### 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) = ∫0tgt'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) = ∫0t Ftc2dt/(m2c4 + (Ft)2c2)½ = (1/F)∫0Ftc ydt/(m2c4 + y2)½
= (1/F)(m2c4 + y2)½|0Ftc = (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 (vx, vy) = (0.6 c, 0) and acceleration (ax, ay) = (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:
Fx = dpx/dt,  Fy = dpy/dt,  px = γmvx, py = γmvy.
dpx/dt = γmax + (dγ/dt)mvx,  dpy/dt = γmay + (dγ/dt)mvy.
dγ/dt = d(1 - vx2/c2 - vy2/c2)/dt = [(vxdvx/dt + vydvy/dt)/c23.
At t = 0 we have dγ/dt = [vx(dvx/dt)/c23, since vy = 0.
Therefore
dpx/dt = γmax + γ3max[vx2/c2] =  γmax[1 + γ2vx2/c2] = γ3max.
dpy/dt = γmay.
(Fx, Fy) = m(γ3ax, γay) , where γ = [1/(√{1 - vx2/c2 - vy2/c2})] = 5/4 .
Thus (Fy, Fx) = ((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 px' = -γ'β'E/c + γ'px,  py' = py.
Here β' and γ' are constants whose magnitude depends on the relative velocity of the two inertial frames.
dpx'/dt = d(-γ'β'E/c + γ'px)/dt,  dpy'/dt = dpy/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.
Fx' = dpx'/dt' = γ'd(-γ'β'E/c + γ'px)/dt.
We write -γ'β'E/c + γ'px = γ'γm(vx - v'),
since E = γmc2, β'E/c = γmv', and px = γmvx.
Here γ = (1 - v2/c2), where v is the speed of the particle in the lab frame.
dγ'γm(vx - v')/dt = mγ'[(dγ/dt)(vx - v') - γ(dvx/dt)]
At t = 0 we have (vx - v') = 0, and
dγ'γm(vx - v')/dt = mγ'γax = mγ2ax,
Fx' = mγ3ax = Fx.
For a frame such as the particle's, moving along with speed 0.6c with respect to the lab we have Fx' = Fx and Fy' = γFy . Thus the force in the instantaneous inertial rest frame of the particle is (Fx', Fy') = (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 = py > 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) = ∫0t (p0 - Ft)c2dt/(m2c4 + (p0 - Ft)2c2)½.
Let t' = p0 - Ft, dt' = -Fdt.  Then
y(t) = -(c/F)∫p0p0 - Ft t'/(m2c2 + t'2)½ = -(c/F)(m2c2 + t'2)½|p0p0 - Ft.
y(t) = -(c/F)(m2c2 + (p0 - Ft)2)½ + (c/F)(m2c2 + p02)½.

(b)  Let p0 and Ft 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.