I understand the idea that, when spacetime is warped by a large mass, objects that try to travel in straight lines instead move along curved geodesics, something Newton would describe as an accelerating force acting on the object in some direction other than the direction of motion, which makes it deviate from its path. That part I get. What I don't understand is how that carries over for stationary objects. If I throw a ball across the room, I can see its path curving as it moves through curved spacetime, but if I hold it in my hand, I can still feel its "weight" pulling down, even though it's not moving along any geodesic. Where does this "force" come from?
[Physics] How does the curvature of spacetime exert a force on a stationary object
general-relativitygravity
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).
You're using the wording "curved spacetime", but you're still only thinking "curved space" with an independent, linear time.
In your curvature model, you're assuming that moving through some 3D spatial point in one spatial 3D direction will experience the same 3D path curvation independent on speed (as if you'd shoot a ball through a curved tube). You'd certainly agree that a different initial 3D direction will result in a different path.
Now we are in 4D, meaning that two different initial speeds are two different 4D directions, and as time cannot be treated as an independent component, but is curved together with space, this easily results in a different path.
Best Answer
A "fictitious force" occurs when your coordinate system is not "natural," in the sense that it is not the right sort of coordinate system to express how Nature works. Basically, if your coordinate system is "simple enough" then you can pretend that your "unnatural" system is really a "natural" one but everything experiences a mysterious "force" which seems to come from nowhere.
In classical mechanics, we already have a great example: a rotating reference frame is not "natural" and therefore has two fictitious forces --centrifugal and Coriolis -- which we can invent. If we invent those then we can pretend that things are standing still! Perversely, the geocentrists were right: you can by all means put the Earth at the center of the Solar system, if you invent the right centrifugal and Coriolis forces to account for its orbit about the Sun. Their only problem was to describe these with "epicycles" -- circles upon circles -- but actually that is now a trick we use all the time, we just call it "Fourier decomposition." (And in any case it's no more physically complicated than a satellite orbiting a Lagrange point.)
Another great example of this is when your car brakes sharply and you seem to be flung "forward": this force is a fictitious force created by the "unnaturalness" of the car's braking motion. Or if you're on a train or subway and you feel suddenly tossed to one side or another, it's hard to remember that actually you are standing still and it is the floor and walls which suddenly tossed themselves at you! We're built to think of the train-car as a "room" and to locate ourselves within it, and if that's our reference frame then our "natural" motion will appear to be acted on by these fictitious forces tossing us side-to-side.
In general relativity, Nature wants to guide particles along geodesics. This means that the appropriate coordinates for describing Nature are coordinates which are in free-fall. Your coordinates are not in free-fall, so they are therefore "unnatural". If you are in free-fall, then the ball in your hand does not feel like it has any weight at all! However your unnatural coordinates are very closely related to the "natural" coordinates that Nature uses, so that you can easily pretend that your coordinates are "natural" too, if you just invent a fictitious force that's pulling everything down. This fictitious force is the gravitational force.