I've had a look at the answers to these sorts of questions already, but feel like I'm still missing something. Starting with this question, and this one and even this one here.

I'm looking at this answer mainly, the others are there to show this is a consistent approach being used to explain the effect.

Unless I'm mistaken it shows that if you construct a metric in polar form, and assume uniform circular motion, that the time dilation is equivalent to just taking the magnitude of the velocity (which remains constant) and deriving a gamma term. However I can't help but notice that you get the same result with uniform linear motion as well, ignoring the acceleration term. In short, it (mathematically) suggests that uniform circular motion in flat space is equivalent to constant relative motion in flat space when asking about time dilation.

This is entirely unsurprising to me so maybe I've missed something? But to me it says that the arc-length and the line-length are equally contracted due to velocity along those paths regardless of the direction of velocity, which means it willfully ignores what, if any, contribution a change of direction (acceleration despite constant speed) has to the metric? My guess is that this is in flat space so general relativity hasn't been considered, ergo no explanation how it affects space-time.

Assuming I've done something wrong, I then looked at this answer... It shows that the SR polar-form arises from the Schwarzschild metric when mass is negligible in magnitude or too far to be of much effect locally. At least that's what I've gathered reading that answer... And now I have more questions than before...

  1. The circular motion component remains unchanged in S.Metric - does this imply that space is only stretched along $r$ direction (under the same assumptions, uncharged non-rotating $M$)?

  2. The factor for $-dt^2$ has changed from $1$ to $1-\frac{2M}r$ - wouldn't this be the actual "gravitational time dilation", as it describes a quantity only affecting $ds^2$ through a local time component and only dependent on (radial) position?

  3. The factor for $dr^2$ has changed from $1$ to $(1-\frac{2M}r)^{-1}$ - I assume this term shows how space-time contracts due to motion away / towards $M$ at a given distance?

  4. How does one arrive at the $1-\frac{2M}r$ term for $-dt^2$, and why does it appear as an inverse factor for $dr^2$? Is there a deeper meaning / significance to this? Is it because a free-falling object with only radial motion must appear locally contracted to a stationary observer it falls past only because of its downward motion relative to the stationary observer?

  5. More broadly, can the equivalence principle explain why acceleration due to gravity causes time dilation separate from motion within a gravitational field, while the same does not hold for uniform circular motion in flat spacetime? Ie, how does the EP hold when comparing gravitational time dilation to uniform circular motion?


1 Answer 1


First off, traveling at constant velocity in flat spacetime is not the same as traveling g in a uniform circular motion. Quite the contrary, free falling towards the gravitational source is actually equivalent to moving with constant velocity in flat spacetime. This is so because the objects are following a geodesic path defined by the geodesic equation. I must add that the following explanations are done under the assumption that the test particle is a point particle.

For your first question, no. Spacetime is also deformed in the temporal components, in addition to the radial term. I understand that the question asked for the space component only, but it is useful not to make a distinction between the two, space and time.

Next, the terms in front of -dt are coincidentally equal to the inverse of the dr term. They arise naturally without any prior assumptions imposed in the derivation other than satisfying the vacuum equations. There are many solutions where indeed the coefficient of the -dt term is not the inverse of the dr term, the Kerr metric is such an example.

As for your questions on time dilation and length contraction in curved space, they can be written in a similar manner to the Lorentz transformation in flat space by replacing the Lorentz factor with its curved space counterpart. Simply, the Lorentz factor γ is replaced with

$\gamma$ = $\frac{dt}{d\tau}$

Whereby τ is the proper time. This then makes the $\gamma$

$\frac{1}{\gamma}$ = $(-g_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt})^{1/2}$

which is then used to calculate time dilation or length contraction of a test particle in the vicinity of a massive object.

I have a little bit of trouble comprehending your fifth question. The effects of moving through space, be it flat or curved, can be deduced entirely from the metric itself. Acceleration of the particle due to gravity is zero, following the equivalence principle.

  • $\begingroup$ Starting to get the idea, have some followups though... re #1 - would space-time contract along the surface of a sphere of constant r (measured from afar) due to a gravitational field being present inside it or if it just seems that way from geodesics converging? re #2-4 - "coincidentally equal" is what makes me unsure; "They arise naturally without any prior assumptions imposed in the derivation other than satisfying the vacuum equations." I want to press this, because S.Metric does have assumptions that the solution is uncharged and non-rotating... re #5 I will add another comment shortly $\endgroup$ May 18, 2015 at 23:46
  • $\begingroup$ re #5 - thinking on it, I don't think answering it will add much without asking an entirely different question; your answer covers the gist of what I was trying to ask. $\endgroup$ May 19, 2015 at 1:38
  • $\begingroup$ Yes I agree. It was a mistake on my part to say that there were no other prior assumptions. Other than that though, my answer should basically cover what you asked for. $\endgroup$
    – Horus
    May 20, 2015 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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