Curvature of Manifolds: Understanding Relative Concepts

[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to say spherical coordinates the path of the elastic relative to spherical coordinates would surely be curved and no longer represent the geodesic relative to spherical coordinates.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.

 

[pervect]

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.

No

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to say spherical coordinates the path of the elastic relative to spherical coordinates would surely be curved and no longer represent the geodesic relative to spherical coordinates.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.

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

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.

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.

 

[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.

 

[Nugatory]

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.

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.

 

[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.

 

[micromass]

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.

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

 

[WannabeNewton]

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?

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.

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.

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

 

[Benjam:n]

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

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.

 

[WannabeNewton]

Because surely that is the straight line between those two points relative to those coordinates.

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.

 

[Benjam:n]

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.
.

But how do you define a straight line without coordinates?

 

[micromass]

But how do you define a straight line without coordinates?

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.

 

[jtbell]

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.

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.

 

[pervect]

But how do you define a straight line without coordinates?

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.

Example

In order to illustrate this process, consider the problem of finding the extremal function y = f (x) , which is the shortest curve that connects two points (x1 , y1) and (x2 , y2) . The arc length of the curve is given by

 

[Nugatory]

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.

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.

 

[Benjam:n]

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!

Ill try and say what I understand from this. If we have some space let's 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 can't 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.

 

[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.

 

[Benjam:n]

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.

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.

 

[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).

 

[Benjam:n]

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.

I know this sounds childish, but I'm quite serious. How do we know that the elastic band isn't only a valid description of a geodesic relative to the Cartesian coordinate system?

 

[nitsuj]

But how do you define a straight line without coordinates?

Despite Euclid Geometry being mentioned numerous times already in this thread...here is what is referred 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.

 

[martinbn]

I think it is not general relativity that confuses you, it is geometry. May be you should read up on it.

 

[Benjam:n]

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).

Ok, so in the little world, which is two dimensional, were we have the same number of points as in R2, i.e. we could set up a mapping which maps every point in our manifold to a point on R2. But at the moment I haven't defined anything, so there's no order to the points - its just a set of points. Even without any order I could still define a path, as a continuous sequence of these points. Then I could construct tangent vectors and derivatives in the way I described in the last post. I really cannot see from what I've created so far, that you could decide whether those tangent vectors were all the same. What condition are you using?

 

[Benjam:n]

I think it is not general relativity that confuses you, it is geometry. May be you should read up on it.

I really don't like books.

 

[WannabeNewton]

All I need is the fact that $\mathbb{R}^{n}$ is a normed vector space; there are no coordinates being invoked here. As a normed vector space, I can use the norm to define a metric (in the analysis sense). Using the metric I can define what a limit is and using the notion of a limit I can define what a derivative is. Then the rest follows.

 

[Benjam:n]

All I need is the fact that $\mathbb{R}^{n}$ is a normed vector space; there are no coordinates being invoked here. As a normed vector space, I can use the norm to define a metric (in the analysis sense). Using the metric I can define what a limit is and using the notion of a limit I can define what a derivative is. Then the rest follows.

But wouldn't a vector space have an order to the points?

 

[WannabeNewton]

There is no a priori order relation imposed on vector spaces; there is nothing in the definition of a vector space that requires an order relation. Regardless, I don't see what that has to do with coordinate-free definitions.

 

[Nugatory]

I know this sounds childish, but I'm quite serious. How do we know that the elastic band isn't only a valid description of a geodesic relative to the Cartesian coordinate system?

The geodesic is a description of the elastic band and not the other way around. The elastic band and jtbell's cat in post #12 don't care what kind of coordinate axes you draw around them.

 

[nitsuj]

I really don't like books.

Yes, it's not the books per-se, it's the concepts within the books. So you really don't like concepts, specifically reading about them...there is some irony here somewhere

 

[Russell E]

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

Yes, but that's not the function you need to minimize. What you've written there is minimizing the integral of the square root of (dr)^2 + (dtheta)^2, as if that represents the incremental distance ds along a path of dr and dtheta. But that's wrong, because we can actually perform measurements of the distance ds along a path for a given dr and dtheta, and we find that the actual relationship (called the metric) is (ds)^2 = (dr)^2 + r^2 (dtheta)^2. Notice the r^2 multiplying the dtheta term. So to find a geodesic path we need to find a path along which the integral of the square root of this expression is minimized. The solutions are the same paths you get if you minimize the integral of the square root of (dx)^2 + (dy)^2 in Cartesian coordinates. For any given system of coordinates there is a corresponding metric which gives the incremental path length in terms of incremental changes in the 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.

Sure, and the measurements give you the ds corresponding to any incremental changes in your coordinates, which represents the metric for those coordinates, and you can then determine the curvature at any location from that metric, which answers your original question.

 

[Benjam:n]

Can I ask someone to really clearly define what the notion of two parallel vectors is in curved spaces. Straight lines no longer exist in curved spaces, except infinitesimal ones - so we changed the useful concept to geodesics. Parallel vectors (infinitesimal vectors) surely also won't exist in curved spaces, except at the same point - so what concept is it replaced with?

Also with regards to christoffel symbols/symmetric portion of connection, I know how to calculate them, but all my understanding of them is that they are about a coordinate system (y) as seen relative to another coordinate system (x) i.e. d2xi/dyndym, where I, n and m are indexs. However I know the formula by which they are given in terms of the metric so are they deeper than that, i.e. is it possible to define them on a manifold without coordinates and if so why, what really are they describing about that space?

 

[nitsuj]

Can I ask someone to really clearly define what the notion of two parallel vectors is in curved spaces. Straight lines no longer exist in curved spaces - so we changed the useful concept to geodesics.

It's hand waving, but those "curved" lines are actually geometrically straight in 4D (or at least temporally & spatially).

It depends on what the illustration of "curved" geodesics is illustrating, but the curve maybe just in time, just in space, or a combination of the two...that said if it's geodesic...it's straight...

For parallel I'd assume the space time interval between the lines is constant.

Re-reading your post The Big words make me think you know better then me...

 

[WannabeNewton]

Can I ask someone to really clearly define what the notion of two parallel vectors is in curved spaces. Straight lines no longer exist in curved spaces - so we changed the useful concept to geodesics. Parallel vectors surely also won't exist in curved spaces, except at the same point - so what concept is it replaced with?

Precisely, the classical Euclidean notion of parallelism only exists at a single point on curved manifolds. There is simply no notion of vectors at different points of space-time being parallel in the Euclidean sense because they live in different vector spaces which have different inner products. What we can do however is parallel transport a vector along some curve from one point to another and then compare the transported vector at the terminal point to vectors that live at the terminal point. What does it mean to parallel transport a vector along a curve? Well let's say $M$ is a smooth manifold and $\nabla_{a}$ is a derivative operator on $M$. If $\gamma$ is a curve and $V^{a}$ is a vector at some point $p$ that the curve passes through, we say it is parallel transported along $\gamma$ if $\xi^{a}\nabla_{a}V^{b} = 0$ where $\xi^{a}$ is the tangent vector to $\gamma$.

...i.e. is it possible to define them on a manifold without coordinates and if so why, what really are they describing about that space?

Christoffel symbols only have meaning if you are given a coordinate chart.

 

[Nugatory]

Can I ask someone to really clearly define what the notion of two parallel vectors is in curved spaces. Straight lines no longer exist in curved spaces, except infinitesimal ones - so we changed the useful concept to geodesics. Parallel vectors (infinitesimal vectors) surely also won't exist in curved spaces, except at the same point - so what concept is it replaced with?

Google around for "parallel transport" and "tangent space".

We can attach a flat surface (two-dimensional plane if you're working with a curved two-dimensional surface like the surface of a sphere, which you've been doing throughout much of this thread) to each point on the curved surface, work with straight lines and non-infinitesimal vectors in this tangent space. The trick is that each point on the curved surface has its own tangent space, so if we want to compare two vectors at two different points for parallelism, we have to use a mathematical procedure called parallel transport to properly transform one of them into the other one's tangent space. (WannaBeNewton will be gagging at this point, but you did ask for a qualitative answer here).

 

[WannabeNewton]

(WannaBeNewton will be gagging at this point, but you did ask for a qualitative answer here).

lmfao! Well we both said the same exact thing so yey !

 

[pervect]

Can I ask someone to really clearly define what the notion of two parallel vectors is in curved spaces. Straight lines no longer exist in curved spaces, except infinitesimal ones - so we changed the useful concept to geodesics. Parallel vectors (infinitesimal vectors) surely also won't exist in curved spaces, except at the same point - so what concept is it replaced with?

Instead of parallel vectors, in curved space you have parallel transport. Parallel transport transports a vector along a curve. There is a geometric interpretation of this in the absence of torsion called Schild's ladder.

See https://www.physicsforums.com/showthread.php?p=4214084#post4214084 for a brief description.

In a flat space, if you parallel transport a vector along two different paths, you get the same vector. Hence you have a general and path-independent notion of what it means to be parallel. In a curved space, parallel transporting a vector along two different paths does not give the same result.

Also with regards to christoffel symbols/symmetric portion of connection, I know how to calculate them, but all my understanding of them is that they are about a coordinate system (y) as seen relative to another coordinate system (x) i.e. d2xi/dyndym, where I, n and m are indexs. However I know the formula by which they are given in terms of the metric so are they deeper than that, i.e. is it possible to define them on a manifold without coordinates and if so why, what really are they describing about that space?

Christoffel symbols rely on coordinates for their definition AFAIK. I believe that the coordinate independent notion equivalent to a Christoffel symbol is called a connection and/or a fiber bundle.

 

[WannabeNewton]

Christoffel symbols rely on coordinates for their definition AFAIK. I believe that the coordinate independent notion equivalent to a Christoffel symbol is called a connection and/or a fiber bundle.

Indeed if $\nabla$ is a connection on a smooth manifold $M$ and $(U,(x^{i}))$ is a coordinate chart, the Christoffel symbols of the coordinate basis $(e_i) = (\frac{\partial }{\partial x^{i}})$ are defined by $\nabla_{e_{i}}e_{j} = \Gamma ^{k}_{ij}e_{k}$. Connections can be defined on fiber bundles, which are vast generalizations of tangent bundles (for the purposes of classical GR, one could think of the connection as a map on the space of smooth sections of the tangent bundle satisfying certain properties).

 

[Benjam:n]

Precisely, the classical Euclidean notion of parallelism only exists at a single point on curved manifolds. There is simply no notion of vectors at different points of space-time being parallel in the Euclidean sense because they live in different vector spaces which have different inner products. What we can do however is parallel transport a vector along some curve from one point to another and then compare the transported vector at the terminal point to vectors that live at the terminal point. What does it mean to parallel transport a vector along a curve? Well let's say $M$ is a smooth manifold and $\nabla_{a}$ is a derivative operator on $M$. If $\gamma$ is a curve and $V^{a}$ is a vector at some point $p$ that the curve passes through, we say it is parallel transported along $\gamma$ if $\xi^{a}\nabla_{a}V^{b} = 0$ where $\xi^{a}$ is the tangent vector to $\gamma$.
.

Can I just run past you my understanding of the covariant derivative and you can correct any misunderstandings if they're there. Take some vector field in a 2D space Vi(x1,x2), we can then quite easily take its derivative with respect to x1 and x2. Now produce a new coordinate system (y1,y2), such that the two are connected by a smooth transition function yi(x1,x2). Then if we want to see the covariant derivative relative to that new frame what we do is transform both the vector at the point and the vector infinitesimally close to that point as though they were both at that same original point (I think this is bringing them both into the same tangent space, but precisely what tangent space means I'm not 100% sure ). This means that we transform the vector into y, dyj/dxiVi(x1,x2) and then differentiate, but the transformation bit is now treated as though it were constant. So it becomes dyj/dxid/dxn(vi(x1,x2)) and then you've just contracted it with xi n, to get the change in the direction of the tangent vector. I am not to sure on the significance of this then being equal to zero. In the example above what it seems to mean is that there is some coordinate system in which the vectors will be seen as actually being parallel? But how does this work on the surface of a sphere. let's do the typical example of taking a pen at the north pole, following a great circle down to the equator moving along the equator and then going back to the north pole and the pen will point in a different direction. How do you actually calculate this derivative here and show for that path it was equal to zero?

 

[Benjam:n]

Because its different to in my example. In my example I just had two coordinates in the same space. Now we just have a curved space embedded in flat 3d space and I don't really understand how the concept can be used?

 

[WannabeNewton]

I'm afraid I don't quite understand what you're saying. The covariant derivative in and of itself is a coordinate independent map defined for smooth manifolds. Only the Christoffel symbols are relative to a coordinate system.

See here for a detailed calculation of parallel transport on $S^{2}$: http://www.scribd.com/doc/57524972/Parallel-Transport

 

[Benjam:n]

Another question. Why do we need the Riemann curvature tensor to measure curvature It took me a good few hours just to calculate a couple of components of it for a curved 2d space. It seems as if its just overly complex. Also the christoffel symbols alone could do the same job surely, since if your in flat space then the christoffel symbol of one coordinate system relative to itself is zero, which isn't true in curved space?

Also I've just about got an intuitive sense for the Riemann tensor. Could anyone shed some light on the Ricci tensor and the curvature scalar and what they mean intuitively (I know there maths definitions)

 

[WannabeNewton]

The Christoffel symbols don't tell us anything about curvature by themselves; they aren't even coordinate independent! How can you make any geometric statement about a manifold using Christoffel symbols as such? They are rather meaningless quantities as far as geometry goes. The Riemann curvature tensor is the most natural curvature tensor you can construct and it can be related to the Gaussian curvature. For 2d-surfaces, you have arguably the most beautiful theorem in geometry to aid you along: http://en.wikipedia.org/wiki/Theorema_egregium

As for the Ricci curvature, see here: https://www.physicsforums.com/showthread.php?t=698693&highlight=ricci+curvature

 

[WannabeNewton]

By the way, I really think you could learn a lot more efficiently if you picked up a textbook on Riemannian geometry. There are a couple of texts that assume the minimal amount of topology knowledge (e.g. Do Carmo) but I don't know your math background.

 

[Benjam:n]

The way I understand the christoffel symbol is as the derivative of the basis vectors. If your in a curved space only certain vectors are allowed at each point (those in the tangent space) ergo my conclusion was that the basis vectors must change and that the christoffel symbol would not be zero?

 

[WannabeNewton]

Even in curved space I can find a coordinate system in which the Christoffel symbols vanish identically at a point (these are called Riemann normal coordinates) but the Riemann curvature tensor will still not necessarily vanish at that point. Like I said, they are coordinate dependent quantities so they by themselves can't be used to say anything unequivocal about the geometry of a (pseudo) Riemannian manifold and curvature is one of the most fundamental geometric properties of a (pseudo) Riemannian manifold.

From the physics side of things, in GR we want to abide by general covariance but explicitly incorporating Christoffel symbols into a physical equation does not abide by general covariance (see Wald 4.1)

 

[stevendaryl]

The way I understand the christoffel symbol is as the derivative of the basis vectors. If your in a curved space only certain vectors are allowed at each point (those in the tangent space) ergo my conclusion was that the basis vectors must change and that the christoffel symbol would not be zero?

There are two different notions: curved space (or spacetime) and curvilinear coordinates. If the Christoffel symbol is nonzero, that means that you are using curvilinear coordinates, but it doesn't mean that your space is curved. For example, on the flat 2D plane, you can use the curvilinear coordinates r and \theta, and have nonzero Christoffel symbol.

The other way around is true, also. In curved spacetime, the Christoffel symbols may all be zero at a particular point, but that doesn't mean that there is no curvature.

The actual test for curvature is a particular quantity, the Riemann tensor, which is constructed from taking derivatives of the Christoffel symbols. (I should clarify that the components can be computed in terms of derivatives of the Christoffel symbols). In flat spacetime, the Riemann tensor is identically zero, no matter what coordinate system you use to compute the components.

 

[pervect]

There are two different notions: curved space (or spacetime) and curvilinear coordinates. If the Christoffel symbol is nonzero, that means that you are using curvilinear coordinates, but it doesn't mean that your space is curved. For example, on the flat 2D plane, you can use the curvilinear coordinates r and \theta, and have nonzero Christoffel symbol.

The other way around is true, also. In curved spacetime, the Christoffel symbols may all be zero at a particular point, but that doesn't mean that there is no curvature.

The actual test for curvature is a particular quantity, the Riemann tensor, which is constructed from taking derivatives of the Christoffel symbols. (I should clarify that the components can be computed in terms of derivatives of the Christoffel symbols). In flat spacetime, the Riemann tensor is identically zero, no matter what coordinate system you use to compute the components.

The Riemann tensor measures the failure of parallel transport to commute.

To put it a bit more simply suppose you go "north" for 100 miles, and "east" for 200, in that order. Do you wind up at the same spot if you reverse the order, go east 200 , THEN go north 100?

If the answer is no, you are definitely on a curved surface. For instance, if you do that on the Earth's surface, you'll find that you don't wind up at the same spot - because it's curved.

To be able to pull this off, you need to know what "north" and "east" are. I've worded my above example for "ease of understanding", which means it'll be easier for some, and less clear and rigorous for others. The more rigorous approach would be to say that you travel north, parallel transporting an east vector , then go east along the vector you just transported. Similarliy, on the other leg of the trip, you parallel transport the north vector while you go east, then you go north. So in my example, I was envisioning the Earth, and starting out at a point on the equator.

Winding up in different spots when you change the order mean that the "north" and "east" vectors, do not commute.

North and east here are actually defined locally, then parallel transported as needed.

If all vectors commute where you're at, you are on a flat spot of whatever surface you're on (and all the components of the Riemann tensor are zero at wherever you're at).

IF all vectors commute everywhere, you're on a flat surface.

 

[Benjam:n]

I'm struggling with why covariant differentiation doesn't commute in curved spaces. To define a covariant derivative - you pick a coordinate system in which this new derivative is just going to equal the normal derivative in that coordinate system. Then from that you can define christoffel symbols of all other coordinate systems and the covariant derivative follows. From this definition I can see that what it effectively is is you take your infinitesimally close vector transport it keeping its components the same to the original point and then transform it into the new coordinates as though it were at that original point and then take the difference between the two vectors. I thought then that it would be this odd way of transforming the vectors that causes the non-commutatively, but then flat space would have the same problem. I presume then that it must be the problem that the vector that your transporting doesn't actually exist in the tangent space of your original point. So how do you actually pick that new vector - is it just the one with the greatest component in the same direction, i.e. that maximises the dot product?

 

[WannabeNewton]

You should really stop thinking in terms of coordinates. Geometry is much more important than coordinates; the latter are nothing but computational tools. The geometric concepts are much more fundamental; I'm not sure if this is a bias on my part from reading so many of Geroch's notes and papers on GR but still. Anyways, see the following screenshot from Wald's GR text: http://postimg.org/image/58tua25td/ The failure of the commutativity of the covariant derivative is related to the presence of curvature in the above manner.

 

[russelljbarry15]

A manifold is a local area that that looks Euclidean in a small area. When you drive around your neighborhood, mountains and hill not included, you only think about two dimensions. So locally, you consider it flat. Manifolds and curvature are two different things. Manifolds are really a fancy word for a surface. A manifold is nothing more than a continuous space of points that may be curved globally, but locally look like plain old flat space.

When you work with the standard vectors i, j, k, they do not change when you move around. But for example in spherical coordinates the basis vectors change when you move around. So in Euclidean geometry the Christoffel symbols do not change they are zero. But in spherical coordinates the Basis Vectors change and this is what the Christoffel symbols are. In spherical coordinates the basis vectors change so when you differentiate the vector you also have to differentiate the basis vector. The Christoffel symbols are the change in the basis vectors. So using the standard i,j,k vectors the Christoffel symbols are zero. Remember in spherical coordinates the basis vectors change, so they also need to be differentiated.

Curvature is something else. If you draw a triangle on a flat piece of paper the sides add up to 180 degrees. If you bend the paper a little, the triangle sides no longer add up to 180 degrees. Depending on the curvature it will be either less or more than 180 degrees. So the space is no longer flat. I hope this helps.

 

[WannabeNewton]

A manifold is a local area that that looks Euclidean in a small area. When you drive around your neighborhood, mountains and hill not included, you only think about two dimensions. So locally, you consider it flat. Manifolds and curvature are two different things. Manifolds are really a fancy word for a surface. A manifold is nothing more than a continuous space of points that may be curved globally, but locally look like plain old flat space.

A manifold is a much, much more general notion than this. A manifold is not necessarily a surface (a surface is a topological 2-manifold) nor does the notion of curvature even need to have any meaning for a manifold.