Goldstein conservation of Hamiltonian example problem

Question

In Goldstein's Classical Mechanis section 8.2 (page 345 in the third edition) there is as example that should illustrate how the Hamiltonian can be conserved without energy being conserved. It goes like this:

Suppose a point mass $m$ is attached to a spring, of force constant $k$, the other end of which is fixed on a massless cart that is being moved uniformly by an external device with speed $v_0$ if we take as the general coordinate the position $x$ then the Lagrangian of the system is obviously $$L=\frac{m\dot{x}^2}{2}-\frac{k}{2}(x-v_0t)^2.$$

Why is this true?

if the coordinate is given by $x-v_0t$ then ''$\dot{x}$'' should be $\dot{x}-v_0$ but that would mean that the Lagrangian would be

$$L=\frac{m(\dot{x}-v_0)^2}{2}-\frac{k}{2}(x-v_0t)^2=L=\frac{m(\dot{x}^2-2\dot{x}v_0+v_0^2)}{2}-\frac{k}{2}(x-v_0t)^2,$$

which is obviously not the same as the above Lagrangian.

What is wrong with this picture?

Answer 1

The kinetic energy is related to the speed of the particle in a fixed frame, not relative to the cart. $x(t)$ is the position of the particle with respect to a fixed axis, so its speed is $\dot x$. $(x-v_0 t)$ appears in the potential energy because the extension of the spring is the position of the particle with respect to the cart.