# Classical Mechanics/Lagrange Theory

This section contains several theoretical developments of the Lagrangian formalism that are not directly necessary for solving problems. However, these considerations help understand the theory more deeply and answer certain important questions.

## Why does the extremum of a functional determine motion?

[edit | edit source]In the Lagrangian formulation of mechanics, the trajectory is determined from the condition that the action functional should have an extremum. (It is not always the case that the trajectory is the *minimum* of the action; in some cases it might be merely an extremum, i.e. a point where the functional derivative vanishes.) This condition is known as the **action principle**. By now, you should be familiar with the mathematical procedures used to derive the equations of motion from the action principle.

So, at this point, you should be well used to the fact that the correct equations of motion for each mechanical system indeed follow from the action principle, if the Lagrangian is chosen appropriately. However, it might still feel like a mystery to you that Newton's laws are equivalent to the condition for the extremum of some functional. You might be asking yourself: why is this possible *at all*?

Here is one explanation that may help. Let us consider a simple mechanical system: a point mass moving in one dimension, with coordinate , in a potential . (The same considerations can be easily generalized to the case of more than one dimensions and more than one coordinate.) Suppose that is the *correct* trajectory according to Newton's law,

How can we use a functional to express the condition that the trajectory is the correct one? One way is to demand that the deviation of from is everywhere zero. This can be expressed using the functional

It is clear that the functional has the minimum value (obviously the minimum is 0) if and only if for all . This is an example of how to use a functional to express some condition on functions: the functional measures the deviation of from all along the way. The smallest possible deviation is no deviation at all; thus, the minimum of the functional is at the trajectory that does not deviate at all from .

Another similar way to specify the trajectory is to use the functional

This functional, together with the boundary conditions , has the minimum value if and only if for all .

Admittedly, the functionals do not help us to formulate the laws of mechanics, because they already contain the correct trajectory explicitly. We shall now construct another functional, , starting from and trying to eliminate the explicit dependence on .

Let us rewrite as

The third term, , is a fixed function and does not vary when we vary . Therefore we may omit that term from . Furthermore, we would like to have rather than , since we could then use Newton's law for the correct trajectory. So let us integrate the second term by parts:

The boundary term does not vary with since the boundary values of are fixed. Therefore we may omit that term. Finally, we use Newton's law to replace by :

If we now *assume* that the trajectory deviates very little from the correct trajectory , then we may approximately write

The term quadratic in can be omitted under the above assumption. The terms and can be omitted since they are independent of . Thus we find that the functional is equivalent, up to inessential terms that do not vary with , to the following functional:

It is clear that is equivalent to the usual Lagrangian up to the coefficient .

In this way, we obtained a functional which has a minimum when is very close to ; i.e. it is a *local* minimum. The new functional does not depend explicitly on , just as we wanted. The price to pay is that this functional works only for *small* deviations from the correct trajectory. Indeed, the functional may have other minima or maxima which the original functional does not have. The only real justification for the correctness of is that the equations of motion coincide with Newton's law.

## Why can we use arbitrary coordinates to write the Lagrangian?

[edit | edit source]In simple cases, the Lagrangian is equal to the difference of the kinetic and the potential energy terms. However, one needs to select some *coordinates* to describe these terms. Then it is completely unimportant *which* variables are chosen as coordinates; these variables could be lengths, angles, or any functions of lengths and angles (but not velocities!). In other words, one can use any coordinate systems or even just *parts* of some coordinate systems, as long as the possible positions of every mass point is adequately described by the coordinates and the appropriate constraints. For this reason, the coordinates entering the Lagrangian are called **generalized coordinates**. Usually, one chooses generalized coordinates for convenience, to minimize the required computational work, or to decrease the number of necessary constraints.

However, you may be asking yourself: *why* is it that one is allowed to use arbitrary coordinates in the Lagrangian formalism? Certainly, as we know, Newton's laws are not the same in different coordinates: for instance, the mass times the acceleration is equal to the force only if the acceleration is computed as , where is the vector of Cartesian coordinates . This formula will be incorrect if the vector were to consist of, say, the radius , the azimuthal angle in the plane, and the coordinate . However, the Lagrangian formalism will work just fine if we express the kinetic and the potential energy through the variables . The equations of motion will be given by the Euler-Lagrange equation,

as before. One says that the Lagrangian formalism is **covariant** with respect to coordinate transformations.

The reason for this can be explained in two ways: either more formally, by showing that the Euler-Lagrange equations remain the same under an arbitrary change of coordinates; or more visually, by approaching the situation from the geometric point of view.

### Formal derivation

[edit | edit source]For simplicity, we shall only consider a one-dimensional problem with a Lagrangian , where is a generalized coordinate. The same consideration is very easily generalized to the case of multiple coordinates.

Suppose that a new coordinate is chosen instead of . The new coordinate can be a function of the old coordinate. Let us consider an even more general case where the change of coordinates depends on time (i.e. we may choose slightly different coordinates at different times). Then the new coordinate is related to the old one by a formula such as

where is a known function.

Now we need to express the old Lagrangian through the new variable and its derivative . We have

where we denote partial derivatives by subscripts with commas, e.g. . This is a condensed notation frequently used in physics.

The Lagrangian expressed through the new variable is therefore

The new variable is a good variable if it is a nontrivial function of the old one, i.e. if . Then the new Lagrangian will be a nontrivial function that depends on as well as on . So we shall assume that at least within *some* interval of .

Now let us compare the equations of motion (EOM) that we would derive in the old coordinates and in the new coordinates.

The old EOM can be written as

The new EOM is

Let us express this equation through instead of :

Therefore, the new EOM is

Simplifying this expression, we find

We find that the new EOM is indeed equivalent to the old one, under the assumption that .

### Geometric picture

[edit | edit source]The computation presented above is straightforward and explicit, but may leave you wondering *why* it works. Here is a more visual explanation.

The Euler-Lagrange equations express the condition that the functional has an extremum at the trajectory . Let us imagine a space of all trajectories, i.e. some huge space where each "point" represents one entire trajectory . The functional has an extremum at some "point" which is the actual trajectory of the mechanical system. When we change coordinates, , we merely change our description of this space of trajectories. We cannot change the fact that the functional has an extremum somewhere, at some "point" . We may only change our description of this "point". Therefore, after a change of variables the new functional will again have an extremum at some "point" , and this "point" will have to correspond to the "point" after the change of variables. The existence of the extremum is a geometric characteristic of the shape of the functional ; that's why it is independent of the way we choose to describe it with coordinates.

Let us consider a simple example where we use functions instead of functionals. The function has a minimum at . We may change coordinates and use instead of , where e.g. . This is a well-defined change of variables on the interval , where . In the new coordinates, the function looks like . This function has a minimum at where . But geometrically speaking, this is exactly the same function as before, except viewed in different coordinates. Therefore, it is no surprise that the minimum is the old minimum after the change of coordinates.

This equivalence can be seen more formally. The condition for the minimum of the function is

This condition is equivalent to the condition for the minimum of the function , namely , as long as . This is why the position of the minimum in the old coordinates, , exactly corresponds to the position of the minimum in the new coordinates, .

Similarly, when we consider functionals, we may write the condition for the minimum of in new coordinates as

It is clear that the condition for the minimum remains the same under the change of variables, as long as the new variables are well-defined, i.e. .

## Is the Lagrangian unique?

[edit | edit source]Another important question is whether there is only one Lagrangian that yields the correct equations of motion for a given system. The answer is that there are infinitely many different Lagrangians that can be used for any given system.

First of all, one may always multiply the Lagrangian by a constant and also add an arbitrary *fixed* function of time, , to the Lagrangian. The modified Lagrangian is then . The term is "fixed" in the sense that it does not depend on . Then we can integrate this term explicitly and express the modified action as

The last term above is simply a number. Clearly, this modification of the action is irrelevant: if is an extremum of , then it is also an extremum of . Adding a constant to a function does not change the position of the extrema.

More generally, we may add an arbitrary *total time derivative* to the Lagrangian:

The resulting modification of the action is

where are the boundary values of . Since these values are fixed and do not vary when we vary , the extra term in the action is again a constant. Therefore, this modification of the action does not change the equations of motion. One says that two Lagrangians differing by a total derivative are equivalent.

One may even allow functions that depend on derivatives of as well as on . However, in this case one would need to keep fixed also the values of the corresponding derivatives of at the boundary points .

So, as we see, the Lagrangian for a given physical system is not unique. The recipe "kinetic energy minus potential energy" is merely a simple rule that yields a good Lagrangian.

The variety of equivalent Lagrangians is not limited to those that differ by a total derivative or by a constant coefficient. For example, the Lagrangians

lead to the same equation of motion,

even though one obviously cannot find a function and a constant such that . (Such a function would produce at most an extra term in the Lagrangian, but not terms that are nonlinear in derivatives.)