# Relativistic interpretation of magnetic effect on a charge due to a perpendicular current carrying wire

Assume that we have a current-carrying conductor and a $$+v_e$$ test charge moving along the current. In the test charge's rest-frame, the electrons in the wire are length-expanded, and the positive ions of the metal are length-contracted. This makes the length charge density net positive, resulting in a repulsive force on the test charge, perpendicular to the wire. This explanation also concurs with the magnetic force $$\mathbf{F}_L=q({\bf v} \times {\bf B})$$ on the charge in lab-frame.

Now, let's assume that the test charge is moving perpendicular to the wire. We know from elementary magnetism that the magnetic force on test charge is now along the wire.

But how do I get the same result via relativistic length-contraction/expansion in the test charge's rest frame?

From what I understand, the moving test charge should observe no length contraction of any elements of wire, since it is moving perpendicular to the direction of relative motion.

• relativity isn't needed here. Classically it makes no difference whether the charge is moving toward the wire or the wire is moving toward the charge Dec 25, 2020 at 15:39
• @R.Emery I don't understand what specific part of the question are you referring to? Dec 25, 2020 at 15:43
• In the first part of your question you classically expect two different results in two different frames. Relativity is required to explain the difference. Since there is no classical difference in the second part of your question there's no reason to invoke relativity to explain something that doesn't require an explanation. You were trying to solve a problem that doesn't exist Dec 25, 2020 at 21:17
• The answer you've accepted is incorrect (see my comment under it). Please consider accepting stuffu's answer instead (which could be more clear and complete but at least points in the right direction). For another way of explaining it, see my answer to a similar question. Jan 1 at 9:56

Let's say that at first the current is on, but the wire is not moving relative to us. The electric field of each electron is length contracted. For some reason this is easier if the electrons move quite fast, so let's say they do move quite fast.

Then the wire starts moving towards us, quite slowly. So basically what happens is that the velocity vector of each electron turns a few degrees. So also the electric field around each electron turns a few degrees, the fields are not perfectly round so this turning actually means something.

It means that our test charge experiences a different force when the fields are turned.

(I mean the shape of the field turns, not necessarily the field itself)

See chapter "Charge moving Perpendicular to the Wire" here: https://physics.weber.edu/schroeder/mrr/MRRtalk.html

• who is 'us'? Lab-frame or test-charge's frame? Dec 27, 2020 at 8:12
• @Kraken we stand on the floor of the lab, test charge is in our hand, which we keep still. Check the link that I added. Dec 27, 2020 at 15:35
• Thank you. The link was very helpful. Dec 28, 2020 at 14:23

The 4-vector describing a charged particle is current, $$J$$. $$J$$ contains information about both the charge and the motion of the particle. For a stationary charge the 3-current is zero, so the 4-current is $$(q,\mathbf 0)$$. The general form of $$J$$ is found by Lorentz transformation and has the form $$J=(\gamma q, \gamma q\mathbf v)$$. Since covariance requires that physical laws are represented by tensors, to describe the electromagnetic force acting on a charged particle we need to contract $$J$$ with a tensor, $$F$$, representing the electromagnetic field. $$F$$ is the Faraday tensor. The result of contracting $$J$$ with $$F$$ is a vector, force. So, Faraday is a rank-2 tensor. Faraday should express both the force acting on a charged particle, and the equal and opposite reactive force exerted by the particle on its environment. If Faraday is an antisymmetric tensor, contracting with one index will give the force on the particle and contracting with the other will give the reactive force. We write down the 4-vector law of force,

$$(\mathrm {Force})^i = F^{ij}J_j.$$ Although it does not at first look like it, this is precisely the Lorentz Force law, which is usually expressed in terms of two 3-vector fields, the electric field $$\mathbf E$$ and the magnetic field, $$\mathbf B$$, $$\mathbf {Force} = e(\mathbf E + \mathbf v \times \mathbf B),$$ where we have $$F^{ij} = \begin{bmatrix} 0 & \mathbf E_x & \mathbf E_y & \mathbf E_z \\ -\mathbf E_x & 0 & \mathbf B_z & -\mathbf B_y \\ -\mathbf E_y & -\mathbf B_z & 0 & \mathbf B_x\\ -\mathbf E_z & \mathbf B_y & -\mathbf B_x &0 \end{bmatrix}.$$ To see this, we need simply consider a static electric field. In this case the 3-vector force is $$q\mathbf E = (q\mathbf E_x, q\mathbf E_y, q\mathbf E_z)$$ We can then write the Faraday tensor, using antisymmetry to determine the other components, $$F^{ij} = \begin{bmatrix} 0 & \mathbf E_x & \mathbf E_y & \mathbf E_z \\ -\mathbf E_x & 0 & 0 & 0 \\ -\mathbf E_y & 0 & 0 & 0\\ -\mathbf E_z & 0 & 0 &0 \end{bmatrix}.$$ For a field due to a moving charge, Faraday is found from Lorentz transformation, acting on both indices, $$F^{m'n'} = k^{m'}_i k^{n'}_j F^{ij}.$$ This introduces the magnetic fields for a moving charge in terms of the electric field of a static charge. Anti-symmetry of $$F^{m'n'}$$ follows directly from antisymmetry of $$F^{ij}$$. Generally, Faraday is not represented simply through boosting a static field, but it is the result of summing the fields generated by many particles, each one of which could be regarded as static in the rest frame of that particle. This would lead to a complicated expression, retaining antisymmetry, and summarised by the resultant electric and magnetic fields, $$\mathbf E$$ and $$\mathbf B$$.

• thank you for your detailed response. I am afraid I do not have sufficient training in mathematics to understand your response. My cursory understanding is that this is some form of mathematical proof that extracts Lorentz force from a relativistic construct (tensor?). I would appreciate if you could dumb it down to how no length contraction results in force on the perpendicular moving test case. Or point out flaws in the entire premise I drew up. Dec 25, 2020 at 20:24
• The relativistic length "contraction" isn't actually a contraction. It is an apparent effect, better compared to a ruler appearing shorter because it is at an angle to view. Proper lengths don't change; objects do not get shorter in a reference frame in which they are not moving. This idea, which is simple enough in itself, is described by the mathematics of Lorentz transformation. I don't really know how to explain it without the maths, except in the simplest cases, and tbh I am suspicious of the argument you cited, which as you say does not work for a single charge with no length. Dec 25, 2020 at 20:40

But how do I get the same result via relativistic length-contraction/expansion in test charge's rest frame?

From what I understand, the moving test charge should observe no length contraction of any elements of wire, since it is moving perpendicular to them.

You have answered your own question. You don’t get length contraction with this setup. Length contraction is directionally dependent and only is important for the parallel configuration.

The perpendicular configuration has no length contraction. The current density and the charge density of the wire are unchanged in the test charge’s frame. Per Faraday’s law, the electric field that pushes the test charge is induced from the time varying B field as the wire moves.

The parallel configuration is of interest because for that case a relativistic explanation is simple and interesting. It should not be taken as a general approach.

• This is incorrect. Length contraction actually does happen here (to the electrons in the wire in a diagonal direction) in the test charge's reference frame and it does allow the force on the test charge (in a direction parallel to the wire) to be explained without invoking magnetic fields. See stuffu's answer to this question and my answer to a similar question. Jan 1 at 9:51
• @GumbyTheGreen that is incorrect. The length contraction that happens for the electrons in the wire happens equally for the protons. There is no net charge produced that way. Assume a boost in the x direction and a current in the y direction then the four-current is $(\rho,j_x,j_y,j_z)=(0,0,j,0)$ which gives $(\rho’,j’_x,j’_y,j’_z)=(0,0,j,0)$
– Dale
Jan 1 at 13:17
• If the electrons are moving upward and the test charge is moving to the left toward the wire, then from the test charge's reference frame, the electrons are moving up and to the right (while the protons are only moving to the right). So their electric fields get contracted diagonally (while those of the protons only get contracted horizontally) and the vertical component of that contraction creates a net charge. See pages 265-267 of Electricity and Magnetism, 3rd Edition and/or the link in stuffu's answer. Jan 1 at 22:22
• Yes, that is indeed how the E field looks in that frame, but the question is what is the source of that field? There is no charge density, so the source of the field is not Gauss’ law. The source of the E field comes from the varying B field. It is simply false that the contraction creates a net charge in that frame. Read your sources carefully, they do not claim the wire is charged in this configuration. On p. 266 Purcell even explicitly states that the wire is uncharged in both frames
– Dale
Jan 1 at 23:03
• I have read it carefully and it does not say or imply that the E field comes from a varying B field. It makes it clear that the E field comes from the charges of the particles. In fact, on p. 277, it says "The sources of magnetic fields are currents, in contrast with the sources of electric fields, which are charges". It doesn't even really get into B fields or induction until the following chapters (except for occasional brief mentions). The effect it describes is the underlying cause of the mathematical construct that we call the magnetic field. Jan 2 at 21:11