According to special relativity, the rate, in an inertial coordinate system Int, of a clock moving with speed $v$ in Int, is reduced from the clock’s rate in a system in which it is stationary (its proper rate), by (is divided by) a factor $$T_{rs} = \frac{1}{\sqrt{1-v^2/c^2}}$$ $c$ = speed of light.

According to general relativity, the rate at s-t point B, in a stationary, time independent space- time in which relative gravitational potentials can be defined, of a clock located at s-t point A, if the gravitational potential of B w.r.t. A (the line integral along any path L from B to A of the gravitational acceleration’s vector’s inner product with a unit tangent vector to L) is P, is reduced by a factor $$T_{rg} = \exp(P/c^2) = e^{P/c^2}$$

Wikipedia’s gravitation time dilation article, including the formula, is at https://en.wikipedia.org/wiki/Gravitational_time_dilation

My derivation of the gravitation time dilation formula, without using full GR, is at https://sites.google.com/view/logic-physics-settheory-math/entries-for-items-1-10

A seeming difference occurs in this thought experiment:

Consider a disc of radius $R$ rotating with speed $v$ at its rim. Its angular velocity is $w = v/R$, & its acceleration at radius $r$ is $rw^2 = r(v/R)^2$, so a test particle of mass $m$ at radius $r$, in the frame rotating with the disc about the disc’s center, experiences a centrifugal force
$$\left[mr(v/R)^2\right]\left[\frac{1}{\sqrt{1-(vr/Rc)^2}}\right]$$, so, by the GR equivalence of acceleration with gravitation, the
disc’s center’s gravitational potential P w.r.t. the disc’s rim is $$[(v/R)^2][\int_0^R (r/\sqrt{1- (r^2)(v/Rc)^2)}\mathrm{d}r] = [(v/R)^2][-(Rc/v)\sqrt{(Rc/v)^2 - r^2}, at (r = R) – at (r = 0)] = (c^2)[1 - \sqrt{1-(v/c)^2}]$$ so its gravitational time rate reduction factor w.r.t. the disc’s rim is
$$\exp[1 - \sqrt{1-(v/c)^2}]$$ so at, e.g., $v = c/2$, it is 1.143…, and its limit as $v$ approaches $c$ is $$\exp(1)= 2.718…$$

But the SR time rate reduction factor for the disc’s center w.r.t. its rim is $$\frac{1}{\sqrt{1-(v/c)^2}}$$ so at $v = c/2$ it is 1.154…, and its limit as $v$ approaches $c$ is infinity.

I am more confident that the SR result is correct than I am that the GR one is correct. I think that either my calculation of the gravitational potential difference between the disc’s center and its rim is incorrect, or that the GR formula for gravitational time dilation is simply not applicable to the rotating disc situation, but I don’t know which is the case, or why. Can anyone enlighten me?

  • $\begingroup$ Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! $\endgroup$
    – jng224
    Apr 29, 2021 at 16:05
  • $\begingroup$ Please do not edit questions that have received answers in a way that invalidates the received answers. You should ask the new material as a separate question $\endgroup$
    – Dale
    May 16, 2021 at 12:53
  • $\begingroup$ I have rolled back the big edit $\endgroup$
    – Dale
    May 16, 2021 at 12:56
  • $\begingroup$ @Dale Dale, what are you talking about? How did I edit my question in a way that invalidated the received answer (Eric Smith's)? I didn't touch the original question, just added material after "Added 2021-5-16". Is it forbidden to add material that disagrees somewhat with someone's answer? The new material mostly didn't consist of questions, rather it was material I had thought of since Eric's answer, partially because of that; the few questions were just incidental to the new material, except my added comment asking Eric whether he agreed with my addition, & 1 other question. $\endgroup$ May 16, 2021 at 14:03
  • $\begingroup$ @MichaelFox said "the few questions were just incidental to the new material". Precisely, this rendered Eric Smith's answer invalid because it did not address any of the new material. If you disagree with my action I would encourage you to open a thread in Meta. The admins can clarify the policy for you, but it is pretty clear. Furthermore, a question is supposed to ask a single question, so adding additional questions is a bad idea even if there are no answers that would be rendered invalid. $\endgroup$
    – Dale
    May 16, 2021 at 14:34

2 Answers 2


The SR calculation is the correct one. Deriving it from the equivalence principle is indeed tricky. I think your expression for acceleration is incorrect; the Wikipedia article on Born coordinates gives $\frac{-\omega^2 r}{1 - \omega^2 r^2}$, which does integrate to give the SR formula. When I tried to derive this myself I ran into difficulty, but I think the difference is that Coriolos forces need to be taken into effect -- an object dropped from the hub does not fall to the rim in a straight line as measured from the rim.


The acceleration formula in the Wikipedia article on Born coordinates which is linked to from
Eric Smith's answer & given in that answer, when it is used to calculate the gravitational
potential of the center of the disc wrt its rim, and this potential is put into my & Wikipedia's
exponential gravitational potential relative time dilation formula, does give the correct, SR,
formula, which depends on only the rim's velocity in the non-rotating frame (the acceleration has
an effect only thru affecting the rim velocity.) (Some people claim that acceleration, in addition to
velocity, does affect relative time dilation, even in flat, Minkowski, space, but they agree that the
effect goes to zero as the acceleration goes to zero. By letting $R$ → ∞, $v$ fixed, $a = v^2/R$
shows that this effect can be ignored in my (defective) gravitational time dilation formula for the
rotating disc, which, for each R, depends on only v.) (Added later: Perhaps not. As $R$ → ∞, the acceleration everywhere goes to 0, but the path length over which the potential is accumulated increases, so, depending on how the time dilation changed with decreasing acceleration, the potential change might not go to 0.) However, although the Born article's acceleration formula for the rotating disc gives the right answer when used in the time dilation equation, I can't see how that acceleration formula can be correct. Eric Smith apparently also had a problem seeing this. Eric says he thinks the difference between the Born article's formula and mine occurs because the Coriolis force needs to be taken into effect (account?). I think this can't be the reason, since the gravitational potential between A and B, just as any potential, is calculated from the force on a stationary particle at each point along a path from A to B, and the Coriolis force is zero for a stationary particle. Even if we used the force on an object dropped from the hub (center), the Coriolis force would not affect the potential, since it is always at right angles to the velocity of the particle. However, the Coriolis force is due to the rotation of the disk, which is directly related to the tangential speed of points on the disk, which is the source of the major time dilation, the SR one, and thus also to the difficulty with defining a global time for the rotating coordinates, discussed below.

One point: I stuck in the gamma, $\Gamma = \frac1{\sqrt{1-v^2/c^2}}$, into my acceleration
formula for the rotating disc to account for a particle's mass increase as its radial position
increases, but this was an error, since, while its mass in the non-rotating frame does increase
with gamma and its radial distance, in the rotating frame, in which the gravitational potential is
calculated, its tangential velocity is zero, so gamma in that frame is always 1. Also, the
gravitational potential rate of change with arc length is the potential energy change per unit
mass at that point, so whatever mass increase wrt some other point has occurred is irrelevant. Using the correct formula for the force due to radial acceleration, without the gamma, my &
Wikipedia's gravitational time dilation formula gives the dilation to be
exp$\left(\frac{v^2}{2c^2}\right)$, which gives an even smaller time dilation than my original, incorrect formula.

The reason my & Wikipedia's gravitational potential time dilation formula gives an incorrect
answer in the rotating disc problem is not, I think, that my calculation of the gravitational
potential between the center and the rim of the rotating disc, in the rotating frame, is incorrect.
My modified formula for radial acceleration, without the gamma, is correct, I think. (So the Born
article formula for acceleration in the rotating disc frame must be incorrect.) The reason that our
time dilation formula which is based on the gravitational potential difference gives the wrong
dilation factor is the second possible one I gave in my original question above: the GR formula
for gravitational time dilation (as an exponential function of the gravitational potential difference
between the two space-time points in question) is simply not applicable to the rotating disc

The reason for this is that the rotating disc situation doesn't satisfy one of the two
requirements that I gave for the derivation on my website of the time dilation formula to apply to
a space-time, which is the requirement that the space-time be non-time-varying in the
coordinate system used in the derivation. (By a non-time-varying space-time, in a coordinate
system Coord, I meant, without stating this explicitly on my website, that Coord could be written
as a product of a 3-dim. spatial part which doesn't depend on the time t, with a 1-dim. temporal
part giving t, which would serve as a global time coordinate for the space-time.) I knew that
some possible GR space-times didn't satisfy this requirement, but when I wrote the initial PSE
question, I didn't realize that the rotating disc s-t was such a s-t. I assumed the rotating
coordinate system could be such a non-time-varying coordinate system, with the time rate at the
center being the same as that in the non-rotating system, and the time rate factor between a
spatial point x in the rotating system and any point, in particular the center, in the non-rotating
system depending only on the radial distance of the point x, as required if it is to be a function of
the gravitational potential difference, so it could be the same for all points in the rotating system
at the same radius r, and so the rotating system coordinate time for all points with the same r
could be the same, as required by time synchronization by light pulses sent from the center.

However, as the Born article points out, there is a severe problem with defining a s-t coordinate
system for the rotating disc in which, for each radius r, all the points at radius r have the same
time. This is because the non-rotating frame times of spatial points which are stationary in the
rotating frame & which have a given r must be an increasing function of their polar angle in the
direction of rotation, because of the Lorentz transformation time relations between relatively
moving frames. However, because the angle eventually reaches 360 degrees, one eventually
comes back to the original point, and so has the requirement that the non-rotating frame time for
each point (except the center of the rotating disc) must be bigger than itself. Thus a
non-time-varying coordinate system for the rotating disk space-time of the type my website's
derivation of the relative time rate formula depends on doesn't exist. The other requirement of
my website's article, that a gravitational potential be definable on the s-t, is also probably not
met for the rotating disk space-time, although I cannot now prove this. (The Wikipedia Born
article says that the s-t is not even Riemannian, apparently on the grounds that no global time
can be defined for it. However, this doesn't prevent it from being Riemannian, since being
Riemannian requires that a time be definable locally, on the tangent space at each point of the
s-t manifold, but doesn't require that it be definable globally.) What a fully general relativistic
analysis of the problem of relative time rate variation as a function of gravitational potential
differences which is applicable to all GR space-times in which a gravitational potential can be
defined would say, I don't know.


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