# Why are stationary observers accelerating in the Schwarzschild metric, intuitively?

First, an introduction.

Consider the surface of the Earth (whose mass is $$M$$): the geometry of the spacetime nearby can be described to a first approximation by the Schwarzschild Metric.

If we wish to compute the 4-acceleration of an observer which is stationary with respect to the $$(r, \theta, \varphi)$$ coordinates, we must do so with the relativistic expression $$a^\mu = u^\nu \nabla_\nu u^\mu,$$ where $$u^\mu \propto (1, 0, 0, 0)$$ is the four-velocity of the observer. Doing the computation, we get that the only nonzero component of the acceleration is

$$a^r = \frac{GM}{r^2},$$ where $$r$$ is the (fixed) value of the radial coordinate, so the magnitude of the acceleration is $$\left|a\right| = \frac{1}{\sqrt{1 - 2GM / r}} \frac{GM}{r^2}.$$

This may be used to explain why we perceive a force downwards: we are in a frame which is accelerating upwards, as per the equivalence principle.

In the computation the relevant Christoffel symbol is $$\Gamma^r_{tt} = \left( 1 - \frac{2GM}{r}\right) \frac{GM}{r^2},$$ which describes how the perturbation to the metric component $$g_{tt} = - (1 + 2 \Phi)$$ varies, that is, how much faster time passes at different elevations.

Now, to my question: is there an intuitive way to explain how this result for the acceleration comes about?

Mathematically it is clear to me how the fact that time passes at different rates at different elevations implies that stationary observers with respect to the coordinates are accelerating, but I would have a hard time explaining it without introducing the whole differential-geometric framework.

Ideally, I'm looking for a graphical explanation which could be understood without a lot of math, but a different approach to the computation could also be useful.

Edit: Such an argument might be only possible in the weak field limit: it is definitely fair for an answer to only apply in the limit $$r \gg GM$$.

• Do you mean the exact result? Or just the general idea of a "static" yet accelerated observer? – Javier May 13 at 17:50
• Well, either would be interesting really. I suspect that getting the exact expression out of a simple argument is quite hard, but maybe there is a way to show how the connection between the gravitational potential and the metric perturbation comes about... – Jacopo Tissino May 13 at 17:55
• I realize I was vague on that point, I wanted to leave the door open for different approaches. – Jacopo Tissino May 13 at 17:55

Now, to my question: is there an intuitive way to explain how this result for the acceleration comes about?

Yes. This was actually Einstein's starting point for general relativity, formulated from the equivalence principle in 1907, which he found by thinking of the twin paradox. The acceleration of the twin causes changes in the rate of time. It took him eight years to fit the maths to the intuition, but the intuition remains, imv, the best way to approach general relativity.

It is as well to start with the general principle, that the laws of physics are locally the same for all observers. General relativity is based on a simple insight. If you, the room you are in, and everything in your immediate environment, were to increase in size, and clocks were to go slower such that the speed of light remains the same, there would be no way that you could tell the difference without looking outside of your environment. There is thus no reason why a clock on a GPS satellite should keep time with an identical clock on Earth. In practice we find that it does not.

In the weak field limit we ignore variations in space components of the metric and considers only the time component. Changes in energy with position follow because energy is the time component of the energy-momentum vector. An increase in the rate of a clock corresponds directly to a loss of (kinetic) energy, which classically is understood as a gain in potential energy.

The Pound-Rebka experiment shows directly that the redshift of a photon rising in a gravitational field is due to differences in the apparent rate of clocks at different heights. The frequency of a photon is proportional to its energy, so a drop in frequency corresponds to a loss of energy.

An identical effect on frequency should be expected for the wave function of any quantum particle, meaning that the net drop in (kinetic) energy due to motion in a gravitational field is accounted by differences in clock rates. Thus in the weak field limit the classical potential energy of a gravitational field is directly dependent on clock rate.

• The energy argument was precisely what I was looking for! For future reference I'll add the explicit expressions here: the energy is given by $E = p^0 = m u^0$, but since we must normalize the 4-velocity we have $u^0 = 1 / \sqrt{1 - 2GM/r}$, so in the weak field limit we can approximate $E \approx m + GMm / r$. – Jacopo Tissino May 13 at 20:13
• Neat! I hadn't done this so simply in formulae. What I have done, is shown the complete consistency of gr with qm, in my books and in academia.edu/40217917/Mathematical_Implications_of_Relationism. – Charles Francis May 13 at 20:22

The general idea that a static observer experiences a proper acceleration is just the equivalence principle, of course. You can easily show this to yourself by downloading an accelerometer app to your phone, and placing it on a table. You will see that the app registers an acceleration of around $$10\, \mathrm{m/s^2}$$ in the upwards direction, instead of zero as you might expect. Of course, this is easily explained in Newtonian mechanics: whatever mechanism the accelerometer uses, it cannot possibly detect a force that affects every piece in the same way - it can only make measurements relative to gravity.

I don't think the exact relativistic form of the acceleration can be easily explained, though. It depends on the Schwarzschild metric, which in turn comes from the Einstein equations. Any small change to the equations will (in most cases) give a different metric, and in turn a different acceleration. Just consider a general spherically symmetric metric:

$$ds^2 = -f(r)\, dt^2 + \frac{dr^2}{g(r)} + r^2 d\Omega^2.$$

Following you calculations we arrive at a proper acceleration

$$|a| = \frac{\sqrt{g}}{f} \frac{f'(r)}{2},$$

which in general will be different from the Schwarzschild result. For example, if we take the usual weak field metric with $$f(r) = 1 - 2M/r$$ and $$g(r) = (1 + 2M/r)^{-1}$$, we arrive at

$$|a| = \frac{1}{\sqrt{1+2M/r}} \frac{1}{1-2M/r} \frac{M}{r^2},$$

which is certainly different from the Schwarzschild result. So while you might be able to come up with some simple geometric derivation, it would have to start from the Schwarzschild geometry - you can't do it from first principles.

• I see, so you are saying that the result holds as long as the field equations do, and we must consider the fact that they imply $f=g$: the variation of $g_{tt}$ alone cannot account for the expression of the acceleration. Still, in the weak field limit $r \gg M$ the difference vanishes, since $f \approx 1 \approx g$: maybe an argument could be made with this additional assumption. – Jacopo Tissino May 13 at 18:39
• @JacopoTissino, In the weak field limit, the variation in $g_{tt}$ does indeed explain the acceleration. However, I think you are asking for an answer without maths, and I will give one. – Charles Francis May 13 at 19:22