On the proof that $4$-velocity transforms like vector

Question

Let $U$ and $U'$ be the $4$-velocities associated to the coordinates $(t,x)$ and $(t',x')$ related through the Poincaré transformation $P:\mathbb R^4\to\mathbb R^4$, i.e. $(t',x')=P(t,x)$.$^1$

Of course the Jacobian $\Lambda\in\mathbb R^{4\times 4}$ of $P$ is a Lorentz transformation. I extracted the following derivation of the tranformation rule for $U$ from this question: \begin{equation} U'=\frac{\mathrm dX'}{\mathrm d\tau}=\frac{\mathrm d(P\circ X)}{\mathrm d\tau}=\Lambda\cdot\frac{\mathrm d X}{\mathrm d\tau}=\Lambda\cdot U\in\mathbb R^{4\times 1} \end{equation} As far as I understand, we used the fact that \begin{equation} \forall \tau:X'(\tau)=(P\circ X)(\tau), \end{equation} but this is not trivial, is it? I will explain my reasoning and I hope for a confirmation/verification:

We have $X(\tau):=X(t(\tau))$, where $I\ni t\mapsto X(t)$ is the $4$-position and $\tau\mapsto t(\tau)$ is the inverse of proper time, i.e. the inverse of the function $\newcommand{\d}{\mathop{}\!\mathrm{d}}$ \begin{align} I\ni t\mapsto \tau(t)=\int_{t_0}^t\sqrt{1-\frac{v(\widetilde t)^2}{c^2}}\d\widetilde t+c \end{align} for some $t_0\in I$ and $c\in\mathbb R$. So what we are really assuming is the following: \begin{equation}\tag{1} X'\circ t'=P\circ X\circ t \end{equation} Let $\Pi:\mathbb R^4\to\mathbb R$ be the projection to the time component, then $X'=P\circ X\circ(\Pi\circ P\circ X)^{-1}$ and hence $(1)$ follows from the fact that \begin{equation} t'=\Pi\circ P\circ X\circ t \end{equation} which is equivalent to \begin{equation} \tau=\tau'\circ\Pi\circ P\circ X \end{equation} and which can be proven through a change of variables.$^2$ Am I right?

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$^1$ The reader familiar with manifolds will note that $(t,x)$ is a chart $\phi: M\to\mathbb R^4$ and that $P=\phi'\circ\phi^{-1}$.

$^2$ We are exploiting the fact that $\tau$ and $\tau'$ are only defined up to a constant when we assume that $t$ and $t'$ have the same domains.

Answer 1

I think you are making this unnecessarily difficult. Here is a proof. $$ U = \lim_{\delta\tau \rightarrow 0} \frac{X(t+\delta\tau) - X(t)}{\delta\tau} $$ Now use that $\delta\tau$ is invariant and the difference of two 4-vectors evaluated at a given event is itself a 4-vector (which is easy to prove). It follows that $U$ is a 4-vector multiplied by a scalar invariant, hence it is a 4-vector.

Answer 2

As @Frobenius said in the comments, this is a bit too complex to understand a relatively easy equation. I would like to make your equations bit clearer here.

As you said, $X'(t)=(P\circ X)(\tau)$ should actually be:

$$(X'\circ t')(\tau)=(P\circ X)(\tau)$$

Then, applying inverse function on the right:

$$(X'\circ t' \circ (t')^{-1})(\tau)=(P\circ X \circ (t')^{-1})(\tau)$$

$$(X')(\tau)=(P\circ X \circ (t')^{-1})(\tau)$$

Using the projection operator:

$$(\tau)=(\Pi \circ P\circ X \circ (t')^{-1})(\tau)$$

Why? Because $(X' \circ t')(\tau) \ne (X \circ t')(\tau)$. Therefore applying $(\Pi \circ X')(\tau) \ne (t')(\tau)$. By the way, in my definition projection operation is something that takes out the first component of a vector. And $X'$ is just a function that is different from $X$.

So I would summarize this redundant set of operations as the following:

1. Take proper time and apply an inverse transform on it which is just the inverse of the operation that a Lorentz boost would do on the proper time.
2. Use this as an argument of your 4-position vector.
3. Apply the Lorentz boost.
4. Take the projection in time.

Voilà! You are at the very same time where you started.