has partial derivatives
where by V'(x) we mean dV/dx. That gives, by the chain rule,
Putting these all together in the Euler-Lagrange equation,
Thus the Euler-Lagrange equation implies that one or both of the
following is true: 1) the particle's energy is zero; 2) Newton's
second law is true. Of course, requiring only the first case to be
true requires that the particle's zero energy remain constant
throughout the motion, which itself implies Newton's second law. Thus,
the Euler-Lagrange equation for this particle gives Newton's second
law. The simpler Lagrangian, 
, with conserved energy 
, would give the same results.
In the reference frame of the railroad car, the bob has position 
, 
(taking the point of support to be at
the origin). In an inertial (external) reference frame, the constant
x-acceleration of the railroad car is superposed as well, so that
(taking x(0) = 0, 
)
and
Thus
and
Taking derivatives,
and
Putting these terms into the Euler-Lagrange equation, the terms with
cancel, leaving
which we rearrange to obtain
We then solve for the pendulum's motion at and near equilibrium. If we
treat 
as an ordinary coordinate, we can construct a potential
so that obeys the equation of motion
. We know that can undergo
oscillations about an equilibrium point 
defined by
, which implies that
. Here the equation of motion
(1) gives when
Near , we expand 
to get, from (1),
Taylor expanding to linear order in 
:
Substituting back in,
where we have grouped O(1) and 
terms in lines 1 and 2
respectively. The first line is just 
(see
(1)), which vanishes. Thus
which we can rewrite in Cartesian coordinates as
where 
is the equilibrium position of the pendulum bob in
the railroad frame.
To proceed further, we must consider (2) more carefully. By
construction (assuming that the pendulum bob remains inside the
railroad car), 
. Equation
(2) implies that has a negative tangent. In this
range of , 
is positive so 
must be negative,
i.e. 
. Thus
Knowing 
and 
gives
Thus
Plugging these back into equation (3) gives
the equation of motion for a simple harmonic oscillator with frequency
.
, we eliminate y in writing our Lagrangian.
We have the usual
which here gives
Note that
and
The E-L equation for X thus gives
Similarly
and
The E-L equation for x thus gives
Note that adding the two E-L equations together just reproduces conservation of horizontal momentum,
times the E-L equation for X to 
times the E-L
equation for x. (Note that all terms with 
would cancel in
this addition.) This complicated addition gives
In the limit where M ;SPMgt;;SPMgt; m, this gives
We could have gotten the same result looking at our original E-L equations. In the E-L equation for X, all terms are negligible except for 
, giving 
. Plugging into the E-L equation for x gives the limit ODE above for x.
Why does this agree with our expectations for motion on an incline
plane? Well, since , the incline plane remains at rest
at X=0. Thus our constraint becomes 
. This gives horizontal and vertical acceleration
of the block
This makes sense since the only acceleration of the block is parallel
to the incline plane, driven by the parallel component 
of the gravitational force. That parallel component of the gravitational force has component 
in the -x-direction, and 
in the -y-direction.
The time elapsed falling down the wire is
The integrand
is a function of y(x) and y'(x) so the curve of least time obeys the Euler-Lagrange equation
Note that
and
The E-L equation for y thus gives (after gathering everything over the common denominator 
)
or more simply,
as given.
