# Thread: How does spacetime curvature cause gravity?

1. From my understanding, objects like to take straight-line paths through spacetime when there is a zero net force acting on them.
If you make a spacetime diagram with three space axes and one time axis, then an object with zero net force will travel at zero acceleration, which will trace a straight line through spacetime.
But when objects with mass exist, they bend spacetime, and the straight paths are ones which fall toward a massive object.
So how, exactly, do we define a straight line through spacetime? Someone in another thread mentioned that it maximizes the proper time between the points. Proper time is defined as the time elapsed between two events from a frame where the events occured at the same time.
Let's go to special relativity for a second. We know that for any two events, the expresion
(dx)^2 - (dt)^2
does not depend on the frame of reference, where dx is the distance between the events, and dt is the time between them. That means that as dx increases, we must increase dt as well, and vice versa. So for an object travelling with zero acceleration, the proper time between any two points along its path is minimized, not maximized.
Can someone help?  2.

3. deleted  4. Proper time in special relativity is (dt)²–(dx/c)².

A straight line in general relativity is such that the covariant directional derivative along the line is zero.

In a coordinate system which is locally Minkowskian at a given point, a straight line has the equation of a straight line at that point. In other words, a straight line in curved spacetime is genuinely straight, even from a naïve viewpoint.  5. Originally Posted by KJW A straight line in general relativity is such that the covariant directional derivative along the line is zero.
What exactly is that? What are you taking the derivative of, and with respect to what?  6. Originally Posted by anticorncob28  Originally Posted by KJW A straight line in general relativity is such that the covariant directional derivative along the line is zero.
What exactly is that? What are you taking the derivative of, and with respect to what?
Suppose one has a spacetime trajectory described parametrically as: where is an arbitrary parameter.

Then this trajectory is a straight line if and only if: where is an arbitrary function of the parameter. is the connection object (a.k.a. Christoffel symbols) that specifies the underlying spacetime as well as the coordinate system over the spacetime: In the case where the parameter is the arc-length (which doesn't exist for light-like trajectories), the above becomes: It should be noted that the curvature vector of the trajectory: is always perpendicular to a tangent vector of the trajectory. Thus, for a straight line: is always parallel to In this case, the trajectory can be re-parameterised such that: This allows one to define a straight light-like trajectory for which the arc-length parameter doesn't exist.  7. My two cents. Don't mind me. People are arguing over the taste of cheese on the moon. The mathmatical representation of the standard model from the 50's. Could it be wrong. Probably. Someone should update it or something. Some kind of math that includes current physics. Einstein left out quantum. Russia left out peasants. And America included witchcraft. The curve is a flaw and shouldn't be there. You're right.  8. Stop posting shit!  9. Ok Teacher  Bookmarks
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