If gravity isn't a force, then how are forces balanced in the real world?

Question

Consider a simple situation like this- an object is sitting on a table. In classical mechanics, we say that the net force on the object is zero because gravity (treated as a force) and normal reaction force are equal and opposite to each other, and hence, it's acceleration is zero. But according to Einstein's General Theory of Relativity, gravity isn't a force at all, but instead curvature created in spacetime by a massive object, and objects near it tend to move towards it because they are just moving along the geodesic paths in that curved spacetime. So if an object kept on a table gets acted only by the normal reaction force (as gravity ain't a force), how is the net force on it zero?

Answer 1

So if an object kept on a table gets acted only by the normal reaction force (as gravity ain't a force), how is the net force on it zero?

I've quoted what I think is the key part of your question, and it's key because the net force is not zero. The object on the table experiences a net force of $mg$ and as a result it is experiencing an upwards acceleration of $g$.

The way you can tell if no force is acting on you is by whether you are weightless or not. If you were floating in space far from any other objects then there would be no forces acting upon you and you'd be weightless. If we fixed a rocket to you and turned it on then you'd no longer be weightless because now the rocket is exerting a force on you. Technically you have a non-zero proper acceleration.

In general relativity your acceleration (your four-acceleration) has two components. We write it as:

$$ a^{\mu}= \frac{\mathrm du^\mu}{\mathrm d\tau}+\Gamma^\mu_{\alpha \beta}u^{\alpha}u^{\beta} $$

The first term $\mathrm du^\mu/\mathrm d\tau$ is the rate of change of your (coordinate) velocity with time, so it is what Newton meant by acceleration, and the second term is the gravitational acceleration. The key thing about general relativity is that we don't distinguish between the two - they both contribute to your acceleration.

If you're falling freely then the two terms are equal and opposite so they cancel out and you''re left wit an acceleration of zero:

$$ a^{\mu}= 0 $$

This is when the net force on you is zero. For the object on the table the coordinate bit of the acceleration is zero but the second term is not and the acceleration is:

$$ a^{\mu}= \Gamma^{\mu}_{\alpha \beta}u^{\alpha}u^{\beta} $$

So the object sitting on the table has a non-zero acceleration and the net force on it is not zero.

Maybe this sounds like I'm playing with words a bit, by defining what I do and don't mean by acceleration. But this is absolutely key to understanding how general relativity describes the motion of bodies. The key point is that gravitational and coordinate acceleration are treated on an equal footing, and if you are stationary in a gravitational field that means you are accelerating.

If you're interested in pursuing this further there is a fuller description in How can you accelerate without moving?. There is more on why spacetime curvature makes you accelerate in How does "curved space" explain gravitational attraction?

A footnote

Given the attention this answer has received I think it is worth elaborating on exactly how relativists view this situation.

The question gives an example of an object sitting stationary on a table, but let's start with an object a few metres above the table and falling freely towards it.

It seems obvious that the apple is accelerating down towards the table. It seems obvious because we are used to taking the surface of the Earth as stationary because that's our rest frame (even though the surface of the Earth is most certainly not at rest :-).

But if you were the apple then it would seem natural to take your rest frame as stationary, and in that case the apple is not accelerating downwards - the table is accelerating upwards to meet it.

So which view is correct? The answer is that both are correct. Whether it's the apple or the table that is stationary is just a choice of rest frame, i.e. a choice of coordinates, and it is a fundamental principle in general relativity that all coordinates are equally good when it comes to describing physics.

But if we can randomly choose our coordinates it seems hard to say anything concrete. We could choose frames accelerating at any rate, or rotating, or expanding or all sorts of bizarre frames. Isn't there something concrete we can say about the situation? Well there is.

In relativity there are quantities called invariants that do not depend on the coordinates used. For example the speed of light is an invariant - all observers measuring the speed of light find it has the same value of $c$. And in our example of the apple and table there is an important invariant called the proper acceleration. While the apple and the table disagree about which of them is accelerating towards the other, if they compute their respective proper accelerations they will both agree what those values are.

In Newtonian mechanics acceleration is a vector $(a_x, a_y, a_z)$, but in relativity spacetime is four dimensional so vectors have four components. The four-acceleration is the relativistic equivalent of the three dimensional Newtonian acceleration that we are all used to. While it's a bit more complicated, the four acceleration is just a vector in 4D spacetime, and like all vectors it has a magnitude – in relativity we call this quantity the norm. And the norm of the four-acceleration is just the proper acceleration that I talk about above.

The proper acceleration can be complicated to calculate. There's a nice explanation of how to calculate it for an object like our table in What is the weight equation through general relativity? It turns out that the proper acceleration of the table is:

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

where $M$ is the mass of the Earth and $r$ is the radius of the Earth.

But hang on – that tells me the proper acceleration of the table is non-zero. But ... but ... isn't the table stationary? Well, this takes us back to where we started. The table and the apple disagree about who is accelerating, but they both agree that the table has a non-zero proper acceleration. And in fact if we calculate the proper acceleration of the apple it turns out to be zero so both the apple and the table agree the apple has a proper acceleration of zero.

There is a simple physical interpretation of the proper acceleration. To measure your proper acceleration you just need to hold an accelerometer. Suppose you're floating around weightless in outer space, then your accelerometer will read zero, and that means your proper acceleration is zero. If you're standing on the surface of the Earth (alongside the table perhaps) then your accelerometer will read $9.81\ \mathrm{m/s^2}$, and indeed your proper acceleration is approximately $9.81\ \mathrm{m/s^2}$ not zero.

To summarise, a comment asks me:

So, let's just get this straight. The book sitting on the table in front of me is accelerating upwards all the time? But when I push it off the table and it falls down, then as it falls down it is not accelerating? Is that what you're saying?

What I'm saying, and what all relativist would say, is that:

• the book on the table has a non-zero proper acceleration

• the falling book has a zero proper acceleration

And this is all we can say. The question of which has a non-zero three-acceleration (Newtonian acceleration) is meaningless because that quantity is not frame invariant. The question of which has a non-zero proper acceleration is meaningful – even if the answer isn't what you expected.

Answer 2

You can describe the physical situation either in a Newtonian way or according to GR (general relativity), but you have to be consistent in each description.
In a curved spacetime a free-falling body follows a geodesic, that is a path with nil acceleration. In the example the object on the table does not follow a geodesic, in fact its four-acceleration is not zero. So the body is acted upon.