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# "Unraveling the Mystery of Gravity: Exploring its True Nature as an External or a Fictitious Force"

Updated: Feb 15

Gravity---on the strength of Einstein's equivalence principle---has be likened to a fictitious force because it can be nullified by an acceleration in the same direction like a freely falling elevator. For if everything falls at the same speed then you get the feeling of weightlessness.

If gravity is fictitious, then it has a different aspect than the other fictitious forces like the centrifugal force, the Coriolis force, and the Euler force which appear only in non-inertial frames that rotate. In terms of wind patterns, the centrifugal force determines its curvature while the Coriolis force determines its direction. In neither case can we determine our position because the rulers change along with the curvature and our compases with the direction of the flow.

The non-Euclidean model which springs to mind is the Poincare disc model where the boundary is determined by the rotational speed, wr, with w=d theta/dt. Transforming from the half-plane model to the disc model, the metric is

ds^2= (dr^2+r^2(d theta)^2)/(1-r^2w^2)^2, (1)

where v=dr/dt. If consider gravity, along with the centrifugal force to be fictitious, then we would add twice the gravitational potential to (1) to obtain

(ds/dt)^2= (v^2+r^2w^2)/(1-r^2w^2-2m/r)^2 (2)

where m=GM, the gravitational parameter. As such, (2) is conformally equivalent to the Schwarzschild exterior solution [cf Eq. (79), p. 325 of Moller, "The Theory of Relativity", 1st ed]. Taking the Lagrangian as 1/2 of (2), the Euler-Lagrange equations give the geodesic equation

dv/dt- rw^2 +[(rw^2-m/r^2)/(1-r^2w^2-2m/r)]v^2 =dv/dt-rw^2+Gamma_{rr}^r v^2=0 (3) where the coefficient of the square term can be written as a Christoffel symbol of the second type. In (3) we have left out terms of quartic order terms in the angular speed, w. So. according to (3), gravity, as given by Newton's law, is a fictitious force on the same level as the centrifugal force, rw^2.

Rather, had we considered gravity as potential energy, V, in the Lagrangian, L=T-V, the gravitational force would have appeared on the right-hand side of (3), as dV/dr= -m/r^2, and not in the Christoffel coefficient, Gamma_{rr}^r. In the first instance, gravity modifies the boundary of the disc, while in the second, it acts as an impressed force. It is noteworthy that the vacuum solution to the static, exterior Schwarzschild solution considers such a modification of the Poincare disc model.

The Coriolis force enters when we transfer to a non-inertial frame where the inertial velocity v is related to the rotational velocity v' by v = v' + wXr, in the numerator of the metric. Therefore, fictitious forces modify the non-Euclidean character of the metric, while inertial ones appear as impressed forces.

The metric found in Moller's book, which, as we have mentioned, is related to ours conformally, is completely static. In GR, the classical potentials appear in the coefficient of the time component, dt, of the metric, and the vector potential in the cross term involving dt. This is odd, to say the very least, since the Newtonian potential is completely static. If the metric is given by

ds^2= -A dt^2+ B dr^2 +... (4)

where A and B can only dependent on r, the gravitational force would also appear in the B coefficient on account of the condition AB=1. Without the time component in (4), meaning that the metric is indeed static, we are allowed to treat gravity either as an impressed force by incorporating the potential V in the lagrangian, or as a fictitious force by including it in the conformal term of the metric.

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