Gravity, according to the General Theory of Relativity, is simply the curvature of space-time. Objects in the universe move through space-time in geodesic paths. Also, the most interesting part is that it is impossible to curve/ warp space without having an effect on time. They are intricately connected. Space warps can be notably seen near black-holes (gravitational lensing) and time dilations are so significant that even GPS systems on Earth have to adjust for it.

But my main concern is the difference in the way objects and light behave when subjected to curved space-time. Projectiles follow parabolic paths in uniform gravitational fields. This can be shown using Newton's law of gravitation, but time dilation can also be used to prove this. But doesn't curved space also be needed to account for? Why is time dilation the only significant factor here?

And what about light? I know that light bends when subject to curved space-time, but which part of space-time curvature is more responsible for this phenomenon?

I guess that since light travels at the max limit, time is effectively not running for light from our frame of reference, so light shouldn't be affected by time dilations. Does this mean that light is only affected by space curvature?

Any help to rid me of these confusions is greatly appreciated :)

  • 1
    $\begingroup$ What do you mean by "This can be shown using Newton's laws, but time dilation can also be used to prove this."? How exactly can time dilation be used to derive parabolic trajectories (which are only an approximation that has Newtonian, as well as tidal, corrections)? $\endgroup$ – Bob Knighton May 13 '19 at 8:50
  • $\begingroup$ What I meant was that if we were to try and figure out the path of an object thrown at an angle in a uniform gravitational field not using arguments of kinematics but rather using the principle "objects always move from one place to another so that a clock carried on it gives the longest possible time as compared to any other trajectory (with, of course, the same starting and finishing conditions)," then the answer would still be the same - a parabola. $\endgroup$ – Apekshik Panigrahi May 13 '19 at 9:03
  • $\begingroup$ So in this alternative proof, we must consider how time slows down in a lower point in a gravitational field (and so we must account for the time dilation). @BobKnighton $\endgroup$ – Apekshik Panigrahi May 13 '19 at 9:05
  • $\begingroup$ I also realized I had written Newton's Laws instead of Newton's Law of Gravitation. Sorry for that (and thanks for pointing that out!) @BobKnighton $\endgroup$ – Apekshik Panigrahi May 13 '19 at 9:08
  • $\begingroup$ There are some major misunderstandings in the question. GR replaces the gravitational force field with space-time curvature. $\endgroup$ – ggcg May 13 '19 at 11:09

Newton limit is approximation of GR in weak fields and SMALL velocities.

Small velocities means, that whole 4-velocity of a particle is basically in time component. So you can imagine, that if spacetime is curved same in all directions, than the time component is most significant simply because the particle almost doesn't move in space at all.

To be more precise:

The spacetime around spherically symmetrical field is given by Schwarzschild metric (in natural units): $$ ds^2=-\left(1-\frac{r_s}{r}\right)dt^2+\left(1-\frac{r_s}{r}\right)^{-1}dr^2+r^2d\Omega\approx ds^2_{flat}+\frac{r_s}{r}(dt^2+dr^2) $$ where $r_s$ is Schwarzschild radius and $ds^2_{flat}$ is Minkowski part (flat spacetime part) of the metric.

As you clearly see, the perturbation of flat spacetime metric has same magnitude in time component as in space component in natural units.

But now, let us compute geodesics. The geodesics equation is given by: $$ a^\mu=-\Gamma^\mu_{\alpha\beta}v^{\alpha} v^{\beta} $$ where $a^\mu$ is 4-acceleration of a particles, $v^\mu$ its 4-velocity and $\Gamma^\mu_{\alpha\beta}$ is Christoffel symbol. Now, the relevant Christoffel symbols for radial motion are $\Gamma^t_{\alpha\beta}$ and $\Gamma^r_{\alpha\beta}$ of which nonzero are only: $$ \Gamma^t_{tr}=\Gamma^t_{rt}\approx -g_{tt,r}/2 $$ $$ \Gamma^r_{rr}\approx g_{rr,r}/2 $$ $$ \Gamma^r_{tt}\approx -g_{tt,r}/2 $$ and all of them are of same order since perturbations of metric components $g_{tt}$ and $g_{rr}$ are of same order (in fact $g_{tt,r}=g_{rr,r}$). So the geodesic equation for radial motion in weak field of spherically symmetric source is: $$ a^t=-\Gamma^t_{\alpha\beta}v^{\alpha} v^{\beta}\approx g_{tt,r}v^{t} v^{r} $$ $$ a^r=-\Gamma^r_{\alpha\beta}v^{\alpha} v^{\beta}\approx g_{tt,r}v^{t} v^{t}/2-g_{rr,r}v^{r} v^{r}/2=g_{tt,r}/2 $$ Where I have used $g_{tt,r}=g_{rr,r}$ from the metric and $v^{t}v^{t}-v^{r}v^{r}=1$ from normalization.

Having 4-acceleration we can get radial 3-acceleration component ($a^r_3$) using: $$a^r=a^t v^r/\gamma+\gamma^2 a^r_3$$ where $\gamma$ is Lorentz factor.

Now this doesn't lead to Newtonian gravitation law without assumption, that velocities are small. With this assumption $\gamma\approx 1$, $v^t\approx-1$, $v^r\ll 1$ and $v^\mu\approx (-1,\vec{v})$ and the equation simplifies further: $$a^r\approx a^t v^r+a^r_3 => a^r_3 \approx a^r - a^t v^r$$ Substituting from geodesic equation: $$ a^r_3\approx g_{tt,r}/2 - g_{tt,r}v^{t} (v^{r})^2=g_{tt,r}/2+o((v^{r})^2)\approx r_s/(2r^2)=-GM/r^2 $$ with $M$ being mass of the source, as Newton gravitation says. So the approximation is not that space-components of curvature can be neglected, it is in the fact that space-components of 4-velocity can be neglected.

| cite | improve this answer | |

Light is affected by both effects of gravity, you can count for time dilation and curvature too, this is the Shapiro effect.

When light passes next to the Sun, its speed measured from Earth will be less then c because:

  1. it moves in curved spacetime

  2. clocks near the Sun tick slower (compared to clocks on Earth)

Please see here:


| cite | improve this answer | |
  • $\begingroup$ That link is gold for me. Thanks! But what I don't seem to understand is why does time dilation also affect light. I think the time component of the 4-velocity of light is 0, so won't light be unaffected by time dilation? $\endgroup$ – Apekshik Panigrahi May 14 '19 at 3:25
  • $\begingroup$ @ApekshikPanigrahi you are correct, light moves in the time dimension with speed 0. Now, when in a gravitational field, the space component changes, due to curvature, but the magnitude of the four vector needs to be c always. So the time component needs to compensate, thus the time component will change. Light will seem to move in the time dimension, which will mean, that when viewed from far away, light will seem to start experiencing time as we do, and thus time will seem to tick slower next to the Sun (relatively, when compared to clocks on Earth) where the light passes. $\endgroup$ – Árpád Szendrei May 14 '19 at 16:55
  • $\begingroup$ Thank you yet again :) That explanation helped me get a much clearer intuition of the subject matter. One last request - Could you cite some online article(s)/books regarding this particular aspect so that I can also have a quantitative grasp of this theory? $\endgroup$ – Apekshik Panigrahi May 14 '19 at 18:02
  • $\begingroup$ Could u answer this question (it is very much related to what u have answered)? physics.stackexchange.com/questions/479946/… $\endgroup$ – Apekshik Panigrahi May 14 '19 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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