Clarification about use of homogeneity and isotropy to determine Lagrangian

Question

I have been trying to review some parts of classical mechanics using Landau and Lifschitz and Arnolds Mathematical Methods of Classical Mechanics and have run into a question regarding how the Lagrangian is determined in the opening chapter of Landau and Lifschitz.

Let's take the following statement from Landau and Lifschitz,

The homogeneity of space and time implies that the Lagrangian cannot con­tain explicitly either the radius vector.

My understanding is that a motion, $ x(\cdot) $, should be an extremal of a given action, $ S[x(\cdot)] = \int_{0}^{T} \mathcal{L}(x(t), \dot{x}(t), t) dt $. From this we find that a motion should satisfy the Euler-Lagrange equations,

$$ -\frac{d}{dt} \left( \frac{d\mathcal{L}}{d\dot{x}}(x(t), \dot{x}(t), t)\right) + \frac{d\mathcal{L}}{dx}(x(t), \dot{x}(t), t) = 0.$$

The way I interpreted homogeneity of space is that all points are equivalent so mathematically, $y(t) = x(t) + r$, for fixed $r$, is another valid measurement which in turn means it should satisfy the Euler-Lagrange equations, or mathematically that,

$$ -\frac{d}{dt} \left( \frac{d\mathcal{L}}{d\dot{x}}(x(t) + r, \dot{x}(t), t)\right) + \frac{d\mathcal{L}}{dx}(x(t) + r, \dot{x}(t), t) = 0. $$

Writing the Euler-Lagrange equations as $F(x(t), \dot{x}(t),\ddot{x}(t), t) = 0 $ space is homogeneous tells us that $$F(x(t), \dot{x}(t),\ddot{x}(t), t) = F(x(t) + r, \dot{x}(t), \ddot{x}(t), t).$$ From here I think we can determine that $F$, or the Euler-Lagrange equations, is independent of the argument for $x(t)$ if there is only a single particle(Arnold uses this same argument to show forces must depend on only relative positions in his opening chapter). Analogously shifting time instead of spatial positions shows $F$ must be invariant of time

From this I have the following questions,

1. Was there any misunderstanding in what I have stated above?

2. How does one go from $F$ is independent of $x$, to deducing information about the Lagrangian being independent of $x$? To add some context to the second question I have found that some other sources such as this attempt to show that homogeneity and isotropy can lead to deductions about the Lagrangian although no source I have found attempts to show any detail about how to do this.

Edit: To address the comment about specificity, I am unclear about how the Euler-Lagrange equations being independent of position and time implies the Lagrangian must also be independent of those properties as implied by Landau. Is it possible to complete this argument? If not, can other concrete information be gained about the Lagrangian from the Euler-Lagrange equations being independent of position or time? If someone has specific feedback about why the current question scope is not specific enough or lacks enough detail I would appreciate it.

Answer 1

Well, rather than having an answer I think I want to argue that the endeavour that you describe is not a fruitful thing to do.

In general: in any logical system there is great freedom to interchange theorem and axiom without changing the contents of the system.

A related thought is expressed in the following quote from a post by physicsSE contributor Kevin Zhou

[...] in physics, you can often run derivations in both directions: you can use X to derive Y, and also Y to derive X. That isn't circular reasoning, because the real support for X (or Y) isn't that it can be derived from Y (or X), but that it is supported by some experimental data D. This two-way derivation then tells you that if you have data D supporting X (or Y), then it also supports Y (or X).

In an earlier version of this question you linked to a document by Vincent Icke. Symmetries and the form of the Lagrangian

It seems to me that the steps that Vincent Icke goes through do not reveal information that wasn't already available. The connection with symmetries is known.

It seems to me: one can start with the symmetries and obtain expressions for physical quantities and their relations, or one can start with the physical quantities and relations, and recognize symmetries.

I submit there is no way of pronouncing one representation as deeper than the other. More abstract does not necessarily imply more fundamental.

Other than that, I notice that Vincent Icke uses more constraint than necessary.

For Hamilton's action the criterion is that you are looking for the point in variation space where the derivative of Hamilton's action is zero. Whether that point-of-zero-derivative is a minimum or a maximum is immaterial.

In fact, there are classes of cases such that in variation space the true trajectory corresponds to a maximum of Hamilton's action. So: a restriction to trajectories that minimize Hamilton's action is incorrect.

It is sufficient to find the point in variation space such that the derivative of Hamilton's action is zero.

As a demonstration of interderivability in classical mechanics: it is possible, without introducing additional assumption, to go from F=ma to Hamilton's stationary action

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[LATER EDIT]

Some additional remarks:
As we know, historically the connection between symmetries and conserved quantites was recognized in the context of applying calculus of variations in physics.

At this point I want to take a step back, and take stock. In my opinion, being able to recognize that does not depend on using the concept of Hamilton's action. I expect that a procedure can be devised, using the same type of operations as in the Noether procedure, to demonstrate the connection between symmetry and conserved quantity directly.