- #1

HubertP

- 9

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I've always loved physics in general, but recently came to conclusion that in order to really understand it I have to get into details and learn the formalism. So, I started with classical mechanics (where else could I start?). I'm making my way through calculus of variations and Lagrangian formulation, which has been great experience for me so far. However, when trying to get my head around the conservation laws and how they arise from symmetries I got stucked at one point.

Let me focus on conservation of (linear) momentum, which is the consequence of the homogeneity of space. This is how I understand it. Our real-life experience is that no part of the physical space is special, meaning that the laws of physics (for isolated system) are everywhere the same. We can move the system one meter to the right, start it with the same boundary conditions (positions and velocities) and it will perform the same motion. This makes sense.

What follows is the tricky part. All papers deriving the conservation law from homogeneity of space say more or less the following:

"Because space is homogenous the (infinitesimal) translation of coordinates will not change the Lagrangian."

The problem is that I can't convince myself this is necessarilly the case! To me the only conclusion from homogeneity of space is that:

"The new Lagrangian (being the result of translation) is such, that the equations of motion (expressed in new coordinates) are the same as the original ones"

In other words, the translated system trajectory is the same as that of the system in its original location.

These two statements don't seem to be equivalent to me. Of course, if Lagrangian is unchanged the trajectory won't change either. However, the opposite is not true. It seems to me that I can easily come up with

**different**Lagrangian which results in

**the same**equations of motion. For example, one can add a term which is time derivative of a function (of positions and velocities) to the Lagrangian:

[itex]\mathcal{L}^{'} = \mathcal{L} + \frac{d}{dt} F(q,\dot{q})[/itex]

The change in the action over the trajectory from such Lagrangian change is:

[itex]\delta{S} = \delta \int_{t1}^{t2} \mathcal{L}dt = \int_{t1}^{t2} \frac{d}{dt} F(q,\dot{q}) dt = F(q,\dot{q}) \bigg|_{t1}^{t2}[/itex]

Which is constant and only depends on the endpoints (and not the trajectory itself). So, the new trajectory with such new Lagrangian will be

**the same**as the original (because it minimizes the action - adding a constant does not change the condition for stationarity).

So, the question is, isn't the conclusion (arising from homogeneity of space) that the Lagrangian does not change under translation too strong? All we observe in nature is that the motion of the translated system is the same. However, we can get the same motion with different Lagrangian! Where do I make mistake?