I Time-dependent mass and the Lagrangian

  • I
  • Thread starter Thread starter Mr rabbit
  • Start date Start date
  • Tags Tags
    Lagrangian Mass
AI Thread Summary
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.
Mr rabbit
Messages
26
Reaction score
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} ##
 
Physics news on Phys.org
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}##
 
  • Like
Likes Mr rabbit
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.
 
  • Like
Likes Mr rabbit
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Dear all, in an encounter of an infamous claim by Gerlich and Tscheuschner that the Greenhouse effect is inconsistent with the 2nd law of thermodynamics I came to a simple thought experiment which I wanted to share with you to check my understanding and brush up my knowledge. The thought experiment I tried to calculate through is as follows. I have a sphere (1) with radius ##r##, acting like a black body at a temperature of exactly ##T_1 = 500 K##. With Stefan-Boltzmann you can calculate...
Back
Top