# GR is confusing me

1. Jul 15, 2013

### Benjam:n

The main problem is when we say a manifold is curved surely this is a relative concept - curved relative to what? I've looked at lots of threads asking similar questions and the answer which keeps coming up is that you can tell if a manifold is curved if you are within that manifold, for instance by constructing a triangle and the angles adding up to more than 180degrees.

My problem with this is surely geodesics are coordinate dependant. For instance if I take a piece of elastic and stretch it between two points that will give the geodesic relative to normal Cartesian coordinates. But if I change coordinates to say spherical coordinates the path of the elastic relative to spherical coordinates would surely be curved and no longer represent the geodesic relative to spherical coordinates. So the elastic band only works for he Cartesian coordinate system.

Why do we trust the elastic band basically, when we look at say a paraboloid and say its curved, how do we know its not us which is curved and the paraboloid which is flat.

2. Jul 15, 2013

### pervect

Staff Emeritus
No
No, the elastic band works for all coordinate systems, as long as it minimizes(**) the distance between the two points (more on this later).

The description in terms of coordinates of the path followed by the elastic band will vary depending on your coordinates, but the unique(*) path that minimizes(**) the distance will be the unique(*) geodesic between two points

There formal statement is that we want the curve that minimizes(**) the distance. The elastic band tends to go to the state of lowest energy(***), and since the energy of the band goes up the longer it is stretched, the state of lowest energy and the state of lowest distance are the same.

* The geodesic is actually only unique if the two points are "near enough". The details of "near enough" are technical. At the moment I don't think it's warranted to get too distracted by these technicalities, so I won't spell them out.

** Minimize is somewhat inexact for reasons related to the point about uniqueness. The details of why "extremized" is preferred over "minimized" are again, technical and distracting.

*** We need to assume the system is static - or at least close enough to being static - so that the inertia of the elastic band is irrelevant for the elastic band to work.

3. Jul 15, 2013

### WannabeNewton

Geodesics are not coordinate dependent. You are thinking of the coordinate representation of a geodesic in a given coordinate system as opposed to the geometric definition of a geodesic. More precisely, if we have a manifold $M$ and a derivative operator $\nabla_{a}$, we say a curve $\gamma$ with tangent $\xi^{a}$ is a geodesic if $\xi^{a}\nabla_{a}\xi^{b} = 0$ i.e. the curve parallel transports its own tangent vector. Clearly this is a coordinate independent statement! What does it mean intuitively? All we're saying is that the tangent vector to a geodesic remains constant along the geodesic which is the natural generalization to manifolds of the primitive statement that a curve has no acceleration if it has a constant tangent vector. Note that some GR texts may define a geodesic as an arc-length extremizing curve; this is only true for manifolds that have a metric. On the other hand, the definition I wrote above holds for any manifold in general; in the general case we refer to such geodesics as affine geodesics.

Let me give a simple example to help try and clear up your possible confusion. Let $M = \mathbb{R}^{2}$. Notice that even before I introduce coordinates, it is obvious that since this space is flat the geodesics will just be straight lines in the plane Euclidean sense. This is a geometric fact independent of any appeal to coordinates.

Now, in cartesian coordinates $(x,y)$ the Christoffel symbols $\Gamma ^{i}_{jk} = 0$ identically so the above geodesic equation simply becomes $\ddot{x}^{i} + \Gamma ^{i}_{jk}\dot{x}^{j}\dot{x}^{k} = \ddot{x}^{i} = 0$ i.e. we just have a straight line $ax + by = c$, as we would expect. Let's now switch over to polar coordinates $(r,\theta)$. Here there are two indepdendent non-vanishing Christoffel symbol components $\Gamma ^{\theta}_{r\theta} = \frac{1}{r}, \Gamma ^{r}_{\theta\theta} = -r$. We end up with the following differential equations, after some work: $\ddot{r} = r\dot{\theta}^{2}$ and $r^{2}\dot{\theta} = \text{const.}$. You might recognize these from Newton's 2nd law in polar coordinates for a non-accelerating particle. The solution to this set of differential equations is given implicitly by $arcos\theta + brsin\theta = c$ (as you can check yourself) which is of course a straight line, yet again.

Last edited: Jul 15, 2013
4. Jul 15, 2013

### Staff: Mentor

Geodesics are not coordinate-dependent. The equation that describes a geodesic will be different in different coordinate systems, but that doesn't make the geodesic coordinate-dependent; I can pick two points and stretch an elastic band between them without ever choosing any coordinate system.

For example: In flat two-dimensional space, $y=2x+1$ describes a geodesic using Cartesian coordinates. The exact same geometric object is described in polar coordinates by $Rsin\theta=1+2Rcos\theta$. Which one I use is completely my choice, based on whichever is more convenient for the problem at hand, but it's the same line either way.

5. Jul 15, 2013

### Benjam:n

Thanks for the help, but I still don't understand.
If I had say polar coordinates and we had two points, which relative to the Cartesian coordinate system are (1,0) and (0,1). Then in the Cartesian coordinates the geodesic is y=-x+1. But surely in the polar coordinates the geodesic is r=1, i.e. x^2+y^2=1 relative to the Cartesian coordinates. Because this is a straight line relative to the polar coordinates. I thought coordinate systems were all equivalent, so why are preferencing (I'm sure this is a real word but its got a red line) straight lines relative to the Cartesian coordinates?

I really don't understand how you can define the notion of quickest path without coordinates. Surely everything we know about the world comes from making measurements and to make measurements you need coordinates.

6. Jul 15, 2013

### micromass

Staff Emeritus
Why do you think the geodesic is $r=1$ in polar coordinates?

7. Jul 15, 2013

### WannabeNewton

That is not a straight line. $r = 1$ is a circle. Again, a straight line is a geometric notion in plane Euclidean geometry. Coordinates have absolutely no relevance; lines are not different in different coordinate systems on the same geometric space. Coordinates don't change the fact that the underlying space is the Euclidean plane.

We have operational devices for measuring certain physical quantities; we can measure acceleration (or lack thereof) using an accelerometer.

8. Jul 15, 2013

### Benjam:n

Because surely that is the straight line between those two points relative to those coordinates. In the Cartesian coordinates if you stay on a line of constant coordinate say x=3, then you move on a straight line. If all coordinates are equivalent then the polar coordinates are equivalent to Cartesian coordinates so why would it be any different?

Also ∫(1+(dr/dθ)^2)^1/2dθ is minimized for r=1.

9. Jul 15, 2013

### WannabeNewton

Yet again, straight lines are intrinsic to the underlying Euclidean space; they have nothing to do with coordinates. Obviously the arc of a circle doesn't minimize the arc-length between two points; the shortest distance between any two points in Euclidean space is a straight line. This is a coordinate independent fact; I can draw the arc of a circle between two points as well as a straight line between those two points and use basic Euclidean geometry to show that the straight line minimizes the arc-length amongst the two. There are no coordinates involved whatsoever. All I need is a ruler (to measure the length of the line segment), a protractor (to measure the angle subtended by the arc which I can then use to find the arc-length by measuring the radius of the arc using the ruler), and then I just have to compare the two.

10. Jul 15, 2013

### Benjam:n

But how do you define a straight line without coordinates?

11. Jul 15, 2013

### micromass

Staff Emeritus
WbN gave a definition in post $3$.

I know this might be a bit difficult. So I highly suggest that you get some differential geometry book and work through that. Something like Do Carmo's curves and surfaces, or O'Neill's Elementary Differential Geometry is very good.

12. Jul 15, 2013

### Staff: Mentor

Put a cat at point A and a piece of catnip at point B. The cat follows the shortest path between point A and point B, regardless of whether you've drawn a Cartesian grid or a set of circles and radial lines on the floor underneath. Cats don't know anything about coordinate systems. At least, I don't think they do. :uhh:

13. Jul 15, 2013

### pervect

Staff Emeritus
Did you see my remarks about a geodesic being the shortest distance between two points? Or did you miss them?

If so, compute the length of the two versions of the straight line. Are they the same length? If not, which one is shorter?

Wannabe Newton's defintions are more precise, the above is (I feel) more likely to be understandable at what I'm guessing is your level of math.

You can work out the equations of a straight line in any coordinate system knowing "only" the calculus of variations, something that's usually presented welll before you get to differential geometry.

Try http://en.wikipedia.org/wiki/Calculus_of_variations for starters. You can work out the curve that minimizes/ extremizes distance in any coordinate system, as long as you have a metric for it.

14. Jul 15, 2013

### Staff: Mentor

No, it is $Rsin\theta = 1-Rcos\theta$. I got this by using the relationship between polar and Cartesian coordinates: $x=Rcos\theta$ and $y=Rsin\theta$ and substituting into $y=-x+1$

Take a sheet of paper; that's a flat two-dimensional surface. Put two tacks in it; those are your two points. Stretch a taut string between the tacks; that's a geodesic. Now you can grab a pencil and a ruler and draw a Cartesian coordinate grid on the paper. You might draw your grid in such a way that one tack is at (1,0) and the other is at (0,1) and then the quation of the geodesic through the two tacks will be $y=-x+1$; but if you draw a different grid, or draw polar coordinates instead of Cartesian, you'll get a different equation, although it's the same line between the same two points.

The important thing here is that the tacks and the string between them came first. You added the coordinate grid afterwards, and only so that you could write down an equation that described the position of the string.

15. Jul 15, 2013

### Benjam:n

Ill try and say what I understand from this. If we have some space lets keep it 2D. We can define independent of coordinates a path through that space, i.e. a sequence of points through which the path goes one after the other. So then the tangent vector at a point is the vector going from that point to the infinitesimally close point in our continuous sequence of points. This is where I cant go any further if you tried to take a derivative like notion then you'd go forward to the infinitesimally close point take its tangent vector and subtract the tangent vector at your original point. Now how do you combine these abstract objects One is the link from your infinitesimally close point to the original point and the other is the link from the infinitesimally close point to the infinitesimally close point beyond that.

16. Jul 15, 2013

### WannabeNewton

Yes if we are working in Euclidean space, that is a perfectly fine way to look at the tangent vectors. They are just infinitesimal arrows from a point on the curve to an infinitesimally separated point on the curve. Now if the curve is to be a geodesic, it is sufficient to check if the tangent vector remains constant along the curve. If this tangent vector is a kind of velocity vector, then this is the rather intuitive statement that the velocity vector is constant along the curve i.e. there is no acceleration along the curve. Now, what are the only curves in the euclidean plane that have constant tangent vectors along them? Just picture everything geometrically.

17. Jul 15, 2013

### Benjam:n

But how do you know if the vectors are the same if we haven't got coordinates? All we have at the moment is little links between infinitesimally close points in a continuous sequence of points.

18. Jul 15, 2013

### WannabeNewton

We can define the derivative of a tangent vector along a curve. In Euclidean space this is just the ordinary derivative. On general manifolds it is given by the derivative operator $\nabla_{a}$. This is how we check if the tangent vector is changing along the curve or not. A derivative does not require coordinates to define in any sense; in Euclidean space it is defined using the $\epsilon-\delta$ definition of a limit and for general manifolds $\nabla_{a}$ is defined as an abstract map satisfying certain properties similar to what regular derivatives satisfy (e.g. Leibniz rule).

19. Jul 15, 2013

### Benjam:n

I know this sounds childish, but I'm quite serious. How do we know that the elastic band isnt only a valid description of a geodesic relative to the Cartesian coordinate system?

20. Jul 15, 2013

### nitsuj

Despite Euclid Geometry being mentioned numerous times already in this thread...here is what is refered to...the [STRIKE]axioms[/STRIKE] postulates (why not an axiom?). Here is some for straight lines. Basically just think of a straight line as a straight line.

Properly defined a few hundred years before Christ. Pulled today from wiki.

"Let the following be postulated":

"To draw a straight line from any point to any point."
"To produce [extend] a finite straight line continuously in a straight line."

Re iterated in language I can even understand from some website

Look at the link to see the illustration of two points & a straight line between them. I believe GR defines straight lines as the shortest "path"/measurement, however probably as time/length; spacetime. So not just familiar length paths, but time paths also.

When discussing physics, allot of terminology & concepts are presumed to be understood based on the initial question. If asking about straight lines/metrics/manifolds in GR, why not know of straight lines in olden days.

Last edited: Jul 15, 2013