₀ - Ft.  E = (m²c⁴ + (p₀ - Ft)²c²)^½,  v/c = (p₀ - Ft)c/(m²c⁴ + (p₀ - Ft)²c²)^½.
  v = dy/dt,  y(t) = ∫₀^t (p₀ - Ft)c²dt/(m²c⁴ + (p₀ - Ft)²c²)^½.
  Let t' = p₀ - Ft, dt' = -Fdt.  Then
  y(t) = -(c/F)∫_p0^{p0 - Ft} t'/(m²c² + t'²)^½ = -(c/F)(m²c² + t'²)^½|_p0^{p0 - Ft}.
  y(t) = -(c/F)(m²c² + (p₀ - Ft)²)^½ + (c/F)(m²c² + p₀²)^½.

  (b)  Let p₀ and F_t be much smaller than mc.  Then
  y(t) = -(1/F)(m²c⁴ + (p₀ - Ft)² c²)^½ + (1/F)(m²c⁴ + p₀² c²)^½
  ~ (mc²/F)[-1 - (p₀ - Ft)²/(2m²c²) + 1 + p₀²/(2m²c²)]
  2p₀Ft/(2mF)  - F²t²/(2mF) = p₀t/m  - ½Ft²/m = v₀t - ½at².

  (c)  y(T) = 0.  (m²c² + (p₀ - FT)²)^½ = (m²c² + p₀²)^½,  (p₀ - FT) = ±p₀, T = 2p₀/F.
  This is the same as the non-relativistic result.