# How many independent components

1. May 25, 2006

### Ratzinger

How many independent components are there to describe the EM field?

The six in the field tensor, the four in 4-vector potential or the two when Coulomb gauge is being used?

thanks

2. May 25, 2006

### marlon

Well, strictly speaking you need the three components of the vector potential and the one component of the scalar potential. So 4 components in total. One can calculate both E and B from these components. The gauges (Coulomb, Lorentz-) are used because the scalar and vector potentials are not unique. The EM field tensor has six components (3 from E and 3 from B) but to get all E and B components, we need the 4 components mentioned above.

regards
marlon

3. May 25, 2006

### lalbatros

Ratzinger,

What do you mean by "independent components" ?

I could imagine this would mean "not linked by a (differential) equation", but this is not very clear.

Another point a view: the fields are completely determined by their sources. These are the 4 components of the current: spatial current and charge. There is one constraint on the source field because of charge conservation: this gives 3 "degrees of freedom" for the sources. Therefore I guess there are 3 "independent components" for the fields too.

In other words: you can choose 3 arbitrary functions of the coordinates (x0=t,x1,x2,x3) to define all kind of possibles sources over space and time. From this you can get all kinds of possible fields. This means "3 degrees of freedom", considering that 1 degree of freedom means one function in space-time f(t,x1,x2,x3). ... And a degree of freedom, this is what I would like to call an independent component ... is that acceptable?

But now, what happens if you consider that the motion of the particles is determined by the fields? Are all degrees of freedom lost? No independent components, but a nonlinear system?

Still I don't totally see the meaning of the question, but it is interresting!

Michel

Last edited: May 25, 2006
4. May 25, 2006

### Ratzinger

With independent components I mean what kind of vector/ tensor is needed to completly describe an EM field.

Actually I got two problems.

1. When all the information is contained in a four-vector, what do we need six components for (two three-vectors or an antisymmetric tensor of rank two)? Isn't there redundant information?

2. In Lewis RyderQuantum Field Theory I read that the problem of quantizing an EM field is due to fact that EM fields possess only two independent components, but are covariantly described by a 4-vector. In choosing two of these components as the physical ones, and thence quantizing them, we lose covarinace.
What should I make of this? Why two independent components? Has that something to do with the polarization vectors?

5. May 25, 2006

### marlon

No redundant information because the F-tensor also incorporates the behaviour of both E and B under Lorentz transformations. Here is more

I think you are referring to why a photon has no z-spincomponent equal to 0. We discussed this here

marlon

6. May 25, 2006

### samalkhaiat

Last edited: May 25, 2006
7. May 26, 2006

### lalbatros

Dear Ratzinger,
Dear all,

The interpretation from two polarisations makes of course sense to me too.

If I take one given fourier component (k,w), it is clear that two components is all that is needed to get any other (k,w) wave. But such a component represents a wave in the vacuum. Would the conclusion for two components remain true also in the presence of charges and currents? Can static fields also be represented from these two components?

(and what could have been wrong with the reasoning of my own that lead me to conclude for 3 components?)

Thanks for help!

Michel

Last edited: May 26, 2006
8. May 26, 2006

### samalkhaiat

Last edited: May 26, 2006
9. May 27, 2006

### lalbatros

samalkhaiat,

I have no problem to admit that two polarisations is all that is needed,
and that this can be furthermore interpreted as related to spin-1 and Lorentz invariance.

Where I do have a problem, is:

- when I try to figure out how to get the result directly from Mawell's equations
- when I try to understand the meaning of the original question by Ratzinger​

I translated the Ratzinger's question by "how to count the different possible sorts of EM fields".

By a trivial reasoning I got the answer that this number is simply 3 times the number of functions of time and space. This is simply the 4 components of the 4-current minus one because of charge conservation. I guessed that this would also be the number of differents fields because there is a one-to-one relation between fields and currents.

But I do have the feeling there is a mistake there.
If I could understand it, I could probably have a little bit more physical interpretation for the real result: 2 degrees of freedom.

Any suggestion ?

Michel

10. May 28, 2006

### lalbatros

samalkhaiat,

I tried to understand a little bit more this discussion. In other words:

Is the number of independent components of the EM field also
the number of independent components in the EM currents (3)?
And what is missing to get the "other answer" (2)? ​

I have these elements to add:

• The simple reasoning above does take gauge invariance into account: it takes charge conservation into account. Invariance and conservation are known to be equivalent properties.
(Noether, but easy to proof for EM gauge invariance).
• I guess that a second independent component (might) disappears because of an additional assumption: the assumption that E.H = 0. This additional condition is Lorentz-invariant and is satisfied for plane waves.

Therefore, I would be interrested to know a little more:
• Why has this additional condition E.B = 0 been introduced.
• What has really been excluded by this additional condition.
• In which physical situation do we have E.B =/= 0. And, more preciselly, why should these situations be discarded in the current discussion? What was the tacit assumption in our discussion?
• Finally, would I be right to say that -generally speaking- EM fields are described by 3 independent components? And why?

I would appreciate you to make the link with your familiar pov which seems to be closer to particle physics. Thanks for your comments.

Michel

11. May 28, 2006

### lalbatros

Ratzinger,

I just realised that I don't agree with the third possibility that you propose:

Indeed, the Coulomb Gauge condition doesn't leave you with two components, but with three:
- the 'electrostatic' potential V
- the three component of the vector potential (Ax,Ay,Az)
- minus one component because of the Coulomb gauge condition

The addition makes three, unless you restrict the possibilities as explained in my earlier post.

Michel

12. May 28, 2006

### samalkhaiat

Last edited: May 28, 2006
13. May 28, 2006

### samalkhaiat

[
3) The (E,B) 3-vector formulation

Here we have 6 components.
But we also have;

$$\nabla . E = \rho$$

$$\nabla . B = 0$$

so the number becomes 4.

We can easily show** that;

$$B^2 - E^2 = inv.$$

$$B.E = inv.$$

these can be used to further reduce the number to only 2.
Are you happy now?

In the em theory, the sources are treated as "given". They impose no restriction on the number of physically significant components of the EM-field.
Seting;
$$J^{\mu} = (\rho , j) = 0$$
does not change any thing in any of the above 3 proofs.
-----------------------------------------------------------

**Consider the complex vector;

$$V=E + iB$$

The only invariant of a vector with respect to rotation is its square;

$$V^2 = E^2 - B^2 + 2iE.B$$
Thus the real quantities $E^2 - B^2$ and $E.B$ are the only two independent invariants of the EM-field.

What do these invariants mean?

Lorentz trans. can be used to give E & B any arbitrary values, subject only to the condition that E.E - B.B and E.B have fixed values.
We can prove;

1) if E & B are mutually perpendicular in one frame, that is E.B = 0, then they are also perpendicular in every other Lorentz frame.

2) if B>E (or B<E) in any system, then we will have the same in every other system.

3) if they make an acute angle in any frame, then they will make an acute angle in every other frame.

4)we can always find a frame in which E & B are parallel at a given point.

5) if E.B = 0, then we can always find a frame in which E = 0 or B = 0 (depending on [E.E - B.B]< or >0) .

6) if, in any frame the field is purely magnetic (E = 0) or purely electric(B = 0), then they are mutually perpendicular in every other frame.

7) the case where both invariants are zero is excluded $V^2 = 0$. In this case, E & B are equal and perpendicular in all system.

Good Luck to all PF members.

This is my last post on these forums, as I will need the time to concentrate on the book I am writting.

Best Wishes

Sam

Last edited: May 29, 2006
14. May 29, 2006

### lalbatros

samalkhaiat,

I still have some problems (sorry for my back-to-the-basics obsession).

First, I think that the Lorentz (or Coulomb) condition is simply choosing a Gauge.
Therefore, the Lorentz condition, does not bring an additional constraint.
And it doesn't change the physics, fortunately.
If I am wrong here, then why should this additional constraint (on physics) be introduced?
This would answer then one of my last posts.

Second, the fact that E.H and E²-H² are invariants are no additional constraints.
Indeed, these are direct consequences of the Lorentz invariance for the Faraday tensor F.
Starting our count of independent components from the vector potential A, already ensure Lorentz invariance for all derived tensors, so then for F too.

Finally, I still remain convinced that an additional constraint is usually included in QFT, namely E.H = 0.
But I am no expert in the field, and I would simply like to know the reason of this additional choice and its physical meaning.

Hope you still find a while to think about my question,

Michel

Last edited: May 29, 2006
15. May 29, 2006

### samalkhaiat

Last edited: May 29, 2006
16. May 29, 2006

### lalbatros

samalkhaiat,

Thanks for the definition, it helps a lot, of course.
We are close to a closing of this discussion, but I still need the ultimate meaning.

I agree that a=4 , since the field has 4 components, those from the vector potential A.

Then, I see 1 constraint for the 'Lorentz' condition, permitted because of gauge invariance:

$$F_{1}(\mathbf{A}) = \partial_{\nu}A^{\nu} = 0$$ ​

Now I am down to 3 independent components.

Could you give me the expression for $$F_{2}(\mathbf{A})$$ and its justification.
Then, I will see why 2 independent components.

Thanks again,

Michel

Last edited: May 29, 2006
17. May 29, 2006

### samalkhaiat

Last edited: May 29, 2006
18. May 30, 2006

### lalbatros

samalkhaiat,

When I say "permitted by gauge invariance", I don't mean that the additional condition is gauge-invariant. (this condition simply amounts to choosing a gauge) I simply mean that the EM field F is not affected by this choice.
The condition that you write simply shows me how to choose that gauge to satisfy the condition. Therefore, I still cannot reach the final conclusion.

Note that I can live for now without the full understanding, I am working in the cement industry ...
I will go back to read Landau & Lipschitz or Ryder to get in shape !

Michel