I’m a physics dilettante… a little undergrad 101 stuff and some exposure to pop sci. I was mulling over the explanation of gravity as being warped space rather than a force, such that an orbiting body for example is not being held in orbit by the gravitational force exerted between it and the object it’s orbiting but is merely traveling inertially in a straight line in a space that has been warped by a big mass in the midst of it.
Okay, thought I, I can picture that.
But then I tried to apply it to another scenario: hole drilled through the middle of the earth (or some simpler, non-rotating, isolated mass… weight dropped into hole. I imagine the weight oscillating back and forth, speeding up as it approaches the center, slowing down as it approaches the surface, then repeating in the other direction. I can’t seem to grok a curved space that’s so curved that an object can go in what appears to be opposite directions along the same path within it without a force being applied to it to make it do so. Yet I understand that from the POV of the oscillating mass, no force is felt. What am I missing?
What you might be missing is time. Things don’t really “sit still” in spacetime—everything is always moving through time, but things that are moving fast through space are moving slower through time—it’s like you have just one speed, but you can point it “out of time” and “into space.”
When you travel through a curved spacetime, “sitting still” still means moving forward in the time direction. It’s not like the curvature due to gravity creates a potential well that you “fall down.” It’s more like you’re one of those stunt motorcyclists driving around the inside of a steel cage, constantly whizzing forward in a locally straight line, but changing what direction you’re pointing due to the curvature of space.
This seems like a great question to me and I’m bummed I can’t answer it. But here’s a toy model that might help a bit.
Take a 2-dimensional spacetime shaped like the surface of a vertical cylinder, with space being the 1-dimensional equatorial circles, and time going vertically. Some of the straight lines in this space are slanted lines just going around and around the cylinder forever, and objects following those as world lines would sort of appear to oscillate around a point traveling along an exact vertical world line.
Anyway that model’s only 2-dimensional, and the bigger problem is it’s not the right type of geometry (it’s Riemannian not Lorentzian). Also the cylinder is flat, not curved. But maybe it still helps.
There is a paper here which does something like this, and draws pretty pictures. The metric has been “absolutized” by replacing the negative coeffecients with their absolute values, so it becomes Riemannian instead of Lorentizian, but the diagram is then annotated with a bunch of yellow triangles showing “the direction of time”, and together these two things apparently contain all the information of the original spacetime. For the spacetime around an ordinary planet all the triangles point in the same direction, so this Riemannian version seems like a valid representation, I guess.
Anyway, the red line in Figure 5 in the pdf shows something like what OP was asking about: a ball is thrown straight up, turns around, and falls down along the same path again.
I think maybe the key point is that although the ball is retracing its path in space, in spacetime it’s just a long line which never loops back on itself, so it may be easier to believe that it’s going “straight ahead”.
The geodesics aren’t lines in space, but in space-time. For the ball to fall through the Earth and back to its starting point takes about 5000 seconds, during which time light goes about 1.5 billion km. So a graph in space-time will be a sine wave whose period is 1.5 billion km and whose amplitude is 6400 km, a ratio of about 250000 to 1. The graph has very low curvature everywhere.
It is the same for the Earth’s orbit round the Sun. It is not the spatial path of the orbit that is a geodesic, but the helical path it traces out in space-time. In one revolution it travels one year into the future, equivalent to a distance of a light-year. As a handy way of visualising this, the ratio of a light-year to an AU (astronomical unit, the radius of the Earth’s orbit) is about the same as a mile to an inch. So in space-time the orbit can be visualised as a helix formed by wrapping a piece of string around a cylinder two inches thick and a mile long, which makes just a single turn over that distance. The curvature of this path is much lower than the spatial curvature of the orbital path.
Imagine a 2D plane where x is space and y is time. Let’s say the Earth is stationary at x=0, so its trajectory is the y axis, and the metric of spacetime is “curved” near it.
We can represent the metric visually by sprinkling a bunch of sand near the y axis. Then lines of inertial movement (“geodesics”) can be understood in two ways:
Given a pair of points, a geodesic is the line between them with the least sand (this represents the line being shortest according to the metric).
Given a starting point and velocity vector, keep moving so as to keep equal amounts of sand on your nearby left vs. nearby right—in other words, curve toward more sand.
Surprisingly, these two views are equivalent! For example, consider the geodesic from (1,0) to (1,1). It will bulge slightly away from the y axis, to avoid sand, and so at each point it will be curving toward more sand.
Now we can answer your original question. Place an object at (1,0) with velocity vector (0,1) (zero spatial velocity) and let it go. It will keep moving in the positive y direction, but curve toward the y axis where there’s more sand, and eventually cross it at an angle. Then it will curve back by symmetry, and so on, oscillating back and forth in the x coordinate while moving forward in time.
Can that really be a shortest line between two points? Why not. Say the object makes one full oscillation, traveling from (1,0) to (-1,1) to (1,2). If you try to “straighten” the line by pulling on the endpoints, the midpoint will be pulled toward the y axis and catch more sand. So it might well be a local minimum.
I’m a physics dilettante… a little undergrad 101 stuff and some exposure to pop sci. I was mulling over the explanation of gravity as being warped space rather than a force, such that an orbiting body for example is not being held in orbit by the gravitational force exerted between it and the object it’s orbiting but is merely traveling inertially in a straight line in a space that has been warped by a big mass in the midst of it.
Okay, thought I, I can picture that.
But then I tried to apply it to another scenario: hole drilled through the middle of the earth (or some simpler, non-rotating, isolated mass… weight dropped into hole. I imagine the weight oscillating back and forth, speeding up as it approaches the center, slowing down as it approaches the surface, then repeating in the other direction. I can’t seem to grok a curved space that’s so curved that an object can go in what appears to be opposite directions along the same path within it without a force being applied to it to make it do so. Yet I understand that from the POV of the oscillating mass, no force is felt. What am I missing?
What you might be missing is time. Things don’t really “sit still” in spacetime—everything is always moving through time, but things that are moving fast through space are moving slower through time—it’s like you have just one speed, but you can point it “out of time” and “into space.”
When you travel through a curved spacetime, “sitting still” still means moving forward in the time direction. It’s not like the curvature due to gravity creates a potential well that you “fall down.” It’s more like you’re one of those stunt motorcyclists driving around the inside of a steel cage, constantly whizzing forward in a locally straight line, but changing what direction you’re pointing due to the curvature of space.
This seems like a great question to me and I’m bummed I can’t answer it. But here’s a toy model that might help a bit.
Take a 2-dimensional spacetime shaped like the surface of a vertical cylinder, with space being the 1-dimensional equatorial circles, and time going vertically. Some of the straight lines in this space are slanted lines just going around and around the cylinder forever, and objects following those as world lines would sort of appear to oscillate around a point traveling along an exact vertical world line.
Anyway that model’s only 2-dimensional, and the bigger problem is it’s not the right type of geometry (it’s Riemannian not Lorentzian). Also the cylinder is flat, not curved. But maybe it still helps.
There is a paper here which does something like this, and draws pretty pictures. The metric has been “absolutized” by replacing the negative coeffecients with their absolute values, so it becomes Riemannian instead of Lorentizian, but the diagram is then annotated with a bunch of yellow triangles showing “the direction of time”, and together these two things apparently contain all the information of the original spacetime. For the spacetime around an ordinary planet all the triangles point in the same direction, so this Riemannian version seems like a valid representation, I guess.
Anyway, the red line in Figure 5 in the pdf shows something like what OP was asking about: a ball is thrown straight up, turns around, and falls down along the same path again.
I think maybe the key point is that although the ball is retracing its path in space, in spacetime it’s just a long line which never loops back on itself, so it may be easier to believe that it’s going “straight ahead”.
Really neat paper, thank you!
The geodesics aren’t lines in space, but in space-time. For the ball to fall through the Earth and back to its starting point takes about 5000 seconds, during which time light goes about 1.5 billion km. So a graph in space-time will be a sine wave whose period is 1.5 billion km and whose amplitude is 6400 km, a ratio of about 250000 to 1. The graph has very low curvature everywhere.
It is the same for the Earth’s orbit round the Sun. It is not the spatial path of the orbit that is a geodesic, but the helical path it traces out in space-time. In one revolution it travels one year into the future, equivalent to a distance of a light-year. As a handy way of visualising this, the ratio of a light-year to an AU (astronomical unit, the radius of the Earth’s orbit) is about the same as a mile to an inch. So in space-time the orbit can be visualised as a helix formed by wrapping a piece of string around a cylinder two inches thick and a mile long, which makes just a single turn over that distance. The curvature of this path is much lower than the spatial curvature of the orbital path.
Imagine a 2D plane where x is space and y is time. Let’s say the Earth is stationary at x=0, so its trajectory is the y axis, and the metric of spacetime is “curved” near it.
We can represent the metric visually by sprinkling a bunch of sand near the y axis. Then lines of inertial movement (“geodesics”) can be understood in two ways:
Given a pair of points, a geodesic is the line between them with the least sand (this represents the line being shortest according to the metric).
Given a starting point and velocity vector, keep moving so as to keep equal amounts of sand on your nearby left vs. nearby right—in other words, curve toward more sand.
Surprisingly, these two views are equivalent! For example, consider the geodesic from (1,0) to (1,1). It will bulge slightly away from the y axis, to avoid sand, and so at each point it will be curving toward more sand.
Now we can answer your original question. Place an object at (1,0) with velocity vector (0,1) (zero spatial velocity) and let it go. It will keep moving in the positive y direction, but curve toward the y axis where there’s more sand, and eventually cross it at an angle. Then it will curve back by symmetry, and so on, oscillating back and forth in the x coordinate while moving forward in time.
Can that really be a shortest line between two points? Why not. Say the object makes one full oscillation, traveling from (1,0) to (-1,1) to (1,2). If you try to “straighten” the line by pulling on the endpoints, the midpoint will be pulled toward the y axis and catch more sand. So it might well be a local minimum.
Perhaps, it’s not you who is missing something.