The related post was found here Lagrangian formalism application on a particle falling system with air resistance and also Wikipedia's definition on generalized force. Essential

$$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L }{\partial q_i}=Q_i.$$

The case concerned was to calculate/translate an arbitrary force into the original system of a Lagrangian.


Setting generalized coordinates to be $(m,r,\phi)$ $$L=\frac{1}{2}m(\dot r^2 +(r+R_E)^2\dot \phi^2 )+\frac{g_0R_E^2}{R_E+r}m$$ the equation of motion with respect to $r$ and $\phi$ were $$\frac{d}{dt} (m\dot r)= -\frac{g_0R_E^2}{(R_E+r)^2}m +m(r+R_E)\dot \phi^2 $$ $$\frac{d}{dt} (m(r+R_E)^2\dot\phi)= 0 $$

which, for $r$, was identified to be $$F_r= -\frac{g_0R_E^2}{(R_E+r)^2}m +m(r+R_E)\dot \phi^2$$

It's straight forward to see that, for an external force in $r$ direction $F_r^{ext}$, $$Q_r=F_r^{ext}$$

However, for $\phi$ direction it was much more complicated.

Through identification $$v_\phi=(r+R_E)\dot \phi,$$ without external force, $$\dot v_\phi=-\frac{\dot r v_\phi }{r+R_E}-\frac{\dot m}{m} v_\phi$$ or, equivalently, $$m\dot v_\phi+\dot m v_\phi=-\frac{\dot r v_\phi }{r+R_E} m$$ one thus identified $$F_\phi=-\frac{\dot r v_\phi }{r+R_E} m.$$

But if one wanted to write $Q_\phi$, then $$\frac{d}{dt} (m(r+R_E)^2\dot\phi)=\frac{d}{dt} (m(r+R_E)v_\phi)= Q_\phi. $$ Clearly, $$Q_\phi^{ext}=F_\phi^{ext}\cdot (r+R_E) \neq F_\phi^{ext}.$$

Why the generalized force in $\phi$ direction seemed so strange? Is there any other way to actually calculated $Q_\phi^\ext$ rather than making a blind identification? How to calculate/translate arbitrary external force into generalized force in Lagrangian such as resistance?

  • $\begingroup$ Sorry, but I don't quite understand your question. So, you are given that Lagrangian and you want to know what the generalized forces are without having to identify them from the equations of motion? $\endgroup$ – AFG Mar 8 at 9:49

Try to avoid the awkward variation due to the existence of a term of order $dt$ in the Lagrangian, I reformulate the problem using dissipation function and gneralized force.

$$\tag{1} G(r, \dot{\phi}) = \frac{1}{2} \frac{dm}{dt} \left\{(r+R_E) \dot\phi - u\right\}^2 $$ where $ \dot m = \frac{dm}{dt}$ is the mass changed between $t$ and $t+dt$, negative for loss. The additional parameter $u$ denotes the enjected tangential volicity of the running masses.

The general force $$ Q_r = \frac{\partial G}{\partial \dot{r}} = 0; $$ and $$\tag{2} Q_\phi =\frac{\partial G}{\partial \dot{\phi}} =\frac{dm}{dt} \left\{(r+R_E) \dot\phi - u\right\}(r+R_E) = \dot{m} (r+R_E)\left\{ v_\phi - u\right\} $$ Note that the general force $Q_\phi$ is indeed a unit of torque, as it should be.

The Euler-Lagrangian equation for $\phi$ becomes:

$$ \frac{d}{dt} \left\{ \frac{\partial L}{\partial \dot{\phi}}\right\} - \frac{\partial L}{\partial \phi} = Q_\phi $$

Becomes $$\tag{4} \frac{d}{dt} (m(r+R_E) v_\phi) = \dot{m} (r+R_E) \left\{ v_\phi - u\right\} $$

An equation for tangential velocity $v_\phi = (r+R_E)\dot\phi$ $$\tag{5} \frac{d}{dt} \left( (r+R_E) v_\phi \right) = -\frac{\dot{m}}{m} (r+R_E) u = \left|\frac{\dot{m}}{m} \right| (r+R_E) u. $$

Since $Q_r$ is zero, the equation in $r$ is the same as before: $$ \tag{6} \frac{d}{dt} (m\dot r) +\frac{g_0R_E^2}{(R_E+r)^2}m - m\frac{v_\phi^2}{(r+R_E)} = 0. $$

And the $v_\phi$ is solve from the integral:

$$\tag{7} v_\phi(t) = v_\phi(0) + \frac{1}{(r+R_E) } \int_0^t dt' \left|\frac{\dot{m}}{m} \right| (r+R_E) u. $$

  • $\begingroup$ Notes: The eq 4 was the same as my line 5, it's the total force acting on the system, with mass varying or not, was the same. The expression under eq 4 was not correct, that's only possible in information geometry. It's really trying to express $\delta L$ at taylor expansion of $\delta t$ there. A condition$\ddot{m}=0$ was missed from that step. Compare to my line 4, the equation above eq. 5 was incorrect. It make sense since "$\delta L$" got ride of the mass varying term. Your radial part had similar mistake. It's incorrect. $\endgroup$ – ShoutOutAndCalculate Mar 28 at 15:50
  • $\begingroup$ It didn't answer the question anyway. The derivation was simulated to be working. The shift to Newtonian picture was because an absolute relative speed from ejecting mass was assumed. At small time steps at each instance, it's same as pushing the entire object since the term of $\Delta t^2$ could effectively be pushed to measure $0$ with small $dt$. The main treatment was the use lagrangian mechanics and newtonian mechanics together by using the inference of D'Alembert's principle, and it causes trouble by just doing lagrangian alone. $\endgroup$ – ShoutOutAndCalculate Mar 28 at 15:59
  • $\begingroup$ "an absolute relative speed from ejecting mass was assumed" this can be incorporate in the formulation of disspation function. I will later formulate the lagragian with a disspation function to avoid the awkward variation inclding a first order term in the lagrangian. $\endgroup$ – ytlu Mar 29 at 5:52
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    $\begingroup$ The original Lagrangian was done in generalized coordinates $(m,r,\phi)$ already, the effect of changing mass term/rocket was resolved through the inference of D'Alembert's principle in Newtonian physics... It's a nice case where Newton's law and equation of motion was used in a combined fashion, otherwise it won't be so interesting. $\endgroup$ – ShoutOutAndCalculate Mar 29 at 6:33
  • $\begingroup$ Using dissipation function and generalized force, I add another answer to physics.stackexchange.com/questions/531329/… . $\endgroup$ – ytlu Mar 29 at 8:19

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