# Is electric force one dimensional?

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How (or why) does an electron negate all the force of a proton?
Explanation (reason) of question:
1> All forces act uniformly in all directions.
Explanation of 1(above):
a) If you lit a light source (like torch or light bulb) you can see it from every direction. Which means light travels in all directions (3D) uniformly.
b) Similarly (to 'a') a mass (of an object) will interact with all other masses (objects) in all directions.
c) Similarly (to a and b) an electron moves around a proton (nucleus) in 3D space i.e. in all directions.
Conforming to '1' that all forces interact with objects uniformly in all directions.

Q>>> So why is it that when a proton finds an electron then it locks in the force in ONE direction (one dimensional) and ignores all other electrons around it?
q> An other question relating to this, if it (Q) is true then does it mean that if two electron are brought together they will repel each other but if there is an other present there, they will not notice/(interact with) it (I mean simultaneously) like two electrons are "busy" repelling each other and another one comes it can pass through them, or will the force (repulsion) between the three of them be halved? Or will they be switching interactions on and off very briefly so it is hard/impossible to notice/measure?
qE>> Explanation of q:
If an electron and a proton are interacting with each other, and there is another proton, will there still be proton proton repulsion if another proton is shoved in the close proximity(space)?

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How (or why) does an electron negate all the force of a proton?

It doesn't

So why is it that when a proton finds an electron then it locks in the force in ONE direction (one dimensional) and ignores all other electrons around it?

It doesn't.

The electrostatic force is F = kQq/r^2, which only tells you about the force between these two particles. It obeys superposition, and the force vectors add. You have to evaluate each force separately.

Newtonian gravity is similar. You can have more than one moon in orbit around a planet. The planet is not limited to interacting with just one at a time. The complication with electrostatics is the possibility of both an attractive and a repulsive force.

If an electron and a proton are interacting with each other, and there is another proton, will there still be proton proton repulsion if another proton is shoved in the close proximity(space)?

Yes, in keeping with my earlier comment about superposition and evaluating vectors.

# Is electric force one dimensional?

) If you lit a light source (like torch or light bulb) you can see it from every direction. Which means light travels in all directions (3D) uniformly.
b) Similarly (to 'a') a mass (of an object) will interact with all other masses (objects) in all directions.
c) Similarly (to a and b) an electron moves around a proton (nucleus) in 3D space i.e. in all directions.
Conforming to '1' that all forces interact with objects uniformly in all directions.

You don't seem very interested in answers to your question(s) , despite the effort you put into your opening post.

Let us start again.

I think your difficultly is mixing up an electric field and an electric force.

This is very common at the beginning of study.

Are you studying this ?

Swanson mentioned vectors, do you understand what he meant ?

Would you like an explanation of vectors and vector fields (of which electric forces and fields are an example ?

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Your premises are wrong, precisely for the reasons that Swansont has pointed out. A proton exerts a force on all the other electrons in the universe, and all the other protons, as well as on any other charged particles. On any charged particle, the electrostatic force is the sum of all the forces exerted on it by all the other charged particles in the universe, according to the superposition principle.

You could say that the electric force is 6 dimensional in some sense, if you include the magnetic effects. But 1-dimensional? That certainly is not the case.

You also seem to be confusing the number of particles on which a charged particle can act with the number of dimensions. Neither of them is one.

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9 hours ago, joigus said:

You could say that the electric force is 6 dimensional in some sense, if you include the magnetic effects.

I think it is better to look at it as a 4-dimensional geometric object in spacetime (the electromagnetic field), which is mathematically described by a rank-2 object (the EM field tensor and its dual). The electric and magnetic parts of the field aren’t independent, they are actually the same object seen from different vantage points, so the 6D description is a little problematic, as the basis vectors of that state space wouldn’t all be linearly independent. But that’s just me going off on tangents again 😄

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3 hours ago, Markus Hanke said:

I think it is better to look at it as a 4-dimensional geometric object in spacetime (the electromagnetic field), which is mathematically described by a rank-2 object (the EM field tensor and its dual). The electric and magnetic parts of the field aren’t independent, they are actually the same object seen from different vantage points, so the 6D description is a little problematic, as the basis vectors of that state space wouldn’t all be linearly independent. But that’s just me going off on tangents again 😄

(My emphasis.)

Your tangents are always very interesting.

Of course, you're right. I meant "in a sense" in the sense that it is an object that only has 6 non-zero components in general, which can be relevant when there is charged matter around, although in that case the new degrees of freedom can be attributed to matter densities --see below. Let me check the counting. Pure EM field must have identically null Lagrangian,*

$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=\frac{1}{2}\left(E^{2}-B^{2}\right)$

as in the vacuum $$\left|\boldsymbol{E}\right|=\left|\boldsymbol{B}\right|$$, so the Lagrangian is identically null on solutions, as corresponds to massless particles. That leaves 5. If we now throw in the transversality,

$\boldsymbol{E}\cdot\boldsymbol{B}=0$

that leaves 4 DOF, which are the ones you're talking about. Correct me if I'm wrong.

When light goes through matter, we no longer have the $$\boldsymbol{E}$$ and the $$\boldsymbol{B}$$, but also the $$\boldsymbol{D}$$ and the $$\boldsymbol{H}$$, so that,

$\mathcal{L}=\frac{1}{2}\left(\boldsymbol{E}\cdot\boldsymbol{D}-\boldsymbol{B}\cdot\boldsymbol{H}\right)\neq0$

with,

$\boldsymbol{D}=\varepsilon\left(x\right)\boldsymbol{E}$

$\boldsymbol{B}=\mu\left(x\right)\boldsymbol{H}$

so we have two "additional" DOFs, the matter medium, characterised by $$\varepsilon\left(x\right)$$ and $$\mu\left(x\right)$$ that take us back to 6. But it's a fictional 6.

*Edit: $$c=1$$ throughout.

Edited by joigus
setting units

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19 hours ago, joigus said:

Of course, you're right. I meant "in a sense" in the sense that it is an object that only has 6 non-zero components in general, which can be relevant when there is charged matter around

Ok, point taken
Actually, I feel a bit outside of my comfort zone here, since I was never particularly interested in EM theory. However, I am tempted to argue that in actual fact EM fields can have no more than 2 physical DOFs. Given conservation of sources, the EM field arises uniquely from a vector potential and a scalar potential - that’s 3+1=4 “free” quantities. But, both of these potentials are invariant under suitable gauge transformations, so in each instance we can arbitrarily pick a gauge, which eliminates one DOF each - leaving only two physical DOFs in the end. This seems to also be consistent with the fact that EM waves have exactly two possible polarisation states at a 90 degree angle.

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11 hours ago, Markus Hanke said:

Ok, point taken
Actually, I feel a bit outside of my comfort zone here, since I was never particularly interested in EM theory. However, I am tempted to argue that in actual fact EM fields can have no more than 2 physical DOFs. Given conservation of sources, the EM field arises uniquely from a vector potential and a scalar potential - that’s 3+1=4 “free” quantities. But, both of these potentials are invariant under suitable gauge transformations, so in each instance we can arbitrarily pick a gauge, which eliminates one DOF each - leaving only two physical DOFs in the end. This seems to also be consistent with the fact that EM waves have exactly two possible polarisation states at a 90 degree angle.

Let me get back to you when I get more time to think about this and review some literature. It's correct that radiation is generally described as having "2 DOF" in the sense of having two polarisation states. But the term "degrees of freedom" is a bit ambiguous. Some people use it in the sense of counting spin states. But there are also space-time variables, according to,

$\left|\boldsymbol{k},s\right\rangle$

The $$k_x$$, $$k_x$$, $$k_x$$, and $$s$$ would be four variables, more in the sense that I was talking about (DOF as number of variables necessary to describe the states).

As I said, let me get back to you. Also, I think you can do the counting on the E's and B's, which are gauge invariant. You can do it on the A's, but AFAIR there's just one gauge fixing condition, because you're in $$U\left(1\right)$$...

Still thinking. Apparently I'm outside my comfort zone too. LOL

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I appreciate the math guys, but the OP's assertion is easily disproven by molecules, where each proton doesn't 'bind' to only one electron, the residual Coulomb attraction is responsible for intramolecular bonds, and some even makes molecules 'polar', resulting in measurable intermolecular effects.

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17 hours ago, MigL said:

I appreciate the math guys, but the OP's assertion is easily disproven by molecules, where each proton doesn't 'bind' to only one electron, the residual Coulomb attraction is responsible for intramolecular bonds, and some even makes molecules 'polar', resulting in measurable intermolecular effects.

Of course...that’s why I was saying I was going off on tangents
But you have to admit that oftentimes tangents are fun.

17 hours ago, joigus said:

You can do it on the A's, but AFAIR there's just one gauge fixing condition, because you're in U(1) ...

There’s only one gauge condition in spacetime, because there’s only one potential field and one electromagnetic field.
But once we decouple this into E and B fields, the gauge condition also decouples into a constraint on the temporal part of A (which is the scalar potential), and a constraint on the spatial part of A (which is the Newtonian vector potential). The U(1) symmetry applies to the electromagnetic field in spacetime, but not to the E and B fields in Euclidean 3-space.

I think the conclusion is the same though, because of this: consider the Lorentz gauge (e.g.)

$\partial_{\mu}A^{\mu}=0$

This is a second order partial differential equation, so to obtain a unique solution, you need to supply precisely two boundary conditions. Since by definition in general field theory (omitting any constants)

$dA=F_{\mu \nu } dx^{\mu } \land dx^{\nu }$

the A’s should be fully determined, given Maxwell’s equations (which now essentially become relations between the components of A). So the boundary conditions above should be the only physical DOFs in the theory.

I have not researched this, I’m just pulling this out of my hat as I go along, so it’s possible - likely even - that I’m missing something. This is strictly my own two cents.

EDIT: Sorry, the above is of course nonsense, this isn’t a second order PDE at all. What I meant to say is that this conditions leads to the constraint equation for A, which is

$\square A^{\alpha } =\mu _{0} J^{\alpha }$

This requires exactly two boundary conditions for a unique solution.

Edited by Markus Hanke

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3 hours ago, Markus Hanke said:

There’s only one gauge condition in spacetime, because there’s only one potential field and one electromagnetic field.

I don't know much about gauge conditions, but having 'one'  EM field assumes homegenity does it not.

Otherwise epsilon and mu become tensors in their own right.

I hope this discussion has not frightened off the OP as the level is way above his presentation as I have already pointed out.

Edited by studiot

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6 minutes ago, studiot said:

I hope this discussion has not frightened off the OP as the level is way above his presentation as I have already pointed out.

The OP has yet to return to see any of these responses...

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4 hours ago, studiot said:

I hope this discussion has not frightened off the OP as the level is way above his presentation as I have already pointed out.

Not exactly frightened, just lost the login information (and could not find the topic and hence answers).

Thank you all for responding, though I did not expect Spanish inquisition arising from a silly question.

4 hours ago, swansont said:

The OP has yet to return to see any of these responses...

I know OP refers to me, here, but what does it stand for?

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Original Poster / Opening Post (depending on context)

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Sorry about the 'Spanish Inquisition' ( that would be Joigus, he's from Spain ), but we do love our 'tangents'.

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9 hours ago, Markus Hanke said:

I think the conclusion is the same though, because of this: consider the Lorentz gauge (e.g.)

$\partial_{\mu}A^{\mu}=0$

This is a second order partial differential equation, so to obtain a unique solution, you need to supply precisely two boundary conditions. Since by definition in general field theory (omitting any constants)

That would be the Lorenz gauge --different Loren*s --.

It's first order. It becomes second order when substituted in Maxwell's equations, which I think is probably what you meant, Markus.

-----

Maybe a split is in order? I agree with @studiot and @MigL. Although I find the discussion with @Markus Hanke very interesting.

40 minutes ago, MigL said:

Sorry about the 'Spanish Inquisition' ( that would be Joigus, he's from Spain ), but we do love our 'tangents'.

It's OK. If somebody had to be in charge of the Inquisition I'd rather it be 21st-Century Canadians. Much more humane, I'm sure.

Nobody expects... that joke. 😆

Edited by joigus

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13 hours ago, Markus Hanke said:

EDIT: Sorry, the above is of course nonsense, this isn’t a second order PDE at all. What I meant to say is that this conditions leads to the constraint equation for A, which is

Aα=μ0Jα

This requires exactly two boundary conditions for a unique solution.

Sorry, Markus. I hadn't seen this. But reading my comments you'll see I understood what you meant.

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Not exactly frightened, just lost the login information (and could not find the topic and hence answers).

Thank you all for responding, though I did not expect Spanish inquisition arising from a silly question.

Well I hope you got something out of it for you did not ask a silly question at all.

It was a perfectly reasonable question for someone starting to study these matters.

You didn't say whether you read my first post, at least you didn't answer it.

Yes it is one dimensional in that it acts along a specific line.

But there is a lot more to the subject which your other questions suggest you haven't yet understood.

Edited by studiot

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14 hours ago, joigus said:

That would be the Lorenz gauge --different Loren*s --.

Oops, sorry

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I know OP refers to me, here, but what does it stand for?

Stands for either "original poster" or "original post".

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