# Deriving lagrangian of a free particle - How do you arrive at Lagrangian independency conclusions

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?


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:

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

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

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.

• Thanks. I understand that in order to have energy conservation, you need the Lagrangian to be (explicitly) independent of time. In the case of the free particle, this would be a reasonable assumption. However, does the homogeneity of time itself have a direct consequence on the independence of the lagrangian of the free particle on time? Also, I checked out that link. That question addresses how you arrive at the statement that the velocity is constant given that the lagrangian is only a function of square of the speed of the particle. – IanDsouza May 12 '15 at 16:51
• I'm not sure if I understand your question correctly but I didn't assume anything about the potential in my derivation so this applies to all (reasonable) potentials. BTW If this post helped you you may consider upvoting or accepting it as your answer. – Gonenc May 12 '15 at 16:55
• In the text, the author uses the argument that since time is homogenous, the lagrangian of the free particle can't be explicitly dependent on time. I take it that homogeneity of time means that the nature of the time evolution of a system doesn't change if you started the system at one time opposed to another with the same positions and velocities. The future positions and velocities are themselves 'translated' in time. Does this by itself mandate the lagrangian of the free particle to be independent of time? – IanDsouza May 12 '15 at 17:28