# Physical meaning of frames in general relativity

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 $1/\sqrt{1-r_s/r}$.

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!