# Some doubts about Field Lines [duplicate]

Take electric field lines as example. In EM textbooks it is stated that

1. The tangent of a field line gives the direction of the electric field

2. The density of field lines is proportional to the field strength at that point

My understanding is that field lines are not rigorous mathematically. So I doubt whether the above can always be true.

For instance, if we draw field lines according to rules 1 and 2, will it happen sometimes that you need to create new field lines or annihilate field lines artificially so that rule 2 is satisfied?

## marked as duplicate by sammy gerbil, Chris♦, stafusa, valerio, Cosmas ZachosFeb 11 '18 at 3:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

To kick things off: this

field lines are not rigorous mathematically

is just plain wrong. It is perfectly easy to come up with a mathematically rigorous definition of a field line: it is a curve $\mathbf c:(a,b)\to\mathbb R^3$ such that $$\frac{\mathrm d\mathbf c}{\mathrm ds} = v(s) \mathbf E(\mathbf c(s)),$$ i.e. such that its derivative $\mathbf c'(s)$ is proportional to the electric field at $\mathbf c(s)$. (The factor of $v(s)$ is there to allow for unit-norm $\mathbf c'(s)$ if so desired.) There is nothing more, and nothing less, to the definition of field lines. It can happen that there are points at which field lines cannot be continued (including e.g. point charges), but that's really all in the game; if $\mathbf E(\mathbf r)$ is regular, then this cannot happen.

OK, so, that settles point 1. Point 2 is harder to deal with, and the answer depends on the configuration; more specifically, on whether the field is being examined in vacuum or in a region with nonzero charge. I've gone over much of this in my answer to Why does the density of electric field lines make sense, if there is a field line through every point?, but the short of it is that the density of field lines takes all of its validity from Gauss's law, $$\nabla\cdot \mathbf{E}=\frac1{\epsilon_0}\rho,\ \text{or equivalently}\ \oint_{\partial\Omega}\mathbf{E}\cdot\text d\mathbf{S}=\frac1{\epsilon_0}Q_\Omega,$$ as well as the concept of a flux tube: a tube-like surface made up of two end-caps $S_1$ and $S_2$, such that all the field lines that start at $\partial S_2$ end at $\partial S_1$, and their union forms the rest of the tube (which is then parallel to the field lines at every point). Image source

(The image has a wonky $S_1$ vs $S_2$ ordering, blame Wikipedia for that.)

The good thing about a flux tube is that there is no electric flux through the 'tube' bit, by construction, which means that Gauss's law reduces to $$\int_{S_1}\mathbf{E}\cdot\text d\mathbf{S} - \int_{S_2}\mathbf{E}\cdot\text d\mathbf{S} = \frac1{\epsilon_0}Q_\Omega$$ in the presence of charges, and to $$\int_{S_1}\mathbf{E}\cdot\text d\mathbf{S} = \int_{S_2}\mathbf{E}\cdot\text d\mathbf{S}$$ in vacuum.

That last equation is the crucial bit: if you draw five streamlines going through $S_2$, then those five streamlines need to go through $S_1$, and that means that if $S_1$ is bigger than $S_2$ then the density of field lines has gone down ─ but so has the field! The flux (like the number of streamlines) is conserved, and if the surface increases, then the field strength needs to go down as well, in exactly the same proportion as the density of field lines.

That brings me, then, to your other key point,

will it happen sometimes that you need to create new field lines or annihilate field lines artificially so that rule 2 is satisfied?

to which the answer is not in a vaccuum, but yes if there are nonzero charges.

• In vacuum, all you need to do is draw your field-line starting points consistently, and then the field-line density will be a good indicator of field strength everywhere; moreover, in the limit where you draw your start-points closer and closer together, then you completely capture all the relevant information about the vector field and you'd be able to reconstruct it just from the diagram.

• In the presence of charges, on the other hand, you will very often be able to draw field lines through every point, but if you want to make an accurate diagram (such that the field-line density is reflective of the field strength) then you will need to kill or start field lines at a rate dictated by the local charge density through Gauss's law as stated above. Doing that correctly is a hardish problem but there's nothing stopping you from doing a full formalization of the procedure, all the way to producing diagrams that encode (in the limit of infinitely tight spacing) the full information of the vector field.

So, why don't we talk about that formalization much? Frankly, because it's not very useful. Streamline plots are useful qualitative aids in both situations, but if you want to use them in earnest for a quantitative or semi-quantitative analysis, then that's only really going to work in the divergenceless case. And, in any case, streamline views of electric fields are extremely limited since they handle superpositions of field very poorly, and in most situations you're better off seeing electric fields as vector fields rather than as collections of streamlines, which is the core place where the lack of focus comes from.

• In two dimensions, there is the technique of conformal mapping in the complex plane to make drawings of field lines. Vector fields may be easier in numerical calculations. – Pieter Feb 6 '18 at 20:35