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

Question

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?

Answer 1

Unfortunately I do not know German. However based upon what you have said, it gives the direction that it will start moving at each point if it is NOT moving at that point. Lets look at the example of a dipole. An image of one can be found here, https://readingpenrose.files.wordpress.com/2015/08/dipole-field-2.gif. Feel free to take a look at this page which I will reference for my explanation. Let us concentrate on the electric field lines in the left lobe, and let us assume that we have a positive particle moving to left and at the top of the left lobe. This particle would be moving in the direction of the electric field line, and would be consistent with the conjecture. Lets further assume the conjecture is correct, and the particle is now at the left extreme of the lobe, and traveling directly downward. This particle would have accelerated from v₀=-v_xi to v₁=-v_yj. Notice that our force is always in the direction of electric field which is always to the left and/or downward. This implies the acceleration is always to the left or downward; however, based upon our velocities, we would need an acceleration to the right and downward. So the electric field lines give us an acceleration to the left (and down); however to keep our trajectory on the electric field lines we would need an acceleration to the right ( and down). This is a contradictions; therefore the conjecture is false. I hope this helps to clarify things.

Answer 2

If you are given an initial acceleration vector (given by qE/m), then if you integrate it over a short enough period of time, which is what I assume the question is asking, then the velocity and the path must be in the same (but not precisely the same) direction. That is assuming they have the same charge.