[Physics] Why would spacetime curvature cause gravity

curvaturedifferential-geometrygeneral-relativitygravityspacetime

It is fine to say that for an object flying past a massive object, the spacetime is curved by the massive object, and so the object flying past follows the curved path of the geodesic, so it "appears" to be experiencing gravitational acceleration. Do we also say along with it, that the object flying past in reality exeriences NO attraction force towards the massive object? Is it just following the spacetime geodesic curve while experiencing NO attractive force?

Now come to the other issue: Supposing two objects are at rest relative to each other, ie they are not following any spacetime geodesic. Then why will they experience gravitational attraction towards each other? E.g. why will an apple fall to earth? Why won't it sit there in its original position high above the earth? How does the curvature of spacetime cause it to experience an attraction force towards the earth, and why would we need to exert a force in reverse direction to prevent it from falling? How does the curvature of spacetime cause this?

When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve. So we cannot just say that it fell to earth because its geodesic curve passed through the earth. Why did the spacetime curvature cause it to start moving in the first place?

Best Answer

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).

Related Question