According to Einstein's general relativity, a free object traces a geodesic in the 4d curved spacetime. But what we can observe is the projected path of the object onto the 3d space, which is itself a Riemannian manifold, with the inherited Riemannian metric from the spacetime.

Is the projected geodesic onto the 3d space still a geodesic?

We know in a flat space, a straight line projected to a subspace remains straight. But I'm not sure whether this is true for a geodesic in a curved space. I guess my question is how the geodesic equation in the spacetime $$\frac{d^2x^{\mu}}{ds^2} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{ds}\frac{dx^{\beta}}{ds}=0$$ changes when we only look at components $x^1,x^2,x^3$ (with $x^0 $ being the time $t$) in the 3d space.

Edit 1:
I forgot to mention my background. I am familiar with Riemannian geometry, but I haven't studied general relativity yet besides some casual reading here and there. But this question was stuck in my head long enough so I guess this is the time I let the demon come out. I will appreciate if you can explain your answer in a level suitable for a beginner in general relativity.

  • $\begingroup$ What is the spatial metric? If $g_{ab}$ depends on time, there is no definite spatial metric. In this case your question is meaningless. If $g_{ab}$ does not depend on time, what do you mean for spatial metric? $g_{ab}$ with $a,b=1,2,3?$. It is disputable if $g_{0,a}\neq 0$... $\endgroup$ Nov 19, 2019 at 14:37
  • $\begingroup$ @ValterMoretti I'm not sure if I understand your concern. But we can always pullback (i.e.,restrict) the Riemannian metric to a submanifold. Essentially, we are saying an inner product on a (tangent) space is still an inner product when restricted to a subspace. $\endgroup$ Nov 19, 2019 at 18:19
  • $\begingroup$ The term "projection", is somewhat of an "umbrella term". Can you please clarify what you mean by "projection". E.g.: "Stereographic projection", "Cylindrical projection" etc. $\endgroup$
    – Matrix001
    Nov 20, 2019 at 6:08
  • $\begingroup$ @Matrix001 This is a valid question. What I had in mind was a neighborhood of a point where the spacetime has coordinate $(t,x,y,z)$ with a Riemannian metric defined there, then the projection is the restriction to the 3d space $(x,y,z)$ with the restricted Riemannian metric. Please let me know if there still is some confusion or any conceptual mistake in general relativity. I forgot to mention that I haven't seriously studied general relativity yet. So I may have made some silly mistake or assumption in my question. $\endgroup$ Nov 20, 2019 at 9:29
  • $\begingroup$ Thankyou for the clarification! In that case, it is not a "projection", but rather a "3-dimensional slice" of the 4-Dimensional curved space-time, at specific point in time. By finding the first-order partial derivative (in respects to x,y and z) of the function describing the 4-Dimensional curved space-time, you get the function for the "sliced" three-dimensional space. You can compare these functions, and plot them, to see the differences. I hope this extra information is useful! $\endgroup$
    – Matrix001
    Nov 20, 2019 at 9:47

2 Answers 2


But what we can observe is the projected path of the object onto the 3d space, which is itself a Riemannian manifold, with the inherited Riemannian metric from the spacetime.

Not true. To get a 3d space, you have to pick a surface of simultaneity. GR doesn't in general have a preferred surface of simultaneity.

Is the projected geodesic onto the 3d space still a geodesic?

No. For example, the earth orbits the sun, and the sun's field is approximately static, so that there is a preferred time-slicing. With this time-slicing, space is nearly flat, and the earth's orbit is an ellipse, which is not a geodesic.

  • 1
    $\begingroup$ What do you mean by surface of simultaneity? In your example, the surface of simultaneity is the 3d space near the sun? $\endgroup$ Nov 19, 2019 at 18:21
  • $\begingroup$ @YuhangChen: A surface of simultaneity is a set of points that are all simultaneous. In special relativity, a surface of simultaneity is a 3-dimensional hyperplane that is orthogonal to the observer's four-velocity. $\endgroup$
    – user4552
    Nov 19, 2019 at 20:05
  • $\begingroup$ I'm not sure what you mean by "a set of points (in 3d space?) that are all simultaneous". The term "four-velocity" is also alien to me. Is there any introductory book in GR that explains this and your example in details that I can refer to? Anyway, I accepted this answer. It cleared my doubt (maybe not fully but I guess that's due to my gap in the understanding of GR) in a simplest possible way. $\endgroup$ Nov 20, 2019 at 21:16

The spatial metric associated to some familly of local observers is the following (in some coordinates system. I'm using signature $\eta = (1, -1, -1, -1)$): $$\tag{1} h_{\mu \nu} = u_{\mu} \, u_{\nu} - g_{\mu \nu}, $$ where $u^{\mu}$ are the components of the 4-velocity of the local observer. Components (1) define a projector: $$\tag{2} h_{\mu \lambda} \, h^{\lambda}_{\; \nu} = h_{\mu \nu}. $$ Also: $h_{\mu \nu} \, u^{\nu} \equiv 0$ and obviously $h_{\mu}^{\; \lambda} \, g_{\lambda \kappa} \, h^{\kappa}_{\; \nu} \equiv h_{\mu \nu}$. The 3D spatial section defined with this lower metric depends on the familly of observers you select.

You could then compute the 3D Riemann tensor on that lower space. The spacelike geodesics of that lower space have nothing to do with the timelike geodesics of the full 4D metric, despite that (1) is a projector.

Notice that the lower metric (1) isn't compatible with the full connection: $$\tag{3} \nabla_{\mu} \, h_{\lambda \kappa} = \nabla_{\mu} (u_{\lambda} \, u_{\kappa}) \ne 0. $$ To define your lower Riemann tensor and also the 3D geodesics, you'll need a new connection $\tilde{\Gamma}_{\mu \nu}^{\lambda}$ from $h_{\mu \nu}$ such that $$\tag{4} \tilde{\nabla}_{\mu} \, h_{\lambda \kappa} = 0. $$


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