study the motion of a particle of mass m in a central potential U(r) = γr^2 with
γ > 0.
a) write the equations of motion in cartesian coordinates (x, y, z). are they coupled or independent with each other?
b) solve the equation of motion for x, y, and z directions.
c) is the angular momentum conserved? can you select a special coordinate system to simplify the problem?
d) can the orbit be an ellipse? if yes, what conditions should the parameters in your solution satisfy?
a) The equations of motion in Cartesian coordinates (x, y, z) for a particle of mass m in a central potential U(r) = γr^2 are:
mx'' = -dU/dr * (x/r)
my'' = -dU/dr * (y/r)
mz'' = -dU/dr * (z/r)
These equations are coupled with each other since the derivative of the potential U with respect to r appears in the equations of motion for each coordinate.
b) To solve the equations of motion in each direction, we begin by expressing the potential U(r) in terms of the Cartesian coordinates:
U(x, y, z) = γ(x^2 + y^2 + z^2)
In the x-direction, we have:
mx'' = -(2γ/m) * x
This is a simple harmonic oscillator equation, and its solution is given by:
x(t) = A * cos(ωt + φ)
where A is the amplitude, ω = sqrt(2γ/m) is the angular frequency, and φ is a phase constant.
By similar reasoning, the solutions for y(t) and z(t) can be written as:
y(t) = B * cos(ωt + φ')
z(t) = C * cos(ωt + φ'')
where B, C, φ', and φ'' are constants.
c) The angular momentum L of a particle is given by:
L = m * (r x v)
where r is the position vector and v is the velocity vector.
In this central potential, the angular momentum is conserved since the potential U(r) is spherically symmetric. The magnitude of the angular momentum remains constant, but its direction may change as the particle orbits around the central point.
To simplify the problem, it is convenient to choose a coordinate system where the motion is planar, meaning the particle moves only in the xy-plane or xz-plane. This can be achieved by considering the conserved angular momentum vector and aligning it with the z-axis.
d) Yes, the orbit can be an ellipse. For an elliptical orbit, the condition is that the total mechanical energy E of the particle must be negative. The mechanical energy is given by:
E = (1/2) * m * (x'^2 + y'^2 + z'^2) + U(x, y, z)
where the prime denotes differentiation with respect to time.
In this central potential, E can be expressed as:
E = (1/2) * m * (x'^2 + y'^2 + z'^2) + γ * (x^2 + y^2 + z^2)
For an ellipse, the condition is that the total mechanical energy E satisfies:
E < 0
This condition will impose constraints on the parameters A, B, C, and the constants.
a) The equations of motion in Cartesian coordinates (x, y, z) can be derived using Newton's Second Law. The central potential U(r) = γr^2 can be written as U(x, y, z) = γ(x^2 + y^2 + z^2). The equations of motion are:
mx'' = -∂U/∂x = -2γx
my'' = -∂U/∂y = -2γy
mz'' = -∂U/∂z = -2γz
These equations are coupled, meaning that the motion in each coordinate direction depends on the values of the other coordinate directions.
b) To solve the equations of motion for x, y, and z directions, we can assume solutions of the form:
x = Xcos(ωt + ϕ)
y = Ycos(ωt + ϕ)
z = Zcos(ωt + ϕ)
Substituting these expressions into the equations of motion, we get:
-mω^2Xcos(ωt + ϕ) = -2γXcos(ωt + ϕ)
-mω^2Ycos(ωt + ϕ) = -2γYcos(ωt + ϕ)
-mω^2Zcos(ωt + ϕ) = -2γZcos(ωt + ϕ)
Simplifying, we obtain:
ω^2 = 2γ/m
Therefore, the solution for the motion in x, y, and z directions is given by:
x = Xcos(ωt + ϕ)
y = Ycos(ωt + ϕ)
z = Zcos(ωt + ϕ)
c) The angular momentum (L) is defined as L = m(r × v), where r is the position vector and v is the velocity vector. To determine if the angular momentum is conserved, we differentiate L with respect to time:
dL/dt = m(d(r × v)/dt) = m(d/dt)(r × v)
Using the product rule for differentiation, we get:
dL/dt = m(dr/dt × v) + m(r × dv/dt)
Since the motion is in a central potential, the force is always directed along the radius vector (r). This implies that dr/dt and dv/dt are parallel to each other and to r. Therefore, the cross product r × v is zero, and the angular momentum is conserved.
We can select a special coordinate system called spherical coordinates (r, θ, φ) to simplify the problem. In this coordinate system, the motion depends only on the radial coordinate (r) and the two angular coordinates (θ, φ).
d) The orbit can be an ellipse. For the orbit to be an ellipse, the parameters in the solution should satisfy the following conditions:
- The amplitudes X, Y, and Z should be non-zero.
- The phase ϕ should be non-zero.
- The angular frequency ω should be positive, which requires γ > 0.
- The masses m should be positive.
These conditions ensure that the particle undergoes periodic motion and traces an elliptical path.
a) To analyze the motion of a particle in a central potential, we can start by writing the equations of motion in Cartesian coordinates (x, y, z). The central potential U(r) = γr^2 implies that the potential only depends on the distance from the origin (r) and has rotational symmetry.
Let's assume the particle's position as (x, y, z) at time t. The forces acting on the particle are the gravitational force (due to the central potential) and any other external forces. We can write the following equations of motion in Cartesian coordinates:
For x-axis: m * d^2x/dt^2 = -dU(r)/dx + F_ext(x)
For y-axis: m * d^2y/dt^2 = -dU(r)/dy + F_ext(y)
For z-axis: m * d^2z/dt^2 = -dU(r)/dz + F_ext(z)
Here, U(r) is the central potential, and F_ext(x), F_ext(y), F_ext(z) are external forces acting on the particle (if any). The terms dU(r)/dx, dU(r)/dy, dU(r)/dz represent the partial derivatives of the potential with respect to x, y, and z.
b) Solving the equations of motion for each direction (x, y, z) is a mathematical process that involves integrating the equations or finding their solutions. The process can vary depending on the exact form of the central potential U(r) and any external forces.
c) In this scenario, where the potential U(r) = γr^2 is rotationally symmetric, the angular momentum is conserved. This means that the magnitude and direction of the angular momentum vector remain constant throughout the motion.
To simplify the problem further, we can choose a special coordinate system called the spherical coordinate system. In this system, the position of the particle is described by its radial distance (r), azimuthal angle (θ), and polar angle (φ). Choosing this coordinate system makes the equations of motion decoupled, meaning the equations for the radial, azimuthal, and polar directions can be solved independently.
In the spherical coordinate system, the angular momentum vector (L) is aligned with the polar axis (z-axis), simplifying the mathematical analysis.
d) Yes, the orbit of the particle can be an ellipse. For the orbit to be an ellipse, the energy of the particle should be negative. The condition for a particle of mass m to have a bound (closed) orbit is given by:
E < 0
where E is the total mechanical energy of the particle given by the sum of kinetic and potential energies.
The total mechanical energy E can be written as:
E = (1/2) * m * (v^2) - U(r)
where v is the magnitude of the velocity vector and U(r) is the central potential.
For an elliptical orbit, the parameters in the solution should satisfy the condition that E < 0.