Landau vol.1 Mechanics: Expansion of L' and Dependence on Velocity

[zhuang382]

Let $K$ and $K'$ be two inertial frame, If K is moving with infinitesimal velocity relative to $K'$ , then $v' = v + \epsilon$.
Note that $L(v^2) - L(v'^2)$ is only a total derivative of a function of coordinate and time. (I understand this part)
Because $L' = L(v'^2) = L(v^2 + 2v\cdot\epsilon + \epsilon^2)$, then we use power series of $\epsilon$ to expand the equation and neglect the second order term:
$$L(v'^2) = L(v^2) + 2\frac {\partial f} {\partial v^2}v\cdot\epsilon$$
Then the author argues that it is only when the second term on the right hand side is a linear function of $\vec{v}$, it is a total derivative of time; therefore $\frac{\partial f} {\partial v^2}$ does not depend on velocity. Can someone help me on the detail of $\epsilon$ series expansion and how the conclusion drawn from this?

 

[PeroK]

Has Landau covered Noether's theorem and, in particular, the concept of divergence invariance?

 

[zhuang382]

Actually, this is chapter one, so I don't know yet. I am just beginning using this book to review the content I learned last semester. I am mainly confused about the mathematical technique he uses, ( expand the Lagrangian as a power series of $\epsilon$) when he analyze the difference of two Lagrangian functions.

 

[PeroK]

Actually, this is chapter one, so I don't know yet. I am just beginning using this book to review the content I learned last semester. I am mainly confused about the mathematical technique he uses, ( expand the Lagrangian as a power series of $\epsilon$) when he analyze the difference of two Lagrangian functions.

That's just a Taylor series expansion to first order in $\vec \epsilon$.

The rest of what he does needs some knowledge of Noether's theorem.

 

[PeroK]

PS Can you explain what he's actually trying to achieve with that calculation? That's the really important thing.

 

[vanhees71]

The ingenious feature of this book is that he starts from the fundamental content of Newtonian mechanics, i.e., invariance under the full Galilei group and derives why the laws look as they look. For a free particle you get uniquely $L=m v^2/2$.

From tranlation invariance in space and time and rotation symmetry you get that $L=L(\vec{v}^2)$. That's all trivial, because under this part of the group $L$ transforms as a simple scalar. For the invariance under Galilei boosts you need a non-trivial term, i.e.,
$$\tilde{L}(\tilde{v}^2)=L(\tilde{v}^2)+\mathrm{d}_t \Omega(\vec{x},t).$$
As a general analysis of Noether's theorem shows, you need it only for "infinitesimal transformations", i.e.,
$$\tilde{\vec{x}}=\vec{x}-\vec{\epsilon} t \; \Rightarrow \; \tilde{\vec{v}}=\vec{v}-\vec{\epsilon}.$$
Then you get
$$L(v^{\prime s})=\vec{\nabla}_{\epsilon} \vec{v}^{2} L'(v^2)=-2 \vec{\epsilon} t \cdot \vec{v} L'(\vec{v}^2) \stackrel{!}{=}-\mathrm{d}_t \Omega(\vec{x},t).$$
Now
$$\mathrm{d}_t \Omega (\vec{x},t)=\vec{v} \cdot \vec{\nabla}_x \Omega + \partial_t \Omega.$$
Comparing the left and the right-hand side shows that
$$\vec{\nabla}_x \Omega = 2 L'(\vec{v}^2) \vec{\epsilon} t, \quad \partial_t \Omega=0.$$
Since the left-hand side doesn't depend on $\vec{v}$ and the right-hand side doesn't depend on $\vec{x}$ we must have
$$L'(\vec{v}^2)=\frac{m}{2}=\text{const}, \quad \Omega=m \vec{\epsilon} \cdot \vec{x}.$$
This implies that the Lagrangian is Galilei-boost invariant in addition to the other symmetries of Galilei-Newton spacetime if and only if $L$ is equivalent to
$$L=\frac{m}{2} \vec{v}^2=\frac{m}{2} \dot{\vec{x}}^2.$$
This implies Newton's Lex II, according to which a free particle moves with constant velocity.

Noether's theorem tells us that the generators for boosts is
$$\vec{K}=m \vec{x}-m t \vec{v}.$$
Since this is the conserved quantity indeed it follows again that $\vec{v}=\text{const}$ along the trajectory of the particle.