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Can we say that an object at rest on the surface of the Earth accelerates in the direction of the ct-basevector, as viewed by an observer standing on Earth? Obviously, there is no acceleration in the space direction.

Il Guercio
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2 Answers2

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In general relativity the four-acceleration has two terms:

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

The first term is the coordinate acceleration i.e. how the position is changing with time in whatever coordinate system we have chosen. This is what Newton meant by acceleration.

The second term is the curvature term. Loosely speaking we can call this the gravitational part of the acceleration. The symbols $\Gamma^\alpha{}_{\mu\nu}$ are the Christoffel symbols that collectively describe how the coordinates we are using are curved.

The important thing to understand about this equation is that neither of the terms on the right are tensors so they are dependent on the coordinates we are using. Either of the terms on the right can be made zero by a suitable choice of coordinates, so the four acceleration can appear to be all coordinate, or all gravitational, or a mixture of both.

You are asking about the form of the acceleration in the rest frame coordinates of an observer standing motionless on the surface of the Earth. In these coordinates the position is constant and time flows steadily at on second per second, so the coordinate acceleration is zero. That means equation (1) reduces to:

$$ A^\alpha = \Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu \tag{2} $$

That is, in these coordinates the acceleration is purely gravitational, and as Dale says in his answer, if we calculate the four acceleration we will get a four-vector in the $\hat r$ direction. In fact the calculation is done in What is the weight equation through general relativity? and the result is:

$$ \mathbf A = \left(0, \frac{c^2r_S}{2r^2}, 0, 0 \right) $$

This seems unintuitive because it seems that we are accelerating without moving, but this is unintuitive only because we are so used to Newtonian mechanics where the curvature term does not exist.

John Rennie
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  • Thanks! Great answer! So for a freely falling observer the object on Earth has a coordinate acceleration, while the second part is zero (because space seems flat for that observer)? Will an object that accelerates upward undergo an acceleration through time also (since it passes positions with different rates of time)? Is this the reason one uses the "falling space"model? – Il Guercio Aug 24 '23 at 09:44
  • What does the four-acceleration of the object on Earth look like for a freely falling observer? The same? – Il Guercio Aug 24 '23 at 09:52
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You speak of acceleration in the time direction. So this is referencing the four-acceleration which is a covariant measure of acceleration. The magnitude of the four acceleration is the magnitude of the proper acceleration, which is the acceleration measured by an accelerometer.

An object at rest on the surface of the earth has a proper acceleration of $g$ in the upward direction. The four acceleration is spacelike, not timelike.

Dale
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