I Time-dependent mass and the Lagrangian

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The discussion centers on the formulation of the Lagrangian for a particle with time-dependent mass, expressed as L = 1/2 m(t) dot{r}^2 - U(r). It explores whether the kinetic energy needs to be rewritten when considering variable mass systems. The first method is deemed appropriate when mass changes do not create additional forces, such as in symmetrical mass ejection. However, it is noted that this approach is invalid for systems like rockets, where mass is expelled in a preferred direction, introducing significant forces. The second method is criticized for incorrectly applying the relationship between external force and momentum change in variable mass scenarios.
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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} ##
 
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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}##
 
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Mr rabbit said:
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} ##
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.
 
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