They say that gravity is technically not a real force and that it's caused by objects traveling a straight path through curved space, and that space becomes curved by mass, giving the illusion of a force of gravity.

That makes perfect sense for planetary orbits, but a lot less sense for the expression of gravity that we are the most familiar with in our day-to-day lives: "what goes up must come down."

Imagine that I hold a ball in my hand, several feet off the ground, with my fingers curled around it, and my hand is above the ball. Then I open my fingers, releasing my grip on it, being very careful to not impart any momentum to the ball from my hand as I do so.

An object at rest remains at rest unless acted upon by an outside force. If the ball is not moving (relative to my inertial reference frame), it has no path to travel that's any different from the Earth's path through space. It should remain at rest, hanging there in the air. And yet it falls, demonstrating that an outside force (gravity) did indeed act upon it.

How does curved space explain this?

  • $\begingroup$ I had the same question too, but while watching one of the lectures on GR, they interpreted it in a very simplified manner, I dont know if it is true but it explained my doubt very well. $\endgroup$ – Ajinkya Naik Jul 8 '17 at 8:14
  • $\begingroup$ If a body is accelerating and we were to plot its motion on a space-time graph, the line representing the motion of the body will be curved and thus depicting acceleration. $\endgroup$ – Ajinkya Naik Jul 8 '17 at 8:15
  • $\begingroup$ We know that Gravity causes acceleration, now since the body itself is not accelerating, and according to principle of equivalence gravity and acceleration are the same phenomenon. $\endgroup$ – Ajinkya Naik Jul 8 '17 at 8:15
  • $\begingroup$ Therefore, in order to represent acceleration caused due to gravity on a graph, we would curve the X and Y or space and time axes, while the line of motion will remain straight. $\endgroup$ – Ajinkya Naik Jul 8 '17 at 8:16
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    $\begingroup$ And so acceleration due to gravity is caused by curved space and time. $\endgroup$ – Ajinkya Naik Jul 8 '17 at 8:17

If you have a look at my answer to When objects fall along geodesic paths of curved space-time, why is there no force acting on them? this explains how on a curved surface two moving observers will appear to exprience a force pulling them together. However two stationary observers will feel no force. The force only becomes apparent when you move on the curved surface.

This is true in general relativity as well, but what is easily forgotten by newcomers to GR is that in GR we consider motion in spacetime, not just space. You are always moving in spacetime because you can't help moving in time. Your speed in spacetime is known as the four-velocity, and in fact the magnitude of the four-velocity (technically the norm) is always $c$. So you can't help moving through spacetime (at the speed of light!) and when spacetime is curved this means you will experience gravitational forces.

You are probably familiar with Newton's first law of motion. This says that the acceleration of a body is zero unless a force acts on it. Newton's second law gives us the equation for the acceleration:

$$ \frac{d^2x}{dt^2} = \frac{F}{m} $$

The general relativity equivalent to this is called the geodesic equation:

$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} u^\alpha u^\beta \tag{1} $$

This is a lot more complicated than Newton's equation, but the similarity should be obvious. On the left we have an acceleration, and on the right we have the GR equivalent of a force. The objects $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols and these tell us how much spacetime is curved. The quantity $u$ is the four-velocity.

Now let's consider the particular example you describe of releasing a ball. You say the ball is initially stationary. If it was stationary in spacetime, i.e. the four velocity $u = 0$, then the right hand side of equation (1) would always be zero and the acceleration would always be zero. So the ball wouldn't fall. But the four velocity isn't zero.

Suppose we use polar coordinates $(t, r, \theta, \phi)$ and write the four-velocity as $(u^t, u^r, u^\theta, u^\phi)$. If you're holding the ball stationary in space the spatial components of the four velocity are zero: $u^r = u^\theta = u^\phi = 0$. But you're still moving through time at (approximately) one second per second, so $u^t \ne 0$. If we use the geodesic equation (1) to calculate the radial acceleration we get:

$$ {d^2 r \over d\tau^2} = - \Gamma^r_{tt} u^t u^t $$

The Christoffel symbol $\Gamma^r_{tt}$ is fiendishly complicated to calculate so I'll do what we all do and just look it up:

$$ \Gamma^r_{tt} = \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) $$

and our equation for the radial acceleration becomes:

$$ {d^2 r \over d\tau^2} = - \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) u^t u^t \tag{2} $$

Now, I don't propose to go any further with this because the maths gets very complicated very quickly. However it should be obvious that the radial acceleration is non-zero and negative. That means the ball will accelerate inwards. Which is of course exactly what we observe. What is interesting is to consider what happens in the Newtonian limit, i.e. when GR effects are so small they can be ignored. In this limit we have:

  • $d\tau = dt$ so $d^2r/d\tau^2 = d^2r/dt^2$

  • $1 \gg \frac{2GM}{c^2r}$ so the term $1 - \frac{2GM}{c^2r} \approx 1$

  • $u^t \approx c$

If we feed these approximations into equation (2) we get:

$$ {d^2 r \over dt^2} = - \frac{GM}{c^2r^2}c^2 = - \frac{GM}{r^2} $$

and this is just Newton's law of gravity!

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    $\begingroup$ although I understand that comments are not supposed to be used for congratulations, I will congratulate you anyway. I have never seen this derivation before now and am thoroughly impressed by it, especially the part where you show how newtonian gravity is recovered when GR effects become negligible. Thanks for posting this. $\endgroup$ – niels nielsen Mar 13 '18 at 6:30

Imagine two cars a mile apart on the equator, driving to the north pole. When they reach it, they are 0 miles apart, and during the entire trip, the distance between them decreases.

What is 'pulling together' the cars? Nothing. The earth is curved, so as they travel, they come together.

Spacetime is the same way. We are always 'travelling' in a forward (time or timelike) direction, and if space is curved, some things accelerate together as if a force had been applied.

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The heart of General Relativity lies in The Equivalence Principle which says: The effect of gravity is completely equivalent to an accelerated frame. There exists no physical process or experiment by which you can distinguish these two. So any question regarding a gravitational region can be thought of using an accelerated frame.

Then, let's consider your problem. When you are holding the ball in earth, you can equivalently think it as if you are holding the ball in a lift which is going upwards. Now think what happens in the lift scenario when you open your grip. Because no force is acting on the ball now, it remains in a const. velocity and actually in lift frame it accelerates downwards. So this is what should happen in a gravity field also. Hence it follows.

If you want a space-time curvature picture you must remember that general relativity talks about gravity as a curvature of space-time not space alone. The argument you gave their is not valid. Earth moves in space-time under the influence of curvature created by Sun. But locally earth itself creates a curvature in space-time. The ball moves not only in the curvature created by Sun but also by Earth. Thus along with moving in the same way as earth is moving around sun it has one extra motion. This extra motion is due to the curvature created by Earth which in earth's frame looks like the force of gravity towards earth center.

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  • $\begingroup$ The best illustration I've seen of the equivalence principle and the basic curvature intuition of GR is in "The Mechanical Universe": youtu.be/w0SeFfV6J3k?t=13m $\endgroup$ – amr Dec 5 '15 at 22:21

Consider astronauts in the space station. They are freely floating in space without any apparent gravitational force acting on them. If they stood on a scale, their weight would be zero. But, the gravitational field in the general area of the International Space Station is just about the same as it is on the surface of the Earth. The reason for weighing zero is that these astronauts are already freely falling. There are no external forces acting on them.

When you are holding that ball in your hand while standing on the surface of the earth, you are applying an external force that keeps it in that particular location. Then when you let go of the ball, that force no longer exists. If a ball is moving, and you apply a force to stop it, and then remove that force, the ball will move again.

Another experiment to perform is to eliminate the forces acting on you by jumping off of a tall building (imagine in your head only) while holding the ball, then let go of the ball. You and the ball will fall at the same rate.

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I like Ari's analogy to an accelerated frame.

If you want your answer in terms of curvature, remeber that GR says that an object always follows a geodesic. If space-time were flat, then this would mean moving at a constant veloctiy or being at rest, which corresponds to what you said with "An object at rest remains at rest unless acted upon by an outside force. ".

This is only true in a flat spacetime. If space and time are curved, then being at rest is not a geodesic anymore.

A geodesic is a world line (trajectory in space and time) that minimises the 'spacetime' interval between two events, $s^2 = g_{\mu \nu}dx^{\mu}dx^{\nu}$.

Being at rest means moving in time only, but if time is curved, then staying at x=0 does not minimise the geodesic! If you moved in space so as to balance the curvature in time, you'd have a minimum again.

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