I would like to describe the powerful method of Langrange-Euler equation in classical physics as opposed to directly applying Newtons laws. It is inherently an energy formulation but the results contain a few nuances. In the Lagrange formulation, the energy used is not the total energy, rather the delta of forms of energy. Let T denote the kinetic energy and V the potential energy. Then, we define L = T-V. This quantity is called the Lagrangian and it is not yet clear what we do with it. But, let us take a deep breath before we discuss a few simple but powerful observations now.
- The quantity defined here is a difference of two scalar quantities. So, the Lagrangian is (a function which can be evaluated to ouput) a scalar quantity that can be meaninfully referred to as smaller or larger, positive or negative etc. (just like numbers we encounter in everyday life; this is part of what makes it easier than dealing with forces as in the Newtonian formulation)
- Energy is a state function and it depends on the current state of the system only, independent of the history. This holds not just for the total energy but for the forms of energy. So, it cannot directly depend on the value of the time. In fact, such a time dependence on the clock reading would violate the conservation of energy. In other words, L(x, x’, t) can be reduced to L(x(t), x'(t)). Of course, the kinetic energy and potential energy do indirectly vary with time t, as the energy depends on the position and velocity of the components of the system and the positions in turn depend on the trajectory that is traced out in time.
- The Lagrangian cannot be re-formulated using the conservation of total energy i.e, T+V = constant (E) . For example, L = 2T – E would be equivalent but we cannot claim E is constant in this case as the Lagrangian is defined not only for all possible trajectories but also all physically impossible trajectories where E is not conserved. The domain of definition needs to be this vast for solving the possible solutions finally, as we shall see.
The principle is that the system will make an effort to reduce the time averaged value of L, as much as possible, according to physical constraints. This is encapsulated by defining the action, A = integral[L(x(t),x'(t)) dt], over the time interval considered. In order to minimize A, we have to lower L over time. Now, to minimize L over the entire interval is not trivial, but the principle does not need us to do that. Rather the principle is that L tends to a local minimum or stationary point. So, all we have to do from a starting L (x,x’) is to change x(t) to nearby values (which will also affect x'(t)), until the wiggling around of the shape of the curve x(t) does not reduce L(x,x’) anymore. And remember that we can wiggle around the curve x(t) as locally as we want, since the integral is just a sum of the contributions from every local interval. So, when we reach the condition that we cannot decrease L(x,x’) by local wiggling, we satsify the condition that any infinitesimal change of x(t) results in a zero corresponding change in the Lagrangian := D(L) = 0.
Let us approximate everything as a set of discrete points using a uniform time step Ts over the interval and then take the limit of the step size going to zero, in the spirit of calculus. Consider three consecutive points of x(t), to form a local picture. If we increase the middle point of x but leave the other two points unchanged, we have done a “local wiggle”. This naturally results in the slope of x increasing at the first point and then decreasing at the middle point by the same amount (compared to the slope values before the wiggle). In other words, if we change x at only a particular time step, leaving the value of x at other time steps unchanged, then x'(t) changes at the previous time step due to the altered value of x. There is an equal and opposite change in x'(t) at the current time step (assuming the forward step differentiation rule, but our conclusions will be independent of the rule used).
If we consider the value of L(x(t),x'(t)) plotted over the time axis, then this plot moves around the spot where t=t0, as we change x(t) at t=t0. Since x’ has equal and opposite changes at two time steps the integral of L (action) does not have any change in the first order resulting from x’. But in the second order, if the dependency of L on x'(t) is getting stronger/weaker with time, the equal and opposite changes in x'(t) will not have the same impact on L since these affect L at different time steps.
D(L) = change due to increase of x at current point + change due to increase of x’ at previous point – change due to decrease of x’ at current point.
Let the change in x at the current point/time-step be dx. Setting v= x’ and denoting derivatives using subscripts, the change is:
D(L) = L(t)x dx + L(t-Ts)v dv – L(t)v dv
Using the definition of velocity we get:
D(L) = L(t)x dx + L(t-Ts)v * dx/Ts – L(t)v * dx/Ts
But from the definition of derivative,
L(t-Ts)v – L(t)v = – d/dt (Lv(t))* Ts
Using this fact in the prev equation,
D(L) = L(t)x dx – d/dt (Lv(t)) * Ts * dx/Ts= [L(t)x – d/dt (Lv(t))] dx
But, D(L) = 0, from our earlier discussion! This yields, the Euler-Lagrange equation for mechanics,
Lx = d/dt (Lv)
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