# In general relativity, why is Earth able to accelerate?

I was told and convinced that gravity is not a force, and in free fall you're an inertial frame and experience no force, and when on the surface of Earth you would be accelerating upwards.

What I admit I can't understand is Earth accelerating, why so? what property of mass explains this? Is there a relationship between an object's mass and its acceleration?

This is quite a common confusion for students new to general relativity, and it's because in GR the term acceleration means something slightly different from its everyday usage.

In everyday usage we say an object is accelerating if its position is changing in a non-linear way with time. In GR we call this the coordinate acceleration because it's the change in the object's coordinates $$(x, y, z)$$ with time. However in GR we use the four-acceleration. In the rest frame of the accelerating object this is equal to the proper acceleration, which I describe in my answer to your previous question: What does it mean that a falling mass in space doesn't sense any force? But more generally it is given by:

$$a^{\mu}= \frac{\mathrm d^2x^\mu}{\mathrm d\tau^2}+\Gamma^\mu_{\alpha \beta}u^{\alpha}u^{\beta} \tag{1}$$

This looks scary, but it's simpler than it appears. The first term $$\mathrm du^\mu/\mathrm d\tau$$ is the coordinate acceleration i.e. the change in the position with time. But note that we have that second term $$\Gamma^\mu_{\alpha \beta}u^{\alpha}u^{\beta}$$ and that tells us the contribution to the four acceleration due to the curvature of spacetime.

In our previous discussion I said you had a proper acceleration of $$1g$$ even when you were standing stationary on the surface of the Earth. Since you are stationary your coordinate acceleration is zero, but that second term $$\Gamma^\mu_{\alpha \beta}u^{\alpha}u^{\beta}$$ is non-zero, and that means your four-acceleration is $$1g$$ even though you and the surface of the Earth are not moving.

This may seem like a silly way of defining acceleration but there is an important principle at play here. The coordinate acceleration depends on your choice of coordinates. For example suppose you leap out of an aeroplane, then the observer on Earth would say the Earth is stationary while you are accelerating downwards at $$1g$$. But in your rest frame you would say you are stationary and it's the Earth that is accelerating up towards you at $$1g$$.

So who is correct? Well both of you are correct! Although you and the guy on the Earth have different opinions about your coordinate acceleration, when you use equation (1) you will both get the same four-acceleration - in fact both you and the guy on Earth will calculate your four-acceleration to be zero. This happens because the two of you will calculate the curvature term $$\Gamma^\mu_{\alpha \beta}u^{\alpha}u^{\beta}$$ differently and the difference in the curvature term balances out the difference in the coordinate term to give $$a = 0$$ for both observers.

And the same applies to the surface of the Earth. In your rest frame the Earth is accelerating towards you so its coordinate acceleration is $$1g$$. For the guy on Earth the coordinate acceleration of the Earth is zero but the curvature term $$\Gamma^\mu_{\alpha \beta}u^{\alpha}u^{\beta}$$ comes out as $$1g$$. So both of you calculate the four-acceleration of the Earth to be $$1g$$ even though you disagree about whether the Earth is moving or not.

• Your answers are nothing short of perfect, and I'm very grateful Jun 2 at 9:15
• Great answer. I think one more thing we can note is that, in Newtonian gravity, we have a choice in modeling Gravity as a force or as geometry, which leads to two possible definitions of "geometry of spacetime", which leads to two possible notions of "co-ordinate invariant proper acceleration". If we choose to model gravity as a force, then we define the geometry as flat, and so we must describe freely falling objects as having non-zero proper acceleration. But in GR, we are compelled to model gravity as geometry, which leads to only one notion of "co-ordinate invariant proper acceleration" Jun 2 at 16:50
• But one thing is true even in Newtonian mechanics regardless of our choice of definition of geometry : the equivalence principle. An observer freely falling in uniform gravity cannot detect acceleration. So I think defining gravity as geometry is still the "right" thing to do, even if there are two choices in defining geometry in Newtonian mechanics Jun 2 at 16:57
• @ThePhoton The meaning of 4-acceleration is, roughly speaking, the acceleration that would be picked up by an accelerometer, just like the ones everyone have in their phones nowadays. It is a measure of how much you're deviating from free-fall. Jun 2 at 18:57

You asked in a comment for a physical meaning.

So, you are accelerating away from the Earth because there is a pressure on you, the force on your feet if you're standing, which is not balanced by any other force. Gravity is no longer considered a force, therefore you are accelerating.

Where does this force come from? Well, this is not really a property of mass, but rather a property of electrons “wanting” to take up space because they cannot be in the exact same state as each other. This is called the Pauli Exclusion Principle, when you bring electron clouds too close they repel.

Gravity is not a force per se in general relativity but it does still bring matter into close proximity, from which it can repel other matter.

“OK but isn't this just a difference in a mathematical description of who accelerates and who stays stationary?”—no, it does make a difference, because of relativity. Special relativity makes one core claim about reality from which all of its weirdnesses can be derived, and it is a claim about acceleration, which you can phrase like this:

It is a universal property of acceleration that whenever you accelerate in some direction “forwards” by some acceleration $$\alpha$$, in addition to all of the normal Doppler shifts that you expect classically, there is an anomalous Doppler-like effect tied to acceleration: you will see any clock ahead of you by coordinate $$x$$ (negative if behind you) ticking at a rate of $$1+\alpha x/c^2$$ seconds per second.

Concretely, imagine I stand under a geosynchronous satellite. If we are both in the Newtonian state of force balance with no acceleration, then I will say that its clocks and my clocks are ticking at the same rate. But if gravity is a fictitious force and we are both actually accelerating upward constantly, I will say that the satellite's clock ticks faster, and the satellite will say that my clock ticks slower ($$x$$ being negative for the satellite), and we will therefore agree that there is a “gravitational time dilation.”

(This also causes lensing but you really want to make this more geometrical rather than phenomenological to describe that.)

This has been observed in experiments and actually needs to be constantly corrected for in the GPS satellites, which work by broadcasting times to the Earth and then your phone triangulates where it is based on microsecond differences in the times it receives at once, plus models of where those satellites should be in relation to Earth's surface. The tick factor $$\alpha x/c^2$$ is about two parts per billion for these satellites ($$\alpha = g$$, $$x$$ = 20,000 km) so it can cause offsets of microseconds after an hour or two. So their atomic clocks actually are calibrated to tick slower by two parts per billion to avoid falling out of sync with Earth's surface.

• +1 for attributing the acceleration to the Pauli force. Jun 3 at 13:06

By general relativity gravitation is an attribute of curved spacetime instead of being due to a force propagated between bodies -- it is a fictitious force. See: https://en.wikipedia.org/wiki/Gravitational_acceleration

Another interesting moment: the vacuum may attract or repel -- try to find something about the Great Attractor.