How to use Lorentz's Transformation properly? I have a series of doubt about how Lorentz's Transformation are written and used commonly:
First doubt:
It's really common to see Lorentz transformation written this way:
$$x'=\gamma x -\gamma \beta t$$
$$t'=\gamma t -\gamma \beta x$$
where, $c=1$ in natural units. However, this way of writing Lorentz's Transformation is valid only if at time $t'=t=0$, the origin of both coordinate systems are in the same position i.e. equal to zero. It seems to me that it is much more natural to write Lorentz's Transformation, as in Morin's special relativity textbook, as follows:
$$\Delta x'=\gamma \Delta x -\gamma \beta \Delta t$$
$$\Delta t'=\gamma \Delta t -\gamma \beta \Delta x$$
In this way we don't have to assume the same zero position at the time zero. I really don't understand why in textbooks and wikipedia pages the first way of writing L.T. is the most popular.

*

*Does it have some advantages I don't understand?

Second doubt:
Let's suppose we want to derive time dilation from the L.T.; my professor does it in the following way:
He starts by stating that the clock in motion with the reference frame $O'$ has spatial coordinate: $x'=0$; from here we see that:
$$0=\gamma x -\gamma \beta t \ \Rightarrow \ x=\beta t=vt$$
(we have $\beta t = vt$ because $c=1$). Then:
$$t'=\gamma t -\gamma \beta x \ \Rightarrow \ t'=\gamma t -\gamma \beta^2 t \ \Rightarrow \ t'=\frac{t}{\gamma}$$
However, this smells fishy to me. Because, if we consider the problem from the point of view of the reference frame $O$ then we are not allowed to state anything about the quantities measured by $O'$ if not by using Lorentz's Transformation. Hence, I find that starting the proof by stating that $x'=0$ is not formally correct. A better way I think is to say that $O$ sees $O'$ moving an amount equal to $\Delta x =\beta \Delta t= v \Delta t$, and so we can say that:
$$\Delta t'=\gamma \Delta t -\gamma \beta \Delta x \ \Rightarrow \ \Delta t'=\gamma \Delta t -\gamma \beta^2 \Delta t$$
and then we get the same result. Also my way of doing it seems faster to me.

*

*Is my reasoning correct?

Third doubt:
And at last let's talk about proving length contraction: I have seen all sorts of lengthy proofs for this one, but seems to me that we can simply state:
Suppose that I (in the O frame) measure the distance ($\Delta x$) between the ends of a rod at rest in my frame. The $\Delta t$ between the events (the ends of the rod) is obviously $0$, because them have their position $x_1,x_2$ at the same time. Then we can simply say:
$$\Delta x'=\gamma \Delta x -\gamma \beta \Delta t \ \Rightarrow \ \Delta x'=\gamma \Delta x$$
and length contraction is proven. Is there something I am missing? (Maybe is not correct to state that the temporal distance of the ends of the rod for $O$ is zero, because the rod is at rest in $O$ and so we could measure the two ends at different times and still get the correct length of the rod; I do not know)
 A: First doubt: Lorentz transformations are specific types of transformations that are consistent with homogeneity, isotropy and universal speed postulates. They are typically defined for the case when spacetime origins of the two frames coincide. More specifically, these include mutually boosted and rotated frames as long as their spacetime origins coincide.
Translation of one frame w.r.t. another is typically considered separately from Lorentz transformations. The overarching group of all possible transformations consistent with special relativity is called the Poincare group, which includes Lorentz transformations and translations. Quoting from Wikipedia:

Poincaré symmetry is the full symmetry of special relativity. It includes:

*

*translations (displacements) in time and space ($P$), forming the abelian Lie group of translations on space-time;

*rotations in space, forming the non-Abelian Lie group of three-dimensional rotations ($J$);

*boosts, transformations connecting two uniformly moving bodies ($K$).

The last two symmetries, $J$ and $K$, together make the Lorentz group (see also Lorentz invariance); the semi-direct product of the translations group and the Lorentz group then produce the Poincaré group. Objects which are invariant under this group are then said to possess Poincaré invariance or relativistic invariance.

Morin's book does account for mutual translation of the two frames' origins, so that's correct (even though not standard). But I hope the above answers your actual question on why we usually don't account for translation in Lorentz transformations - because it's considered separately as another symmetry.
I had the same doubt as yours, so it might be worthwhile checking out Derivation of the Lorentz transformation without assuming that clocks are synchronized when origins align and its answer if you like.

Second doubt: A rephrasing might help. If $E$ is an event whose location in $O'$ is given by $x'$, then yes, you can't talk about its location w.r.t. $O$ without first using the Lorentz transformation. I think the confusion is because $x'=0$ is used to denote the origin of $O'$.
Look at it in this way: for an event $E$, what should be the relation between its $t$ and $x$  coordinates in $O$, for us to call it the origin of $O'$? i.e. How should $t$ and $x$ be related so that $x'=0$? I think that perspective should make it clearer.

Third doubt: You measured $\Delta x$ when $\Delta t=0$, which is fine, but did you measure $\Delta x'$ when $\Delta t'=0$? You have to make sure that if the coordinates of the rod ends are $(t'_1,x'_1)$ and $(t'_2,x'_2)$, then $|x'_2-x'_1|$ represents the rod length only if $t'_1=t'_2$. Length contraction proofs are lengthy to ensure that this is the case.
A: 
I really don't understand why in textbooks and wikipedia pages the first way of writing L.T. is the most popular

Because L.T. are supposed to transform coordinate frames. From your expression, how would you know what are the new coordinates of a point? Any shift of origin cancels out from your formula, so you are loosing information. The assumptions on having same origin of both frames is just for simplicity. Whenever you wish to pick another origin, you can easily adjust the formula yourself and there is no need to bother with it in general.
If you are little more mathematically inclined, the common L.T. formula transform coordinates, while Morin's formula transforms tangent vectors, so the two formulas are concerned with different mathematical objects. The convention could then be given by which mathematical object we are considering to be more central to our analysis.

we are not allowed to state anything about the quantities measured by O′ if not by using Lorentz's Transformation

First of all, $x'$ is not strictly speaking measured quantity, it is a coordinate. Second, you are indeed using L.T. so I do not see the problem.
You are starting with the knowledge of coordinate transformation between two frames $O$ and $O'$. If the event P has coordinates $(x',t')$ in coordinates $O'$ then you already have a formula for coordinates of this event in $O$ coordinates. In particular, if you pick a curve $(0,t')$ - which is by definition the curve along which origin of $O'$ moves in $O'$ coordinates , then you have L.T. to compute $O$ coordinates of the same curve, that is $(v\gamma t',\gamma t')$. To interpret this result simply note, that the event of clocks of $O'$ showing time $t'$ (that is the event $P=(0,t')$), is simultaneous w.r.t $O$ with the event of clocks of $O$ showing time $t=\gamma t'$ and to note, that at the origin the two clocks are at the same place.

...and length contraction is proven. Is there something I am missing?

Indeed you are. Assuming the measurement happened at $t=0$, your $\Delta x'$ is spatial distance w.r.t $O'$ between events $P_1=(x_1,0)$ and $P_2=(x_2,0)$ written in $O$ coordinates. Writing these two events in $O'$ coordinates you get $P_1=(\gamma x_1,-\gamma \beta x_1)$ and $P_2=(\gamma x_2,-\gamma \beta x_1)$. But since $x_1\neq x_2$ the events $P_1$ and $P_2$ are not simultaneous w.r.t to $O'$ frame. The length however needs to be measured simultaneously at both ends of the rod, since the rod is by your own assumption moving in $O'$ frame. Note also, that your rod from $O'$ point of view is not contracted, but extended so your proof gives wrong formula.
