- #1
Mr rabbit
- 26
- 3
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} ##
##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} ##