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?
Newtonian Mechanics – If Gravity Isn’t a Force, How Are Forces Balanced in the Real World?
forcesgeneral-relativitynewtonian-mechanicsspacetime
Related Solutions
To really understand this you should study the differential geometry of geodesics in curved spacetimes. I'll try to provide a simplified explanation.
Even objects "at rest" (in a given reference frame) are actually moving through spacetime, because spacetime is not just space, but also time: apple is "getting older" - moving through time. The "velocity" through spacetime is called a four-velocity and it is always equal to the speed of light. Spacetime in gravitation field is curved, so the time axis (in simple terms) is no longer orthogonal to the space axes. The apple moving first only in the time direction (i.e. at rest in space) starts accelerating in space thanks to the curvature (the "mixing" of the space and time axes) - the velocity in time becomes velocity in space. The acceleration happens because the time flows slower when the gravitational potential is decreasing. Apple is moving deeper into the graviational field, thus its velocity in the "time direction" is changing (as time gets slower and slower). The four-velocity is conserved (always equal to the speed of light), so the object must accelerate in space. This acceleration has the direction of decreasing gravitational gradient.
Edit - based on the comments I decided to clarify what the four-velocity is:
4-velocity is a four-vector, i.e. a vector with 4 components. The first component is the "speed through time" (how much of the coordinate time elapses per 1 unit of proper time). The remaining 3 components are the classical velocity vector (speed in the 3 spatial directions).
$$ U=\left(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau}\right) $$
When you observe the apple in its rest frame (the apple is at rest - zero spatial velocity), the whole 4-velocity is in the "speed through time". It is because in the rest frame the coordinate time equals the proper time, so $\frac{dt}{d\tau} = 1$.
When you observe the apple from some other reference frame, where the apple is moving at some speed, the coordinate time is no longer equal to the proper time. The time dilation causes that there is less proper time measured by the apple than the elapsed coordinate time (the time of the apple is slower than the time in the reference frame from which we are observing the apple). So in this frame, the "speed through time" of the apple is more than the speed of light ($\frac{dt}{d\tau} > 1$), but the speed through space is also increasing.
The magnitude of the 4-velocity always equals c, because it is an invariant (it does not depend on the choice of the reference frame). It is defined as:
$$ \left\|U\right\| =\sqrt[2]{c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2} $$
Notice the minus signs in the expression - these come from the Minkowski metric. The components of the 4-velocity can change when you switch from one reference frame to another, but the magnitude stays unchanged (all the changes in components "cancel out" in the magnitude).
If a ball is kept on a table then there is gravity acting on it as well as a normal reaction force by the table on the ball.
This is correct.
But as both the forces are being exerted on the same object i.e. the ball, then it cannot be called a- action reaction pair under Newton's third law of motion because it requires the forces acting on two different bodies.
This is also correct. The force of gravity and the normal force from the table on the ball are not action-reaction pairs as described by Newton's third law.
So, if it is not an action reaction pair, then how is any force being exerted by the table on the ball? Also, is the ball applying any force on the table?
As you have recognized, both forces in action-reaction force pairs do not act on the same object, so why would you think that in order to have forces acting on an object they have to be part of an action-reaction pair?
There are two action-reaction force pairs here.
- The force of gravity: Earth on the ball and ball on the Earth
- The normal force: Table on the ball and ball on the table
As you can see, we have our two forces acting on our ball: gravity and normal force. Each of these forces has a corresponding force that it forms an action-reaction pair with, as stated by Newton's third law.
As a small aside, the more I see questions on this site about Newton's third law and actions/reactions, the more I realize how confusing this terminology is to new students to physics. If this is all still confusing to you, may I suggest a different view of Newton's third law?
All forces arise from interactions
If you want to pick out your action-reaction pairs, just pick out your interactions. Are the ball and Earth interacting? Yes, through gravity. Are the ball and the table interacting? Yes, through the normal force (electrostatic interactions).
Best Answer
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:
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.