# Lagrangian density of the EM field

1. Feb 25, 2007

### Ahmes

Hello,
I took an Electrodynamics course this semester, where we derived Maxwell's equations from the field's Lagrangian density.
As a motivation, we "looked" for a scalar (in the relativistic sense) having something to do with EM fields - and had we found one we would have declared it a candidate to be our Lagrangian density.
We found two such scalars (in the characteristic polynomial of the EM tensor): $|\mathbf{B}|^2-|\mathbf{E}|^2$ and $\mathbf{E}\cdot\mathbf{B}$

The professor hand-wavingly claimed that the latter "does not preserve parity" and therefore invalid, so we ignored it for the rest of the course. But now (after the test...) I came back to this dot product, trying to see what "new Maxwell equations" we can theoretically get from it.
I wrote it using the 4-vector-potential and its derivatives, put it in the Euler Lagrange equations - and surprisingly - this element contributes nothing! Putting it in the EL equation just produces zero.

So what is the story of the parity and how does it relate to E dot B?

Last edited: Feb 25, 2007
2. Feb 25, 2007

### lalbatros

Personally I don't really like using E and B in such a derivation.
The most natural way to do that, with a relativistic point of view, is to use the Faraday tensor.
The result is the same, of course since the same invariants can be obtained from the Faraday tensor.

It is true that E.B is a pseudo scalar and can be dropped from the start.
It would be interresting to get the physical reason why it disappears anyway from the least action principle.

See on wiki also:

http://en.wikipedia.org/wiki/Electromagnetic_tensor
http://en.wikipedia.org/wiki/Pseudoscalar

Michel

3. Feb 26, 2007

### dextercioby

Let's ignore parity for a while and suppose the $\mathcal{L}=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$

Is the resulting theory still a gauge theory ??

4. Feb 26, 2007

### ObsessiveMathsFreak

Code for "There is a proper fundamental reason we ignore it but I've either forgotten it or never came across it." I would suggest caluculating the demsional units of the product to see if anything can be gleaned from that.

5. Feb 26, 2007

### StatMechGuy

If you use CGS units, like it seems that he did, then E.B would have the units of energy density. However, as has been pointed out above, E.B does not preserve spacial inversion (parity) because E is a polar vector and B is a pseudovector.

6. Feb 26, 2007

### Ahmes

Oops... I meant to convert to SI. I dislike CGS too :yuck:
But you can anyway get units of energy density if you multiply the thing by some power of c. The point is that both these are relativistic scalars (and a c will surely not change it). Now you say E·B is a "pseudoscalar" rather than a normal scalar because B is a pseudovector.

I read the Wikipedia articles about pseudoscalar & vectors, but I don't understand why there is a difference between B and E (which is a normal "polar" vector)

Maxwell equations in my opinion do not expose a difference, as they were completely symmetric had magnetic monopoles existed. And if we assume them to exist, but we're happen to be at an area were they aren't found - B and E should still act the same anywhere (at least in an electric charge free zone).
So what is the source of the difference?

[Sorry dextercioby, I didn't study about gauge theories. And according to Wikipedia, it's too complicated for me to relate to at the moment]

7. Feb 26, 2007

### Parlyne

Parity inversion is the operation of taking $$\vec{x} \rightarrow -\vec{x}$$. A polar vector is one which transforms under parity inversion like the position vection, while pseudovectors remain unchanged. It should not be too hard to convince yourself that the cross-product of two polar vectors does not change under parity inversion. This means that such a cross-product is actually a pseudovector.

Now, consider the Bios-Savart law. This shows that the vector character of B comes from a cross-product between electrical current and position ($$\vec{I} \times \vec{x}$$). If invert parity, we know that the position vector changes sign by definition. The current vector must also change sign because it measures the actually physical flow of charged particles, whose actual direction of motion doesn't change when we switch the direction of our coordinates. But, if current and position both change sign under parity, B cannot. Hence, B is a pseudovector.

8. Feb 27, 2007

### Ahmes

But still, the Biot-Savart law comes from the fact that all the Bs around come from currents of electric cherges. I didn't want to make this a discussion about magnetic monopoles (which is an interesting subject) but if they exist, I guess there will be a "Biot-Savart law" for the electric field - and then it will be a pseudovector too (unless from some reason $mathbf{I}_m$ is a pseudovector and than the cross product will be a polar vector, but as you said - it makes no sense).

Last edited: Feb 27, 2007
9. Feb 27, 2007

### Parlyne

Actually, my understanding is that $$\vec{I_m}$$ is a pseudovector. The best argument I have (off the top of my head) is as follows. When we perform parity inversion on vectors, the change in sign comes because the physical arrow keeps pointing in the same direction it always did, but the coordinates have reversed on it. This means that the physical direction of a pseudovector actually reverses, so that its sign remains the same.

Now, picture the magnetic field of a standard bar magnet. What do we need to do to make the field switch orientation without changing its shape? We need to switch the poles of the magnet. This means that under parity inversion, the poles of a magnet reverse. What was a north pole becomes a south pole and vice-versa. But, a magnetic monopole is just a lone N or S. So, under parity, it must change sign. This means that magnetic charge is actually a "pseudoscalar" quantity. This is exactly unlike electric charge which is unchanged under parity reversal.

Since magnetic charge is a pseudoscalar, when we consider the action of $$\vec{I_m}$$ under parity, we have to include one factor of -1 for the physical direction of the monopoles' velocities to remain unchanged, and a second -1 for the switched magnetic charge. This means the magnetic current vector will be unchanged under parity, making it a pseudovector.

10. Feb 28, 2007

### Ahmes

Hi Parlyne,
It appears to me that your argument:
is flawed. You can look in the same way on an electric dipole and divide its poles to N and S, but that doesn't mean the electric monopole is a pseudoscalar.

I did see on the Wikipedia article about pseudoscalars that the magnetic monopole ('s mathematical definition) is given as an example to such a creature. And as to yet I can't see way. I might have been wrong in eliminating your argument, but I surely want to know why.

11. Feb 28, 2007

### Parlyne

The key point wasn't that a monopole is, in a sense, half a dipole. The key point was that, under parity, the poles of a magnetic dipole must change sign in order to correctly generate the (now flipped) magnetic field. Since the electric field doesn't change physical direction under parity, there is no need for the poles of an electric dipole to change sign.

12. Mar 1, 2007

### Ahmes

I can't see why there is a difference...
The electric field of an electric dipole is:
$$\vec{E}=\frac{1}{4\pi\epsilon_0} \frac{3\hat{n}(\hat{n}\cdot\vec{p})-\vec{p}}{x^3}$$
The magnetic field of a magnetic dipole is:
$$\vec{B}=\frac{\mu_0}{4\pi} \frac{3\hat{n}(\hat{n}\cdot\vec{m})-\vec{m}}{x^3}$$

So why would there be a difference in how they behave under parity transformation? I guess the only answer left is that p and m are fundamentally different, the question is why...

13. Mar 1, 2007

### Parlyne

The fundamental difference between $$\vec{p}$$ and $$\vec{m}$$ is that $$\vec{m}$$ is the result of a current loop and, thus, has uses a cross-product to define the direction perpendicular to it, while $$\vec{p}$$ comes from a physical separation of two electric charges and, hence needs no cross-product to define the direction along the dipole.

14. Mar 1, 2007

### Ahmes

But that's a circular argument

- Why is B a pseudovector?
- Beacue it is a cross product.
- What if a magnetic monopole exists?
- Then it acts like half a dipole and therefore a pseudoscalar.
- But there's no fundamental difference between E dipole and B dipole
- Yes there is, m is a cross product (because it is made out of currents) and p is not.

Again, if magnetic monopoles exist, their current will produce an electric dipole field. And then your argument could be applied to ordinary electric charges being pseudoscalars.

15. Mar 1, 2007

### Mentz114

Has the asymmetry between E and M got anything to do with negative and positive energy solutions ? The invariant of the EM field E^2-B^2 looks like energy but does the minus imply negative energy for B ?

Dirac is unambiguous in stating that the longitudinal magnetic component of the field has negative energy.

16. Mar 1, 2007

### Parlyne

The argument is not at all circular. It relies on the empirical knowledge that electric charge in a scalar quantity, making a current of electric charges a polar vector. The magnetization (and, hence, the magnetic field it generates) of a physical magnetic dipole involves the cross-product between this current and the position vector, making it a pseudovector. The statement about the poles switching under parity follows from this. A magnetic monopole is effectively defined to be a particle that acts like a lone pole of a magnetic dipole. So, it's sign must change under parity, making a monopole current a pseudovector.

17. Mar 2, 2007

### Ahmes

Yes, you were right about the magnetization m (that we know from our daily lives, which is produced by currents only) being a cross product. But I'm still not convinced that if magnetic monopoles exist, they act like a single pole. I don't see why they shouldn't be exactly symmetric to electric charges. Maybe there is a reason to it, I want to know what it is (or just that it exists if it's too complicated).

Also...
How can one observe something being a scalar rather than a pseudoscalar?
Can one do improper rotations "for real" and see what happens?

Last edited: Mar 2, 2007