# Alternative Physics - Attractive Electromagnetism / Vector Gravity

If you flip the sign of the term containing the field strength tensor (e.g. change $$-\frac{1}{4}F^{\alpha \beta}F_{\alpha \beta}$$ to its negative) in the Lagrangian for electromagnetism, you get a theory where like charges attract and opposite charges repel instead of the other way around. Imagining a universe where only special relativity applied (and not general relativity), this attractive force would look like Newtonian gravity at low relative velocities if the charge of normal matter was roughly proportional to inertial mass (e.g. protons and neutrons have unit charge and electrons have no charge).

However, the energy of the "gravitational" field in this theory would be negative. Does this cause problems if one were to try to make this into a quantum field theory? (The theory would be unstable, since there would be no ground state?) If the strength of the "gravitational" force were low enough, would the theory at least be effectively metastable?

• Commented Aug 13, 2020 at 5:32

If I write down the Lagrangian $$\mathscr{L} = -\frac14 F^2\,,$$ and change its sign, nothing happens, since the derived equations of motion are the same. However, if I introduce a current $$\mathscr{L} = -\frac14 F^2 + A_\mu\bar e \gamma^\mu e + \bar e({\rm i}\partial_\mu\gamma^\mu - m) e\,,$$ now changing the sign of $$-\frac14F^2$$ has the effect of changing the relative sign of the electron kinetic term and the photon kinetic term. In other words, the lowest energy state has photons with infinite kinetic energy. I don't think this is the desired effect.
• Let's restore a canonical kinetic term by taking $A_\mu\to {\rm i} A_\mu$. With this field re-definition, we see that this new theory is equivalent to electrons with complex charge. In this sense, what you are saying is correct, in that like charges appear to attract. However, as pointed out in the post linked by @qmechanic, because this is equivalent to a negative kinetic term, it breaks unitarity, which is a very (very very) strong constraint on a theory.
• Also, I should add that $A$ is by definition a real vector field (like every gauge field), so the redefinition $A\to{\rm i} A$ is unusual in this regard as well.