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?
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.
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.)
To transform to another inertial frame we use:
(p0, p) = (E/c, p) is a 4-vector.
At t = 0
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)-½, 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.
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.
One-dimensional problem: F = dp/dt, E = (m2c4 + p2c2)½, v/c = pc/E
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 = 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)∫p00p0 – 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.