Lagrangian
First of all, let's stipulate that the Laws of Newton are based on experiment, they are not derived from some more fundamental theories.
Lagrangian mechanics presents a different approach to analyze the motion than Newtonian mechanics.
In many cases it presents a simpler, more universal way to describe the motion of a mechanical system than Newtonian one.
Let's start with an example where both methodologies lead to the same result.
Spring Oscillation
Consider an ideal spring with one end fixed and a point-mass attached to another end.
The oscillations will occur along the length of a spring that coincides with X-axis.
Position of a point-mass on the spring's end will be described by it X-coordinate x(t) as a function of time t with initial position at time t=0 being an origin of X-coordinate, that is x(0)=0.
According to the Hooke's Law, the force F of a spring applied to an object attached to its end is proportional to a length x by which this spring is stretched or squeezed from its neutral position and directed always towards a neutral point x=0 of no stretching or squeezing.
F = −k·x
where k is a coefficient of elasticity that characterizes physical properties of a spring.
According to the Newton's Second Law, the acceleration a of an object is proportional to a force F applied to it
F = m·a
where m is the object's mass being a coefficient of proportionality.
A linear acceleration a(t), as a function of time, is a derivative of a linear speed v(t) by time t
a(t) = dv(t)/dt = v'(t)
A linear speed v(t) is, in turn, a derivative of a position of an object x(t) by time
v(t) = dx(t)/dt = x'(t)
Therefore, an acceleration is a second derivative of a position by time
a(t) = d²x(t)/dt² = x"(t)
Equating the value of force by Hooke's Law to that of Newton's Second Law, we get a differential equation that defines a motion of the object
−k·x(t) = m·a(t)
or, equivalently,
−k·x(t) = m·x"(t)
Solution to this differential equation of the second order is a trajectory of our object.
Let's approach the same problem from another side.
An object attached to a spring's end that has mass m and linear speed v has kinetic energy that is equal to
Ekin = ½m·v²
Since speed v(t), as a function of time t is just a derivative of a position x(t) by time, we can express kinetic energy in terms of position, as in the case of potential energy above
Ekin = ½m·[x'(t)]²
NOTICE:
Ekin depends explicitly only on speed x'(t) and
d/dt[∂(Ekin)/∂x'] =
= d/dt[m·x'] = mx"(t) = F
A stretched or a squeezed spring has potential energy equal to the amount of work needed to stretch or squeeze it against the force of its elasticity (you can refer to a lecture Physics 4 Teens - Energy - Potential Energy - Spring on UNIZOR.COM).
Thus, a potential energy of a spring squeezed or stretched by the length x(t), as a function of time t, equals to
Epot = ½k·[x(t)]²
where k is the same coefficient of elasticity as above that characterizes the physical properties of a spring.
NOTICE:
Epot depends explicitly only on position x(t) and
∂(−Epot)/∂x = −k·x(t) = F
Based on two NOTICEs above, it is IMPORTANT to see that
∂(−Epot)/∂x =
= d/dt[∂(Ekin)/∂x'] = F
Since Epot depends explicitly only on position x(t), not on speed x'(t), and Ekin depends explicitly only on speed x'(t), not on position x(t),
∂(Ekin−Epot)/∂x =
= ∂(−Epot)/∂x =
= d/dt[∂(Ekin)/∂x'] =
= d/dt[∂(Ekin−Epot)/∂x']
Φ[f(x)] = ∫[a,b] F[x,f(x),f '(x)]dx
where F[...] is some known smooth real function of three arguments - real variable x, real value of function f(x) and real value of derivative f '(x).
This Euler-Lagrange differential equation looks like this:
(∂/∂f)F [x,f0(x),f '0(x)] =
= (d/dx)(∂/∂f ')F [x,f0(x),f '0(x)]
Let's change more abstract symbols x and f(x) to those applicable to our task.
The argument will be time t instead of abstract x. The function will be a position x(t) instead of abstract f0(x).
Now the Euler-Lagrange equation looks like
(∂/∂x)F [t,x(t),x'(t)] =
= (d/dt)(∂/∂x')F [t,x(t),x'(t)]
Compare this to the equation above that equates partial derivative from Ekin−Epot by x with its partial derivative by x'.
Obviously, L=Ekin−Epot satisfies the Euler-Lagrange equation
(∂/∂x)L[t,x(t),x'(t)] =
= (d/dt)(∂/∂x')L[t,x(t),x'(t)]
which in this case is exactly the same as the equation obtained from the Newton's Second Law
−k·x(t) = m·x"(t)
Expression
L[x(t),x'(t)]=Ekin(x')−Epot(x)
is called Lagrangian.
At any moment it has certain kinetic and potential energy, so we can constract a Lagrangian
L(t) = Ekin(x'(t)) − Epot(x(t))
Consider an integral of this Lagrangian by time
S = ∫[t1,t2]L(t)·dt
This integral is call action.
The trajectory that minimizes or maximizes this action is a solution to an Euler-Lagrange equation
(∂/∂x)L[x(t),x'(t)] =
= (d/dt)(∂/∂x')L[t,x(t),x'(t)]
which has the same solution as Newtonian F=m·a.
Therefore, the trajectory obtained using a Lagrangian approach is the same the one from Newtonian mechanics. BUT IN SOME CASES IT MIGHT BE MUCH MORE CONVENIENT.
The equivalence of a differential equation obtained from the Newton's Second Law and the Euler-Lagrange equation is not just a coincidence peculiar for springs.
In general, kinetic energy always depends on mass and speed
Ekin = ½m·v²
In general, derivative of Ekin by speed v is a momentum of motion p
p = m·v = ∂/∂v(Ekin) =
= ∂/∂v(½mv²)
In general, derivative of momentum p by time t is the force F
dp/dt = d(m·v)/dt = m·a = F
In general, potential energy is, actually, an amount of work W.
Since dW=F·dx, its derivative by x is the force, and a derivative of potential energy by coordinates gives the force as well.
So, the Newton's Second Law and Euler-Lagrange equation are equivalent. Why do we need both?
Practical mechanical problems are rarely as simple as we are taught at high school.
It appears that the more complicated problems with more than one object involved are easier to solve using Lagrangian L=Ekin−Epot than to deal with complicated forces and their interaction constructing the equations of F=m·a type.
The next lectures will be dedicated to a few important physical problems and their solutions using Euler-Lagrange equation.
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