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I want to know if, in general, is it true that the Lagrangian describing the energies of the center of mass of a system of particles is the same as the Lagrangian written for the energies corresponding to every particle in the system.


The specific problem I am trying to solve is that of two particles of mass $m$ constrained to move inside a sphere of radius $b$, with the restriction that the two particles are always a distance $2a$ apart from each other (say, due to a massless rigid rod).

When I write the kinetic energy for the center of mass of the system, I don't get the same expression as the one corresponding to the two individual particles:

$T_{CM}=\frac{1}{2}(2m)R^{2}(\dot{\phi}^{2}\sin^{2}\theta+\dot{\theta}^{2}) \qquad \text{where}\qquad \theta=\frac{\theta_2+\theta_1}{2}, \quad R^2=b^{2}-a^{2}$

$T_{\text{individual}}=\frac{1}{2}mb^{2}(\dot{\phi}_{1}^{2}\sin^{2}\theta_1+\dot{\phi}_{2}^{2}\sin^{2}\theta_{2}+\dot{\theta}_{1}^{2}+\dot{\theta}_{2}^{2})$

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    $\begingroup$ If I have understood your notation correctly, your $T_CM$ only includes the kinetic energy of the centre of mass and is missing a contribution from motion within the centre of mass frame (i.e. kinetic energy due to the rotation of the particles around the centre of mass) $\endgroup$ – By Symmetry Sep 4 '16 at 15:13
  • $\begingroup$ $$ \left(m_{1}+m_{2}\right)\mathbf{r}_{\text{cm}}=m_{1}\mathbf{r}_{1}+m_{2}\mathbf{r}_{2} \quad \nRightarrow \quad \tfrac{1}{2}\left(m_{1}+m_{2}\right)\Vert \dot{\mathbf{r}}_{\text{cm}}\Vert^{2}= \tfrac{1}{2}m_{1}\Vert \dot{\mathbf{r}}_{1}\Vert^{2}+\tfrac{1}{2}m_{2}\Vert \dot{\mathbf{r}}_{2}\Vert^{2} $$ $\endgroup$ – Frobenius Sep 4 '16 at 17:13
  • $\begingroup$ @BySymmetry that is true, that solves the problem that my equation only appears $R^2$. Introducing the kinetic energy including the rotation of the particles from the center of mass I obtain $T_{CM}=\frac{1}{2}(2m)b^{2}(\dot{\phi}^{2}\sin^{2}\theta+\dot{\theta}^{2})$. Still I can't recover the expression for the kinetic energy of the individual particles $\endgroup$ – Saavestro Sep 4 '16 at 17:40
  • $\begingroup$ @Frobenius you mean $\frac{1}{2}(m_{1}+m_{2})||\dot{\boldsymbol{r}}_{cm}||^2\neq \frac{1}{2}m_{1}||\dot{\boldsymbol{r}}_{1}||^2+\frac{1}{2}m_{2}||\dot{\boldsymbol{r}}_{2}||^2$ ? that kind of makes sense, so it would mean that the lagrangian written for the center of mass is not the same as the lagrangian written as the sum of the energies corresponding to every particle? $\endgroup$ – Saavestro Sep 4 '16 at 17:44
  • $\begingroup$ @Saavestro I think you have overloaded a symbol in your derivation there. Your system has 3 degrees of freedom. (4 variables minus 1 constraint) Your expression for the centre of mass energy only has 2 variables. You need a 3rd one to describe the motion of the particles within the CM frame. $\endgroup$ – By Symmetry Sep 4 '16 at 17:51
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No. The answer to the general question is NO. $L=T-V$. This is not true for either $T$ or $V$, although in this particular problem there is no potential energy. The total kinetic energy (sum for each particle) is equal to the CM energy plus the kinetic energy "relative to the CM" for each particle.

In this particular problem, your $T_{individual}$ is correct, but you did not enforce the constraint. There are only 3 degrees of freedom, instead of 4. Your $T_{CM}$ is incorrect even if considered as CM energy. The constraint is not modeled correctly.

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  • $\begingroup$ yeah, I realized that some days ago, I wasn't considering the energy relative to the center of mass. Anyway, thank you for your answer. $\endgroup$ – Saavestro Sep 12 '16 at 22:00

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