Galilean invariance and the Lagrangian

Question

My textbook says that in a time invariant space with translational and rotational symmetry the Lagrangian only depends on the magnitude of the velocity. The galilean invariance says that a Lagrangian is mechanical equivalent to this Lagrangian after a Galilean transformation. With a Taylorexpansion we get:

$$L(v'^2)=L((\vec{v}+\vec{\epsilon})^2)=L(v^2)+\dfrac{\partial L}{\partial v^2}2\vec\epsilon\cdot\vec v$$

since these two are mechanical equivalent, partial derivative part should be a total differential. All the above I understand. But then my textbook says that $L$ can only be of the form $\dfrac{1}{2}mv^2$ or else we can not bring it to the form of a total differential.

I don't understand why $L$ can not be something else. Could someone please explain this?

Answer 1

You are possibly confused by the fact that your textbook, when it says we need this to be a total differential, means with respect to time, not speed.

We know that, if we perform a change of ${\cal L}$ into ${\cal L}+\delta\!{\cal L}$ such that $\delta\!{\cal L} = df(\vec x)/dt$, the dynamics is unaffected, where $f$ is allowed to be a function of position only, not of speed as well. The reason is that, when we use the least action principle, we are comparing paths which start and end simultaneously, at the same positions, but we allow them to have different speeds at the start of the path, at the end of the path, and everywhere else as well.

In other words, we have $\delta\!\vec r_0 = \delta\!\vec r_1 = 0$, i.e. the paths must start and arrive at the same positions, simultaneously; no one ever mentioned the velocities being the same at the initial and/or final positions.

Thus, if $\partial {\cal L}/\partial v^2 = $ a constant, then we can write

$$\delta\!{\cal L} = k \vec\epsilon\cdot\vec v = \frac{d}{dt} (k\vec\epsilon\cdot\vec r)$$

which is indeed a total differential. But, if $\partial{\cal L}/\partial v^2$ is not a constant, then you cannot write your $\delta\!{\cal L}$ as a total differential with respect to time, such that $f$ is a function of position only.

Answer 2

If the Lagrangian is linear in $v^2$, then $\partial_{v^2} L = k$, $k$ some constant. So the difference becomes

$$\delta S \propto \vec \varepsilon \cdot \int dt \frac{d}{dt} \vec x(t) = \vec \varepsilon \cdot \int d \vec x(t) = \varepsilon \cdot (\vec x_b - \vec x_a)$$

That term does not depend on the path $x(t)$, since the endpoint of the variation are fixed, so that

$$\frac{\delta S[x(t)]}{\delta x(t)} = 0$$

for that term, hence not contributing to the Euler-Lagrange equation.