# Spacetime curvature

1. Sep 12, 2007

### PhanthomJay

That famous experiment during a solar eclipse, which showed the curvature of light from a star as the light rays passed by the sun, pretty much I gather confirmed Einstein's space time curvature theory.

Question: Does spacetime curve into one of those hidden spatial dimensions that M-theory talks about, or is it an 'ordinary' curvatire in 3D space? I find either scenario difficult to comprehend.

2. Sep 13, 2007

### PhanthomJay

Bump? (Wow, my first bump!)

You know, if space is curved in 3D, why can't I take a shortcut along the straightline 'chord' of the curve? And how can it curve into the 5th dimension which is curled up perhaps in the order of the planck size of 10 to the minus 33 centimeters??

3. Sep 13, 2007

### Wallace

In General Relativity gravity is not a force, rather mass alters space-time such that if a particle travels in a straight line according to their local frame then a distant observer sees them travel on a curved path. Note that you do not actually feel a gravitational acceleration, so if you were orbiting the Earth in a space ship with no windows you wouldn't know you were moving in a circular path.

All this is described in General Relativity which is a four dimensional theory, it doesn't need string theory or other hidden dimensions. In the future these ideas may all be linked together but GR at present deal quite happily with four dimensions.

To answer your question about taking the 'shortcut along the chord' of a curved space-time region, you can do that but it requires an additional (non-gravitational) force to achieve. If you just coast along you will follow the path that the space-time dictates. The same is true for light and hence the bending of light around massive bodies like the Sun.

4. Sep 13, 2007

### PhanthomJay

Thanks, I think that does make it a lot clearer. Am i correct in interperting your answer about curvature, that spacetime is apparently curved to the distant observer (he sees the light 'apparently' bending as it passes the sun), but to a person hypothetically riding along witht the ray, that person would view spacetime as 'straight' at that point, and GR would argue that both observers are correct in their observation?

5. Sep 14, 2007

### Wallace

Light rays really do bend around the sun, it's not just an 'apparent' effect. The point I was making is that you don't 'feel' like you are being moved on a curved path if you are the one following it. For instance, if you go around a corner in a car you can feel the acceleration moving you in that path, this is because the acceleration is provided by the force of the car pushing on you.

However if you were on a spaceship coasting past a star, your path would bend around it yet you would not 'feel' this happening, though you could determine that it is the case by seeing that your bearings are changing with respect to distant objects (i.e. the distant stars in front of you). To travel in a 'straight' line would require you using rockets to counteract the effect of the star and you would then feel that acceleration.

To put it another way, Newton's first law states that an object will continue its state of motion if no external force is acted on it, i.e. that if the force on an object is zero then acceleration will be zero. In General Relativity the same thing holds except that the effect of gravity (through the curvature of space-time) alters to direction of the unaccelerated path. In technical terms we call the path in which an observer travels if they are not accelerated a 'geodesic'.

In even more technical terms to geodesic equation (put into words) goes something like this:

Acceleration = Force applied + Effects of spacetime curvature

So for a region with no curvature (and hence no matter in that Universe) or when the Force is much much greater than the effects of curvature on the distance scales of the experiment (true for almost all experiments performed on Earth) this reduces simply to Newtons second law.

Last edited: Sep 14, 2007
6. Sep 14, 2007

### PhanthomJay

Thanks for the clarification. But it appears that one can circumvent the spacetime curvature by simply applying a non gravitational force, such as a rocket thrust, to take the shortcut along the chord. In which case, theoretically, someone travelling at near lightspeed could take the shortcut, using a thrust force, and arrive at his destination faster than the light beam, which is confined to the longer path curve, and thereby, he exceed the speed of light? What is wrong in my thinking?

7. Sep 14, 2007

### genneth

I believe that is actually further, not shorter. Essentially, you end up going slower due to time dilation effects. Welcome to the curvature of spacetime... :grin:

8. Sep 14, 2007

### PhanthomJay

OK, I'll buy it, but let's get away from this relativistic stuff and get back down to earth, if I may, and discuss what happens when I jump down from a roof, using the GR theory of gravity instead of the Newtonian version. Presumably, when i jump, and in the absence of any other forces, am i actually jumping in a curved path, albeit a very miniscule curvature; and if someone else jumped with me at the same time, using a small hand held rocket thrust which applied a force perpendicular(?) to her motion that would be just enough to offfset the curvature such that she fell in a perfectly straight line, would she hit the ground before me, after me, or at the same exact time as measured by a neutral observers clock? (Oops, i'm back into relativity again, sorry).

9. Sep 14, 2007

### Wallace

I'm not sure what you are asking here? If you are dropping towards the Earth you are doing so because of the curvature of space-time caused by the Earth, but you do so on a perfectly straight radial path (unless you add some sideways motion by pushing off in that direction, then you'd be following a parabola). If you used an added force to 'counterattack' the curvature you would be stationary with respect to the Earth. You don't need a rocket to do this, you're doing it right now! Whatever you are standing/sitting on is providing a force upwards.

Note that this is a fundamental difference in interpretation in Newtonian and Relativistic gravity. In Newtonian gravity when you are on the roof you are at rest and when you step of you begin to accelerate due to the force of gravity. In GR the geodesic path is one that moves to the centre of the Earth and hence when you are on the roof you are being accelerated upwards (you can feel this acceleration, stand up long enough and your feet get sore) but once you step off you are now following the unaccelerated geo-desic path.

10. Sep 14, 2007

### PhanthomJay

Thank you, I think i'm running with you now, so bear with me a bit longer. When I step off the roof, I follow the natural spacetime geodesic, which is the radial path toward earth's center, and an external force, such as the normal force, or presumably a thrust force applied during myfall equal and opposite to my Newtonian weight, would allow to me to circumvent the 'curve' and remain stationary or floating, as the case may be, and such force would be applied opposite to my natural radial path, not perpendicular to it as I had incorrectly assumed. But where is this shortcut? If I'm staionary or floating, I'm not going anywhere, so that's not much of a shortcut thru spacetime, I would say. Further, at some point distant from the earth, say on that sun hidden star of that solar eclipse, if I fell from there toward earth, my path is no longer radial toward earth, correct?, it has so many arc seconds of curvature , and in what direction would an external force have to be applied to 'straighten' me out, if you know what I mean? Thanks.

11. Sep 18, 2007

### PhanthomJay

Anyone? If I bypass the geodesic with some sort of external force to get me off that curve and take a shortcut along the chord of the curve, then how can the geodesic be the shortest distance between 2 points if my path is shorter? Are we talking wormholes? I think this curvature must be into a higher order dimension than 3 space, maybe its a curvature into the time dimension?

12. Sep 20, 2007

### pervect

Staff Emeritus
The geodesic path through space-time is the one that maximizes proper time.

Usually we think of a straight line as the shortest distance between two points, but in the Minkowskian geometry of relativity, a straight line is the longest time between two points.

Consider for example the twin paradox of SR. The "stay at home" twin is the one that ages the most, not the travelling twin. So the twin that follows a geodesic path ages the most, rather than the least.

If you want some background reading on this, try some of the sample chapters from Taylor's book "Exploring black holes" available for download here, for instance this one.

I'll give a short quote from one of the chapters which you could download:

So, what does this have to do with your example? Suppose you stand on the ground, and throw one of your watches up in the air, and catch it. The watch follows a geodesic path, but you don't, because you stand on the ground. As a consequence of this, the thrown watch has a maximum (technically, and extremum) of elapsed time. This means that the watch you throw up and catch reads longer than the watch you keep with you.

This is one of the simpler ways to describe what makes a geodesic path "special" - it extremizes proper time.

There is some fine print here on the difference between extremizes and maximizes that I won't get into - there are a few situations where this becomes important, but it would be too much information to present at once, and besides, it's dinner time....

13. Sep 20, 2007

### PhanthomJay

Thank you for the response, but just when i thought I had understood it, you hit me with two apparently conflicting statements:
and then

So in your first statement, it is the stationary twin who ages the most, but in the latter, the traveling twin (i.e, the travelling watch) reads longer, and thus, it ages the most, right? Can you clarify?; thank you for your time.

14. Sep 20, 2007

### pervect

Staff Emeritus
In the first example, I neglected gravity, because it introduces many extra complications. Imagine that your "stay at home" twin is out on a small space station, well away from any planet, star, or any other massive body. This makes the problem much simpler to analyze. I think you can then see that the "stay at home" twin ages the most? You might want to read some of the links I mentioned as well.

In the second example, I introduced gravity. In this case, we see that the person standing on the ground ages slightly less than one following a geodesic path.

In all cases, the twin that ages the most will follow a geodesic path. However, it is not necessarily true that any one particular geodesic path will be the path of maximal aging. (The path will be an "extremum", but it may not be a global maximum).

This is a rather confusing point, so it's best to make sure you understand the simple case first, the case without gravity.

The confusions introduced by gravity are very simlar to the coufusions involved in doing geometry on a curved surface. The shortest distance between two points is always a straight line, but when you have multiple straight lines connecting two points, if you just pick one particular "straight line" at random, that particular line is still "straight", but it may not actually be the shortest distance between two points. This is where the technical defintion of "extremum" comes in.

Imagine two towns with a mountain between them - there is a "straight line" path over the mountain, but it's not the shortest path!

Gravity complicates things in space-time in the same manner as curved geometry complicates distances and the concept of the "shortest distance".

Things are much simpler without gravity. Without gravity, you can regard space-time as flat, and there will only be one geodesic path between any two events, just as there is only one straight line between any two points in a plane.

Last edited: Sep 20, 2007
15. Sep 21, 2007