I was watching veritasium's video about the gravity not being a force. The Equation in the image states that you have to have accelaration in order to stay in your position (like being on the earth's surface). But then I thought of a man which is not moving in space where no mass is near him. He should not be accelerating because he is in a inertial frame of reference but since he is not moving he needs to have an acceleration (according to the equation). What did I thought wrong?enter image description here

  • $\begingroup$ How is the "acceleration" $a^r$ in the image defined? $\endgroup$
    – Krup'a
    Jan 16, 2021 at 12:38

1 Answer 1


If the man in space is moving only under the influence of gravity, he should move on what's called a geodesic (the analog of a straight line in curved spacetime). The equation governing this trajectory is the so-called geodesic equation, which can be thought of as the analog of $ma=0$ (the governing equation for a free particle) in Newtonian mechanics. It is given by

$$ {d^2x^\rho\over d\tau^2} + \sum_{\mu=0}^4 \sum_{\nu=0}^4 \Gamma^\rho_{\mu\nu} {dx^\mu\over d\tau} {dx^\nu\over d\tau} = 0.$$

In this equation $\tau$ is the proper time and $x^\mu(\tau)$ symbolises the coordinates of the man in spacetime at proper time $\tau$. To simplify the discussion, we can simply think of a 1+1-dimensional spacetime, where the coordinates $x^\mu$ are simply $x^0 = t$ and $x^1=r$, and the indices run from zero to one. The $\Gamma$'s are called Christoffel symbols, and they depend on both the geometry of the spacetime, as well as the coordinate system used.

If you set $\rho=1$ in the equation above, you get the equation on the whiteboard if you assume $a^r=0$ and if you only keep the term that is dominant in the nonrelativistic limit.

If I understand it correctly, the $a^r$ in the expression on the whiteboard actually refers to the right hand side of the equation $ma=F$, and we should think of $a^r$ as the relativistic counterpart of $F/m$. In any case, since the man you're talking about is not under the influence of forces except gravity, it should be set to zero.

And as I said, the Christoffel symbols depend on the geometry of the spacetime. In flat spacetime they are zero, at least if we are working in a cartesian coordinate system. This means that the right hand side of the equation on the whiteboard is actually zero, so ${d^2r\over d\tau^2}=0$, as you would expect.

Conversely, if the man were close to the earth, the Christoffel symbols would be nonzero and the equation on the whiteboard (again with $a^r=0$) would precisely give Newton's law of gravity, plus a relativistic correction.


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