# The proof of conservation of momentum in Mechanics by Landau and Lifshitz

In reading the first chapter of Mechanics by Landau and Lifshitz, there is one point on which I consistently get stuck. This regards the proof in $$\S 7$$ that space homogeneity implies conservation of momentum. To paraphrase Landau and Lifshitz, the proof begins as follows.

1. Consider a system whose Lagrangian (in a certain inertial frame) is $$L(\mathbf{r}, \mathbf{v}, t)$$, where $$\mathbf{r}$$ represents the particle coordinates and $$\mathbf{v} = \frac{d\mathbf{r}}{dt}$$ represents the velocities.
2. Now consider a parallel inertial frame, with coordinates $$\mathbf{r}' = \mathbf{r}+ \varepsilon$$ for some infinitesimal $$\varepsilon$$. Changing coordinates, the laws of motion in this frame must be governed by the different Lagrangian $$L' = L(\mathbf{r}' - \varepsilon, \mathbf{v}, t)$$ homogeneity of space, and yet the laws of motion must be the same by the homogeneity of space.
3. Since $$L' = L(\mathbf{r}' - \varepsilon, \mathbf{v}, t)$$ and $$L(\mathbf{r}', \mathbf{v}, t)$$ are two Lagrangians describing the same system (they have the same equations of motion) it must hold that $$L' = L$$.

However, it was shown in $$\S 2$$ that another possibility by which $$L'$$ and $$L$$ could yield the same equations of motion is if $$L' = L + \frac{d}{dt} f(\mathbf{r}, t)$$. How does one deduce in this case that momentum is conserved?

• L&L do not assume (3), and they do not prove that every Lagrangian system has to conserve momentum. The underlying solid point is that if the Lagrangian does not change value on shifting the system in space (or shifting values of the coordinates), then value of $\sum_a \frac{\partial L}{\partial \mathbf v_a}$ is conserved. It is the same kind of argument as with "energy" in §6: if the Lagrangian does not change value on shifting $t$, then value of $\sum_i \frac{\partial L}{\partial \dot{q}_i}\dot{q}_i - L$ is conserved. Commented Aug 2 at 18:00
• Of course, there is no law or principle requiring that no Lagrangian can change value when shifting the system or changing coordinate system; there are such Lagrangians, and they describe systems which may not conserve momentum (e.g. particle in external field). Commented Aug 2 at 18:03

Unfortunately that proof is simply wrong as it stands. (I do not have LL’s book here and I cannot check the actual statements, I’m just referring to what asserted in the OP). Maybe it is possible to add further physical hypotheses for achieving the wanted result. But, at the end of the day, it would be better and more didactically honest, in my view, to assume as a postulate that the Lagrangian is translationally invariant.

If one assumes from scratch that in inertial reference frames the Lagrangian of a free point is $$mv^2/2$$, then translational invariance (invariance under all isometries of the space more generally) arises automatically and it does need independent postulates.

The passage

1. Since $$L' = L(\mathbf{r}' - \varepsilon, \mathbf{v}, t)$$ and $$L(\mathbf{r}', \mathbf{v}, t)$$ are two Lagrangians describing the same system (they have the same equations of motion) it must hold that $$L' = L$$.

is even more wrong, the fact that two Lagrangian produce the same equations if their difference is a total derivative is just a sufficient condition, in general not necessary. ($$L= mv^2/2$$ and $$L’= e^{L/k}$$, for any constant $$k$$ with the dimensions of energy, give rise to the same equations of motion but their difference is not a total derivative.)

What is physically true is that the equations of motion of an isolated physical system in an inertial frame are invariant under general Galilean transformations, in particular spatial displacements. However this much more physical notion of dynamic symmetry is not equivalent to the notion of Lagrangian symmetry (contrarily to what happens for the notion of Hamiltonian symmetry). Therefore the outlined approach is hopeless from scratch, it is difficult (perhaps impossible) to pass from the invariance of equations of motion to the invariance of the Lagrangian.

• you state that the difference being a total derivative is sufficient, but not necessary - could you please provide an example, where this is not happening, yet the EOM are the same? ; could you provide the theorem, which is both sufficient and necessary? Commented Aug 1 at 20:47
• If $L= mv^2/2$ and $L’= e^L$, these Lagrangian produce the same equation of motion for a free particle, but their difference is not a total derivative. As far as I know the problem has not yet a solution as it is entangled with the one of the existence of a Lagrangian for given EOM, which has only partial solutions. However I am not an expert on these issues. Commented Aug 2 at 6:11
• Right. Thank you. I think some partial answers exist, under the title "The inverse problem of calculus of variations". Commented Aug 2 at 8:58

Technically point 3 is incorrect. The correct statement would be that since $$L'=L(\mathbf{r}'-\epsilon, \mathbf{v}, t)$$ and $$L(\mathbf{r}', \mathbf{v}, t)$$ are two Lagrangians describing the same system, it must hold that $$L'=L+\frac{df}{dt}$$ for some function $$f$$. Then the rest of the argument would go through, although the total derivative term $$f$$ would enter into the momentum.

This part is just speculating. But it may be that when Landau and Lifschitz derive conservation of momentum explicitly for some examples, $$f=0$$, so they don't bother bringing up that point.

• I tried going this route. But starting with $L' = L + \frac{df}{dt}$ leads to the time derivative of momentum being $\frac{df}{dt}$ instead of $0$, which needless to say doesn't look like a conservation law any more. Commented Jul 31 at 22:58
• @jnhnum1 An example of a place where you can see this kind of approach worked out is Tong's QFT notes, Sections 1.3.1 and 1.3.2. (see especially Eqs 1.35 and 1.38). damtp.cam.ac.uk/user/tong/qft.html Commented Jul 31 at 23:45
• Another way to say it is that if you've found $\frac{dp}{dt}=\frac{df}{dt}$, then you can redefine $p'=p-f$, then $\frac{dp'}{dt}=0$. Commented Jul 31 at 23:47

Hoping my English is clear,if anywhere improper,please point it out.

Before answering your question,if you are a beginner,I strongly recommend you don't read Landau's textbook,this is completely not for a beginner.When I read Landau,I found some sentenses are very hard to understand,like

A second conservation law follows from the homogeneity of space. By virtue of this homogeneity, the mechanical properties of a closed system are unchanged by any parallel displacement of the entire system in space. Let us therefore consider an infinitesimal displacement $$\epsilon$$ , and obtain the condition for the Lagrangian to remain unchanged.

We didn't discussed terms are:homogeneity,mechanical properties,closed system,unchanged,parallel displacement,entire system,remain unchanged.When I was a beginner,I can only understand them intuitively,but I can't speak what they exactly are!These terms are so abstract and concise,and we also don't have some examples in Landau's book to understand them,which is quite unfriendly for beginners(at least all people I meet in real,but not online..)

I guess the previous answer by Andrew is incorrect,seems Andrew treated the $$f$$ as force so he claim $$f=0$$ in the text.But this is not true.

Let's consider $$L=\frac{1}{2}mv^2-\frac{1}{2}kx^2$$,let $$x_n =x+v_0t$$ and $$v_n=v+v_0$$,which is an inertial coordiante transformation.sub n for new.

Then as an exercise you will find $$L_n=\frac{1}{2}mv_n^2-\frac{1}{2}kx_n^2$$ differ a total derivative term and will give you the same equation of motion,which is intuitively obivious $$ma=-kx$$ because we just did an inertial transformation,which doesn't creat a new force.

First,your understanding seems need to be changed a bit.

How does one deduce in this case that momentum is conserved?

The philosophy is:If we assume the so called homogenity of spacetime,then some certain lagrangian in this space there will offer some conserved quantity.The conservation of momentum is asserted both by the homogenity(property of the backgroud) and the expression of Lagrangian(property of the system).

So Landau indeed mixed the two things.Let's consider a uniform force F acting on a particle $$L=\frac{1}{2}mv^2+Fx$$,which is clearly a legal Lagrangian but it is impossible to derive the conservation law of momentum!

The lack of homogenity of background space is more subtle and I think explain it seriously is not so easy,but intuitive I guess its understandable.

So second,what if $$L$$ and $$L_n$$ differred by the total derivative term?

The answer is we have the same conserved quantity but in different value.

For another question,if we consider the motion in polar coordinate(like the motion of planet).What transformation should we do to keep the conservation of angular momentum?

The answer is an uniformly rotating transformation $$\phi '=\phi +\omega _0 t$$,and after the transformation $$r'=r+v_0t$$,the two Lagrangian will not differred by a total derivative!

1. For what it's worth, L&L do not change horses midstream: They work with one and the same Lagrangian $$L$$, and ask when the Lagrangian is invariant $$\delta L=0$$ under an infinitesimal translation $$\delta\vec{r}_a=\vec{\epsilon}$$, and derive the condition $$\sum_a\frac{\partial L}{\partial\vec{r}_a}~=~0, \tag{7.1}$$ and use Lagrange equations (5.2) to show that the total momentum (7.2) is conserved.

2. If we instead ask for invariance $$\delta L=\frac{d(\vec{k}\cdot\vec{\epsilon})}{dt}$$ of the Lagrangian up to total time derivative terms, then the Noether current (in this case the total momentum) may receive an improvement term $$\vec{k}$$.