I'm reading Landau's Book.

He tries to conclude the law of inertia from the Lagrange equations.

For that, he argues (by nice suppositions about space and time), that the lagrangian must depend only on the velocity. More specifically, only on the square of the velocity.

The point is, since the lagrange equations is:

$$\nabla_x L+(\nabla_{\dot{x}}L)' \equiv 0$$

He gets that $(\nabla_{\dot{x}}L)'\equiv 0$, which implies $\nabla_{\dot{x}}L=constant$ (in time, along the trajectory).

Now here is my problem: he concludes that $\dot{x}$ is constant. How? He doesn't know anything about $L$, besides the "symmetry" properties. For instance, $L\equiv 0 $ satisfies all of the properties required from such $L$, and we would not be able to infer that $\dot{x}$ is constant. In fact, any curve would be extremal with respect to the action.

What is his reasoning, then?


2 Answers 2


You're correct.

To find the equations of motion, we have:

\begin{align*}c_i&=\frac{\partial L(v^2)}{\partial v_i}\\ &=L'(v^2) 2 v_i \end{align*}

so that $L'(v^2) v_i$ is constant for all of time.

Firstly, you could imagine a world in which all paths ${\bf x}(t)$ are valid mechanical paths. Then the Galilean transform of a valid mechanical path is also a valid mechanical path, and so respects Galilean relativity. (Trivially, because all paths are valid) This is what happens when $L'=0$ identically.

Another degenerate case is, for example, $L(y)=\sin(y)$. Then if $v^2=\frac{pi}{2}$, $L'(v^2)=0$ and the path is allowed to change direction at will.

Yet another degenerate case is when $L(v^2)=\sqrt{v^2}$. Then $\frac{\partial L}{\partial v_i}=\frac{v_i}{\sqrt{v^2}}$ and, for example, the one dimensional motion $x(t)=t^2$ satisfies $\frac{2t}{\sqrt{4 t^2}}$ constant (for $t>0$).

So that begs the question: what DOES that passage of Landau prove? Fortunately, though I haven't looked thoroughly at it, the next section (which proves that $L\propto v^2$) does not seem to depend on the assumption that $v={\mathrm{const}}$ for a free particle.

  • $\begingroup$ But there is also another problem: it does not uniquely determine the path. We have a family of pathes (namely, all the ones with constant $\textbf{v}$ satisfy the principle). How can I be sure that no other path will also satisfy? $\endgroup$ Jun 17, 2015 at 0:16
  • 1
    $\begingroup$ @AloizioMacedo Sorry, I was actually thinking about fixing that. The goal of classical mechanics is to uniquely determine the equations of motion. I'll update my post. $\endgroup$
    – user12029
    Jun 17, 2015 at 0:20
  • $\begingroup$ I'm sorry, I didn't understand. What do you mean by the lagrangian "uniquely determining the equations of motion"? $\endgroup$ Jun 17, 2015 at 0:40
  • 1
    $\begingroup$ @AloizioMacedo turns out this is a very annoying question you asked! Let me know if my most recent edit "helps". $\endgroup$
    – user12029
    Jun 17, 2015 at 1:23
  • 1
    $\begingroup$ Certainly helps! Thanks for your attention, really : ). I am afraid that the next section depends on the assumption that $v=\text{const}$ for a free particle. I just found this question: physics.stackexchange.com/questions/23098/… which is similar to mine (and even includes other doubts I was going to ask). The answer seems to respond it, but since I don't know Noether's Theorem and the other tools involved in that answer, I'll refrain from this matter for some time. I'm accepting your answer, since you answered my question. Thank you again. $\endgroup$ Jun 17, 2015 at 2:45

It follows from $L$ being a function $\propto\dot{x}^2$. With this at hand, you are left with two choices:

  1. $\left(\nabla_{\dot{x}}L\right)'\sim\left(\dot{x}\right)'=0$ implies $\dot{x}=\rm const.$
  2. $L=0$ implies $\dot{x}=0=\rm const.$

Either way, you get that the velocity is constant in time (for this particular, free-particle case).

  • 1
    $\begingroup$ $L=0$ doesn't imply $\dot{x}=0$. Actually, it implies absolutely nothing! $\endgroup$
    – user12029
    Jun 16, 2015 at 23:59
  • $\begingroup$ @NeuroFuzzy: If $L$ is only a function of $\dot{x}$ (as the author claims), then saying $L$ is zero means that $\dot{x}$ is also zero (which is constant). $\endgroup$
    – Kyle Kanos
    Jun 17, 2015 at 0:00
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    $\begingroup$ If $L=0\cdot \dot{x}^2$, $\dot{x}$ can be anything and satisfy $L=0$, and that's the issue the OP is getting at. $\endgroup$
    – user12029
    Jun 17, 2015 at 0:01
  • $\begingroup$ @NeuroFuzzy: I don't see how one can rationally argue that $L=0\cdot\dot{x}^2$ is a function of $\dot{x}^2$ (or anything) since your constant coefficient is necessarily zero. $\endgroup$
    – Kyle Kanos
    Jun 17, 2015 at 0:11
  • $\begingroup$ @Kyle Kanos $L=0$ is a function, and depends only on $\dot{x}$ simply by definition: it doesn't depend on the other variables. Even if you disconsider this case, you could define $L=\sqrt{v^2}$. This depends only on $v$ and, as exposed by NeuroFuzzy, gives rise to solutions with non-constant $v$. $\endgroup$ Jun 17, 2015 at 2:47

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