# Is the Configuration of Magnetic Fields Only a Convention?

• I
If I understand correctly, the concept of electric and magnetic fields originated with Faraday and was developed by reconceptualizing forces acting at-a-distance.

For example, the electric field concept was developed by looking at the force on a test charge in the presence of a source charge rather than looking at the force between two charges (i.e. Coulomb’s law). The magnetic field was similarly developed by looking at the force on iron filings or compass needles in the presence of a magnet or electric current.

But could Faraday have equally developed the concept of magnetic fields by looking at the force on one (test) electric current in the presence of another (source) electric current, per Ampere’s discovery of the attraction/repulsion between currents? If so, wouldn’t that make the configuration of magnetic fields only a convention, in somewhat the same way that positive/negative signs are just a historical convention due to Franklin?

anuttarasammyak
Gold Member
You may change convention of magnetic field to inverse one if you change also B to -B in Maxwell's equations. If you want to keep Maxwell equations as they are, you should keep current convention.

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Dale
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If so, wouldn’t that make the configuration of magnetic fields only a convention, in somewhat the same way that positive/negative signs are just a historical convention due to Franklin?
The magnetic field direction is only a convention anyway. You can flip the direction of all magnetic fields either by changing the right hand rule to a left hand rule, or by flipping the order of the cross products.

• vanhees71 and Delta2
The magnetic field direction is only a convention anyway. You can flip the direction of all magnetic fields either by changing the right hand rule to a left hand rule, or by flipping the order of the cross products.
True, but that's only with respect to whether the direction of the field line is positive or negative. But if the field lines were based on the force between currents, the field lines would diverge from currents rather than curl around them.

Dale
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But if the field lines were based on the force between currents, the field lines would diverge from currents rather than curl around them.
Oh, I see what you are saying.

I have not considered this in detail, but I don’t think it would work. If the field lines radiate out from the current then you will have the same field regardless of which direction the current flows.

Staff Emeritus
I am having a hard time visualizing the direction you want this to point, other than "not in the direction of the B field". Well, there are really only two choices.

One is the magnetic vector potential. One can always use this, since ∇×A= B. However, I suspect that had things evolved this way, it would not be long before we introduced an auxiliary field that was the curl of the A field to ease calculations.

If that's not it, it would be some field like B×A so that it was perpendicular to both. I am pretty sure that this would work, but it would be computationally horrible.

• cg0303
If the field lines radiate out from the current then you will have the same field regardless of which direction the current flows.
Why? We could choose one direction to be 'outwards' and one 'inward', the same way we chose the positive charge to be outward and negative to be inward. The only difference is that that with electric charges opposites attract, while with currents opposites repel.

I am having a hard time visualizing the direction you want this to point, other than "not in the direction of the B field". Well, there are really only two choices.

One is the magnetic vector potential. One can always use this, since ∇×A= B. However, I suspect that had things evolved this way, it would not be long before we introduced an auxiliary field that was the curl of the A field to ease calculations.

If that's not it, it would be some field like B×A so that it was perpendicular to both. I am pretty sure that this would work, but it would be computationally horrible.
The field line would point away from an electric current.

I'm not sure what the magnetic vector potential of a current would look like. Could you post a diagram that would help?

Dale
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Why? We could choose one direction to be 'outwards' and one 'inward', the same way we chose the positive charge to be outward and negative to be inward.
By symmetry. The laws of physics are invariant under rotation, including specifically a 180 degree rotation. Start with a current ##\vec J## that produces an outward field. By symmetry a 180 degree rotated current ##R_{180}(\vec J)## must also produce an outward field. But ##R_{180}(\vec J)=-\vec J##. So ##\vec J## and ##-\vec J## produce the same field.

• cg0303
Staff Emeritus
Could you post a diagram that would help?
I could, but I posted a link with diagrams.

• Dale
Staff Emeritus
By symmetry. The laws of physics are invariant under rotation, including specifically a 180 degree rotation.
While what he proposing is still pretty vague, I think he means that clockwise current gives an outward field and counterclockwise gives an inward field. (Or the reverse)

Dale
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While what he proposing is still pretty vague, I think he means that clockwise current gives an outward field and counterclockwise gives an inward field. (Or the reverse)
Which is the same issue since a clockwise current rotated 180 degrees becomes counterclockwise.

I could, but I posted a link with diagrams.
I was after something different than what was contained in the link, but I think I was able to get what I needed from it.

If that's not it, it would be some field like B×A so that it was perpendicular to both. I am pretty sure that this would work, but it would be computationally horrible.
If I understand you correctly, it would be this.

Staff Emeritus
Then it would be computationally horrible.

Then it would be computationally horrible.
Fair enough

Staff Emeritus
Which is the same issue since a clockwise current rotated 180 degrees becomes counterclockwise.
Take a look at the Wikipedia page, which has a toroid. B is drawn, A is drawn, and their cross-product is outward from the current. So I think the direction is - or at least can be - well-defined. The OP's description still kind of confuses me (sometimes it sounds like A, sometimes perpendicular to A) so I am not sure that it is well-defined. But I think something like he wants could be well-defined.

Dale
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B is drawn, A is drawn, and their cross-product is outward from the current. So I think the direction is - or at least can be - well-defined.
Agreed. The direction can be well defined and with your formula it is always outward.

The problem is here:
could Faraday have equally developed the concept of magnetic fields by looking at the force on one (test) electric current in the presence of another (source) electric current
So we have a test current and a source current. The field always points outward from the source current regardless of which direction the current is flowing. So the field at the test current is the same regardless of which direction the current flows. But the force on the test current is not the same. So this field is not and cannot be directly related to the force on the test current.

In order to distinguish the direction of the force would indeed be computationally horrible and would probably wind up solving for B at some point anyway.

• vanhees71
Gold Member
I am having a hard time visualizing the direction you want this to point, other than "not in the direction of the B field". Well, there are really only two choices.

One is the magnetic vector potential. One can always use this, since ∇×A= B. However, I suspect that had things evolved this way, it would not be long before we introduced an auxiliary field that was the curl of the A field to ease calculations.

If that's not it, it would be some field like B×A so that it was perpendicular to both. I am pretty sure that this would work, but it would be computationally horrible.
Well, in fact I think historically there was a lot of confusion about the potentials in the beginning. It's a tricky concept as we know today, because it's not the potentials which uniquely describe an electromagnetic field but only an equivalence class, i.e., "the potentials modulo a gauge transformation". That is, because the Maxwell equations for the potentials do not uniquely determine these potentials given some physical initial (and maybe boundary) conditions, while of course the Maxwell equations for the field(s) ##(\vec{E},\vec{B})## have unique solutions under given initial/boundary conditions, and indeed ##\vec{E}## and ##\vec{B}## are what's operationally defined via the forces on test charges.

• Delta2
Staff Emeritus
computationally horrible and would probably wind up solving for B at some point anyway.
I agree. I believe that the thing that is most likely being discussed works out to be ∇(A2/2)−(A⋅∇)A. One would need to solve for A and take it's curl to get anything that looks like a force.

@vanhees71 is right that A is determined by B only up to a gauge transformation. I think the monster above is also determined by A up to a gauge transformation, and some care might be needed to make sure that one chooses a consistent set. I haven't worked this out, but can see how one might screw this up.

• vanhees71 and Dale
Dale
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can see how one might screw this up.
Excellent understatement!

• vanhees71
Staff Emeritus
My beautiful Latex was destroyed. (Greg, is there a Purple Heart badge for this?) I tried to write:

$$\nabla (A^2/2) - ({\bf A} \cdot \nabla){\bf A}$$

What's a vector and what is not is important.

All you have to do is invert this. • cg0303
vanhees71
Gold Member
What do you want invert here? It's not an equation. So far the great feature of classical electromagnetism was that you get very far with linear laws. What should this non-linear expression help (whatever the missing right-hand side of the equation reads).

Staff Emeritus
I have no idea why anybody would want to add this much complexity.

berkeman
Mentor
Greg, is there a Purple Heart badge for this?)
$$\heartsuit (A^2/2) - ({\bf A} \cdot \heartsuit){\bf A}$$
There, I fixed it for you. And you're welcome. • 