An intuitive approach to inertial forces and the centrifugal force paradox in general relativity

For the purposes of this article we may consider the event horizon to be an (invisible) sphere. If one ventures inside of this sphere, one cannot come back out again.

There is a radius where free photons, that is, photons whose motion is determined only by gravity, can move on circular orbits around a black hole. The circumference of this circle is 1.5 times the circumference of the surface of the black hole (the event horizon).

The proper acceleration of an object is the acceleration measured relative to an inertial system (a freely falling system) momentarily comoving with the object.

Strictly speaking we are using geometrized units in which $c=1$⁠. In these units, time has the same dimensions as distance. If we want to express distances and times in terms of standard units, we should replace any instance of $v$ by $v∕c$⁠, where $c$ is the velocity of light in standard units.

The embedded geometry (⁠$t=0$ and $θ=π∕2$ in Schwarzschild coordinates) corresponds to a section of a parabola (see Ref. 25), $z=2RGr−RG$⁠, revolved around the vertical $(z)$ axis (⁠$RG$ is the radius at the event horizon).

In the derivation we divided by $v$⁠, thus assuming $v≠0$⁠. For $v=0$ any $n̂∕R$ will do.

Strictly speaking the argument only holds exactly as long as the velocity is purely horizontal.

Consider a $1+1$ dimensional scenario. Let $u$ be the velocity of the reference frame relative to an inertial system in which the reference frame is momentarily at rest, and let $ημ=(γ(u),γ(u)u)$ be the corresponding four-velocity. Let $vs$ be the velocity of the test particle relative to the inertial system in question, and let $uμ=(γ(vs),γ(vs)vs)$ be the corresponding four-velocity. Let $v$ be the velocity of the test particle relative to the reference frame. We have $γ(v)=−ημuμ$ (using the (−, +, +, +) convention). If we differentiate both sides of this expression by $d∕dt$ and use the fact that $u=0$ and $v=vs$ momentarily, we find $v(dv∕dt)γ4=−γ2v(du∕dt)+v(dvs∕dt)γ4$⁠. This result corresponds to Eq. (17) (multiply Eq. (17) by $γ4$⁠, divide by $δt$⁠, and take the limit where $δt$ is infinitesimal). Also we know that the proper acceleration $α$ is given by $α2=−(duμ∕dτ)(duμ∕dτ)$⁠. If we differentiate $uμ=(γ(vs),γ(vs)vs)$ with respect to $τ$⁠, we find $duμ∕dτ=γ3(dvs∕dτ)(v,1)$⁠. It follows that $α=γ3(dvs∕dt)$⁠, which is Eq. (18).

It is true in the Newtonian limit that accelerations of the reference frame perpendicular to the direction of motion do not affect the local speed derivative. We can also reason this strictly relativistically knowing a little about time dilation and simultaneity. We could also understand it using four-tensors knowing that $γ=ημuμ$⁠, where $ημ$ is the reference frame four-velocity and $uμ$ is the particle four-velocity. Consider a particle moving in the $x$ direction and consider the reference frame to reach a velocity $δvy$ in the $y$ direction after a time $δt$⁠. To first order in $δt$ we have $ημ:(1,0,δvy,0)$ and $uμ:(γ0,vx,0,0)$⁠. Here $γ0$ is the value of $γ$ at $t=0$⁠. We see that to first order $γ=ημuμ=γ0$⁠, and thus a perpendicular acceleration of the reference grid does not affect the local speed derivative. Similarly an infinitesimal perpendicular velocity of the particle will have no first-order effect on the speed. Any one-dimensional reasoning of how the speed derivative is related to force parallel to the direction of motion thus holds also when there are perpendicular effects.

The difference between the proper rotation and the rotation as observed from outside of the merry-go-round for the non-central points is not only due to time dilation, but also to relativistic precession (rotation) effects. We can also use the formalism of this paper for these points assuming that we correctly express the proper local reference frame rotation $ω$⁠, that is, the rotation as experienced by an observer at rest relative to the reference frame at the points in question.

It is easy to make a formal proof of how the factors should enter. For contravariant vectors we can reason it out instead. Consider two coordinate points separated by an infinitesimal vector $dxk$⁠. Assume that we stretch space by a factor $e−Φ$⁠. A coordinate vector that has the same norm (length) relative to the stretched space that $dxk$ had relative to the standard space would have to be component-wise smaller by a factor $e−Φ$⁠. Hence given a normalized vector relative to the standard space, we obtain a corresponding normalized vector relative to the rescaled space by dividing by the stretching factor $e−Φ$⁠.

Here we consider the perpendicular part of the spatial part of Eq. 44 of Ref. 10 and set $θ̃αβ=0$ as is appropriate for this case.

Set $θ̃αβ=0$⁠, $η̃α∇̃αΦ=0$ and identify $τ̃0=tc$ in Eq. (44) of Ref. 10.

Actually they are not all exact geodesics. They cannot be in general (while at the same time being orthogonal) when the surface is curved. But at the center of coordinates they are orthogonal, and the curvature of the coordinate lines vanishes, which is sufficient for the type of arguments made in this article.