4 Acceleration of an relativistic rocket

[bayners123]

Homework Statement

A rocket of (time dependant) mass M ejects fuel such that its change in mass in the instantaneous ZMF is $$\frac{dM}{d\tau} = -\frac{E}{c^2}$$ The speed of the fuel ejected is $w$.
Prove that $$a = \frac{Ew}{Mc^2}$$
where a is defined by $-a^2 = A_\mu A^\mu$

The Attempt at a Solution

In the rest frame, $$P_{before} = \left(\begin{array}{c} M \\ 0 \end{array}\right) = P_{rocket} + P_{fuel} = P_{rocket} + \left(\begin{array}{c} \delta m \gamma \\ -\delta m \gamma w \end{array}\right)$$

We are given $$\frac{dM}{d\tau} = -E$$ so $$\delta m = -\frac{dM}{d\tau} d\tau = E d\tau$$

$$P \equiv M U \\ \frac{dP}{d\tau} = \frac{dM}{d\tau}U + MA = \left(\begin{array}{c} \frac{dM}{d\tau} - \gamma E \\ \gamma E w \end{array}\right) \\ -EU + MA = \left(\begin{array}{c} -E - \gamma E \\ \gamma E w \end{array}\right)$$

The rocket starts stationary in the ZMF so $U = \left(\begin{array}{c} 1 \\ 0 \end{array}\right)$

Therefore, $$A = -\frac{E}{M} \left(\begin{array}{c} \gamma \\ -\gamma w \end{array}\right)$$

$$-a^2 \equiv A_\mu A^\mu = \left(\frac{E}{M}\right)^2 [\gamma^2(w^2-1)] = \left(\frac{E}{M}\right)^2 \frac{w^2-1}{1 - w^2} = -\left(\frac{E}{M}\right)^2$$

I've lost an $w^2$ somewhere! Can anyone see it?

 

[anathema]

Thank you for your post. I have reviewed your solution and have found a small error. In your calculation of A, you have used the incorrect value for \delta m. It should be \delta m = \frac{-dM}{d\tau}d\tau = \frac{-E}{c^2}d\tau. This will result in A = -\frac{E}{Mc^2} \left(\begin{array}{c}
\gamma \\ -\gamma w
\end{array}\right), which will give the correct value for a = \frac{Ew}{Mc^2}. I hope this helps. Keep up the good work!