Form of the spin 1 equation

1. Apr 15, 2010

RedX

The equation for a massive spin 1 particle is given by the Proca equation:

$$\partial_\mu(\partial^\mu A^\nu - \partial^\nu A^\mu)+ \left(\frac{mc}{\hbar}\right)^2 A^\nu=0$$

My question is why this equation? In particular, why can't it be like this instead:

$$\partial_\mu(\partial^\mu A^\nu +2 \partial^\nu A^\mu)+ \left(\frac{mc}{\hbar}\right)^2 A^\nu=0$$

The Lagrangian is given by:

$$\mathcal{L}=-\frac{1}{16\pi}(\partial^\mu A^\nu-\partial^\nu A^\mu)(\partial_\mu A_\nu-\partial_\nu A_\mu)+\frac{m^2 c^2}{8\pi \hbar^2}A^\nu A_\nu$$

and again, why can't the derivative terms instead be:

$$X\partial_\nu A_\mu \partial^\nu A^\mu+Y\partial_\nu A^\nu \partial^\mu A_\mu +Z\partial_\nu A^\mu \partial_\mu A^\nu$$

for arbitrary real numbers X,Y, and Z?

The answer can't be that there is a U(1) symmetry, because the mass terms don't obey the U(1) symmetry.

2. Apr 17, 2010

Avodyne

First, note that your Y term can be converted to your Z term (or vice versa) by integrating each derivative by parts to move it onto the other A. But there are still two possible inequivalent terms, so your question remains.

The answer in given in Weinberg vol I. If you choose any relative coefficient other than the QED one, there is an extra propagating spin-zero field, essentially the 4-divergence of A. Choosing the QED kinetic term eliminates this spin-zero field as a propagating degree of freedom. See Weinberg for the details.

3. Apr 17, 2010

RedX

Thanks for the information! I'll have a look at Weinberg's book, but real quick, does the reason have to do with quantum field theory, or does it hold true even in relativistic quantum mechanics?

Also, I was thinking about spin-2. The Lagrangian is given by:

$$S=\int d^4x\sqrt{-g}R$$

where $$R=R(g_{\mu \nu})$$

is the scalar curvature, a function of the metric $$g_{\mu\nu}$$ or graviton field.

Would a different spin 2 particle, say $$z_{\mu\nu}$$, have the Lagrangian:

$$S=\int d^4x\sqrt{-g}R$$

where this time the scalar curvature $$R=R(z_{\mu \nu})$$

is of the same form as previously, but replacing a z for every g?

Because if this is true, then spin-2 particles don't all obey the same equation. Every spin 0 particle obeys Klein-Gordan. Every spin 1/2 the Dirac equation. But the graviton seems special.

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