1
$\begingroup$

I have recently learned briefly that an accelerating frame is equivalent to a gravitational field in the equivalence principle. I'm sitting on a chair and I am experiencing the force of gravity and its gravitational field. I am in an accelerating frame right?

However, the net force on me is zero because the force of the ground and the force of gravity cancel each other out, thus I am not accelerating. However, there still is gravity constantly acting and the gravitational field is all around me. Shouldn't I still be accelerating? And if so, what am I accelerating towards?

$\endgroup$
  • $\begingroup$ If you don't consider a gravity to be a force - which many well-tested theories do not - then the only force on you is the normal force from the ground, so you do accelerate upwards. If the ground vanished, you would follow a geodesic, non-accelerating path. $\endgroup$ – Asher Oct 23 '15 at 7:24
3
$\begingroup$

The equivalence principle can be framed in lots of ways, but in this context it says there is no local experiment that can tell the difference between acceleration and a gravitational field. Suppose I put you in a small box that you can't see out of and I give you an accelerometer that you find reads 9.81 m/sec$^{2}$. You won't be able to tell whether you are stationary on the surface of Earth or blasting through outer space on a rocket accelerating at 9.81 m/sec$^{2}$.

Note the use of the term local. The gravitational acceleration on Earth varies with height because it's given by $GM/r^2$, so on Earth the accelerometer will have a slightly lower reading at the top of the box than it does at the bottom. When we say a local experiment we mean that the scale of the experiment is small enough that the variations in the gravitational field across your box are too small to be measured.

Anyhow, in relativity (both special and general) we measure the acceleration using a quantity called the four-acceleration. This is a frame independant quantity i.e. all observers in all frames will agree on its value (though its representation will be coordinate dependant). The four-acceleration measured in the accelerating observer's rest frame is just the value shown on an accelerometer held by the accelerating observer i.e. in your case it is 9.81 m/sec$^{2}$ (strictly speaking the norm of your four acceleration is 9.81 m/sec$^{2}$ - it is the norm that is an invarient).

The four-acceleration is defined relative to an observer in an inertial (non-accelerating) frame. Suppose you drop the accelerator I gave you earlier. The accelerator is now in free fall so it will read zero. Your four-acceleration is defined relative to a freely falling object like the dropped accelerometer, and since you observe it to be accelerating away at 9.81 m/sec$^{2}$ that means your four acceleration must be 9.81 m/sec$^{2}$.

If you want to make this quantitative, your four acceleration is calculated using the geodesic equation:

$$ {d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0 $$

The quantities $\Gamma^\mu_{\alpha\beta}$ are called the Christoffel symbols and they describe the curvature of spacetime. In flat spacetime the Christoffel symbols are zero$^1$ and the equation becomes:

$$ {d^2 x^\mu \over d\tau^2} = 0 $$

which is basically just Newton's first law i.e. the acceleration is zero if no forces are acting.

Calculating your four-acceleration when sitting in your chair is straightforward as general relativity goes, and twistor59 does this in his answer to What is the weight equation through general relativity?. It turns out that (the norm of) your four-acceleration is:

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

which you should recognise as Newton's equation for the gravitational force with a correction factor due to relativitic effects.


$^1$ someone is bound to point out that the Christoffel symbols in flat spacetime are non-zero if you're using curved coordinates e.g. polar coordinates. OK, OK, the Christoffel symbols in flat spacetime are zero when you're using Cartesian coordinates.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.