Euler-Lagrange Equations for Gravitation: UNIZOR.COM - Physics+ 4 All - Lagrangian

Notes to a video lecture on UNIZOR.COM

Euler-Lagrange Equations
for the Gravitational Field

The following is an illustration of using Lagrangian Mechanics to analyze the movement of a planet in a central gravitational field of the Sun.
It will also show how the Kepler's laws of planetary movements are derived from Euler-Lagrange equations.

Let the Sun be modeled as a point mass M and a planet be modeled as a point mass m.
In the lecture about a central field we proved that the orbit of a planet lies in a plane. That allows us to choose polar coordinates r(t) and θ(t) with the Sun at the origin as generalized coordinates.

To apply Euler-Lagrange equation, we have to express kinetic energy T and potential energy U in terms of generalized coordinates r and θ.

Kinetic energy depends on the square of the magnitude of velocity v. In polar coordinates this is expressed as a sum of squares of radial speed v_r and perpendicular to it tangential speed v_θ:
v_r = r'
v_θ = r·θ'
v² = v_r²+v_θ² = (r')²+(r·θ')²
T = ½mv² = ½m[(r')²+(r·θ')²]

Potential energy of a planet in the gravitational field (you can refer to lectures in
Physics 4 Teens → Energy → Energy of Gravitational Field) is
U = −G·M·m/r
where G is the universal Gravitational Constant

Lagrangian of a planet is
L = T − U =
= ½m[(r')²+(r·θ')²] + G·M·m/r

The Euler-Lagrange equations for generalized coordinate are
∂L/∂r = d/dt ∂L/∂r'
∂L/∂θ = d/dt ∂L/∂θ'

From the equation for r:
∂L/∂r = m·r·(θ')² − G·M·m/r²
∂L/∂r' = m·r'
d/dt ∂L/∂r' = m·r"
Resulting equation is
m·r·(θ')² − G·M·m/r² = m·r"
Cancelling m produces
r·(θ')² − G·M/r² = r"

From the equation for θ:
∂L/∂θ = 0
(because L does not explicitly depend on θ)
∂L/∂θ' = m·r²·θ'
Resulting equation is
0 = d/dt m·r²·θ'
from which follows that
m·r²·θ' = const = |L|
The expression on the left side of the above equation is the magnitude of a angular momentum vector (usually it's denoted by symbol L, but we will use |L| to separate it from the Lagrangian).
Since the Lagrangian does not depend explicitly on θ, this coordinate is cyclic, and therefore the corresponding angular momentum is conserved.
So, this equation expresses the angular momentum conservation law.
At the same time the constancy of the expression r²·θ' means that equal areas are swept by the radius vector per unit of time. Thus, Kepler's Second Law follows directly from the Euler–Lagrange equations.
From this follows that r²·θ' is also a constant of motion (angular momentum per unit of mass), let's call this constant k. Now the second Euler-Lagrange equation looks like this
r²·θ' = |L|/m = k

The system of both Euler-Lagrange equations for two generalized coordinates is

r·(θ')² − G·M/r² = r"
r²·θ' = k

From the physical standpoint, the second equation determines how fast the planet moves along the orbit, while the first determines the shape of the orbit.

Let's use the expression for θ'=k/r² from the second Euler-Lagrange equation and substitute it into the first term r·(θ')² of the first equation
r·(θ')² = r·(k/r²)² = k²/r³
Now the first equation looks like
k²/r³ − G·M/r² = r"
This is exactly the equation we derived in the lecture Motion in Polar Coordinates of the Laws of Kepler part of this course, though the derivation in that part was much longer and involved a lot of vector manipulations in Cartesian coordinates.

Both Euler-Lagrange equations define r(t) and θ(t) as functions of time.
In a tradition of using polar coordinates, let's derive r as a function of θ.

Differentiation by time of an expression X can be done according to the following procedure
[A1] dX/dt = (dX/dθ)·(dθ/dt)
or, equivalently,
[A2] dX/dθ = (dX/dt)/(dθ/dt) = X'/θ'

Apply rule [A2] for X=r:
dr/dθ = r'/θ'
and, therefore,
⇒ r' = (dr/dθ)·θ' =
use the second Euler-Lagrange equation above resolving it for θ'=k/r²
= (dr/dθ)·k/r² = −k·d(1/r)/dθ

Consider the equality we've got
r' = −k·d(1/r)/dθ

Let's differentiate it by time. On the left side we will get r".

To differentiate d(1/r)/dθ on the right side by time, we will use the rule [A1] above getting
d/dt[d(1/r)/dθ] =
= d/dθ[d(1/r)/dθ]·(dθ/dt) =
= [d²(1/r)/dθ²]·θ' =
= [d²(1/r)/dθ²]·k/r²

Equating derivatives of the left and the right sides, we get
r" = −(k²/r²)·[d²(1/r)/dθ²]

This expression for r" we substitute as the right side of the first Euler-Lagrange equation getting
r·(θ')² − G·M/r² =
= −(k²/r²)·[d²(1/r)/dθ²]

Multiplying both sides by r²:
r³·(θ')² − G·M =
= −k²·[d²(1/r)/dθ²]
Again substitute θ'=k/r² getting
r³·k²/r⁴ − G·M =
= −k²·[d²(1/r)/dθ²]
or
(1/r) − G·M/k² = −[d²(1/r)/dθ²]

Assign x=1/r. In terms of x as a function of θ our equation becomes
d²x/dθ² + x = G·M/k²
This is a known type of differential equation, in a simplified form its solutions are
x(θ) = a + b·cos(θ)
More precisely, the general solution is
x(θ) = G·M/k² + C·cos(θ−θ₀)
from which follows
r(θ) = 1/[G·M/k²+C·cos(θ−θ₀)]

The last equation in polar coordinates represents second degree curves (called conic sections - circle, ellipse, parabola, hyperbola) with eccentricity e=C·k²/(G·M).
Geometrically,
if e is smaller than 1, a curve is an ellipse;
if e is equal to 1, a curve is a parabola;
if e is greater than 1, a curve is a hyperbola.

This equation describes an orbit of an object flying around the Sun along a trajectory that is a conic section. This is in compliance with Kepler's laws of planetary movement.

Thus, using the methodology of Lagrangian Mechanics, we were able to derive the Law of Conservation of Angular Momentum, Kepler's Second Law and the shape of a planetary orbit - a curve of the second degree (conic section).