# Why would spacetime curvature cause gravity?

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?

• I have always wondered about this (and related). This is so brushed aside in populist explanations! – GreenAsJade Mar 11 '14 at 12:06
• This exact question was puzzling me on my drive home yesterday, and my kids were wondering why I was tracing curves in the air over the steering wheel :) – Charlweed Jan 20 '18 at 22:57
• Space-time curvature doesn't "cause" gravity. Space-time curvature is gravity. – EvilSnack Apr 13 at 16:35

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:

$$\left\|U\right\| =\sqrt{c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2}$$

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).

• Can you comment on why the four-velocity is always the speed of light? – GreenAsJade Mar 11 '14 at 12:07
• @GreenAsJade: The four-velocity is commonly defined as $u^i = \frac{dx^i}{d\tau}$. Written as a four vector, it looks like $\vec{u} = \gamma (c, \mathbf{v})$, with $\gamma = (1-v^2/c^2)^{-1/2}$. Its length squared is then $\gamma^2 (c^2-v^2)$, which is equal to $c^2$. – Javier Mar 11 '14 at 14:19
• Does this mean that particles passing through space in the speed of light does not pass through time? Do electrons not have an age? – Pål GD Mar 11 '14 at 23:35
• @PålGD: Correct, particles moving at the speed of light do not age. Thus the "time delation" mentioned in space travel things, where they travel a long way but not much time passes. – Mooing Duck Mar 12 '14 at 0:36
• @AlanSE The apple is moving through time at light speed only in the reference frame where the apple is at rest (spatially). In some other reference frame (where the apple has some spatial velocity) its speed through time is slower. The four-velocity is a vector which has 4 components. All these components can vary between frames, but the magnitude of this 4-vector remains unchanged (always equals c). – mpv Mar 12 '14 at 6:00

When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve.

Even when at rest in space, the apple still advances in space-time. Here is a visualization of the falling apple in distorted space-time:

• I've always visualized it as the apple being held by the stem at the top of a valley, the bottom of which is the Earth's gravity well / center of mass. The stem breaks and the apple rolls down the hill into the "valley's" gravity. The animation is very good, but I'm left wondering why space-time is curving away from the direction of earth's gravity. (I think I'm interpreting that correctly because the stem/branch force arrows point opposite from the center of the earth, and that direction is shown to be the center of the curve of the spacetime graph.) – Patrick M Mar 12 '14 at 16:56
• This is a great video... – Earth is Donut Mar 13 '14 at 2:49
• @PatrickM It's because you've been misled. The gravity well picture is good for showing (classical) gravitational potential and geodesics. It does not represent how spacetime is curved which would be difficult to show for any spatial dimension > 1. Hence the apple video is the only legitimate example I've seen of what curvature really means. Notice here the spatial dimension is not curved at all, but the time dimension is -- that is close to how it is in reality under GR. – Nimrod Sep 16 '18 at 22:20

As to the first paragraph, gravity shows up as geodesic deviation; initially parallel geodesics do not remain parallel.

Since, for a freely falling particle, the proper acceleration (the reading of an accelerometer attached to the particle) is zero, it is correct to say that a particle whose worldline is a geodesic has no proper acceleration.

But it is not correct to say that a freely falling particle has no coordinate acceleration.

Regarding the second paragraph, if a particle's wordline is not a geodesic, the particle will have a proper acceleration, the particle's accelerometer will not read zero. Two particles that are preventing from falling towards one another will have weight.

Regarding the third paragraph, I think you need to sharpen your conception of worldines and geodesics. If a particle exists, it has a worldline and the worldline of a particle that is free to fall is a geodesic even if the particle is momentarily stationary.

Not everything needs to follow geodesic Spacetime curvature available to it. With external force, you can prevent a particle from following Spacetime curvature. Only "freely" falling particles follow Spacetime curvature available to them. So, when you see a stationary object not following Spacetime curvature, it's because an external force is preventing it from going to it's inertial trajectory... Means, it's not in "Free Fall".

Come to Apple: In terms of Spacetime, nothing is in rest. An Apple, when attached with tree, is also in motion. But, the motion exist fully in time with zero space component. This motion is NOT according to Spacetime curvature available to it because external forces holding root of Apple oppose it at microscopic level. When these external forces stop working, Apple starts to follow Spacetime curvature which converts time component of motion to space component. That's why Apple's acceleration is merely inertial motion. You can see removal of time component of motion in Gravitational Time Dilation.

Imagine you are on the northern hemisphere on Earth (assuming it's a perfect sphere).

Now go north with constant speed: you can just go straight north you don't need to steer.

Now go east with constant speed: this is something different now, in order to remain on the same latitude circle you must steer northwards constantly. If you don't see why, try to imagine you are doing it on the 89° latitude circle. If you stop steering, you begin to go "straight" along a geodesic and "fall" towards the equator.

This correction force depends on where you are and which direction you are going (and wish to remain on a "coordinate-straight" path), it's a linear map that maps your velocity into force. It's called the Christoffel-symbols. It's a property of your chosen coordinate system and the geometry of the space-time.

Now in reality on Earth you are in a coordinate system where the coordinates given by the latitude, longitude and altitude and time. Your 4-velocity in space-time is constant $c$. If you stand still, you go straight in the time direction. But in order to keep this four velocity, you feel an upwards force from the floor, this is the effect of the Christoffel-symbols. If you lose the floor your trajectory in space time will be a geodesic and you will fall.

• This is a very nice example! – Vojtěch Dec 5 '18 at 18:03