# Motivation for electromagnetic field tensor

1. Apr 1, 2009

### Bobhawke

Is there any deep reason for introducing the electromagnetic field tensor other than the fact that Maxwell's equations can be written in a very succinct form in terms of it? Would it be possible to write down a lagrangian involving a normal kinetic term for $$A^{\mu}$$ that reproduces the physics of electromagnetism correctly?

2. Apr 2, 2009

### genneth

No, because a kinetic term for A would not be gauge invariant.

I know of two ways to get Maxwell's theory of electromagnetism from asking for the "simplest" theory of some class of theories. One to to ask for the simplest possible gauge theory --- then you would choose the simplest compact Lie group U(1) and minimal coupling; the other is to ask for the simplest Lorentz covariant field theory that couples to a 4-current linearly.

The need for F can be boiled down to this: E and B are observed to mix under Lorentz transforms, thus we can hypothesize that they are elements of a tensor. It cannot be the case that they are a 4-vector, because there are not enough degrees of freedom to express E and B. Thus we can go for the next simplest thing, which is a 2-form (antisymmetric 2-tensor); this happens to have exactly the right number of degrees of freedom.

3. Apr 2, 2009

### quZz

Equations of motion can be written in terms of $$F$$ alone:
$$\frac{dp^{\mu}}{ds} = q F^{\mu\nu}u_{\nu}$$.
But anyway $$F_{\mu\nu} \equiv \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$$ so $$A$$ is fundamental.

4. Apr 2, 2009

### Hans de Vries

You can do this like in the following way, to obtain the Lagrangian density
of the massless or massive $$A^\mu$$ vector field:

$$L ~~=~~\left\{\begin{array}{l} ~~+\Big(~~~\frac{1}{2}{\dot A_0}^2 ~-~ \frac{1}{2}\nabla A_0\cdot\nabla A_0 ~-~ \frac{1}{2}~m^2 A_0^2 ~\Big)\\ \\ ~~-\Big(~~~\frac{1}{2}{\dot A_x}^2 ~-~ \frac{1}{2}\nabla A_x\cdot\nabla A_x ~-~ \frac{1}{2}~m^2 A_x^2 ~\Big)\\ \\ ~~-\Big(~~~\frac{1}{2}{\dot A_y}^2 ~-~ \frac{1}{2}\nabla A_y\cdot\nabla A_y ~-~ \frac{1}{2}~m^2 A_y^2 ~\Big)\\ \\ ~~-\Big(~~~\frac{1}{2}{\dot A_z}^2 ~-~ \frac{1}{2}\nabla A_z\cdot\nabla A_z ~-~ \frac{1}{2}~m^2 A_z^2 ~\Big) \end{array}\right.$$

Which is the scalar Lagrangian for each vector component with the metric chosen so
that the whole still transforms like a scalar.

This is correct but it's only one half of the standard Lagrangian. The other half
term induces the same U(1) phase changes in a charged field and is therefor
indistinguishable from the Lagrangian above.

See section 22.9 of my book in progress:

http://physics-quest.org/Book_Chapter_Lagrangian.pdf

The same holds for the non-Abelian Lagrangian:

"The simplest derivation of the non Abelian field tensor and Lagrangian density"
http://www.physics-quest.org/Non_Abelian_Lagrangian.pdf

For a step by step treatment of the derivation of the Lorentz force from the U(1)
symmetry of the Klein Gordon equation using Wilson loops see:

http://www.physics-quest.org/Book_Lorentz_force_from_Klein_Gordon.pdf

Regards, Hans.

Last edited: Apr 2, 2009
5. Apr 2, 2009

### Bobhawke

Cool, thanks for the replies guys

6. Apr 13, 2009

### Bobhawke

Is there any reason why we couldnt get EM in terms of either

1) A rank 2 field strength tensor $$F_{\mu \nu}$$ with no symmetry/anti-symmetry requirements, and then choose a gauge to get rid of all the redundant dofs

or

2) A field tensor of rank>2 $$A_{\mu\nu\rho...}$$ with again the redundant dofs eliminated by fixing the gauge

For 2), Im reading something by t'hooft that says such a tensor field could not be introduced for 2 reasons: the energy density has to be bounded from below, and the theory has to be renormalisable ie all interactions have to have sufficiently low dimensions that their couplings have mass dimension greater or equal to zero. But I cant figure out why such a tensor would cause the energy density to be unbounded from below or why it would cause the theory to be unrenormalisable

7. Apr 14, 2009

### genneth

If you choose to use a gauge theory, then your physical variable is a connection on some G-bundle over spacetime. The F-tensor is the curvature of this connection, and so will be antisymmetric by decree of mathematics --- nothing physical there. It so happens that for EM, if we choose a U(1) bundle, then F has 6 dofs and exactly express the electric and magnetic fields that we see. You can do a bit of group theory to figure out all the possible combinations of the connection (in coordinates, A_mu), which would not depend on the choice of basis, i.e. be gauge invariant. Then you find that F^2 is the only non-trivial possibility.

8. Apr 14, 2009

### Hans de Vries

The simplest derivation possible of the Lorentz force:

$$p_c^\mu ~~=~~ p^\mu+eA^\mu ~~=~~ -\partial^\mu\,\phi$$

Where p is the inertial momentum depending only on the velocity
and $\phi$ is the phase of the field. The combination of p and eA can
not have any curl since $\phi$ is a scalar. So any curl in eA must
be compensated by an opposite curl in p.

$$-(\partial^\mu p^\nu-\partial^\nu p^\mu) ~~=~~ e(\partial^\mu A^\nu-\partial^\nu A^\mu)$$

One obtains the Lorentz force by using $U_\nu=\partial x_\nu/\partial \tau$ to turn all
the spatial derivatives into time derivatives.

$$-\frac{\partial p^\nu}{\partial x_\mu}\,\frac{\partial x_\nu}{\partial \tau} ~~+~~\frac{\partial p^\mu}{\partial x_\nu}\,\frac{\partial x_\nu}{\partial \tau}~~=~~ eF^{\mu\nu}\,U_\nu$$

The first term cancels, it represents the derivatives of the
invariant mass $p_\nu p^\nu$, so we obtain.

$$\frac{\partial p^\mu}{\partial \tau}~~=~~ eF^{\mu\nu}\,U_\nu$$

Regards, Hans

Last edited: Apr 14, 2009