Show that the vector C = p × l − mkê_r is a conserved quantity for the motion in a central potential U(r) = −k/r. Here p is the linear momentum, and l is the angular momentum.
Note: C, p, and l are supposed to have arrows on top to indicate they are vectors; ê_r is a unit vector in the direction of r.
C = p × l − mkê_r is the Laplace-Runge-Lenz (LRL) vector.
Ah, vectors and conservation, two of my favorite topics! Let's have some fun exploring this.
We want to show that the vector C = p × l - mkê_r is conserved. To do this, we need to show that its derivative with respect to time, dC/dt, is zero. So, let's dive in!
First, let's break down the vector C step by step. The cross product p × l can be calculated using the formula:
(p × l)_i = ε_ijk * p_j * l_k
where ε_ijk is the Levi-Civita symbol and it takes values of +1, -1, or 0 depending on the permutation of i, j, and k. Now, since we are working in spherical coordinates, our i, j, and k components are r, θ, and φ, respectively.
So, let's calculate it. For the r component:
(p × l)_r = ε_rjk * p_j * l_k
= ε_rφk * p_φ * l_k (since j ≠ r)
= (δ_rφl_θ − δ_rθl_φ) * p_φ * l_θ (using the Levi-Civita symbol)
= −l_θ * p_φ (since δ_rθ = 0 and δ_rφ = 1)
So, the r component of C is −l_θ * p_φ.
Now, let's calculate the -mkê_r part. Since ê_r is a unit vector in the direction of r, we can write ê_r = (1/r) * r̂, where r̂ is the unit vector in the direction of r.
Therefore, -mkê_r = −mk * (1/r) * r̂.
Now, let's combine the components we found and see what we get:
C = −l_θ * p_φ − mk * (1/r) * r̂
Okay, now that we have C, let's differentiate it with respect to time and see what happens:
dC/dt = d/dt (−l_θ * p_φ − mk * (1/r) * r̂)
Now, since p, l, m, and k are constants (in this context), their derivatives with respect to time vanish. Therefore, the derivative of C simplifies to:
dC/dt = −d/dt (mk * (1/r) * r̂)
The derivative of (1/r) with respect to time is zero, and the unit vector r̂ only depends on the spatial coordinates, not time. Hence, dC/dt = 0.
Voilà! We've shown that the vector C = p × l - mkê_r is conserved for the motion in a central potential U(r) = −k/r. It's always fun to have a conserved quantity around to keep things stable, doesn't it?
To prove that the vector C = p × l - mkê_r is conserved for the motion in a central potential U(r) = -k/r, we need to show that its time derivative is equal to zero, i.e., dC/dt = 0.
Let's calculate the time derivative of C:
dC/dt = dp/dt × l + p × dl/dt - m(dkê_r)/dt
The first term, dp/dt × l, can be rewritten as the cross product of the force and the distance vector:
dp/dt × l = F × r
By Newton's second law, F = d(p/m)/dt, so we can substitute it into the equation:
dp/dt × l = (d(p/m)/dt) × r
The second term, p × dl/dt, represents the angular momentum changing with time. In central potential motion, angular momentum is conserved, so dl/dt = 0. Therefore, the second term is zero:
p × dl/dt = 0
The third term, m(dkê_r)/dt, is zero because the potential energy U(r) = -k/r does not depend on time. Thus, dk/dt = 0.
Combining all the terms:
dC/dt = (d(p/m)/dt) × r + 0 - 0
Since the force is central and depends only on the distance r, then it is oriented along the r direction. Therefore, we can write F = faê_r, where fa is a scalar. Now, let's express dp/dt as d(mv)/dt = ma. Substituting this into the equation above, we get:
dC/dt = fa * r
Since the force fa is collinear with r, the cross product of fa with r is zero: fa × r = 0. Hence:
dC/dt = 0
Therefore, we have shown that the vector C = p × l - mkê_r is a conserved quantity for the motion in a central potential U(r) = -k/r, as its time derivative is zero.
To show that the vector C = p × l − mkê_r is conserved for motion in a central potential, we need to demonstrate that its time derivative is zero.
Let's begin by finding the time derivative of vector C, denoted as dC/dt.
dC/dt = d/dt (p × l - mkê_r)
Using the product rule for differentiation, we can expand this expression:
dC/dt = d/dt (p × l) - d/dt (mkê_r)
Now, let's evaluate each term separately.
First, consider the term d/dt (p × l). This derivative can be computed using the properties of the cross product:
d/dt (p × l) = dp/dt × l + p × dl/dt
The first term involves the time derivative of the linear momentum p. However, since vector p represents linear momentum, it is given by p = mẋ, where m is the mass and ẋ is the velocity. Therefore, dp/dt = m dẋ/dt = m a, where a is the acceleration.
Similarly, the time derivative of the angular momentum l can be expressed as dl/dt = r × v, where r is the position vector and v is the velocity vector.
d/dt (p × l) = m a × l + p × (r × v)
Next, let's consider the second term, d/dt (mkê_r). Since k is a constant, its derivative with respect to time is zero. This means that d/dt (mkê_r) = 0.
Substituting these expressions back into the original equation, we have:
dC/dt = m a × l + p × (r × v) - 0
Now, utilizing the properties of vector algebra, we can simplify further:
dC/dt = m a × l + p × (r × v)
= m (ẍ × l) + p × (r × v)
= m r¨ × l + p × (r × v)
= -m r × (dr/dt) × l + p × (r × v)
= -m r × v × l + p × (r × v)
= -m (r × v) × l + p × (r × v)
At this point, we can observe that the term (r × v) is common to both terms of the equation. Therefore, we can factor it out:
dC/dt = [(p × (r × v)) - m r × v] × l
Now, consider the potential energy U(r) = -k/r. The force is given by F = -∇U, where ∇ represents the gradient operator. The radial component of the force can be written as:
F_r = -dU/dr = d/dt (-k/r)
= k/r^2
Since the force acts along the radial direction, we have F = F_r ê_r, where ê_r is the unit vector in the radial direction. Thus, the radial force is:
F = k/r^2 ê_r
Now, recall that the linear momentum p is given by p = mẋ. Therefore, the acceleration is:
a = dp/dt = m dv/dt
Using Newton's Second Law, F = ma, we can write:
k/r^2 ê_r = m dv/dt
Since ê_r is in the radial direction, we can express the velocity vector v as:
v = v_r ê_r
Substituting these expressions, we have:
k/r^2 ê_r = m d(v_r ê_r)/dt
= m (d(v_r)/dt) ê_r
Comparing the radial components, we get:
k/r^2 = m d(v_r)/dt
Since the potential energy U(r) = -k/r contributes only to the radial component of the force, we can say that the motion in a central potential U(r) = -k/r implies that the radial acceleration a_r = d(v_r)/dt is zero.
With this information, we can now simplify the expression for dC/dt:
dC/dt = [(p × (r × v)) - m r × v] × l
= [(m r × v) × (r × v)] × l
= [(m r × v) × (r × v)] × l
= (0) × l (since a_r = 0)
= 0
Since dC/dt is equal to zero, we have shown that the vector C = p × l - mkê_r is conserved for motion in a central potential U(r) = -k/r.