In Landau Mechanics, he derived the conservation of momentum assuming that $\delta L = 0$ under infinitesimal translation $\epsilon$. However, one just need the change of Lagrangian to be a total derivative of time to preserve the EoM. Is there any other assumption leading to $\delta L = 0$ ?

A parallel displacement is a transformation in which every particle in the system is moved by the same amount, the radius vector $\:\mathbf{r}\:$ becoming $\:\mathbf{r}\!+\!\boldsymbol{\epsilon} $. The change in $\:L\:$ resulting from an infinitesimal change in the co-ordinates, the velocities of the particles remaining fixed, is \begin{equation} \delta L=\sum\limits_{a}\dfrac{\partial L \hphantom{_{a}}}{\partial\mathbf{r}_{a}}\boldsymbol{\cdot}\delta\mathbf{r}_{a}=\boldsymbol{\epsilon}\boldsymbol{\cdot}\sum\limits_{a}\dfrac{\partial L \hphantom{_{a}}}{\partial\mathbf{r}_{a}}, \nonumber \end{equation} where the summation is over the particles in the system. Since $\:\boldsymbol{\epsilon}\:$ is arbitrary, the condition $\:\delta L=0\:$ is equivalent to \begin{equation} \sum\limits_{a}\partial L/ \partial\mathbf{r}_{a}=0. \tag{7.1} \end{equation} From Lagrange's equations (5.2) we therefore have \begin{equation} \sum\limits_{a}\dfrac{\mathrm d \hphantom{t}}{\mathrm d t}\dfrac{\partial L \hphantom{_{a}}}{\partial\mathbf{v}_{a}}=\dfrac{\mathrm d \hphantom{t}}{\mathrm d t}\sum\limits_{a}\dfrac{\partial L \hphantom{_{a}}}{\partial\mathbf{v}_{a}}=0. \nonumber \end{equation} Thus we conclude that, in a closed mechanical system, the vector \begin{equation} \mathbf{P}=\sum\limits_{a}\partial L/ \partial\mathbf{v}_{a} \tag{7.2} \end{equation} remains constant during the motion; it is called the momentum of the system.
Differentiating the Lagrangian (5.1), we find that the momentum is given in terms of the velocities of the particles by \begin{equation} \mathbf{P}=\sum\limits_{a} m_{a}\mathbf{v}_{a}. \tag{7.3} \end{equation}

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    $\begingroup$ The condition $\delta L=0$ under translation is the assumption. Using this assumption, Landau derives that momentum is conserved, so if the Lagrangian is translation-invariant, that implies conservation of momentum. $\endgroup$ Jun 7, 2018 at 18:47
  • $\begingroup$ Ideally, each conservation law should be independent of the others. Can you think of a counter example of an L that cannot be expressed as a total derivative yet exhibits some conserved quantities? The procedure of looking for invariance upon introducing an infinitesimal change in something is the procedure for finding conserved quantities. $\endgroup$
    – user196418
    Jun 7, 2018 at 19:51

1 Answer 1


There are different types of symmetries we might search for in a Lagrangian theory. When setting out to prove identities, we assume that the theory is invariant under the change we are making. With this assumption, we then look at what the implications are for conservation laws. In this case, Landau assumes $L$ is invariant under translations. When this is true, linear momentum is conserved.

Noether's theorem searches for those symmetries that arise from a continuous global transformation of a coordinate. These symmetries have a general proof that relies on the fact that a perturbation in coordinate $q^k$ leaves the conjugate momentum unchanged: $$ \frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot q^k}\bigg) = 0 $$ Symmetries of this kind lead to the conservation of {angular} momentum under a translation {rotation} or perhaps conservation of $h(q,\dot q)$ under time translations, where: $$ h(q,\dot q) = \frac{\partial L}{\partial \dot q^k} \dot q^k - L $$

We can however, search for symmetries in the action, Lagrangian, the equations of motion or the solutions of the equations of motion.

Amongst other things, this leads us to gauge symmetries. Two Lagrangians that differ by a total time derivative: $$ L' = L + \frac{d}{dt}f $$ lead to the same equations of motion due to only surface terms in the action (boundary point terms).


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