If gravity is the curvature of spacetime it should bend everything equally. To clarify my point I would like you to imagine two scenarios. Think of a bird flying in the storm while the wind is blowing sideways. Because the wind is an active force and the bird is fighting against it, the amount by which it's trajectory is bent should depend on it's momentum for a given strength of wind. Qualitatively speaking, more the momentum the less it bends. Now think of a train moving on a curved track. It should always bend by a fixed degree regardless of it's velocity or mass because the path itself is bent by that precise degree.

I am aware that in my examples I have only considered curvature in SPACE but the theory of relativity talks about curvature in SPACETIME and I am guessing that every object's trajectory through spacetime is indeed equally bent due to gravity. Can someone explain?

EDIT: After reading some comments, I have realized that perhaps my choice of words wasn't that great in the original question. I know that different object will follow different geodesic trajectories based on their initial conditions. What I was interested in is the CHANGE in trajectories due to spacetime curvature. The term I should have used in the question is acceleration. For example, the gravitational acceleration an object will experience near the surface of the earth is g regardless of it's mass or initial velocity. So it's natural to expect that every object will experience the same acceleration when they are moving sideways with respect to a gravitational body. In other words the "bending" should be equal. Hope it's a little clearer now.

  • $\begingroup$ Sorry I forgot to mention in the question I was thinking about a beam of light moving tangent to earth's orbit and wondering why doesn't it follow the same trajectory as the earth and start orbiting the sun. $\endgroup$
    – user58110
    Aug 30, 2014 at 7:21
  • $\begingroup$ The gravity is not strong enough....what you are trying to know happens only in the case of event horizons and photon spheres of black holes... $\endgroup$
    – GRrocks
    Aug 30, 2014 at 7:28
  • $\begingroup$ The bending of light was just an example. What I wanted to know was why should an object's momentum affect how much it's trajectory is curved if the spacetime itself is curved and the object is only following a straight line in that curved spacetime. $\endgroup$
    – user58110
    Aug 30, 2014 at 7:39
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    $\begingroup$ An object's momentum determines part of its trajectory even without gravity. Why should this change with gravity? $\endgroup$
    – CuriousOne
    Aug 30, 2014 at 7:40
  • $\begingroup$ Another way to think about it is if the earth slows down for some reason will it move along the same orbit or will it spiral inwards? My intuition tells me that it should spiral inwards but I can't explain why. $\endgroup$
    – user58110
    Aug 30, 2014 at 7:41

2 Answers 2


First the somewhat misleading rubber sheet analogy:

You've probably seen the bending of spacetime described as the deformation of a rubber sheet. Be careful taking this too literally as the sheet bending doesn't illustrate the bending of time, only space, and in any case it's not that good a mathematical model. Anyhow, it should be obvious that the trajectory of a ball rolling on the sheet depends on its initial velocity. A ball moving very fast won't be deflected much while a slow moving ball will be deflected a lot. In both cases the ball follows a geodesic, but the geodesic depends on the initial conditions.

Now the tl;dr version:

The geodesic is calculated using the geodesic equation.

$$ {d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0 \tag{1} $$

The $\Gamma^\mu_{\alpha\beta}$ are called Christoffel symbols (of the second kind) and they depend on the spacetime curvature. Although this looks fearsome it's just a second order differential equation (well, four such equations) and you solve it just like any other second order differential equation to give the equation for the trajectory. However, like all such differential equations, the final equation depends on the initial conditions.

To see this consider flat spacetime. If we use the usual $(t, x, y, z)$ Cartesian coordinates all the Christoffel symbols are zero. If we further assume that all motion is in the $x$ direction so we can ignore $y$ and $z$ equation (1) simplifies to:

$$\begin{align} {d^2 t \over d\tau^2} &= 0 \\ {d^2 x \over d\tau^2} &= 0 \tag{2} \end{align}$$

which is just telling us that the acceleration is zero, which it would be in a flat spacetime. However equation (2) is satisfied by all trajectories of the form:

$$ x = x_0 + vt $$

where $x_0$ is the position at $t = 0$ and $v$ is a constant velocity, and $x_0$ and $v$ can have any values. So even in this simple case objects with different initial velocities don't travel on the same geodesic trajectory.

  • $\begingroup$ thank you for your nice and thorough answer. Talking about misleading analogies, I was thinking about an object moving on the earth's surface and how it's curvature doesn't depend on it's momentum as there is nothing "forcing" it off the straight line. From the perspective of the object, it's trajectory is perfectly straight. For an outsider, the curvature is always the same regardless of it's momentum. The geodesic equation explains HOW the curvature depends on the initial conditions and HOW MUCH it should bend, but does it explain WHY? $\endgroup$
    – user58110
    Aug 30, 2014 at 8:31
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    $\begingroup$ @Rahat: in general why questions don't have an answer in physics. General relativity is a mathematical model that gives the right answers if we compare its predictions with experiment. But as to why GR should correctly describe the universe - well you'd have to ask a philosopher. $\endgroup$ Aug 30, 2014 at 9:36

No, every object's trajectory doesn't curve equally...for instance, light travels a null geodesic, which is very different from an orbiting planet. Your first example is quite correct, and the the path does depend on the momentum.But this applies to the second example as well...if the train is moving very fast, it won't bed at all. Since compared to gravity, the curvature of track is very trivial, the speed required for a train (running without any external forces or engine) will be very less to curve as the track does...IN OTHER WORDS, IF YOU TRAVEL FAST AND FAR ENOUGH, GRAVITY WILL NOT AFFECT YOU. THE CLOSER AND SLOWER YOU MOVE, THE GREATER THE BENDING OF TRAJECTORY YOU WILL EXPERIENCE. In you example, the train will curve at that precise degree if it travels at a speed slow enough to not be able to break off its trajectory. If it doesn't, the train will continue moving in a straight line(provided no external forces act). Again, centripetal and centrifugal forces play their part, but they have nothing to do with gravity. (Anyway, on a turn, you do slow down the vehicle,right??)

  • $\begingroup$ So in my train example I make it go faster and faster. At a point it will leave the track. It will go from a fixed curvature (the curvature of the track) to another fixed curvature (no curvature at all). The curvature is still not changing in line with the momentum UNLESS there is another force (friction for example). $\endgroup$
    – user58110
    Aug 30, 2014 at 7:30
  • $\begingroup$ But remember that it has nothing to do with the original trajectory....because it is being changed... $\endgroup$
    – GRrocks
    Aug 30, 2014 at 7:32
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    $\begingroup$ "IF YOU TRAVEL FAST AND FAR ENOUGH, GRAVITY WILL NOT AFFECT YOU". Nothing travels faster than light and yet light is affected by gravity. $\endgroup$ Aug 30, 2014 at 8:54
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    $\begingroup$ @Brandon Enright, please read the whole thing...I also wrote "Far enough"... $\endgroup$
    – GRrocks
    Aug 30, 2014 at 11:33
  • $\begingroup$ That still makes no sense. Gravity will still affect you. $\endgroup$
    – HDE 226868
    Aug 30, 2014 at 16:58

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