Despite studying the general theory for quite some time, this still eludes me.

The geodesic equation can be cast in the form $$ m\frac{d^2x^\mu}{d\tau^2}=-m\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}, $$ so the connection coefficients play the role of a 4-force. The nontensorial nature of this expression is due to the fact that this 4-force essentially contains all inertial "pseudo-forces", including gravity, so it is frame dependent.

It is clear that this equation relates "coordinate-acceleration" to forces as seen by some observer. The essential question is what kind of observer sees "gravitational force" this way?

I mean, in special relativity, for a global Lorentz-frame $(t,x,y,z)$, this frame represents a freely falling observer, whose "space" is described by the cartesian coordinates $x,y,z$.

Switching to a curvilinear coordinate system (in the spatial variables only) makes this less palatable, but I guess we can then employ a local perspective: The coordinate vectors $\partial_i|_p$ are the "yardsticks" of an observer at $p$. But what if we also change the direction of the time coordinate? What does that mean?

Given a general coordinate system in GR, what sort of observer does that coordinate system represent at a certain point $p$? How does he or she experience "space" and "time" from his/her perspective?

What if we use an orthonormal frame instead of a coordinate frame?

Note: I ask this question somewhat more generally but my aim is to be able to tell how some observers experience gravitational force.

For example if I describe Earth with a Schwarzschild-metric (assuming nonrotation) and there is an observer at a fixed point on the surface of the Earth, and a particle is moving freely (for example a projectile fired with some initial conditions), I want to be able to describe mathematically how this observer sees the particle move, and what force does he feel is affecting the particle.


Since my question is apparantly confusing, I interject that I believe my question would be answered incidentally, if someone told me how to handle the following problem:

Let $(M,g)$ be a spacetime where $g=-a(r)dt^2+a(r)^{-1}dr^2+r^2(d\vartheta^2+\sin^2\vartheta d\varphi^2)$ is the Schwarzchild-metric. The Schwarzchild-metric is caused by a planet of mass $M$, whose surface is located at $r=r_p$.

At some point $(r_p,\vartheta_0,\varphi_0)$ and time $t_0$ there is an observer. The observer is motionless with respect to the origin of the coordinate-system, so it's spatial positions are described by $(r_0,\vartheta_0,\varphi_0)$ all the time.

The observer carries three rods of unit length, $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$ satisfying $g(\mathbf{e}_i,\mathbf{e}_j)=\delta_{ij}$ as a reference frame.

Assume a freely falling particle's worldline crosses the observer's worldline at a point (so that the observer can make local measurements), I assume that the 3-velocity the observer would measure is simply $v^i=e^\mu_i\frac{dx^\nu}{d\tau}g_{\mu\nu}$, right? Same for all 4-tensorial quantities.

But what about the gravitational force? To calculate the connection coefficients, one must know not only the frame at a point, but also in a region around the point. So what is the mathematical expression to describe how the observer detects the force acting on the particle? How is it related to $\Gamma^\mu_{\alpha\beta}$?

If I fired a cannonball from the surface of the earth, how could I use GR to find its (3-)trajectory? The 3-trajectory I see?

  • $\begingroup$ Question is not clear. Somewhat reluctant to bring back some first principles, as this may be interpreted as obvious, rather than a request for clarification of your dilema; 1) The equivalence principle is that locally space-time appears the same for any observer in any frame of reference. Resistance to Gravitationally induced velocity changes is experienced as acceleration. Co-ordinate systems are just different ways of describing position - they have no impact on the physical consequences, but rather on how they may be derived or measured. $\endgroup$
    – JMLCarter
    Jan 25, 2017 at 13:12
  • $\begingroup$ 3) A distant observer sees the action of forces as changes (or resistance to changes) in velocity? The see an object on a geodesic to be in free fall, which can be interpreted as gravitational attraction classically, or relativistically as the curvature of space-time. $\endgroup$
    – JMLCarter
    Jan 25, 2017 at 13:15
  • $\begingroup$ @JMLCarter We, as observers, experience time and space as different things. GR phenomena might be coordinate-independent, but when we make measurements, we employ a point-of-view, which is a frame. It bothers me that I know a lot about GR as an abstract theory, but if someone gave me the excercise "Hey, let's fire a particle of mass $m$ at initial 3-velocity $\mathbf{v}$, calculate its trajectory using GR instead of Newtonian gravity, I would not be able to do it. $\endgroup$ Jan 25, 2017 at 13:15
  • $\begingroup$ An observer at rest in a freely falling frame feels no gravity. Another at rest on the surface of the Earth does. Is this a question about what "at rest" means? $\endgroup$
    – mmesser314
    Jan 25, 2017 at 13:16
  • $\begingroup$ @Uldreth I really didn't get that from your question. If you want a link to a worked example I would ask for it. Must be one somewhere. $\endgroup$
    – JMLCarter
    Jan 25, 2017 at 13:24

2 Answers 2


To calculate the gravitational acceleration you simply calculate the four-acceleration:

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

In your question you mention the geodesic equation, but the geodesic equation describes the freely falling observer i.e. the observer for whom the four-acceleration is zero. And indeed setting $\mathbf A = 0$ in (1) immediately gives us the geodesic equation:

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

So by starting with the geodesic equation you are starting by assuming the acceleration is zero and you'll never be able to calculate what you want.

The gravitational acceleration $G$ felt by an observer, i.e. the proper acceleration, is the norm of the four-acceleration for that observer:

$$ G^2 = g_{\alpha\beta} A^\alpha A^\beta $$

and since this is a scalar we can use any convenient coordinate system to calculate it. The example commonly used for students is an observer hovering at fixed distance from a spherical mass i.e. the spacetime geometry is the Schwarzschild geometry. Since $u_r = u_\theta = u_\phi = 0$ the four-acceleration is simply:

$$ \mathbf A = (0, \frac{GM}{r^2}, 0, 0) $$

And the norm is then:

$$ G = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$

Which is just the Newtonian expression modified by that factor of $1/\sqrt{1-r_s/r}$.

  • $\begingroup$ I think this answers my question (not fully but at least it opened a helpful perspective I did not consider before), so thank you, however I'd like to point out I did not say the observer satisfies the geodesic equation. My observer is a shell observer, and it is observing a freely falling particle. I realize a freely falling observer has zero 4-acceleration, however the (nongeodesic) observer will see it accelerating (if I drop a stone I see it accelerating). With that said, the actual 4-acceleration of the observer is obviously the same thing I am looking for, which I did not consider. $\endgroup$ Jan 25, 2017 at 16:46
  • $\begingroup$ @Uldreth: Oops, sorry yes. Anyhow, I'm glad this helped. $\endgroup$ Jan 25, 2017 at 17:12
  • $\begingroup$ What about the perihelion precession? Will this force $ mG = \frac{GmM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$ create the additional perturbation term as we know it with $\frac{2 GmM 3 h^2}{r^4}$? I doubt so $\endgroup$
    – Eddward
    Dec 6, 2019 at 11:40

I like this question. My understanding is that a general coordinate system does not necessarily correspond to the coordinates measured by any physical observer. However, certainly one can ask: given an observer, what coordinate system does she correspond to?

Well the 'time' coordinate of our observer must be the proper time along their wordline – this is what our observer physically measures as time. So her timelike basis vector is simply her 4-velocity. Then at any given point along the observer's worldline, we can give our observer three spacelike vectors orthogonal to each other and to the 4-velocity. The metric at a given event on the wordline is then simply the Minkowski metric.

Of course, at every point along the worldline, we are free to rotate our spacelike basis vectors at will. Which is the physical choice? Well, the wordline of an observer is not a complete description of that observer – it is also necessary to know how the observer is rotating. If we demand that the observer is not rotating, then given a choice of spacelike basis vectors at one point of the worldline, the basis vectors everywhere else are determined by Fermi-Walker transport. The failure of our basis vectors to be Fermi-Walker transported is a measure the extent to which our frame is spinning.

So now we have a basis $e_\alpha$ (which is orthonormal) defined at every point along our observer's worldline, which corresponds to the basis that our observer would physically use. With this, we can determine the value of any tensor-derived quantity our observer would find if she were to measure it. If a particle with momentum $p$ were to come flying past, its measured energy would be $e_0^\mu p_\mu$. If the ripples of an electromagnetic field were to pass through her, the measured electric field in the $x$-direction would be $e_0^\mu e_1^\nu F_{\mu \nu}$ (up to a sign).

But, to come at long last to the question at hand, what about the gravitational field? What would our observer measure for this? The problem is that the gravitational field experienced by an observer depends on derivatives of the metric (in the appropriate coordinate system). However, our choice of basis is only defined on the worldline itself, and hence we do not know what the components of the metric are away from the worldline, according to our observer.

We need to invoke some more physics.

At any instant of time, suppose you are stood next to me on the surface of our planet, experiencing the force of gravity. I, on the other hand, have just jumped into the air and am at the peak of my trajectory, instantaneously at rest; gravity for me has disappeared. We both agree on what to call 'time', since our 4-velocities agree. For me, an inertial observer, there is a physical coordinate system defined in my immediate vicinity, normal coordinates.

Physically, however, we expect that both of us should assign the same spatial coordinates to a given point in space, at that instant of time. Hence I now have a good definition of your coordinates – you assign the same spatial coordinates to a nearby point in space as does a stationary freely-falling observer (whose spacelike basis vectors are aligned with yours).

With a coordinate system defined in the vicinity of your worldline, it is a simple matter to compute the components of the metric tensor in your coordinate system, and thence its partial derivatives, and thence the gravitational field!

References: General Relativity, An Introduction for Physicists; Hobson, Efstathiou and Lasenby; Cambridge; sections 5.13 and 7.5.

If I find time later, I will try to illustrate how this works for an observer at fixed $(r,\theta,\phi)$ in Schwarzschild spacetime. Of course, I'm sure it would be worthwhile to attempt this for yourself!

  • $\begingroup$ Thanks for your answer, I'll surely try some test calculations when I get the time, but of course, I'd be thrilled to read your example if you got the time. Especially, that something is unclear to me. Do you mean that my spatial coordinates are defined by the freely falling observer's normal coordinates, but my time coordinate is different? This is the only sensible thing for me, else all connection coefficients would vanish, but still feel the need to ask, for clarity. $\endgroup$ Jan 25, 2017 at 22:07
  • 1
    $\begingroup$ Yes, that's absolutely right. I should have been more clear: at a given point on our observer's worldline, we can always find an inertial frame such that at this point, the inertial frame basis vectors and observer's basis vectors coincide. The inertial observer can then define a 'constant time hypersurface' in the vicinity of this point, and we take the time as measured by our observer to be constant over this hypersurface also. Then on this hypersurface, we choose our observer's spatial coordinates to be those of the inertial frame. $\endgroup$
    – gj255
    Jan 26, 2017 at 0:30

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