# Why magnetic and electric field

1. Mar 14, 2015

### exponent137

Do anyone know, what is the most primitive mathematical reason that electromagnetic field includes magnetic and electric field, not only one field? One reason is hidden in F tensor:
http://en.wikipedia.org/wiki/Electromagnetic_tensor
but, what is the most primitive mathematical reason ...?

2. Mar 14, 2015

### vanhees71

The question is what you mean by "most primitive". From the modern point of view, the forms of possible physical laws are dictated by symmetries. Starting from space-time symmetry, which is underlying all physics, you can analyze large classes of possible physical laws. Starting from the special-relativistic space-time, Minkowski space, the appropriate symmetry group is the proper orthochronous Poincare group. One particularly elegant way to realize this symmetry in terms of dynamical laws is local field theory, and the symmetry structure of space-time leads to a huge class of possible dynamical laws, among which is that of a massless vector field, which leads (under the additional assumption that there's only a discrete intrinsic spin/polarization degree of freedom) inevitably to the notion of gauge invariance, and gauge invariance then more or less determines the math, which describes the electromagnetic field. There's no electric and magnetic field, but one electromagnetic field, which has, seen from an inertial frame of reference, electric and magnetic field components.

3. Mar 14, 2015

### exponent137

So, according to your correction, more precise question is:
"what is the most primitive mathematical reason that electromagnetic field has two components, not only one?"

(under the additional assumption that there's only a discrete intrinsic spin/polarization degree of freedom)
Do you think spin of photon, 1\hbar, or any quantum spin?

inevitably to the notion of gauge invariance, and gauge invariance then more or less determines the math, which describes the electromagnetic field.
Can this link give new information: http://en.wikipedia.org/wiki/Introduction_to_gauge_theory
Can you tell more?

4. Mar 14, 2015

### Staff: Mentor

5. Mar 14, 2015

### vanhees71

The reason that there are only two polarization states not three as you'd expect from a vector field is that the photon is massless.

It goes back to the representation theory of the Poincare group. Looking for irreducible unitary representations you come to the conclusion, that good states are the momentum eigenmodes. In order to be irreducible, the Lorentz transformations must bring you from any "standard momentum" to any other. This implies that $p \cdot p=m^2$ with $m^2$ a given constant for all states represented by the irrep. Physically relevant are the cases $m^2>0$ (leading to the many-body description of massive particles, when the field is quantized) and $m^2=0$ (massless particles).

For massless particles you can choose the standard momentum as $p=\Lambda(1,0,0,1)$ (with $\Lambda \neq 0$ arbitrary). The irrep. then is further characterized completely by the representation of the subgroup which leaves this standard momentum invariant (the socalled "little group"). The group leaving the light-like standard momentum invariant is not compact since this group is equivalent to ISO(2), i.e., the group built by translations and rotations of the Euclidean plane. Now, to have only discrete intrinsic ("polarization") degrees of freedom, you must represent the translations in this plane trivially and only the rotations non-trivially (if you represent them also trivially, you get massless spin-0 particles, which is mathematically fine but not observed in nature). This leads to the conclusion that the only remaining degree of freedom is the rotation around the $z$-axis, also leaving the light-like standard vector unchanged. This is a U(1), and it's representation is given by multiplication with a phase factor $\exp(-\mathrm{i} \lambda \varphi)$. In principle, $\lambda \in \mathbb{R}$, can take any value, but you want to lift the representation of the little group to one of the entire Poincare group, and there you have the usual rotations in space, and thus you must have $\lambda \in \{0,\pm 1/2,\pm 1,\ldots \}$. So for massless particles you have only two polarization-degrees of freedom, which are represented by the rotations around the direction of the momentum of the massless particle, whose generator is the projection of the particle's angular momentum to the direction of its spin, the socalled helicity. For a massless spin-1 particle you the two helicities, $\lambda=\pm 1$. In classical electromagnetism this describes left- and right-circular polarized electromagnetic waves.

6. Mar 14, 2015

Staff Emeritus
Until someone comes up with a metric of primitiveness, I predict this thread will go around and around in circles.

It also comes at this from the wrong direction. Electric and magnetic phenomena are not a predictions of our mathematical models of electromagnetism. These models were created to mathematically capture the observed phenomena.

7. Mar 14, 2015

### vanhees71

True, but the OP is about the "most primitive mathematical reason", and that's representation theory of the Poincare group. Of course, all natural science is empirical. Sometimes mathematical consistency allows to make nice predictions (like the introduction of Maxwell's "displacement current" into the electromagnetic system of equations, making them to what's know known as "Maxwell equations", which lead to the prediction of electromagnetic waves and the identification of light with such), but the general scheme of the physical laws are indeed all due to observations.

8. Mar 14, 2015

### exponent137

Maybe the answer with Poincare Group is this, what I searched. I need to read more. But as "primitive" I thought, for instance, that McIrvin explained why the same charges reppel and opposite charges attract. http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html
This is more primitive explanation. Weizsacker explained that Pauli matrices are three independent and this is (maybe) cause for three space dimensions.

The next question can be: if only electric field had existed without magnetic one, what would have been special relativity vector form instead of F tensor? If this form is clumsy, this is also reason for two components, E and B.

The basic motivation for the question was: "photon is 3 dimensional (B, E, direction of speed), space is three dimensional. Maybe this is correlated and connected?

In the case I will not find direct answers on basic motivation, I will anyway better understand QFT.

Last edited: Mar 14, 2015
9. Mar 14, 2015

### Staff: Mentor

That's not possible.

The existence of magnetic fields is a relativistic effect from Lorentz contraction.

Here is a link that gives the full mathematical detail of how Maxwell's equations follow from electric fields:

EM is as it is because of the existence of electric fields and space-time geometry.

Although understanding all these different views of EM is worthwhile I am with Vanhees - that is the most fundamental reason, also unfortunately the most advanced.

Thanks
B ill

10. Mar 14, 2015

### Staff: Mentor

There is only one thing: the electromagnetic field.
The definitions of electric and magnetic fields have historic reasons and they can be convenient in many setups, but there is nothing fundamental about this convention.

11. Mar 15, 2015

### exponent137

Yes, very simple answers. The question was a lapse. But, anyway, I still hope to better understand the most fundamental reason.

Thus a photon has three dimensional structure, B, E and direction of waves.
Gravitometric force also exists at gravitational wavings: http://en.wikipedia.org/wiki/Gravitoelectromagnetism
How it is here with B_g direction?

12. Mar 15, 2015

### exponent137

Because of two components, the photon has 3 dimensional structure, B, E and direction of waves. Thus two components are not only virtual? Or am I wrong?

13. Mar 15, 2015

### Staff: Mentor

What a photon is, is only elucidated in Quantum Field Theory which requires, unfortunately, many years of study. Anything you read outside a QFT textbook is likely wrong - certainly the above is wrong.

If you would like to undertake that journey the following is accessible after a first course in QM:

By first course I mean something like Susskind that explains the full theory - not a lay version:
https://www.amazon.com/Quantum-Mechanics-The-Theoretical-Minimum/dp/0465036678

It will however take time and require close attention.

And once that's done you can start to understand why U(1) symmetry is the key idea.

Thanks
Bill

14. Mar 15, 2015

### Staff: Mentor

Or 3, or 4, or whatever number you like if you make arbitrary "components" of the electromagnetic field.
I don't think that question makes sense.

15. Mar 15, 2015

### exponent137

I try still to find clearer connection between U(1), gauge theory and electromagnetism.
U(1) is pretty simple, it is based on basic circle. For instance:
http://uw.physics.wisc.edu/~himpsel/449group.pdf
A, (and Q) are probably included in wave functions?
And transformation is gauge transformation?
What gauge transformations
$A_\mu=A_\mu'+q^{-1} \partial_\mu \varphi$ etc. have to do with the unit circle? Or how to include it in system of explanation with U(1)?

16. Mar 15, 2015

### Staff: Mentor

Have a look at the first link I gave:
http://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html

Its got to do with invariance to complex phase ie invariant to exp^iy - which is basically a circle - recall the phaser representation of complex numbers - as y varies it traces out a circle.

To go any further than this you need to start into much more advanced math - areas I am not even conversant with. Can we know your level of math training eg have you done multi-variable calculus to the level to understand the first link? That's so I can recommend some texts.

Thanks
Bill

Last edited: Mar 15, 2015
17. Mar 15, 2015

### exponent137

Yes I understand your link, as I wrote in PM. But, more I need, why electromagnetism is U(1), except, that wave function describe unit circle.

18. Mar 15, 2015

### Staff: Mentor

As explained in the link its the requirement from local phase symmetry ie if you require the equation to be locally invariant to exp^iy then you need to have a term e/cA to cancel the terms that appear because of that requirement. exp^iy when graphed in the complex plane is a circle. Thus we require the equation to be invariant to the symmetry of a circle which by definition is U(1) symmetry.

I cant explain it better than that. If you still don't get it I will need to leave it to others to explain.

Thanks
Bill

19. Mar 21, 2015

### exponent137

The answers above are now clear.
I want to understand on the next level.

Why names bundles and fibres, where it is similarity with prolonged structures? What is analogy with curvature and with »connections«? Which components of curvature are B and E? Can anyone give any link which visulizes these things? Links like those http://www.math.toronto.edu/selick/mat1345/notes.pdf are not visulized enough.

20. Mar 22, 2015

### Staff: Mentor

I think that this is sufficiently different from the OP to warrant a separate thread. Since the questions in the OP are now answered, lets close this thread.

exponent137, please try to formulate your question about bundles and fibres as clearly and as simply as possible, and avoid a "shotgun" series of questions. You also may be better off simply re-reading the material from this thread for a while until it has had a chance to really sink in.