Time dilation calculation in the relativistic rocket Rindler coordinates are useful for describing objects moving at constant acceleration:
\begin{equation}
\mathbf{x} = \begin{pmatrix}
t \\ x
\end{pmatrix} = \begin{pmatrix} \xi\sinh [\alpha \zeta] \\ \xi\cosh[\alpha \zeta]
\end{pmatrix}.
\end{equation}
This transforms the Minkowski metric into
\begin{equation}
ds^2 = -(\alpha \xi)^2 d\zeta^2 + d\xi^2.
\end{equation}
I think Rindler coordinates are helpful in describing the relativistic rocket. For example, if I fix $\alpha$ then two clocks on two curves with different $\xi$ will have the relation
\begin{equation}
d\tau_1 = \alpha \xi_1 d\zeta, \quad d\tau_2 = \alpha \xi_2 d\zeta \implies d\tau_1 = \frac{\xi_1}{\xi_2} d\tau_2.
\end{equation}
If $\xi_2 > \xi_1$ then this implies that clocks higher up in the rocket tick faster than clocks lower down, which is the standard result. So far so good. 
To justify my assertion that we are looking at the relativistic rocket, both clocks must have the same proper acceleration, as this is its defining characteristic. I have seen it written that $\alpha$ is the proper acceleration, which would be grand, so now I need to prove it. First, let's find the four-velocity:
\begin{equation}
\mathbf{v} \equiv \frac{d\mathbf{x}}{d\tau} =  \frac{1}{\alpha\xi}\frac{d}{d\zeta} \begin{pmatrix} \xi \sinh [\alpha \zeta] \\ \xi \cosh[\alpha \zeta] \end{pmatrix} = \begin{pmatrix} \cosh[\alpha \zeta] \\ \sinh[\alpha \zeta] \end{pmatrix}.
\end{equation}
$\mathbf{v}$ is normalised so so far so good. Taking another derivative provides
\begin{equation}
\mathbf{a} \equiv \frac{d\mathbf{v}}{d\tau} = \frac{1}{\alpha\xi}\frac{d\mathbf{v}}{d\zeta} = \frac{1}{\xi}\begin{pmatrix} \sinh[\alpha \zeta] \\ \cosh[\alpha\zeta] \end{pmatrix}.
\end{equation}
The proper acceleration is
\begin{equation}
\sqrt{a_ia^i} = \frac{1}{\xi}.
\end{equation}
This is not fixed and therefore Rindler coordinates do not help with the relativistic rocket. Is this true or have I made a mistake? I have seen the derivative taken with respect to $\zeta$ rather than $\tau$, in which case the answer comes out as $\alpha$ but the normalisation that makes these two trajectories physical (i.e. actual moving objects obeying $v_iv^i = 1$) implies that $\alpha = 1/\xi$, and so only one of the trajectories is physical if $\alpha$ is fixed. What am I doing wrong? What is the correct way to describe the relativistic rocket? Unless I'm missing something, John Baez' notes on the topic don't include any calculation for the clocks in the rocket. 
 A: Your calculation is correct. What you've discovered is that in Rindler coordinates, every worldline of constant $\xi$ has its own (constant) proper acceleration. As to why, see Bell's spaceship paradox and Born rigid motion. The worldline of the rocket itself is given by $\xi \equiv c^2/\alpha$, if it's treated as a point particle. For more details, see this answer.
For a rocket with a sizeable length, the front and back have to move with different proper acceleration, otherwise it will tear itself apart (that's Bell's paradox). So,
$$
x_\text{front} = \xi_\text{front}\cosh(\alpha\zeta/c),\\
x_\text{center} = \xi_\text{center}\cosh(\alpha\zeta/c),\\
x_\text{back} = \xi_\text{back}\cosh(\alpha\zeta/c).
$$
$$
\xi_\text{center} = \frac{c^2}{\alpha},\\
\xi_\text{front} = \frac{c^2}{\alpha} + L/2,\\
\xi_\text{back} = \frac{c^2}{\alpha} - L/2,
$$
where $L$ is the length of the ship at rest. The corresponding proper accelerations at the front and back are therefore
$$
\alpha_\text{front} = \frac{c^2}{\xi_\text{front}},\\
\alpha_\text{back} = \frac{c^2}{\xi_\text{back}}.
$$
A: Pulsar's answer was helpful but I want to lay out the misconception more clearly. The setup is to consider the inside of the relativistic rocket with clocks placed at different locations. I jumped to Rindler coordinates as the well-known coordinates to describe events within the rocket. From there, I correctly derive the time difference between clocks at different points in the frame. Then I say I must have been wrong because:
Assumption 1: A uniformly accelerating rocket must be characterised by its various points having equal proper acceleration. 
This is not the standard characterisation of the relativistic rocket. The standard characterisation is an observer with uniform acceleration and the coordinate chart that they place on the world around them. This chart is defined by rods that are equidistant from each other and the observer. See a derivation of Rindler coordinates for this. As Pulsar pointed out, this characterisation does not impose equal proper acceleration on all clocks in the frame.
Asumption 2 The rocket must be characterised by various points having equal proper acceleration because this is what one would find in a uniform gravitational, and there should be some equivalence.
This is not so obviously the definition of a uniform gravitational field. Note that, from the observer's point of view, objects moving in the in the frame are going to have a 'Rindler coordinate acceleration' which will not be the same as either the 'Lab coordinate acceleration' or proper acceleration. There is literature on the definition of uniform gravitational fields which is beyond the scope of the question. There is a discussion in Baez's notes (linked to in the question) on falling objects in the relativistic rocket that might be a good starting point.
A: What is the correct way to describe the relativistic rocket?
I use the relativistic equation of motion
$$\frac{d}{dT}\left(\gamma\,M\,v\right)=\gamma\,F\tag 1$$
$\gamma=\frac{1}{\sqrt{1-v^2(T)}}\quad, c=1$
where :
$T$ is the proper time (rocket time)
$M$ is rocket plus fuel mass
$F$ is the Newton force 
for $F=M\,a$ where a is a is the constant rocket  acceleration we get (equation (1)):
$$\frac{dv}{dT}=a(1-v^2)$$
$\Rightarrow$
$$v(T)=\tanh(a\,T)\quad ,v(0)=0$$
to get the earth time $t$ you can use this equation:
$$\frac{dt}{dT}=\gamma(v(T))$$
$\Rightarrow$
$$t(T)=\frac{\sinh(a\,T}{a}$$
the distance from the earth:
with 
$\frac{dv}{dt}=\frac{dv}{d\lambda}\frac{d\lambda}{dt}$
$\Rightarrow\quad$
$\frac{dv}{d\lambda}=\frac{1}{\gamma(v(\lambda))}\frac{d\lambda}{dt}=a(1-v(\lambda))^{3/2}\quad \Rightarrow\quad v(\lambda)$
$x(\lambda)=\int v(\lambda)\,d\lambda=\frac{\sqrt{1+a^2\lambda^2}-1}{a}\quad ,x(0)=0$
so the distance from the earth is:
$$x(\lambda=t(T))=\frac{\cosh(a\,T)-1}{a}$$
These result  are the Rindler equations
