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Javier Gómez Morales

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Landau · Classical Mechanics · Ch. 1

The Equations of Motion

March 5, 2026

A distinctive editorial choice of this book is that it opens with the Lagrangian formulation of mechanics, instead of introducing the subject through Newton's laws. In other words, dynamics is presented from the start as a variational problem: the equations of motion emerge as conditions on the action functional, rather than being postulated as force laws.

First-order expansion (the "first variation")

Consider a trajectory $q(t)$ and a small deviation $\delta q(t)$. The corresponding deviation of the velocity is $\delta \dot{q}(t)=\frac{d}{dt}\delta q(t)$. A first-order multivariable Taylor expansion of the Lagrangian around $(q,\dot{q},t)$ gives

$$ L\!\big(q+\delta q,\ \dot{q}+\delta \dot{q},\ t\big) = L(q,\dot{q},t) + \frac{\partial L}{\partial q}\,\delta q + \frac{\partial L}{\partial \dot{q}}\,\delta \dot{q} + \mathcal{O}\!\big((\delta q)^2\big). $$

Substituting into the action difference $S[q+\delta q]-S[q]$ and keeping only terms linear in the deviation yields

$$ S[q+\delta q]-S[q] = \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q}\,\delta q + \frac{\partial L}{\partial \dot{q}}\,\delta \dot{q} \right)\,dt + \mathcal{O}\!\big((\delta q)^2\big). $$

Definition. The first variation $\delta S$ is the linear part in $\delta q$ and $\delta \dot{q}$:

$$ \delta S = \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q}\,\delta q + \frac{\partial L}{\partial \dot{q}}\,\delta \dot{q} \right)\,dt. $$

Intuitively, if the change in the action were strictly proportional to the deviation, then changing the sign of the deviation would change the sign of the action's change. But a minimum cannot behave that way: a perturbation cannot make the action larger in one direction and smaller in the opposite direction. Therefore, at a minimum the linear contribution must vanish, i.e. $\delta S=0$ at first order for all admissible variations.

Integration by parts

To proceed, focus on the second term in $\delta S$. It has the form $\int_{t_1}^{t_2} A(t)\,\frac{d}{dt}(\delta q)\,dt$. Applying integration by parts gives

$$ \int_{t_1}^{t_2} A(t)\,\frac{d}{dt}\big(\delta q\big)\,dt = \Big[ A(t)\,\delta q(t) \Big]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{dA}{dt}\,\delta q(t)\,dt. $$

In our case, $A(t)=\frac{\partial L}{\partial \dot{q}}$. Hence

$$ \int_{t_1}^{t_2} \frac{\partial L}{\partial \dot{q}}\, \frac{d}{dt}\big(\delta q\big)\,dt = \Big[ \frac{\partial L}{\partial \dot{q}}\,\delta q \Big]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}}\right)\, \delta q\,dt. $$

The boundary term vanishes because the variations are taken with fixed endpoints: $\delta q(t_1)=\delta q(t_2)=0$. This condition expresses the idea that we compare paths that start and end at the same points in configuration space; only the intermediate shape of the path is allowed to vary. In that setting, $\big[ \frac{\partial L}{\partial \dot{q}}\delta q \big]_{t_1}^{t_2}=0$.

Inertial frames

I particularly like how the book motivates the notion of an inertial frame through simple symmetry statements about space.

Homogeneity. Space is homogeneous if there are no privileged points. Put differently: the laws of physics are the same here as one meter away, or at any other position in space.

Isotropy. Space is isotropic if all directions are equivalent. There are no privileged orientations.

Next chapter Chapter 2 · Conservation Laws →

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