When the mass of a planet causes the curvature of space-time we see that an approaching free-falling object deviates its path towards the planet. We also see the amount of that deviation depends on it's speed so that slowly moving objects hit the planet while faster ones pass it by.

Obviously the same curvature of space-time is experienced by both objects and so it would seem that both of them must follow the same geodesic (i.e. shortest path through curved space-time) irrespective of their speed relative to the planet.

Is there a relatively easy way of explaining this apparent contradiction without a deep study of differential geometry (for example)?

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    $\begingroup$ You say "both of them must follow the same geodesic". This is wrong. Even if they start in the same place and go in the same direction, if they have different speeds then they follow different geodesics. $\endgroup$
    – Mike
    Commented Jul 22, 2016 at 10:51
  • $\begingroup$ Thanks Mike. I acknowledge that my statement "both of them must follow the same geodesic" is wrong. Thanks also to the other respondents who have greatly enhanced my understanding of the subject. $\endgroup$ Commented Jul 22, 2016 at 13:11

3 Answers 3


Start by considering the ordinary Newtonian gravity. This tells us that the acceleration of a test mass due to our planet of mass $M$ is:

$$ a = \frac{GM}{r^2} $$

The acceleration is the rate of change of velocity with time. A fast moving object spends less time near the planet than a slow moving object so its velocity changes less. That means fast moving objects are deflected less than slow moving ones.

Since general relativity reduces to Newtonian gravity when the gravitational fields are small (i.e. everywhere that isn't near a black hole) this also explains why fast moving objects deflect less than slow moving ones in GR.

Showing this rigorously does involve some differential geometry, but I think it's possible to grasp the principle without getting too deeply embedded into the maths. The trajectory followed by the freely falling test mass is described by the geodesic equation:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{1} $$

This isn't as complicated as it looks (well, not quite!). The left hand side is sort of an acceleration, and the symbols $\Gamma^\alpha_{\,\,\mu\nu}$ are the Christoffel symbols that describe how curved spacetime is. In flat spacetime using the usual $(t,x,y,z)$ coordinates the Christoffel symbols are all zero and our equation becomes:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = 0 $$

which is just telling us that in flat spacetime the acceleration is zero i.e. the object travels in a straight line.

In curved spacetime the Christoffel symbols are not zero so we get a non-zero acceleration and the trajectory will be curved, but we still need to explain why the trajectory is different for different velocities. That is simply due to the term $U^\mu$, which is the four-velocity.

So the geodesic equation tells (a) that the path isn't a straight line in curved spacetime and (b) that the amount the path curves by is dependent on the four-velocity $\mathbf U$ of the test mass. That's why test masses moving at different velocities follow different paths.

  • $\begingroup$ Thanks for the link explaining geodesic equations. From that I found another useful closely-related link: Geodesics in GR $\endgroup$ Commented Jul 22, 2016 at 11:47

It is not true that there is a unique geodesic through every point. To understand it, imagine a point on a sphere (where geodesics are just great circles) or even on a plane (here geodesics are straight lines). Through that point you can draw infinitely many geodesics, e.g. infinitely many straight lines passing through this point. However, if you restrict to small piece of your geometric space (no need to do this in flat case, though) geodesic will be uniquely specified by specifying two nearby points that it should connect. If you draw a small circle on a ball, choose two points inside then you can choose precisely one trajectory going on a great circle which goes through those two points without leaving this circle. Since any two nearby points specify unique trajectory you can consider a limit in which their separation approaches zero. So a geodesic is specified uniquely by saying what is $\vec x (t) $ and $\vec x (t+ \mathrm d t)$, or equivalently - both position and velocity at initial time.

Moreover, curvature you feel when you make a trip in curved space can, and usually will depend on the direction in which you go (which in space-time terms used in GR is exactly the same as your velocity). To see this, consider a bug going around a saddle. There are some directions which curve "upwards", some direction which curve "downwards" and by continuity there must exist intermediate direction in which saddle is not curved at all. I encourage you to try to draw this on a piece of paper, or just google "saddle point" and go into graphics.


The attractivity effect depends on the time during the two masses are at proximity. You can experiment it in sending a ball through a water fall or a water flow at different speeds and look the direction the ball takes after crossing the water.

The representation of space time curvature by a deformated plane around a planet is a farce. Try to represent a hole in space, just for fun ! Space time has four dimensions : 3 for the space plus one for the time.

You can experiment the space time in measuring a box with a measuring stick : move the object and the stick : 1 : at the same relative speed 2 : at different relative speed When the first limit of the box is facing the zero of the stick, you look the measurement value on the stick where is the second limit of the box.

What is the length of the box in the case 1, and in the case 2 ?


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