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I guess this question has been asked before, but I'm looking at a slightly different aspect.

I'm reading Landau's book on classical mechanics. In deriving the lagrangian for a free particle, I understand the significance of the homogeneity of space and time, and isotropic nature of space with respect to time evolution of the system. But I am a little unclear as to how this affects the lagrangian.

For example, in an inertial frame, the homogeneity of space means that if you translate positions of all particles at an instant of time, the future positions are also translated by that same amount. But what's the mathematical logic behind arriving at the conclusion that the lagrangian does not depend on the trajectories $x(t),y(t),z(t)$?

In the lagrangian formalism, $S = \int_{t_1}^{t_2} L(q,\dot{q},t)\,dt$ , the boundary condition is on the initial and final positions.

Using this view or the Euler-Lagrange equations, could someone explain the mathematics used to arrive at the conclusion here as well as the other two cases - independency of lagrangian on time and direction of velocity?

Matrix23
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$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$ $\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$ $\newcommand{\l}[0]{\mathcal L}$ $\newcommand{\q}[0]{\dot q}$ $\newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$

A word of advice: Landau's book, don't read it if you haven't studied that subject before!

Your first question has already an answer here. For your second question: Nobody is saying that the Lagrangian should be independent of time, unless you are working in a closed system. Take the total derivative of the Lagrangian with respect to time ie:

$$\dv{\l}{t}=\pdv{\l}{q}\pdv{q}{t}+\pdv{\l}{\dot q}\pdv{\dot q}{t}+\pdv{\l}{t} \tag{1}$$

The Euler Lagrange equation reads:

$$\dv{}{t}\pdv{\l}{\q}=\pdv{\l}{q} \tag{2}$$

You immediately see the right hand side of (2) and plug it in into the right hand side of (1).

\begin{align}\implies \dv{\l}{t} &= \dv{}{t} \left( \pdv{\l}{\q} \right) \pdv{q}{t}+\pdv{\l}{\dot q}\pdv{\dot q}{t}+\pdv{\l}{t}\\ &= \dv{}{t} \left( \pdv{\l}{\q} \right) \pdv{q}{t}+\pdv{\l}{\dot q}\pdvt{q}{t}+\pdv{\l}{t} \end{align}

Look what happened to the right hand side of the equation. Noticing the product rule of differentiation you can simplify it as:

$$\dv{\l}{t}=\dv{}{t} \left( \pdv{\l}{\q} \pdv{q}{t} \right) + \pdv{\l}{t}= \dv{}{t} \left( \q \pdv{\l}{\q} \right) + \pdv{\l}{t} $$

$$\implies \dv{}{t}\left( \l- \q \pdv{\l}{\q} \right) = \pdv{\l}{t}$$

Now if $\pdv{\l}{t}$ is zero than we have a conservation law. Something that doesn't change under “translations” in time,which is energy. If you don't believe me and don't understand the time invariance mumbo jumbo you can also work out the units.

If on the other hand you don't have $\pdv{\l}{t}=0$, that would mean that you haven't chosen your system properly (meaning you haven't chosen a closed system), and there is energy being created or destroyed. If you want to have energy conservation in a closed system, as we all love to have, you better make sure that you choose your Lagrangian to be time invariant.

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