Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been reading a book on General Relativity lately (Gravitation and Cosmology, Weinberg), and I was reading about the weak field approximation. It derived the time dilation in a weak gravitational field (spherically symmetric so that $\phi=-GM/R$) as $$\frac{dt}{d\tau}=(-g_{00})^{-1/2}=(1+2\phi)^{-1/2}=\left(1-\frac{2GM}{Rc^2}\right)^{-1/2}.$$ The term itself reminded me a lot of time dilation in Special Relativity. Namely, $$\frac{dt}{d\tau}=\left(1-v^2/c^2\right)^{-1/2}.$$ And then I noticed that if you were to somehow equate these (which I know can't really be done, but just for analogy), then a particle travelling at exactly the escape velocity of the spherical body in question would experience the same dilation as a particle sitting on the surface. My question pertains as to why this is true.

Better put: Is there some symmetry of nature (possibly a consequence of the equivalence principle or something else) that I'm just not noticing, or is this just a coincidence of math (I see this possibility as unlikely)?

Please note that I'm quite the beginner in GR, so if there's something well known that I just haven't learned (read) yet, please let me know.

share|cite|improve this question
related: – Christoph Jun 15 '14 at 7:49
@Christoph That actually seems to conceptually make a lot of sense. If I'm understanding the "river model" correctly, then space is moving relative to the "stationary" observer, and so an inertial coordinate change just puts the stationary observer as moving with respect to space at the escape velocity, correct? – Bob Knighton Jun 15 '14 at 7:55
that would be my understanding as well – Christoph Jun 15 '14 at 8:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.