# Deriving the Lagrangian for a free particle

I'm a newbie in physics. Sorry, if the following questions are dumb. I began reading "Mechanics" by Landau and Lifshitz recently and hit a few roadblocks right away.

1. Proving that a free particle moves with a constant velocity in an inertial frame of reference ($\S$3. Galileo's relativity principle). The proof begins with explaining that the Lagrangian must only depend on the speed of the particle ($v^2={\bf v}^2$): $$L=L(v^2).$$ Hence the Lagrance's equations will be $$\frac{d}{dt}\left(\frac{\partial L}{\partial {\bf v}}\right)=0,$$ so $$\frac{\partial L}{\partial {\bf v}}=\text{constant}.$$ And this is where the authors say

Since $\partial L/\partial \bf v$ is a function of the velocity only, it follows that $${\bf v}=\text{constant}.$$

Why so? I can put $L=\|{\bf v}\|=\sqrt{v^2_x+v^2_y+v^2_z}$. Then $$\frac{\partial L}{\partial {\bf v}}=\frac{2}{\sqrt{v^2_x+v^2_y+v^2_z}}\begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix},$$ which will remain a constant vector $\begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}$ as the particle moves with an arbitrary non-constant positive $v_x$ and $v_y=v_z=0$. Where am I wrong here? If I am, how does one prove the quoted statement?

2. Proving that $L=\frac{m v^2}2$ ($\S$4. The Lagrangian for a free particle). The authors consider an inertial frame of reference $K$ moving with a velocity ${\bf\epsilon}$ relative to another frame of reference $K'$, so ${\bf v'=v+\epsilon}$. Here is what troubles me:

Since the equations of motion must have same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L'$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of coordinates and time (see the end of $\S$2).

First of all, what does same form mean? I think the equations should be the same, but if I'm right, why wouldn't the authors write so? Second, it was shown in $\S$2 that adding a total derivative will not change the equations. There was nothing about total derivatives of time and coordinates being the only functions, adding which does not change the equations (or their form, whatever it means). Where am I wrong now? If I'm not, how does one prove the quoted statement and why haven't the authors done it?

P. S. Could you recommend any textbooks on analytical mechanics? I'm not very excited with this one. Seems to hard for me.

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– Qmechanic Apr 1 '12 at 11:20
I think the best advice one can give you is don't read Landau & Lifshitz. They are great books for reference, but practically impossible to learn from. If you're into analytical mechanics then Goldstein is a good place to start, or Arnold, if you're more interested in the mathematical aspects. – yohBS Apr 1 '12 at 21:54

1. In physics, it is often implicitly assumed that the Lagrangian $L=L(q^i,v^i,t)$ depends smoothly on the (generalized) positions $q^i$, velocities $v^i$, and time $t$, i.e. that the Lagrangian $L$ is a differentiable function. Let us now assume that the Lagrangian is of the form $$L~=~\ell(v^2),\qquad\qquad v~:=~|\vec{v}|,\qquad\qquad(1)$$ where $\ell$ is a differentiable function. The equations of motion (eom) become $$\vec{0}~=~\frac{\partial L}{\partial \vec{q}} ~\approx~\frac{d}{dt}\frac{\partial L}{\partial \vec{v}} ~=~\frac{d }{dt} \left(2\vec{v}~\ell^{\prime}\right) ~=~2\vec{a}~\ell^{\prime}+4\vec{v}~(\vec{a}\cdot\vec{v}) \ell^{\prime\prime}.\qquad\qquad (2)$$ (Here the $\approx$ symbol means equality modulo eom.) We conclude that on-shell $$\vec{a} \parallel \vec{v}.$$ (The words on-shell and off-shell refer to whether eom are satisfied or not.) Therefore by taking the length on both sides of the vector eq. $(2)$, we get $$0~\approx~2a(\ell^{\prime}+2v^2\ell^{\prime\prime}),\qquad\qquad a~:=~|\vec{a}|.$$ This has two branches. The first branch is that there is no acceleration, $$\qquad \vec{a}~\approx~\vec{0},\qquad\qquad (3)$$ or equivalently, a constant velocity. The second branch imposes a condition on the speed $v$, $$\ell^{\prime}+2v^2\ell^{\prime\prime}~\approx~0.\qquad\qquad (4)$$ To take the second branch $(4)$ seriously, we must demand that it works for all speeds $v$, not just for a few isolated speeds $v$. Hence eq. $(4)$ becomes a 2nd order ODE for the $\ell$ function. The full solution is precisely OP's counterexample $$L~=~ \ell(v^2)~=~\alpha \sqrt{v^2}+\beta~=~\alpha v+\beta,$$ where $\alpha$ and $\beta$ are two integration constants. This is differentiable wrt. the speed $v=|\vec{v}|$, but it is not differentiable wrt. the velocity $\vec{v}$ at $\vec{v}=\vec{0}$ if $\alpha\neq 0$, and therefore the second branch $(4)$ is discarded. Thus the eom is the standard first branch $(3)$.

2. Firstly, the definition of form invariance is discussed in this question. Concretely, Landau and Lifshitz mean by form invariance that if the Lagrangian is $$L~=~\ell(v^2)\qquad\qquad (5)$$ in the frame $K$, it should be $$L'~=~\ell(v^{\prime 2})\qquad\qquad (6)$$ in the frame $K^{\prime}$. Here $$\vec{v}^{\prime }~=~\vec{v}+\vec{\epsilon}$$ is a Galilean transformation. Secondly, OP asks if adding a total time derivative to the Lagrangian $$L ~\longrightarrow~ L+\frac{dF}{dt}$$ is the the only thing that would not change the eom? No, e.g. scaling the Lagrangian $$L ~\longrightarrow~ \alpha L$$ with an overall factor $\alpha$ also leaves the eom unaltered. See also Wikibooks. However, we already know that all Lagrangians of the form $(5)$ and $(6)$ lead to the same eom $(3)$. (Recall that acceleration is an absolute notion under Galilean transformations.) Instead, I interpret the argument of Landau and Lifshitz as that they want to manifestly implement Galilean invariance via Noether Theorem by requiring that an (infinitesimal) change of the Lagrangian $$\Delta L~:=~L'-L ~=~2(\vec{v}\cdot\vec{\epsilon})\ell^{\prime}$$ is always a total time derivative even off-shell. In general, how do we know if an expression $\Delta L$ is a total time derivative? Well, one way is to apply the Euler-Lagrange operator on the expression, and check if it is identically zero off-shell. We calculate $$\vec{0}~=~ \frac{d}{dt}\frac{\partial \Delta L}{\partial \vec{v}} -\frac{\partial \Delta L}{\partial \vec{q}}$$ $$~=~4\vec{\epsilon}~(\vec{a}\cdot\vec{v}) \ell^{\prime\prime} +4\vec{v}~(\vec{a}\cdot\vec{\epsilon}) \ell^{\prime\prime} +4\vec{a}~(\vec{v}\cdot\vec{\epsilon}) \ell^{\prime\prime} +8\vec{v}~(\vec{v}\cdot\vec{\epsilon})(\vec{a}\cdot\vec{v}) \ell^{\prime\prime\prime}. \qquad\qquad (7)$$ Since eq. $(7)$ should hold for any off-shell configuration, we can e.g. pick $$\vec{a}~\parallel~\vec{v}~\perp~\vec{\epsilon}.$$ Then eq. $(7)$ reduces to $$\vec{0}~=~ 4\vec{\epsilon} ~(\pm a v) \ell^{\prime\prime}.$$ We may assume that $\vec{\epsilon}\neq\vec{0}$. Arbitrariness of $a$ and $v$ implies that $$\ell^{\prime\prime}~=~0.\qquad\qquad (8)$$ (Conversely, it is easy to check that eq. $(8)$ implies eq. $(7)$.) The full solution to eq. $(8)$ is the standard non-relativistic Lagrangian for a free particle, $$L~=~ \ell(v^2)~=~\alpha v^2+\beta,$$ where $\alpha$ and $\beta$ are two integration constants. For more on Galilean invariance, see also this question.

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Regarding your answer to the first question: $L = v^4$ would be smooth, but $\frac{\partial L}{\partial v} = 3 v^3 = const.$ and so one could not conclude that the vector $v$ is constant. Probably it is easier to argue, that $$\frac{d}{d t} \frac{\partial L}{\partial v}$$ is the acceleration and so one identifies/defines $\frac{\partial L}{\partial v}$ as the velocity/momentum (until this moment $v$ means an arbitrary variable with no physical meaning). Which in turn implies the quadratic dependency of the Langrangian. – Tobias Diez Apr 1 '12 at 14:55
Thank you, the first question is now clear! It's a shame, the authors didn't bother to mention the smoothness there. But I'm still confused with the second part. I didn't find a clear definition of form-invariance here. Does form-invariance mean that the families of solutions will be the same? If so, do there exist second-order DEs that generate the same sets of solutions, but are different (linear-independent)? – Someone Apr 1 '12 at 23:23
Finally, did I get it right that Landau only proved, that $\frac{m v^2}{2}$ will fit the requirements? Is there a proof that this is the only possible expression? – Someone Apr 1 '12 at 23:27
@altertoby, I didn't understand your argument, as $$4v^3=\text{const}$$ seems to imply $v=\text{const}$ (one dimentional case). In 3 dimensions it would be $$4v^2{\bf v}=\text{const},$$ what lead me to the same result ($v_i^3=\text{const}$). Also, how can $v$ (hence ${\bf v}$ as well) have no physical meaning? Would ${\bf v}$ not had the meaning of velocity, how could we apply the Galileo's relativity principle and come to the conclusion that $L=L(v^2)$? – Someone Apr 2 '12 at 1:52
@Someone: Thanks for your valid objection. Lets consider the equation in 2D. Eg $4 v^2 v_x = 2 = 4 v^2 v_y$ has solutions $v_x = 1 = v_y$, but its not the only one. (Admittedly, these extra solutions are complex, but I think one could find a better example with real solutions). So the smoothness of $L$ should not be enough to conclude $v = const.$. Of course you pointed out a flaw in my reasoning. If $v$ is just a placeholder variable in the first place, it has no meaning to apply the Galileo principle to it. So maybe one could skip this point as $L=L(v^2)$ is not needed in my argumentation. – Tobias Diez Apr 3 '12 at 10:18

Qmechanic's answer is good, and I only want to comment (but I can't) on two small mistakes in the logic of the derivation for the first part, but which doesn't affect the final answer.

From $(2)$ we see that $\vec{a} \approx \vec{0}$ solves the equation. Now, assuming that $\vec{a} \neq \vec{0}$, then we get the statement that on-shell $\vec{a} ~||~ \vec{v}$. After all, there is no notion of vectors being 'parallel' if one of them is the zero vector.

The next statement is not gotten by taking the length on both sides of the equation. This is because we don't know if $|\vec{x} + \vec{y} | = |\vec{x}|+|\vec{y}|$ or $|\vec{x}|-|\vec{y}|$. The former is when the two vectors are pointing in the same direction and the latter is when they are pointing in opposite directions.

Instead, dot with $\vec{v}$. Then one gets \begin{align} 2(\vec{a}.\vec{v})(l'+2v^2 l'') \approx 0, \end{align} whereby from the assumption that neither $\vec{a}$ nor $\vec{v}$ are identically 0, the stuff in the other paranthesis vanishes. Then the result that this branch is bad follows, leaving us with $\vec{a} \approx \vec{0}$.

Hope that was useful :)

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Dear @nervxxx. Thanks for providing the full argument, which may be helpful for some. Your two points were not actually mistakes, just shorthand arguments. Specifically, 1) I always implicitly declare that a zero-vector is parallel to any vector. 2) The phrase taking the length on both sides was meant to mean go from a vector eq. to a scalar eq. – Qmechanic Jan 10 '13 at 0:55

Also Qmechanic gave the correct answer, I believe it is overloaded, because there is actually no need to use motion equations (Euler-Lagrange equations) to answer to the second part of the OP question at least.

Actually you can simply just generalize original Landau approach to this issue for an answer, so I will mention it here in details:

Lets suppose that $L\left(\vec{v}^{2n}\right)$($2n$ is to have scalar value) is the Lagrangian of a free particle in inertial frame $K$. suppose another inertial frames of reference $K'$ that moves relative to $K$ with infinitesimal velocity $\vec{\varepsilon}$, the Lagrangian $L'=L\left[\left(\vec{v}+\vec{\varepsilon}\right)^{2n}\right]$ in $K'$ that describes the particle should be same Lagrangian as in $K$ up to a total time derivative.

To show that, we expand $\left(\vec{v}+\vec{\varepsilon}\right)^{2n}$ in first order of $\vec{\varepsilon}$, to find it, we suppose at first that $n=1$, then: $$\left(\vec{v}+\vec{\varepsilon}\right)^{2}\simeq v^{2}+2\vec{\varepsilon}\cdot\vec{v}$$ then to find for $n=2$ we write: $$\left(\vec{v}+\vec{\varepsilon}\right)^{4}\simeq\left(v^{2}+2\vec{\varepsilon}\cdot\vec{v}\right)\left(v^{2}+2\vec{\varepsilon}\cdot\vec{v}\right)\simeq v^{4}+4\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2}$$ repeating this couple times insures you that: $$\left(\vec{v}+\vec{\varepsilon}\right)^{2n}=v^{2n}+2n\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2n-2}+O\left(\varepsilon^{2}\right)$$ then we can write by expanding the Lagrangian that: $$L\left[\left(\vec{v}+\vec{\varepsilon}\right)^{2n}\right] = L\left(\vec{v}^{2n}\right)+2n\left(\vec{\varepsilon}\cdot\vec{v}\right)v^{2n-2}\frac{\partial L}{\partial\vec{v}^{2n}}+O\left(\varepsilon^{2}\right) \simeq L+g\left(v\right)\sum_{i}\varepsilon_{i}\frac{dx_{i}}{dt}$$ Where: $$g\left(\left\Vert \vec{v}\right\Vert \right)\equiv2n\, v^{2n-2}\frac{\partial L}{\partial v^{2n}}$$ because $L\left(v\right)$, it should be clear that $g$ have to be a function of speed only (not of velocity or it's components), also we see that the sum sign, is actually already a full time derivative by it's own, so to keep the second term full time derivative, we see that the only possible option for us it to have $g\left(v\right)=const$, from this it follows immediately that $n=1$ (note that $n>0$) and $L=\alpha v^{2}+\beta$.

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