1

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?

Qmechanic
  • 201,751
Matt Dickau
  • 418

1 Answers1

3

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.

Guy
  • 1,261
  • Hi David, I am aware that the sign change makes the theory unstable when charged particles are included (at least, that was my guess). What I'm wondering is if the theory could be considered to be metastable if the coupling constant is sufficiently small. – Matt Dickau Aug 13 '20 at 14:28
  • 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. – Guy Aug 13 '20 at 17:18
  • 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. – Guy Aug 13 '20 at 17:39
  • How does the negative kinetic term break unitarity? Isn't the Hermiticity of the Hamiltonian a sufficient condition for unitarity? (If H is Hermitian then exp(-iHt) is unitary, no?) – Matt Dickau Aug 13 '20 at 18:21
  • Is the Hamiltonian still hermitian? – Harry Wilson Aug 14 '20 at 14:43