The Tsiolkovsky equation derived via the Lagrange funtion

Question

I wanted to derive the Tsiolkovsky rocket equation using the Lagrange formalism. $$ T = \frac{m(t)}{2}\dot{\xi(t)}^2 + \frac{\mu(t)}{2}(\dot{\xi(t)} - u)^2$$ $m$ - mass of the rocket

$\xi$ - some coordinate along which the rocket is moving

$\mu$ - mass of the thrown out fuel

$u$ - fuel velocity (in the rocket's reference frame).

The standard procedure yields the following equation of motion: $$\ddot{\xi} (m+\mu) + \dot{\xi} (\dot{m} + \dot{\mu}) = \dot{\mu} u.$$ Since $m(t) = m_0 - \eta t$, and $\mu = \eta t$, there follow $\dot{m} + \dot{\mu} = 0$, $m+\mu = m_0$, and $\dot{\mu} = - \dot{m}$.

Therefore we have $$m_0 \frac{d}{dt}\dot{\xi} = m_0 \ddot{\xi} = -u \eta.$$

This is not the anticipated result, the Tsiolkovsky equation is $$\frac{d}{dt}\dot{\xi} = - u \frac{dm}{m},$$ where $m = m(t)$. Specifically, my issue is the fact that $m_0 \neq m(t)$.

Where is my mistake? Is the lagrangian wrong, or am I missing something else here?

Answer 1

Your Lagrangian doesn't accurately describe the energy of the system. You have a term in your kinetic energy $$ \frac{\mu(t)}{2}(\dot{\xi(t)} - u)^2 $$ which is intended (I assume) to describe the kinetic energy $T_\text{fuel}(t)$ of all of the fuel that has been thrown out of the rocket up to time $t$. But the fuel that's thrown out at a time $t_e < t$ will not be moving at speed $\dot{\xi}(t) - u$; it will be moving at speed $\dot{\xi}(t_e) - u$.

We can try to write down an expression for $T_\text{fuel}(t)$ by integrating over the rocket's history. The kinetic energy of the fuel that's thrown out between $t_e$ and $t_e + \Delta t_e$ will be $$ \frac{\Delta \mu}{2}(\dot{\xi(t_e)} - u)^2 $$ where $\Delta \mu \equiv \dot{\mu}(t_e) \Delta t_e$. Thus, the total kinetic energy of the ejected fuel will be $$ T_\text{fuel}(t) = \frac{1}{2} \int_0^t \dot{\mu}(t_e) (\dot{\xi}(t_e) - u)^2 \, dt_e. $$

Note that since your Lagrangian $\mathcal{L}$ will now involve an integral over the past history of the function $\xi(t)$, rather than simply depending on $\xi(t)$ and $\dot{\xi}(t)$. This means that the plain-vanilla Euler-Lagrange equations $d/dt(\partial\mathcal{L}/\partial \dot{\xi}) = \partial\mathcal{L}/\xi$ cannot straightforwardly be applied. You will instead have to return to first principles in the calculus of variations to derive the equations of motion.