- #1

TubbaBlubba

## Homework Statement

In Newtonian mechanics, the rocket equation is derived by solving the simple differential equation -dm U = m dV, where U is the velocity of the expelled material relative to the rocket; a matter of conservation of momentum. In order to get the correct relativistic equation, one wishes to arrive at the differential equation ##-dm U = \gamma^2(V) m dV##. I know of lengthy derivations using velocity addition and so on, but I want to try to understand it in a more intuitve and managable way. My textbook (Rindler) is not exactly helpful by providing the hint of writing M for the rest mass of the rocket and dM for the "relativistic mass" of the expelled material...

## Homework Equations

We wish to arrive at ##-dm U = \gamma^2 m dV## using relativistic mechanics.

## The Attempt at a Solution

My line of reasoning is this: The final velocity V is the velocity relative to Earth (or wherever the rockets starts from). At any given moment, we will observe the rocket of mass m changing its velocity by dV to increase its momentum by ##\gamma(V) m dV##. On the other hand, the expelled material is expelled at a velocity U,

*constant relative to the instantaneous rest frame of the rocket*. So from Earth, the expelled material will be observed to have momentum ##-\frac{U}{\gamma(V)}dm##, giving us the desired differential equation ##-dm U = \gamma^2(V) m dV##..

Is this line of reasoning acceptable, intuitively speaking?