# I Time-dependent mass and the Lagrangian

1. Oct 7, 2017

### Mr rabbit

I was talking to a friend about Lagrangian mechanics and this question came out. Suppose a particle under a potential $U(r)$ and whose mass is $m=m(t)$. So the question is: the Lagrangian of the particle can be expressed by

$L = \frac{1}{2} m(t) \dot{\vec{r}} ^2 -U(r)$

or I need to re-write the kinetic energy? Maybe this way

$\displaystyle T = \int \vec{F} \cdot d\vec{r} = \int \frac{d\vec{p}}{dt} \cdot \vec{v} \: dt = \int \vec{v} \cdot d\vec{p} = \int \vec{v} \cdot (\vec{v} \: dm + m \: d \vec{v}) = \int v^2 \: dm + \int m \: \vec{v} \cdot d \vec{v}$

2. Oct 7, 2017

### hilbert2

It's as in your first equation. If $r$ is a single 1d coordinate, the equation of motion will be

$0 = \frac{\partial L}{\partial r} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right)$
$= -\frac{\partial U}{\partial r} - \frac{d}{dt}( m(t)\dot{r})$
$= -\frac{\partial U}{\partial r} - m(t)\ddot{r} - \dot{m}(t)\dot{r}$

3. Nov 5, 2017

### Jano L.

The first method is appropriate when the process that changes the mass does not result in additional force on the remaining body. For example, when the body ejects mass in two opposite directions with the same rate. It is not valid for a rocket in flight, because mass is thrown away in a preferred direction and as a result, there is a strong force acting on the rocket. The second method is not a valid derivation, since for variable mass systems, external force does not in general equal d(mv)/dt.