What dimension does space-time curve in?

[Ez4u2cit]

Is not space curvature the curving or projecting into a higher dimension? Like a curved sheet of paper perceived by a two dimensional creature? The mystery seems to reside in our ape brains being unable to perceive (but not conceptualize) higher dimensions than three or relativistic, quantized time.

 

[Nugatory]

Is not space curvature the curving or projecting into a higher dimension? Like a curved sheet of paper perceived by a two dimensional creature?.

It is not. The curvature produced by bending a two-dimensional sheet of paper through the third dimension is called extrinsic curvature, and it does not affect the two-dimensional geometry of the surface of the paper. For example, if you draw a triangle on the surface of the paper, the sum of the interior angles of the triangle will be 180 degrees. When you bend the paper or even roll it up into a cylinder nothing will change and the angles will still add to 180.

The curvature of spacetime is intrinsic curvature and it does not need an extra dimension. The effect of intrinsic curvature is directly visible to hypothetical two-dimensional creatures living on the curved surface without any reference to the third dimension. For example, on an intrinsically curved surface the interior angles of a triangle will not add to 180 degrees.

 

[Ez4u2cit]

So to put this in simple terms, what is curving if other dimensions are not involved? How is it possible to perceive a straight line as curved and remain in three dimensions?

 

[PeterDonis]

How is it possible to perceive a straight line as curved and remain in three dimensions?

It isn't. A line cannot have intrinsic curvature. In order to have intrinsic curvature, you have to look at a manifold with at least two dimensions--for example, a 2-sphere. Then your question can be rephrased as: how is it possible to tell that a 2-sphere is curved, without making any use of an embedding of it into a space with more than 2 dimensions? The answer to that is, by looking at geodesic deviation, which can be measured purely within the surface.

 

[Orodruin]

It isn't. A line cannot have intrinsic curvature. In order to have intrinsic curvature, you have to look at a manifold with at least two dimensions--for example, a 2-sphere. Then your question can be rephrased as: how is it possible to tell that a 2-sphere is curved, without making any use of an embedding of it into a space with more than 2 dimensions? The answer to that is, by looking at geodesic deviation, which can be measured purely within the surface.

This might be a bit steep for a B level thread so let me try to put it in more accessible terms.

Let us imagine you start at location A and travel in a straight line for a distance d. After doing this you turn 90 degrees to your right and travel a distance d in a straight line again. Repeating two more times, you will end up back at A if your space is flat. If you do not end up at A your space is therefore curved. To visualise this, I suggest thinking about how this would work on a sphere when d is one fourth of the circumference.

 

[Nugatory]

So to put this in simple terms, what is curving if other dimensions are not involved? How is it possible to perceive a straight line as curved and remain in three dimensions?

Imagine that your two dimensional universe is a sheet of thin rubber stretched out on a flat table. Now I can tug on the corners and stretch it, distorting shapes on the surface, turning circles into ovals, making it so the distance between non-intersecting lines is not constant, and all the other effects of intrinsic curvature - and we're staying in two dimensions.

 

[bahamagreen]

Orodruin and Nugatory,

It seems like all paths that stay on the surface of a sphere are curves; a straight path will leave the sphere's surface and doing so traversing a square path will place you back where you started (the straight line path comprising the square route in a flat plane intersecting the surface of the sphere with one of the route square's corners); travels that stay on the surface of a sphere are curved, not straight).

It also seems like when a flat surface is distorted by displacement within the plane and what were circles become ellipses, etc., those ellipses are not circles anymore; they just need to be redrawn. Walking a square of straight lines will return one to there source point both before and after the distortion - what would throw off the path is if the distortion was ongoing during the walk...

Both analogies seem ineffective at breaking out of referring to all curves with respect to flat space. How does one eliminate flat space as the basis from which curvature is a departure?

 

[PeterDonis]

It seems like all paths that stay on the surface of a sphere are curves; a straight path will leave the sphere's surface

Here you are assuming a definition of "straight" that is only valid for a flat space. On the sphere, a "straight" path (i.e., a geodesic) is a great circle.

How does one eliminate flat space as the basis from which curvature is a departure?

By not using a definition of "straight" that privileges flat space. If you use the correct definition of "straight"--that a "straight" line is a geodesic--then flat space becomes just one option among many.

 

[Nugatory]

It seems like all paths that stay on the surface of a sphere are curves; a straight path will leave the sphere's surface and doing so traversing a square path will place you back where you started (the straight line path comprising the square route in a flat plane intersecting the surface of the sphere with one of the route square's corners); travels that stay on the surface of a sphere are curved, not straight).

When you say "a straight path will leave the sphere's surface" you have to mean a straight path in three-dimensional space, not a straight path on the two dimensional space that is the surface of the sphere - otherwise, it wouldn't be be able to leave the surface. That's fine if you're thinking of the points making up the curved surface of the sphere as a subset of the points making up a flat three-dimensional space, and using the three-dimensional space's definition of "straight" as a relationship between points in that three-dimensional space. But if you're doing that, you're already assuming the existence of points outside of the two-dimensional space so of course you conclude that there must be another dimension for those points to live in.

However, that assumption is unnecessary. The mathematical object that is a set of points that collectively behave like the two-dimensional surface of a sphere is a perfectly good mathematical object in its own right, with its own perfectly good definition of "straight" (intuitively, the path of a string stretched tight between two points, called a "geodesic"). The third dimension is only needed because you and I live in three-dimensional space so if we're going to build a physical object with those mathematical properties we have to build a three-dimensional globe and then say "Hey - only pay attention to the points on the surface". If we were flatlanders (if you haven't read E.A. Abbott's classic book "Flatland", find a copy) we'd understand our world just fine without needing that third dimension; we'd just observe the geometrical behavior of our world, select a metric tensor that matched that behavior, and let differential geometry take it from there. Likewise, as creatures living in four-dimensional spacetime we can describe intrinsic curvature of that spacetime completely without ever needing a fifth dimension.

How does one eliminate flat space as the basis from which curvature is a departure?

You can't. There's flat spacetime in the absence of gravity, and if you add gravity you get curvature and departures from that flatness - no getting away from that. But you don't need an extra dimension to curve spacetime through to produce those departures. You just need a geometry in which the metric tensor, which describes the distance between adjacent points, is different from the flat-spacetime metric tensor.

 

[Dale]

It seems like all paths that stay on the surface of a sphere are curves; a straight path will leave the sphere's surface and doing so traversing a square path will place you back where you started (the straight line path comprising the square route in a flat plane intersecting the surface of the sphere with one of the route square's corners); travels that stay on the surface of a sphere are curved, not straight).

You are thinking about a 3D flat embedding space. A sphere is a 2D manifold and there is an intrinsic geometry entirely within that 2D surface. The fact that the path bends in the radial direction in 3D is utterly irrelevant to this intrinsic geometry because the radial direction does not exist in the 2D geometry.

 

[bahamagreen]

Well, I tried looking around in Wiki...

"For any natural number n, an n-sphere of radius r is defined as the set of points in (n + 1)-dimensional Euclidean space which are at distance r from a central point, where the radius r may be any positive real number."

"It is an n-dimensional manifold in Euclidean (n + 1)-space."

"In particular:...
a 2-sphere is the two-dimensional surface of a (three-dimensional) ball in three-dimensional space."

 

[Nugatory]

Well, I tried looking around in Wiki...

And stuff like this is the reason why wikipedia is not, in general, an allowed source here. The text you've quoted is wrong (it's confusing "X is Y" with "Y is an example of X") in a way that doesn't always matter - but it matters here. You'll find some mention of the problem about two-thirds of the way down the "talk" page for the article on "n-spheres" where those quotes come from.

 

[Ez4u2cit]

Thanks for the thoughtful replies. But I am having a problem moving forward from the theoretical to the perceived reality of my (ageing) ape brain. If I travel in an airplane between two points in three dimensions, taking the shortest possible route (straight line) experiencing gravity the entire trip, where is the curvature in three dimensional space?

 

[jbriggs444]

Thanks for the thoughtful replies. But I am having a problem moving forward from the theoretical to the perceived reality of my (ageing) ape brain. If I travel in an airplane between two points in three dimensions, taking the shortest possible route (straight line) experiencing gravity the entire trip, where is the curvature in three dimensional space?

You are making that trip in four dimensional space-time. The path is not "straight" in four dimensional space time. That is, it is not a "geodesic path". You can tell because you are accelerated -- your seat is pushing up on you all the way through the trip. The problem is that you are using your pre-relativistic picture of three dimensional space with an independent one dimensional time. That picture is not appropriate for explaining gravity in terms of curved space time.

 

[Nugatory]

Thanks for the thoughtful replies. But I am having a problem moving forward from the theoretical to the perceived reality of my (ageing) ape brain. If I travel in an airplane between two points in three dimensions, taking the shortest possible route (straight line) experiencing gravity the entire trip, where is the curvature in three dimensional space?

Gravity is the result of intrinsic (no fifth dimension needed) curvature in four-dimensional spacetime, not three-dimensional space.

Search this forum for member @A.T. 's excellent video showing how an apple falling freely to the ground is following a straight-line path through curved spacetime, while an apple hanging from the branch so not falling is being dragged away from that straight path by its stem which is connected to the branch.

 

[A.T.]

Thanks for the thoughtful replies. But I am having a problem moving forward from the theoretical to the perceived reality of my (ageing) ape brain. If I travel in an airplane between two points in three dimensions, taking the shortest possible route (straight line) experiencing gravity the entire trip, where is the curvature in three dimensional space?

Here the mentioned simpler examples, of hovering and falling radially in a gravitational field:



Tajectories that require more than 1 spatial dimension, are very difficult to visualize though. See this post for explanation on the different meanings of "curvature":

This is my own non-animated way of looking at it:

• Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

 

[alw34]

"But I am having a problem moving forward from the theoretical to the perceived reality of my (ageing) ape brain. "
yeah, you and Einstein have [had] the same problem. Not to worry, I stumbled across this yesterday:

It's not that I'm so smart, it's just that I stay with problems longer.
Albert Einstein

 

[PeterDonis]

If I travel in an airplane between two points in three dimensions

If the airplane is traveling between two points on the Earth's surface, and it travels at a constant altitude, then the path it takes, viewed in the 3-dimensional space in which the Earth is embedded, is not straight; it's curved. But it's the shortest possible path between the two points given the constraints (both points are on the Earth's surface and the plane keeps a constant altitude above the Earth's surface); the constraints basically amount to restricting the plane's path to a 2-sphere, the Earth's surface, and within that restricted domain, the plane's path is "straight"--a geodesic.

Now suppose that we're talking about a spaceship, not an airplane, and it's traveling from Earth to Mars, and it takes the shortest possible spatial path in the 3-dimensional space in which the Earth and Mars are embedded. This path will be (at least to a very good approximation--space in the solar system is actually not perfectly Euclidean because of the Sun's gravity, but the effect is too small to matter here) a Euclidean straight line, the sort of thing you're used to describing as a "straight" path.

However, as others have mentioned, this path will be "straight" only in space. It will not be straight in spacetime. Why? Because the spaceship will not be in free fall; it will have to accelerate (fire rockets) in order to maintain this straight path. The free-fall path between Earth and Mars would be an elliptical orbit that, spatially, looks curved. But in spacetime, it is the free-fall elliptical orbit that is straight (a geodesic), and the accelerated path (the one that looks "straight" spatially) that is curved. This is a manifestation of the fact that, in the presence of gravity, spacetime itself is curved; the free-fall elliptical orbit is a straight (geodesic) path in curved spacetime.

 

[Ez4u2cit]

Thanks for the answers. I have a much better understanding of the concept as a model or map which helps explain and predict outcomes. I won't pretend I understand it beyond that any more than I can visualize four dimensional objects. I take my (small) hat off to those who can.

 

[alw34]

Is it entirely proper to answer the original question,
What dimension does space-time curve in?
by saying:
If you consider space and time as a four dimensional geometry, gravity IS the curvature of those dimensions.

 

[jbriggs444]

Is it entirely proper to answer the original question,
What dimension does space-time curve in?
by saying:
If you consider space and time as a four dimensional geometry, gravity IS the curvature of those dimensions.

While true, it is side-stepping the question. As has been stated in this thread, intrinsic curvature is perfectly meaningful and definable even without an extra dimension to curve through. A direct answer points this out: "no such dimension is needed".

 

[sweet springs]

We do not need extra dimensions to explain curvature of our 1+3 dimension space.

But sometimes hypothetical extra dimension in Euclid geometry helps us to imagine what is going on.

For example our universe is regarded as 3D sphere in 4D Euclid space

$$x_1^2+x_2^2+x_3^2+x_4^2= a^2$$　

ref. (107.5) of Landau-Lifshitz's Classical Theory of Field

 

[PeterDonis]

our universe is regarded as 3D sphere in 4D Euclid space

This is not true for our current best-fit model of the universe; in that model, the universe is spatially flat and infinite in extent. The 3-sphere model is a model of a closed, spatially finite universe.

 

[sweet springs]

Instead, 3D hyperbole in 4D fictional Euclid space
$$x_1^2+x_2^2+x_3^2-x_4^2=a^2$$
works?

 

[pervect]

Thanks for the answers. I have a much better understanding of the concept as a model or map which helps explain and predict outcomes. I won't pretend I understand it beyond that any more than I can visualize four dimensional objects. I take my (small) hat off to those who can.

Trying to visualize a 4d object is not the approach we are recommending at all.

The first step is to understand curvature in 2 dimensions. Orodruin's definition in post #5 is one of the simplest tests for curvature, and you can use it to show that the surface of a globe is curved. I say a globe, rather than the Earth, because we can make our globe perfectly spherical, and don't have to deal with complicating issues of the Earth being not-quite a perfect sphere.

You start off at some point on the surface of the globe, and draw a line on the surface of the globe (the line obviously does not leave the surface of the globe, it's painted on the surface). This is similar to the way we do not leave space-time, we're always in space-time, we never go "outside" space-time. But right now we are trying to describe curvature in only two dimensions, so you know what we mean by curvature, and how you can tell something like the surface of a globe is curved without ever imagining anything that is not on the surface of the globe. Later on, we can use the analogy to better understand space-time being curved.

You draw your lines on the globe on the path that is basically the shortest distance between two points on the sphere. This is perhaps slightly oversimplified, but not by very much. This curve of shortest distance is a great circle. Sometimes this path is called a geodesic rather than a straight line. Because we are trying to keep things simple, the technical term "geodesic" wasn't used, rather the less formal and less precise term "straight line" was used. It appears the attempt at simplificaiton didn't work, so I thought I'd try introducing the concept of a geodesic, which at this point we can regard as being the shortest distance between two points that lies on the surface.

So curvature may be slightly tricky, but the hopefully familiar example of the globe's surface being curved, and applying Odoruin's test of making 4 right angle turns and moving along geodesics (which can be regarded for our purposes as being the shortest path connecting points that lies entirely on the surface without ever leaving it) making 4 90 degree turns is sufficient to determine whether or not our 2d geometry is curved or not.OK - we've started out moving in some direction, for the sake of definiteness let's say we start heading north. Then we might a 90 degree turn (lets say to the right), and start to head East. But a circle of lattitude is not a geodesic path on the globe, because it isn't great circle, and we know that all geodesics on the surface of the globe are great circles. (You may have to look this up for yourself, or just trust us on this point, unless you have calculus of variations to prove it.)

We make another 90 degree right turn, and we start to head south-west. We're not heading exactly south, because we veered from moving due east when we insisted in tracing out a goedisc path. It will be a mildly tricky exercise in spherical geometry to work out all the details, and follow the whole path, but basically when you follow Odoruin's description, you find that you do not wind up exactly at your starting point on the globe, but you do if you follow the description on a flat plane. Therefore Odoruin's procedure provides a test for curvature that doesn't require knowing anything about "higher dimensional spaces", you can do all your geometrical drawings and measurements on the surface of the globe.

 

[PeterDonis]

3D hyperbole in 4D fictional Euclid space

This would describe an open universe, i.e., negative spatial curvature. Our best-fit current model is spatially flat (no curvature).

 

[Ez4u2cit]

Because the spaceship will not be in free fall; it will have to accelerate (fire rockets) in order to maintain this straight path. The free-fall path between Earth and Mars would be an elliptical orbit that, spatially, looks curved. .

If Mars or some other body is stationary, the spaceship is simply falling due to gravity, then the path of the falling body is not a straight line? Did I understand that right?

 

[PeterDonis]

If Mars or some other body is stationary, the spaceship is simply falling due to gravity, then the path of the falling body is not a straight line?

The worldline of the spaceship, assuming it's in free fall, is a straight line in spacetime. Whether or not it looks straight in space depends on how you set up the initial conditions. If it's falling purely radially towards a planet such as Mars, then its spatial path would look straight. But in the scenario I was describing, the spaceship is traveling from Earth to Mars in free fall; its spatial path in that case would be an elliptical orbit, so it would look curved.

 

[Ez4u2cit]

Just to simplify: I am on the spaceship. It is momentarily motionless in reference to the planet which is also motionless. As I fall towards the surface(experiencing gravity) , do I still need to fire rockets to stay on the straight line perceived by myself or someone on the planet?

 

[Nugatory]

Just to simplify: I am on the spaceship. It is momentarily motionless in reference to the planet which is also motionless. As I fall towards the surface(experiencing gravity) , do I still need to fire rockets to stay on the straight line perceived by myself or someone on the planet?

No. In that particular case the straight-line path through spacetime happens to coincide with the straight-line path through space. The general principle is that if you aren't firing your rockets, you are traveling a straight-line path through spacetime; if you're firing your rockets they are pushing you off the straight-line path through spacetime so you aren't traveling such a path.

 

[alw34]

As I fall towards the surface(experiencing gravity) , do I still need to fire rockets to stay on the straight line perceived by myself or someone on the planet?

Just to be clear: you normally can 'perceive' straight line paths in spacetime by whether or not you are feeling acceleration...see the earlier post of about hanging apple on a stem versus free fall for one example. Here is a bit more...

A geodesic is a free fall worldline; A geodesic has zero [proper] 4-acceleration and zero curvature. You don't feel any acceleration.

When you don't 'feel' acceleration, you are in free fall and moving in a straight line [geodesic] in spacetime. This may appear curved, as already described, in space alone. [exactly as described in post #28] Analogous to an apple stem holding an apple in place, sitting in a chair in your home IS properly-accelerating. You feel it on your backside, and a test device called an accelerometer will register acceleration.

Feeling acceleration is the telltale sign you are NOT moving in a straight line in spacetime.

It's sometimes counter intuitive: going down the highway in a straight path through space,a 'flat highway' you still feel gravity on your backside,so you know you are not going in a straight line in spacetime. Then you come to a curve in the road, in space, and now you feel additional acceleration, that's how you know you are moving in a different curve in spacetime. And when you go up or down hills at constant speed you also feel different accelerations which tells you your path through spacetime is curved differently.

When would you be going in straight line in spacetime? When you run off a bridge and are in free fall, just like the Earth to Mars example in the prior post. Of course, it looks curved in space, a parabola is it??. Our puny senses can't detect the very small effect in time.

 

[pervect]

I will post an image of an "easy to visualize" trip on the globe that illustrates the point Odoruin was making. The trip follow four geodesic paths, each of which is at right angles to the other. All segments of the path has the same length, and starting and ending points of the journey are different. This is not possible on a plane.

To recap - the equivalent of straight lines on the globe are called geodesics, they represent the shortest route that you can take between two points on the globe without leaving the surface of the globe. On a spherical globe, these geodesic paths turn out to be great circles. I hope the concept of right angles is reasonably clear, rather than muddy up the exposition by going into details, I'll assume for now that it is. If questions arise, we can address that in a separate post.

Onto the picture of the journey. A short summary - the trip starts at A. Th four legs of the journey are A-B, B-C, C-A, and A-B. Thus the trip starts out at A, and winds up at B.

You start at point A, which is on the equator, at 90 degrees west longitude. You go north until you reach the North pole, at B. Next, you make a right angle turn, and go south down along the line of 0 degrees longitude, back to the equator. You make another right turn (the third), and head west along the equator. At this point you've taken only three legs of the journey, but you're already back at your starting point of your journey! To finish the trip, you make another right turn, and take the last and final leg of your journey, which goes back up to the North pole.

So, following the geodesic lines on the globe, here we have an example of a trip that takes 4 right angles, wherein all legs of the trip have the same length (1/4 of the circumference of the globe). The trip starts on the equator, and winds up on the north pole.

This then illustrates the difference between a curved spherical geometry and a flat plane geometry. On the spherical geometry you can start at a point, travel along what is essentially a squre (a quadrilateral with four right angles and equal sides), and wind up at a spot different than where you started. This is not possible on a plane geometry, and it illustrates how you can determine the existence of curvature via geometrical meansm without ever leaving the curved surface, simply by drawing geodesic lines which can be understood (slightly oversimplified) as the shortest distance between two points.

 

[A.T.]

When you don't 'feel' acceleration, you are in free fall and moving in a straight line [geodesic] in spacetime.

Just want to point out for the OP that "feel acceleration" means that an inertia based accelerometer, that you hold measures non-zero accelration. This is for example the case when you sit at your desk.