# Maxwell's Equations: Mutual consistency

1. May 31, 2010

### Arijit Dutta

The no. of unknowns in the Maxwell's Equations is 6- the components of E and B. But there are 8 partial differential equations if we separate them out component- wise. So when we find the solution to them how do we ensure their mutual consistency?

2. May 31, 2010

### nicksauce

Last edited: May 31, 2010
3. May 31, 2010

### Arijit Dutta

4. May 31, 2010

### Meir Achuz

By working with the potentials and the wave equation, there are 4 source terms and 4 unknowns. These determine E and B.
There are constraints on possible E and B.

5. Jun 1, 2010

### Born2bwire

In addition, there are really three independent Maxwell's equations. I believe that you can derive Gauss' Law for Magnetism from the other three laws. I would imagine that if you are able to generate 8 partial differential equations that some of them may be degenerate. In addition to what Meir Achuz stated, there is of course the implicit requirement that boundary conditions are required for a unique solution.

6. Jun 1, 2010

### Phrak

Of Maxwell's equations, they are all independent until the magnetic vector potential and electric potential are introduced as fundamental fields. In this case, Gauss law for magnetism is a mathematical identity. Really though, all that is needed is to posit a 4-vector field (A, phi) and identify first derivatives with components of the electric and magnetic fields. Charge and charge continuity are second derivatives. We could call these identifications equalities if we wish, but other than that, the equal sign doesn't appear.

To answer Arijit Dutta's question, to some amount, any four continuous, twice differentiable, independent scalar valued fields Ax, Ay, Az and phi is a solution.

Last edited: Jun 1, 2010
7. Jun 2, 2010

### |squeezed>

You are right about the number of unknowns in Maxwell's eqs but you must also take into account the constitutive relations.

8. Jun 2, 2010

### Anamitra

We must not forget the charge density $$\rho$$ and the current density j,which has three components. We have ten variables indeed!. In general all these variables are functions of space and time.

Last edited: Jun 2, 2010
9. Jun 2, 2010

### Phrak

To anyone that follows differential p-forms:

We can cover space and time with a 4-vector field A. Amu=Amu(t,x,y,z). mu = {0...3}. Greek indices will range over spacetime. Latin ranges over space.

A is the so called four vector potential. It is unphysical up to a constant field, phi. A-->A'=A+d(phi). Here, A has upper indices.

Unscripted variables in the following are taken to be antisymmetric tensors with upper indices.

F = dA defines the Faraday tensor. It's elements are the electric and magnetic field components. All together, there are 6 independent components as should be expected.

$$F = F_{[\mu\nu]} = (1/2)(F_{\mu\nu}-F_{\nu\mu}) = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$$

By mathematical identity, ddA=0. (all exact forms are closed)

ddA=dF=0 says there are no magnetic charges. This equation also contains

$$-(\nabla \times E)_{k}+\partial_{t}B_{k}=0_{k}$$

*d*dF=-J is Faraday's law of magnetic induction and defines electric charge as the divergence of the electric field. (*) is the Hodge duality operator. It takes a tensor with lower indices, mulitiplies it by the completely antisymmetic unit tensor or rank k, with indices raised and lower so as to return a tensor of rank n-k, where n=4 for spacetime.

$$J = J_{\mu},\ and \ J^{\mu} = (J^{i}, \phi)$$

J combines charge density and current density.

10. Jun 2, 2010

### Dickfore

The equations are not independent. For example, from:

$$\mathbf{\textup{curl}} \, \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$$

by taking the divergence of both sides and using the vector identity:

$$\textup{div} \, \mathbf{\textup{curl}} \, \mathbf{a} = 0$$

we get:

$$\textup{div} \, \frac{\partial \mathbf{B}}{\partial t} = 0$$

$$\frac{\partial}{\partial t} \left(\textup{div} \, \mathbf{B} \right) = 0$$

$$\textup{div} \, \mathbf{B} = f(\mathbf{r})$$

where $f(\mathbf{r})$ is an arbitrary function of coordinates only. But, according to the Principle of Relativity, this has to hold in any inertial reference frame. Therefore, the only possibility is that this funcion is actually to zero. The second Maxwell equation simply serves as a verification of this, but is not an independent constraint.

Similarly, taking the divergence of the fourth Maxwell equation and using the first one, we get:

$$\frac{\partial \rho}{\partial t} + \textup{div} \, \mathbf{J} = 0$$

This is an additional condition that the external sources (charge and current densities) have to satisfy. It is the continuity equation and is the Law of conservation of charge in a differential form.

11. Jun 3, 2010

### Arijit Dutta

@ Meir Achuz
But what about the guage condition? Isn't it another constraint. So now there r 4 unknowns and 5 equations. Again a question of mutual consistency arises.

@ Phrak
I really don't get all that mathematics. Sorry, but I can't help. I guess I'm not upto ur level in mathematical proficiency.

@Dickfore
You just showed that the Maxwell's equations are mutually consistent as well as consistent with charge conservation and special relativity.

12. Jun 3, 2010

### zeebek

@Dickfore

There can be magnetic monopoles in principle. It does not contradict anything so far, just nobody found them. The equations are independent. I can refer to any famous course on EM. Math does not prove anything .All major physical equations were guessed, not derived.

@Arijit Dutta

ME were not derived, their historical devellopment was just a genious guess based on the experimental facts. There are 2 "dynamic" equations: ampere-maxwell's one and faraday's one, apart from them the conservation of charge should be included ( ampere-maxwell's + gauss's law) and the fact that there are no magnetic monoploes: div B = 0 (anlike electric charges). So they are consistent because they were made consisent. Suggest you to follow Walter Levin's cource on EM
http://www.learnerstv.com/lectures.php?course=ltv008&cat=Physics

Last edited: Jun 3, 2010
13. Jun 3, 2010

### Dickfore

This has nothing to do with the thread. If there is a source term in the second ME, then we must compensate with a convective magnetic current density in the third ME, leading to another continuity equation which is a necessary condition for ME to have a solution, just as the continuity equation for electric charges.

14. Jun 3, 2010

### Antiphon

For completeness let's also include a Lorentz force law on the monopole. F=qH-(vXepsE) where q is the monopole charge, eps is the permittivity of free space, E is the electric field, and v the velocity of the monopole.

15. Jun 4, 2010

### Phrak

This is true up to a constant for all points in spacetime. The constant, is of course, current density.

$$\mathbf{\textup{curl}} \, \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t} + \mathbf{J}$$

There are exactly four independent variables,

$$A^{\mu} = A(t,x,y,z)\ ,$$

at each spacetime event. These will define the electric field strength, the magnetic field strength and the charge and current density. This is simply because each is derivable from the magnetic potential Ai and electric potential phi. So, as you expected there is some interdependence between E, B, J and charge density.

This, very general result, does not include boundary conditions, so that admissible solutions can include every sort of radiation field.

16. Jun 4, 2010

### Phrak

For completeness we'd also have to include the Lorentz force on electric charges as well.

Allowing nonzero magnetic charge, and ignoring the Lorentz force for a moment, the total number of independent variables is eight.

One element of classical electromagnetism that is never explicitly stated, to my knowledge, is that charge is always associated with mass. This is another constraint. We are given Maxwell's equations plus the Lorentz force, but without exception, in my studies, we are expected to know that all charges have mass. Standing alone, Maxwell's equations are perfectly consistent without associating charge with nonzero mass.

But you elude to a good question, "How does the Lorentz force reduce the number of independent variables?"