Orsdxy

English is not my native language, so please forgive my errors.

Consider this example:

This is a classic: an exercise requiring you to calculate the electric field produced by a charged ring on its axis. Here I will expose my reasoning to show you what I can't understand.

1. Every small charge $dq$ on the ring is contributing to the electric field. Its electric field is obviously a vector: $$dvec{E} = frac{dq}{4piepsilon_{0}r^{2}}hat{u}$$
   


2. We know that because of the symmetry, the $x$ components of the field behave "normally" meaning that they add together, but the $bot$ components respectively cancel themselves. So we only consider the $x$ component of the field for our calculations:
   $$dE_{x} = dvec{E},costheta$$
   from what I know, being $dE_{x}$ the component of another vector this should be just a number. However, I can also see that the simmetry facts make only the direction of the field constant, but the $dE_{x}$ field has still a verse depending on the positivity or negativity of the charges of the ring, so it cannot be just a number. My book, however, does like this:


3. It introduces the $x$ field as a function of numbers:
   $$dE_{x}(x) = frac{lambda dl}{4piepsilon_{0}r^{2}}costheta$$
   


4. It then proceeds to integrate, and it considers the final field still as a function of x but as a full fledged vector:
   $$vec{E}(x) = frac{lambda costheta hat{u_{x}}}{4piepsilon_{0}r^{2}}int_{0}^{l} dl$$
   

Now, I get really confused here.

1. First we isolated a component $dE_{x}$ which was not a vector.


2. However that component still is a vector even if its direction is fixed, or maybe the book was just considering its magnitude. We are now considering the field as a function of a vector: $vec{E}(x)$ that is equal to the product between the original field formula $dvec{E}$ and the $costheta$ because of the simmetry.


3. So in the final expression by the book we have the field as a vector, but also the $costheta$ which was isolating the $x$ component and then also the unit vector of the field is called $hat{u}_{x}$!

Isn't this a repetition? As you can see, I am really confused. What, in this process of calculation, keeps being a vector and what not?
If it was to me without getting confused by any book I would just return to the step 2 of the first list, to the expression
$$dE_{x} = dvec{E},costheta$$
and just proceed to the calculus
$$dE_{x} = frac{dq}{4piepsilon_{0}r^{2}}hat{u},costheta$$
but I couldn't say how exactly these vectors interact.

First of all, an expression like
$$dE_{x} = dvec{E},costheta$$
can never be correct. If you have a vector on the right, you will also have a vector on the left. The correct expression would be
$$dE_{x} = dvec{E} cdot hat{u}_x = frac{dq}{4piepsilon_{0}r^{2}} underbrace{(hat{u} cdothat{u}_x)}_{costheta} = frac{dq}{4piepsilon_{0}r^{2}} costheta$$
which is indeed a scaler ("just a number", as you call it).

What is happening here is that we want to know the electric field vector
$$ vec{E} = begin{pmatrix} E_x \ E_y \ E_z end{pmatrix}. $$
However, we know from symmetry, that two of these entries must be zero.
$$ vec{E} = begin{pmatrix} E_x \ 0 \ 0 end{pmatrix}. $$
Now, because we already know which direction the field is pointing (along the $x$-axis), we only need to calculate the magnitude of $E_x$ to get our result. So we extract the scalar $E_x$ from the vector like this:
$$ E_x = hat{u}_x cdot vec{E} = begin{pmatrix} 1 \ 0 \ 0 end{pmatrix}cdotbegin{pmatrix} E_x \ E_y \ E_z end{pmatrix}$$
Now we want to know the contribution of a line element to the total field, which you have already given
$$dvec{E} = frac{dq}{4piepsilon_{0}r^{2}}hat{u}$$
But we only need the $x$-component, which is the projection of $dvec{E}$ onto the $x$-axis, which is already given above
$$dE_{x} = dvec{E} cdot hat{u}_x = frac{dq}{4piepsilon_{0}r^{2}} costheta$$
from the geometry of the problem.

Then calculation for $E_x$ continues as in your post. However, in the end we want the electric field vector, so we need to substitute the magnitude of $E_x$ back into a vector that has only an $x$-component.
$$vec{E} = begin{pmatrix} E_x \ 0 \ 0 end{pmatrix} = E_x cdot begin{pmatrix} 1 \ 0 \ 0 end{pmatrix} = E_x cdot hat{u}_x$$

You cannot have a scalar equal to a vector.

Starting from $dvec{E} = dfrac{dq}{4piepsilon_{0}r^{2}}hat{u}$ to get to the componet in the $hat x$ direction

$dvec{E} cdot hat x = dE_{rm x} = dfrac{dq}{4piepsilon_{0}r^{2}}hat{u} cdot hat x =dfrac{dq}{4piepsilon_{0}r^{2}} cos theta$

A vector's component is also a vector. If we have a vector $vec{v}$ in (for example) two dimensions, we can say that it is the sum of two components, which are its projections on the $x$-axis and $y$-axis. Mathematically:

$$vec{v} = vec{v}_x + vec{v}_y$$

In the problem you are solving:

$$dvec{E} = dvec{E}_x + dvec{E}_y$$

This a vector sum, of course. A vector's component is still a vector, it is not a scalar. If we want to talk about the modulus of the vectors however:

$$dE^2 = dE_x^2 + dE_y^2$$

We may also write this in terms of the angles. We would find:

$$dE_x = dE costheta$$
$$dE_y = dE sintheta$$

So a vector is never equal to a scalar.