2
$\begingroup$

The test particle action in Scalar-Tensor-Vector Gravity (STVG) can be written as

$$ S = -\int( m + q \omega u^{\mu} \phi_\mu ) d\tau $$

where, $\tau$ is the proper time along the world line of the test particle and $m$ denote the test particle mass, $u^{\mu}=dx^\mu / d\tau$ is the four-velocity. Here, $\omega$ is the coupling constant characterises the strength of the coupling between the fifth force vector field $\phi_\mu$ and matter. The fifth force charge $q$ of a particle is proportional to its mass $m$

$$ q = k m = \sqrt{\alpha G_N} m,$$

where, $G_N$ is the Newton's gravitational constant and $\alpha$ is dimensionless parameter.

My question is, is their any way the `charge' appearing in the solution can couple to the test particle's action? For example, in the galileon type theories the parameter can couple to the action of a test particle, eventually breaking Einstein's equivalence principle.

$\endgroup$
1
$\begingroup$

You can generally couple a continuum action $S_{\rm field} = \int_V \mathcal{L}(\psi(x^\mu)) d^4 x $ to a worldline action $S_{\rm particle} = \int_{\Delta \lambda} L(x^\mu(\lambda),\psi(x^\mu(\lambda))...)d\lambda$ through a delta function on the world-line $$S_{\rm tot} = \int_V \left(\mathcal{L}(\psi(x^\mu)) + \int_{\Delta \lambda} L(x^\mu(\lambda),\psi,...)\delta^{(4)}\left(x^\mu - x^\mu(\lambda)\right) d \lambda \right)d^4 x$$ In other words, the total Lagrangian will have the smooth field density contribution and a distributional contribution on the world-line. Now you will see that the variation with respect to $\psi(x^\mu)$ and $x^\mu(\lambda)$ will give you the correct field equations. (Keep in mind that usually you will have to understand this Lagrangian as an effective field theory with divergences on the world-line which will require curing by regularization.)

As for your case of a scalar coupling - once you couple your particle to a Lagrangian field density such as $\mathcal{L} = \sqrt{-g}(R + \partial_\mu \phi \partial_\nu \phi g^{\mu\nu} + 2 V(\phi))$ (up to various factors), you should be able to find out from the resulting field equations that, in fact, the "charge" carried by the particle is determined by the magnitude of the coupling parameter $\omega$. Good luck!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.