# Are the field lines the same as the trajectories of a particle with initial velocity zero

Is it true that the field lines of an electric field are identical to the trajectories of a charged particle with initial velocity zero? If so, how can one prove it?

The claim is from a german physics book from Nolting "Grundkurs theoretische Physik 3 - Elektrodynamik" page 51, let me quote:

Man führt Feldlinien ein und versteht darunter die Bahnen, auf denen sich ein kleiner, positiv geladener, anfangs ruhender Körper aufgrund der Coulomb-Kraft (2.11) bzw. (2.20) fortbewegen würde.

In english:

One introduces field lines and means trajectories along which a small, positively charged, initially resting body moves due to the Coulomb-foce (2.11) resp. (2.20).

2.11 is just the coulomb law, 2.20 is $F = q E$.

(If someone has a better translation, feel free to edit it).

I don't see why this should be true. So it would be great to see a proof or a counterexample with solved equations of motion.

For a magnetic field this claim is obviously wrong since the Lorentz Force depends linearly on the velocity.

Are there other physical fields where the claim is analogously true?

Edit: The answers show that the claim is not true in general but holds in the special case of a highly viscous medium. Is this also the case for moving charged cotton along the field lines in air, as shown in this animation: http://www.leifiphysik.de/web_ph09_g8/grundwissen/01e_feldlinien/01e_feldlinien.htm ?

Do you have any references or more details for this viscous media limit?

Do you have any computational counter example why it doesn't hold in general or a simulation which shows that?

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No, the statement is false even in the electric case. At the very beginning, the acceleration is $\vec a \sim \vec E$ so they have the same direction at $t=0$: the tangents agree.

However, as soon as the particle reaches some nonzero velocity $\vec v \neq 0$, its acceleration is still $\vec a\sim \vec E$, in the direction of the field lines, however its velocity – and it's the velocity that determines the tangent direction of the trajectory – is not proportional to the acceleration.

Again, the field lines have direction corresponding to the acceleration at the given point but the trajectories have directions given by the velocity and $\vec v$ isn't a multiple of $\vec a$ in general.

Imagine a simple example above. If you start with a positive charge, and $\vec v=0$ and very close to the positive-charge source above, they will repel and the moving charge will quickly achieve a huge speed. This speed will act as inertia that will make the trajectories much more straight than the field line and the discrepancy will become more extreme if the initial position of the moving charge will be even closer to the positive source.

You would only get the relationship "field lines are equal to trajectories" if you were stopping the moving test charge at every moment and slowly allowed the field to accelerate from scratch after each infinitesimal amount of time.

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Thanks. Do you have a computational counter-example (physical possible field configuration and a solution to the corresponding equations of motions to get an exact counter example)? –  Anna Sep 16 '12 at 19:00
@Anna you can numerically or analytically solve pretty much any situation where the field lines are not straight, and you will see an example of this effect. –  David Z Sep 16 '12 at 20:37
You may treat them as the path that would be followed by a test charge moving very slowly through a viscus medium because that would be quasi-static and every instant would fulfill the "no initial velocity" condition within whatever tolerances you require.. –  dmckee Sep 17 '12 at 1:34
The statement is true in a highly viscous medium with linear in velocity viscosity, like pollen in water. This is what the author is surely mangling. But +1 anyway. –  Ron Maimon Sep 17 '12 at 9:01

As Luboš points out, in any situation with non-straight field lines the particle's inertia will make it "skid" across field lines and make the trajectory depart from the field line.

If you are intent on a physical analogy or representation for the field lines, there is a relatively similar one which does hold. If you want to eliminate inertial effects from a mechanical analogy, one way to do it is to introduce a damping term proportional to the velocity, instead of the acceleration, which dwarfs the inertia. Thus, given an electric field configuration, consider a particle of mass $m$ subject to the field in addition to a damping force of constant $R$:

$$m\mathbf{a}-R\mathbf{v}=\mathbf{F},$$

and assume that the particle is light enough and subject to a strong enough damping that $m$ can be assumed to be zero in this equation. (Note, though, that you cannot compare $m$ and $R$ due to their differing dimensions. This means you must postulate that $m/R$, which has dimension of time, is much smaller than any time of interest in your system. Even worse, the system's "characteristic times" are so far ill defined and have to be defined in terms of what spatial resolution you consider acceptable.)

Once you do this, though, the particle's trajectory follows the field, since $$\mathbf{v}\approx-\frac{q}{R}\mathbf{E}.$$ This can be rephrased physically as saying that if the damped particle is light enough that it is always going at the local terminal velocity dictated by the force and the damping, then its trajectory follows the field lines.

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Dear Emilio, would you accept that my answer (which contains exactly what you said) remains posted here? I was so happy that –  Lupercus Sep 16 '12 at 21:34
... that I found it. –  Lupercus Sep 16 '12 at 21:44
@Lupercus - I should admit to being a bit upset at an almost exact duplicate to my answer popping up an hour after my post. –  Emilio Pisanty Sep 17 '12 at 12:52
You asked whether I'm OK with it so I told you what I think. I do not see any clear differences apart from the hiding of "haphazard" argumentation under the "it can be shown" rug. For me the crucial physical concept is the notion of a light particle so heavily damped that it will always go at the local terminal velocity. I do not ask that the post be taken down, but since you asked for my opinion - I am upset about it. To be clear and to avoid FWs, this is my last post on this thread. –  Emilio Pisanty Sep 17 '12 at 15:14
I am so sorry. I will retract my post, the single one I can do. –  Lupercus Sep 17 '12 at 15:19

Let me make it simple. What are electric lines of force? They are lines in which a +ve charge move. Let us consider a +ve charge and lines of force(field) are emitted from it. If we keep an another +ve charge on any line of force(field line), it moves away in certain trajectory. What if it kept more closer to the first +ve charge? It also traces the same trajectory(considering you kept it on the same line). So, logically, the field lines of an electric field are identical to the trajectories of a charged particle with initial velocity zero.

To get more insight, i think, you have heard of an experiment in magnetism where an conducting wire is passed through the center of a paper perpendicularly containing iron fillings. The fillings align along the field. Now, consider the same with electric field and the iron fillings are electric charges.

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Usually field lines are the lines where $\vec{E}$ is tangent to at each point. So you start from the wrong definition! –  Anna Sep 25 '12 at 18:03
There are a lot of ways to define Electric Field. I explained it in the lamest way. I am sure i am right. You can refer to Feynman lecture on Physics part 2 for more info (about my answer). –  Fr34K Sep 25 '12 at 18:38
Can you give a page number and a quote to what you refer to Feynman lectures? –  Anna Sep 25 '12 at 18:51
Looking up in the Feynman lectures I find (p.4-11): "We draw lines in the direction of the field - lines which are always tangent to the field, as in Fig. 4-12. These are called field lines" That's exactly what I said, isn't it? –  Anna Sep 25 '12 at 18:57
Yup. I am sure that it is from that book. What explained is same as the path of unit positive charge in an electric field. Put q1=1 in the force equation (force between 2 charges). The electric field can be defined in alot of ways. As you have mentioned 'particle with initial velocity zero', i have taken charges to explain it. Else, to understand the fields, what you have told is true. –  Fr34K Sep 25 '12 at 19:03

Here is a simulation which seems to be reasonable (the argument Lubos-Motl pointed out). However I don't have the source code to verify that the calculations are done correct:

Trajectory of a test charge in electric field

For a reference that your claim is a misconception (except of some special cases) there is for example: "Intermediate Electromagnetic Theory" by J.V.Stewart Page 58:

We note that it is a misconception that particles move along the lines of force. The lines of force do not describe the path a particle takes but the lines of force it experiences.

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If anybody knows how to embed this simulation, feel free to do it! –  student Sep 25 '12 at 18:36