In the general relativity, a local inertial frame is mentioned. The local inertial frame is a notion, which is related to (or represents) the equivalent principle. Here, I understand that, in the Schwarzschild space, description by a free falling observer is actually a collective result, which is made on a series of local inertial frames defined along a trajectory or geodesic line of the free-falling observer, rather than on a single local inertial frame.

[Supplementary explanation]

Here, the expression “a series of local inertial frames” is used to refer to frames, each of which is Euclidean but whose lengths are different from each other by the metric tensor. For example, in the Schwarzschild metric, the length of dx' is constantly changing in a ratio with respect to a particular length dx in a chosen coordinate system, as a radial coordinate r changes. That is, we can define a local inertial coordinate system for each r on a trajectory, but the lengths of dx' in such inertial coordinate systems will change as r changes. The expression “a series of local inertial frames” is used in this sense. (If this expression is inappropriate, please let me know a more appropriate expression.)

(Although, because of ignorance of differential geometry, it is not certain, this question seems to be related to the affine connection in mathematics.)

Is my understanding correct?


The free-falling observer is accelerating with respect to an observer in an inertial frame (e.g., of zero gravity). I understand that, in the general relativity, effects associated with the acceleration of the free-fall observer are not described in an explicit manner, but in an implicit manner (e.g., by specifying one of the series of local inertial frames or by specifying a position of the free falling observer).

[Supplementary explanation]

This second question is related to the above argument that the lengths in the series of local inertial frames change as the radial distance r changes. That is, the expression "specifying a position of the free falling observer" is used to explain that, by specifying the radial distance r, we can specify which of the local inertial frames the free-fall observer is passing through.

Is my understanding correct?

Any suggestions or recommendations are welcome!

  • $\begingroup$ In GR, the frame of the freefalling body is inertial, the frame of the observer at rest on the surface of the Earth is not inertial, even if we neglect the rotation & revolution of the Earth. $\endgroup$ – PM 2Ring Jun 20 '19 at 8:20
  • $\begingroup$ @PM2Ring Thank you for your comment. Do you mean that I should say that the lab frame is positioned at a position infinitely far away from the center of mass, not on the earth inertial. $\endgroup$ – SOQEH Jun 20 '19 at 8:48
  • $\begingroup$ Ok, if the lab frame is in a region of zero gravity, then it's inertial. $\endgroup$ – PM 2Ring Jun 20 '19 at 8:53
  • $\begingroup$ @PM2Ring Thank you. I revised the question, in consideration of your comment. $\endgroup$ – SOQEH Jun 20 '19 at 9:00
  • $\begingroup$ Related: physics.stackexchange.com/questions/3193/… $\endgroup$ – PM 2Ring Jun 20 '19 at 10:16

Suppose we choose some coordinates. These can be any coordinates that are convenient such as Schwartzschild, Gullstrand-Painleve, Kruskal-Szekeres or whatever. If we denote your position in these coordinates $(x^0, x^1, x^2, x^3)$ then we can differentiate twice to get your acceleration in these coordinates:

$$ \frac{d^2x^\alpha}{d\tau^2} $$

This is called the coordinate acceleration and it's basically the same as the acceleration we are used to from Newtonian mechanics except that we differentiate wrt proper time not coordinate time, and we include the time coordinate i.e. we include $d^2t/d\tau^2$.

We can also write down an expression for the curvature of the spacetime in our chosen coordinates, and in particular we can compute the Christoffel symbols:

$$ \Gamma^\alpha{}_{\mu\nu} $$

Then the four acceleration is the sum of the two terms:

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

where $\mathbf U$ is the four velocity expressed in our chosen coordinate system. The point of this is that for a free falling observer, i.e. in an inertial frame the four acceleraion is zero, and substituting this in equation (1) gives us the geodesic equation:

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

And the trajectory of your free fall in my chosen coordinates is just the solution to this equation (2).

But we are free to choose any coordinate system we want, and we can choose coordinates that make your coordinate acceleration zero - this is just your rest frame. Alternatively we can choose coordinates that make the Christoffel symbols zero - these are the Fermi normal coordinates. For a free falling observer equation (2) tells us that the two coordinates are the same i.e. that the Fermi normal coordinates are the rest coordinates of a freely falling observer.

This is the equivalence principle i.e. that by changing our coordinates the four acceleration can be made to appear purely coordinate, purely gravitational or some combination.

So if I understand your question 1 correctly you are alluding to the fact that the locally inertial frame is the Fermi coordinate system and this does indeed change along the path i.e. the transformation between my (stationary) coordinates and your (Fermi) coordinates changes as you fall. But then this is of course true in Newtonian physics.

I'm not sure what you are asking in your second question - perhaps you could clarify it - but it seems to me you're asking the same point in a slightly different form so hopefully the above discussion answers your second question as well.

  • $\begingroup$ Thank you for your kind answer. The Fermi normal coordinates mentioned by you is an unfamiliar word to me, and thus, it is hard to judge for now whether it is directly related to my curiosity. Anyway, from your answer, I feel that my questions was vaguely written to cause confusion. So I added a supplementary explanation to my question. If you would like to comment on this, I would appreciate it. $\endgroup$ – SOQEH Jun 20 '19 at 13:31

Imagine a series of concentric paper thin onion skins (spherical shells) surrounding the earth and centered on the center of mass of the earth. These skins extend all the way to the center of mass point. There is one special shell which is the surface of the earth itself (assume a perfect non rotating sphere). The free falling observer will not be able to pass beyond this shell and his weight will manifest as a force on a weight scale. Each shell above this shell will register a smaller and smaller weight if it were possible to attach a scale there. But without the scale the observer in free fall will feel nothing. In reality, per Einstein, the shells are infinitely thin and merge to a continuum.

The reason the observer will accelerate is because space is convergent on the center of mass and the acceleration will be necessary to maintain momentum. In Newton, space is not convergent, its parallel and so momentum will be maintained at constant speed. Note that we consider only the convergent vector here and not any orbital vector so the motion will be a straight line to the center of mass and not a geodesic. Also note that the acceleration is initiated due to the shell at the observers feet being slightly smaller (greater gravitational pull) than the one at his head.

  • $\begingroup$ What? In GR, a body in freefall is following a (timelike) geodesic. $\endgroup$ – PM 2Ring Jun 20 '19 at 8:16

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