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On p. 5 in $\S$3 pf the book of Mechanics by Landau & Lifshitz, it is claimed that

[...] for a free particle, the homogeneity of space and time implies that Lagrangian cannot depend on position or time, explicitly.[...]

However, in my understanding, Lagrangian of a system is the function that determines the equation of motion, i.e given a the initial conditions and the Lagrangian of the system, we can determine the future configuration of the system, as in the case Newton's second law.

However, we also do know that adding a constant to our Lagrangian, or a time derivative of a function of position and time, the equation of motion does not change, hence, we get an "equivalent" Lagrangian in the sense that both functions lead us to the same conclusion about the dynamics of the system at hand.

Given this, I cannot understand why the Lagrangian of a free particle cannot depend on the position or time, explicitly.

I mean it is clear that if that is the case, we have a simple Lagrangian that satisfy all the properties that you would expect it to have; however, why this is the only case that a free particle can have as a Lagrangian.

Note: I have read this question, but I still cannot understand why does the origin would have a privileged status in that case.

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  • $\begingroup$ Can you show how adding your proposed dependencies on position and time results in the same equations if motion. I'm not sure I'm following. $\endgroup$
    – BioPhysicist
    Nov 23, 2018 at 16:11
  • $\begingroup$ @AaronStevens I'm not claim such a thing. I'm just saying that I cannot see how does having a position or time dependency in our Lagrangian lead to a contraction, or a violation of one of our principles. $\endgroup$
    – Our
    Nov 23, 2018 at 16:13
  • $\begingroup$ @AaronStevens I mean, for example, if we have a differential equation at hand, you can say that $f(x)$ is a solution of this DE, but it does not mean that it is the only solution. Similar to this, I can see that $K(v^2)$ is a possible nice Lagrangian for the free particle, however, why can't another Lagrangian is not possible ? $\endgroup$
    – Our
    Nov 23, 2018 at 16:14

2 Answers 2

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A free particle has no external forces acting on it. Therefore, momentum and energy are conserved. By Noether's Theorem this means the system has spatial and temporal translation symmetry. If the Lagrangian has explicit position or time dependence, then this cannot be the case.

Also note that adding a constant to the Lagrangian is not the same thing as adding in explicit position or time dependence.

Ultimately, it seems like the book is making a physical argument rather than a mathematical one. There do exist transformations that end up making the same equations of motion, but if you want to interpret the Lagrangian as the difference between the kinetic and potential energy, then you don't want explicit position and time dependence. If there was this dependence, then this would mean the particle is no longer free.

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  • $\begingroup$ Dude, I'm in the beginning of the book. According to the theory that I have developed so far, there is no Noether's theory, nor energy, nor momentum. $\endgroup$
    – Our
    Nov 23, 2018 at 15:46
  • $\begingroup$ @onurcanbektas I have added some more information. $\endgroup$
    – BioPhysicist
    Nov 23, 2018 at 18:27
  • $\begingroup$ However, two equivalent Lagrangians can differ by some function $d f(x,t) / dt$, and in such a case, $\frac{\partial L(v)}{dx } = 0$, but $\frac{ \partial L(x,v,t)'}{ dt} = \frac{\partial [L(v) + d f(x,t)/dt]}{ \partial x} \not =0 $. $\endgroup$
    – Our
    Nov 23, 2018 at 20:02
  • $\begingroup$ @onurcanbektas are you saying because $\frac{df}{dt}$ has a $\dot x$ dependence? $\endgroup$
    – BioPhysicist
    Nov 23, 2018 at 20:06
  • $\begingroup$ Not $\dot x$, but $x$. $\endgroup$
    – Our
    Nov 23, 2018 at 20:22
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I will try to avoid jargon: if the Lagrangian is explicitly independent on a coordinate, like position or time, then the corresponding canonical momenta of that coordinate is conserved. So, for a free particle there are no external forces acting on it, so the Lagrangian only has a kinetic energy term which depends only on the velocity of the particle. Thus, energy is conserved because time does not appear in the Lagrangian, and linear momentum is conserved because position does not appear in the Lagrangian.

As you said, you may add a time derivative of a function that satisfies the Euler-Lagrange equations, but this does not grant you the privilege of imposing explicit time or position dependence.

however, why this is the only case that a free particle can have as a Lagrangian.

I think this is a far deeper question than you might realize: as it turns out, Lagrangians are not mathematically unique generally, but in terms of producing the correct equations of motion for some system only certain Lagrangians work. It's an open problem of theoretical physics to answer "why these Langrangians for these systems?" and there's no obvious answer to it currently other than that it works in producing the correct equations of motion (verified experimentally).

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  • $\begingroup$ About your first paragraph, the book tries to understand the form of the lagragian of a free particle, and we at at the beginning of the book; Your argument depends on thing that are going to be developed, so in a sense your argument is circular, because anything that will be derived will depend on this interpretation, so you cannot use those derivations to explain this thing. $\endgroup$
    – Our
    Nov 24, 2018 at 6:50
  • $\begingroup$ I mean, I just said the same thing as AaronStevens but just in simpler languange, and you didnt seem to mind the jargon that he whipped out from near the end of your book...? The first paragraph is just to set some context, but the core of my answer is the last paragraph where I address your "why" question: it's a question that has no answer currently. And the first paragraph is not my interpretation, it's how it is: by definition a free particle only has kinetic energy and so the lagrangian only has a kinetic energy term. It's truly that simple. $\endgroup$
    – Daddy Kropotkin
    Nov 24, 2018 at 13:14

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