- #1
Peeter
- 305
- 3
Problem 1.8 from a Goldstein's mechanics text, is (verbatim since it's hard to describe otherwise) :
"A system is composed of three particles of equal mass m.
Between any two of them there are forces derivable from a potential
[tex]
V = -g e^{-\mu r}
[/tex]
where r is the disance between the two particles. In addition, two of the
particles each exert a force on the third which can be
obtained from a generalized potential of the form
[tex]
U = -f \mathbf{v} \cdot \mathbf{r}
[/tex]
[itex]\mathbf{v}[/itex] being the relative velocity of the interacting particles
and f a constant. Set up the Lagragian for the system, using as coordinates the
radius vector [itex]\mathbf{R}[/itex] of the center of mass and the two vectors
[tex]
\begin{align*}
\boldsymbol{\rho}_1 &= \mathbf{r}_1 - \mathbf{r}_3 \\
\boldsymbol{\rho}_2 &= \mathbf{r}_2 - \mathbf{r}_3
\end{align*}
[/tex]
Is the total angular momentum of the system conserved?"
For the Lagrangian I get:
[tex]
\mathcal{L} =
g \left(
e^{-\mu \left\vert{\boldsymbol{\rho}_1}\right\rvert}
+ e^{-\mu \left\vert{\boldsymbol{\rho}_2}\right\rvert}
+ e^{-\mu \left\vert{ \boldsymbol{\rho}_1 - \boldsymbol{\rho}_2 }\right\rvert} \right)
+ f \left(\mathbf{R} - \frac{1}{3}(\boldsymbol{\rho}_1 + \boldsymbol{\rho}_2) \right) \cdot \left( \dot{\boldsymbol{\rho}_1} + \dot{\boldsymbol{\rho}_2} \right)
[/tex]
For the part about the angular momentum conservation I'm not as sure. Since there is no external torque on the system I think that the angular momentum is conserved.
However, perhaps the idea of the problem is to show this explicitly given the Lagrangian. It's not clear to me an effective approach to do so though, as this looks like a messy calculation, and I haven't actually tried doing so.
Does anybody have any hint (which could be "suck it up, and do the calculations") for me if there is a way to show or disprove the momentum part of the question if the answer isn't just "there is no external torque".
"A system is composed of three particles of equal mass m.
Between any two of them there are forces derivable from a potential
[tex]
V = -g e^{-\mu r}
[/tex]
where r is the disance between the two particles. In addition, two of the
particles each exert a force on the third which can be
obtained from a generalized potential of the form
[tex]
U = -f \mathbf{v} \cdot \mathbf{r}
[/tex]
[itex]\mathbf{v}[/itex] being the relative velocity of the interacting particles
and f a constant. Set up the Lagragian for the system, using as coordinates the
radius vector [itex]\mathbf{R}[/itex] of the center of mass and the two vectors
[tex]
\begin{align*}
\boldsymbol{\rho}_1 &= \mathbf{r}_1 - \mathbf{r}_3 \\
\boldsymbol{\rho}_2 &= \mathbf{r}_2 - \mathbf{r}_3
\end{align*}
[/tex]
Is the total angular momentum of the system conserved?"
For the Lagrangian I get:
[tex]
\mathcal{L} =
g \left(
e^{-\mu \left\vert{\boldsymbol{\rho}_1}\right\rvert}
+ e^{-\mu \left\vert{\boldsymbol{\rho}_2}\right\rvert}
+ e^{-\mu \left\vert{ \boldsymbol{\rho}_1 - \boldsymbol{\rho}_2 }\right\rvert} \right)
+ f \left(\mathbf{R} - \frac{1}{3}(\boldsymbol{\rho}_1 + \boldsymbol{\rho}_2) \right) \cdot \left( \dot{\boldsymbol{\rho}_1} + \dot{\boldsymbol{\rho}_2} \right)
[/tex]
For the part about the angular momentum conservation I'm not as sure. Since there is no external torque on the system I think that the angular momentum is conserved.
However, perhaps the idea of the problem is to show this explicitly given the Lagrangian. It's not clear to me an effective approach to do so though, as this looks like a messy calculation, and I haven't actually tried doing so.
Does anybody have any hint (which could be "suck it up, and do the calculations") for me if there is a way to show or disprove the momentum part of the question if the answer isn't just "there is no external torque".
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