# When does a vector component keep being a vector, exactly?

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

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 $$d\vec{E}$$ and the $$cos\theta$$ because of the simmetry.
3. So in the final expression by the book we have the field as a vector, but also the $$cos\theta$$ 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 $$dE_{x} = d\vec{E}\,cos\theta$$ and just proceed to the calculus $$dE_{x} = \frac{dq}{4\pi\epsilon_{0}r^{2}}\hat{u}\,cos\theta$$ but I couldn't say how exactly these vectors interact.

• When decomposing the vector $(2, 3)$ into a sum of vectors parallel to the axes, $(2, 0)$ is a vector, but $2$ is not. By careless people both of these are called the horizontal component of the vector. And for people who aren't careless, which of them is actually called "the horizontal component", and what to call the other one may vary. – Arthur Feb 18 at 16:18

First of all, an expression like $$dE_{x} = d\vec{E}\,\cos\theta$$ 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 $$dE_{x} = d\vec{E} \cdot \hat{u}_x = \frac{dq}{4\pi\epsilon_{0}r^{2}} \underbrace{(\hat{u} \cdot\hat{u}_x)}_{\cos\theta} = \frac{dq}{4\pi\epsilon_{0}r^{2}} \cos\theta$$ 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: $$E_x = \hat{u}_x \cdot \vec{E} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\cdot\begin{pmatrix} E_x \\ E_y \\ E_z \end{pmatrix}$$ Now we want to know the contribution of a line element to the total field, which you have already given $$d\vec{E} = \frac{dq}{4\pi\epsilon_{0}r^{2}}\hat{u}$$ But we only need the $$x$$-component, which is the projection of $$d\vec{E}$$ onto the $$x$$-axis, which is already given above $$dE_{x} = d\vec{E} \cdot \hat{u}_x = \frac{dq}{4\pi\epsilon_{0}r^{2}} \cos\theta$$ 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. $$\vec{E} = \begin{pmatrix} E_x \\ 0 \\ 0 \end{pmatrix} = E_x \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = E_x \cdot \hat{u}_x$$

• This answer is not correct, as $dE_x=d|\overrightarrow E| \cos\theta$ loses the sign of $q$, or in other words this is the absolute value of $dE_x$, not $dE_x$ itself. I believe this sign is the core of the confusion in the original question. – tobi_s Feb 18 at 22:35
• @tobi_s How does it lose the sign? The sign is captured by the $\cos\theta$ part, since it can take on values between -1 and 1. – Kyle Feb 18 at 22:42
• I doubt that this was at the core of the confusion of the OP, but I agree that it was sloppy notation; it should be clear now. – noah Feb 18 at 23:40
• @Kyle I don't think that a charge-dependent redefinition of angles helps the original poster in any way. But your viewpoint may be consistent with the negative absolute values that you seem to tolerate :-) Anyway, that's moot now that @ noah improved his answer. – tobi_s Feb 19 at 3:04
• @noah It's certainly the point where the OP lost me, but confusion can arise in many and different ways :) Thanks for cleaning up the notation nonetheless. – tobi_s Feb 19 at 3:05

You cannot have a scalar equal to a vector.

Starting from $$d\vec{E} = \dfrac{dq}{4\pi\epsilon_{0}r^{2}}\hat{u}$$ to get to the componet in the $$\hat x$$ direction

$$d\vec{E} \cdot \hat x = dE_{\rm x} = \dfrac{dq}{4\pi\epsilon_{0}r^{2}}\hat{u} \cdot \hat x =\dfrac{dq}{4\pi\epsilon_{0}r^{2}} \cos \theta$$

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:

$$d\vec{E} = d\vec{E}_x + d\vec{E}_y$$

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:

$$dE_x = dE \cos\theta$$ $$dE_y = dE \sin\theta$$

So a vector is never equal to a scalar.

• This is using a different definition of "component" than the other answers. In this answer, the component and the unit vector are not separate entities; in the other answers, the component is the dot product of a vector and a unit vector (and hence is a scalar). The objects you describe as "components" here might be referred to in the other answers as projections of a vector onto a given unit vector. Please make this distinction clear. – probably_someone Feb 18 at 13:42

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