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:
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:
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/s^{2}. 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:
To transform to another inertial frame we use
(p_{0}, 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 = γmc^{2}, β'E/c = γmv', and p_{x} = γmv_{x}.
Here γ = (1 - v^{2}/c^{2})^{-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} = mγ^{2}a_{x},
F_{x}' = mγ^{3}a_{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.
A particle with mass m is launched upward at t = 0 from rest at y = 0 with
relativistic momentum p = p_{y} > p_{0}. 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 = (m^{2}c^{4}
+ p^{2}c^{2})^{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 = (m^{2}c^{4}
+ p^{2}c^{2})^{½}, v/c = pc/E
(a) Equation of motion: F = dp/dt
p = p_{0} - Ft. E = (m^{2}c^{4} + (p_{0}
- Ft)^{2}c^{2})^{½}, v/c = (p_{0} - Ft)c/(m^{2}c^{4}
+ (p_{0} - Ft)^{2}c^{2})^{½}.
v = dy/dt, y(t) = ∫_{0}^{t} (p_{0} - Ft)c^{2}dt/(m^{2}c^{4}
+ (p_{0} - Ft)^{2}c^{2})^{½}.
Let t' = p_{0} - Ft, dt' = -Fdt. Then
y(t) = -(c/F)∫_{p0}^{p0 - Ft} t'/(m^{2}c^{2}
+ t'^{2})^{½} = -(c/F)(m^{2}c^{2} + t'^{2})^{½}|_{p0}^{p0
- Ft}.
y(t) = -(c/F)(m^{2}c^{2} + (p_{0} - Ft)^{2})^{½}
+ (c/F)(m^{2}c^{2} + p_{0}^{2})^{½}.
(b) Let p_{0} and F_{t} be much smaller than mc. Then
y(t) = -(1/F)(m^{2}c^{4} + (p_{0} - Ft)^{2}
c^{2})^{½} + (1/F)(m^{2}c^{4} + p_{0}^{2}
c^{2})^{½}
~ (mc^{2}/F)[-1 - (p_{0} - Ft)^{2}/(2m^{2}c^{2})
+ 1 + p_{0}^{2}/(2m^{2}c^{2})]
2p_{0}Ft/(2mF) - F^{2}t^{2}/(2mF) = p_{0}t/m
- ½Ft^{2}/m = v_{0}t - ½at^{2}.
(c) y(T) = 0. (m^{2}c^{2} + (p_{0} - FT)^{2})^{½}
= (m^{2}c^{2} + p_{0}^{2})^{½},
(p_{0} - FT) = ±p_{0}, T = 2p_{0}/F.
This is the same as the non-relativistic result.