## Newtonian force as a covariant or contravariant quantity

 Quote by WannabeNewton Again this requires some additional structure; your example is not a archetype of the norm. For an arbitrary smooth manifold M and $p,q\in M$ you can't just subtract p from q to get some vector in $T_{p}(M)$ like you could geometrically in euclidean space.
The point of my example was to illustrate how it is possible to have a space with parallel lines, but no perpendicular lines.

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 Quote by stevendaryl The point of my example was to illustrate how it is possible to have a space with parallel lines, but no perpendicular lines.
Yes, but your final claim was:

 Quote by stevendaryl So for a general abstract manifold, you can always make sense of parallel lines
This is false. You need a connection in order to make sense of this, as wbn mentioned.

 Quote by micromass Yes, but your final claim was: This is false.
Fine. I feel that in a discussion, it is important to have the level of the discussion to be appropriate to the topic at hand, rather than in complete generality. The point of using displacements is that displacements give an intuitive idea of the tangent space, where a notion of "parallel" is always defined (in terms of one vector being a linear multiple of another).

The mathematical definition of the tangent space is a little involved. It's not something that people are familiar with just from learning vectors in the Euclidean cartesian context.

So your clarifications are completely correct, but they are a little unfocused. The people who can appreciate what you're saying are the people who don't need to hear it.

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 Quote by stevendaryl Fine. I feel that in a discussion, it is important to have the level of the discussion to be appropriate to the topic at hand, rather than in complete generality. The point of using displacements is that displacements give an intuitive idea of the tangent space, where a notion of "parallel" is always defined (in terms of one vector being a linear multiple of another). The mathematical definition of the tangent space is a little involved. It's not something that people are familiar with just from learning vectors in the Euclidean cartesian context. So your clarifications are completely correct, but they are a little unfocused. The people who can appreciate what you're saying are the people who don't need to hear it.
It is very dangerous to pretend that a topic is much easier than it actually is. I'm not saying that we should treat each topic in its full generality, but at least we should try not to make statements which are factually incorrect. If you want to talk about "parallellism" and tangent spaces in the easier context of $\mathbb{R}^n$, then this is perfectly fine. But you shouldn't say that it is the same in arbitrary manifolds since it is simply not true. I'm not saying we should actually define general tangent space, connections, etc. But at least, let's try to be precise and correct.

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 Quote by TrickyDicky Thanks for the relevant quotes. I guess I fail to see how this procedure is more geometrical, I have always thought of geometry as something related to distances and angles, and for that you need a metric. Sure there are computations such as the one mentioned about volumes of parallelepipeds that seem to not need the metric, but I think this is a good example of the hidden metric case Burke precisely is trying to avoid.
Projective geometry (that of perspective and vanishing points, etc...) has no metric (akin to the Euclidean metric).
"Geometry" (in Felix Klein's view http://en.wikipedia.org/wiki/Erlangen_program and http://arxiv.org/abs/0807.3161 ) is (paraphrasing) a space and group of transformations of the space, seeking invariants under the group of transformations.

 Quote by Klein p2 But metrical properties are then to be regarded no longer as characteristics of the geometrical figures per se, but as their relations to a fundamental configuration, the imaginary circle at infinity common to all spheres. p4 As a generalization of geometry arises then the following comprehensive problem: Given a manifoldness and a group of transformations of the same; to investigate the configurations belonging to the manifoldness with regard to such properties as are not altered by the transformations of the group. ... Given a manifoldness and a group of transformations of the same; to develop the theory of invariants relating to that group.
which predates Special Relativity. I'm certain that Klein's view influenced Minkowski.

 Hmm, maybe it's just me but if perpendicularity is meaningless without a metric(wich probably is), I don't know how parallelism is any more meaningful, I mean one seems to need something resembling Euclidean space locally (actually its affine generalization) to have the notion of parallelism between lines.
As you mention, parallelism is an affine concept.
The notion of two lines being parallel makes sense (in the same way) in Euclidean, Minkowskian, or Galilean [sorry to bring it up again] space[time]s. It is more primitive.
Next, add additional structure [e.g. a more restrictive group--say, a choice of defining what "rotations" are] to get perpendicularilty, in addition to the already existing notation of parallelism. As you know, "perpendicularity" in Euclidean space is generally different from the other space[time]s.

If you take away the parallelism structure (relaxing the group of transformations), then you allow a more general "geometry", like elliptic/spherical geometry. However, you still have even more primitive structures like "incidence" (whether a point is on a line).

With regard to the volume of parallelepipeds, I think the correct statement is that whether you choose the Euclidean, Minkowski, or Galilean metric, you get the same value for the volume [possibly, up to signs]. So, it really doesn't depend on the metric... but instead on something common to those metrics (something describing the affine structure). Does the "determinant" rely on the choice of metric among these three space[time]s... or just the parallelepiped's vectors themselves?

 Quote by micromass You need a connection in order to make sense of this, as wbn mentioned.
Exactly, and I like to think of this affine connection that is needed in order to differentiate vector fields in differentiable manifolds as a structure that takes advantage of the common property of all manifolds, being locally like an affine Euclidean space.
So it is clear that some extra structure is needed, it just seems natural to me that this connection needed to parallel transport is the canonical Riemannian connection in the vast majority of classical physical problems, be it in configuration space (configuration manifold), Euclidean space, Lorentzian manifolds, etc.
My point was that I don't understand Burke's prejudice against metric tensors and what is it "not geometric" about them that might bother anyone. I agree with him that it should be more enphasized when the Euclidean metric is being implicitly used and when it is not.

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 Quote by TrickyDicky Exactly, and I like to think of this affine connection that is needed in order to differentiate vector fields in differentiable manifolds as a structure that takes advantage of the common property of all manifolds, being locally like an affine Euclidean space. So it is clear that some extra structure is needed, it just seems natural to me that this connection needed to parallel transport is the canonical Riemannian connection in the vast majority of classical physical problems, be it in configuration space (configuration manifold), Euclidean space, Lorentzian manifolds, etc. My point was that I don't understand Burke's prejudice against metric tensors and what is it "not geometric" about them that might bother anyone. I agree with him that it should be more enphasized when the Euclidean metric is being implicitly used and when it is not.
How is it goin Tricky Dicky. Just to clarify, a connection is not the ONLY way to differentiate vector fields on a smooth manifold. Take, for example, the lie derivative. Both offer different geometric notions of "parallelism". I think his issue is just that in coordinate - based computations, there is so much blurring of what quantities should naturally be taken as one - forms and what as vectors because people tend to use the metric tensor to go from one to the other readily.

 Quote by robphy Projective geometry (that of perspective and vanishing points, etc...) has no metric (akin to the Euclidean metric).
Yes, that is a good example of not needing the concepts of angle and distance in geometry generalizations. A geometry only based in those concepts would be very poor. But my point was rather that there wasn't anything wrong with metrics if one wants to stress the geometrical side of something.

 Quote by robphy With regard to the volume of parallelepipeds, I think the correct statement is that whether you choose the Euclidean, Minkowski, or Galilean metric, you get the same value for the volume [possibly, up to signs]. So, it really doesn't depend on the metric... but instead on something common to those metrics (something describing the affine structure). Does the "determinant" rely on the choice of metric among these three space[time]s... or just the parallelepiped's vectors themselves?
Of course it doesn't. But I always think of the cross product when talking about the determinant, is there anything more Euclidean than that?

 Quote by micromass It is very dangerous to pretend that a topic is much easier than it actually is.
I'm not sure that I agree. Progress in physics really was only possible because physicists oversimplified in a way that was good enough for the problems that they were interested in. I think it's important to have a feel for what the limitations are of a particular approach, and sometimes going forward means going backwards and redoing what you've already done, but in a more careful way.

Anyway, I was aiming my comments at the level of someone who is familiar with vectors in the context of Euclidean space, but doesn't realize the full implications of lacking a metric. I think going into the full complexities of differential geometry is the wrong level for the discussion.

 Quote by WannabeNewton How is it goin Tricky Dicky. Just to clarify, a connection is not the ONLY way to differentiate vector fields on a smooth manifold. Take, for example, the lie derivative. Both offer different geometric notions of "parallelism".
Yes, the only difference is that the Lie derivative admits torsion. I was jst thinking that most classical physics situations don't need to include torsion. An exception is the non-mainstream Einstein-Cartan theory that relates torsion with QM spin, but QM spin is not a classical concept.
 Quote by WannabeNewton I think his issue is just that in coordinate - based computations, there is so much blurring of what quantities should naturally be taken as one - forms and what as vectors because people tend to use the metric tensor to go from one to the other readily.
Yeah, agreed. But again, sure, you don't need metrics to talk about differential forms and vectors, but I still don't know if I buy the notion that classical physics quantities must be taken naturally as one or the other independently of the geometry of the physical problem at hand.

 Quote by stevendaryl I think going into the full complexities of differential geometry is the wrong level for the discussion.
This looks a bit condescending with the people in this forum. (Not with poor ignorant me but with many others very capable of handling that level).

 Quote by TrickyDicky This looks a bit condescending with the people in this forum. (Not with poor ignorant me but with many others very capable of handling that level).
I don't mean to be condescending, it's just that in an informal setting, as opposed to writing a textbook, it seems to me that pitching at the right level is more important than being perfectly general. Yes, there's the danger that someone will take something more literally than it was intended, and could be led astray, but I think that people generally can recognize the difference between a textbook style formal definition and an intuitive argument, and should know that the latter is only meant to steer someone in the right direction, rather than to be the definitive last word on the subject.

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 Quote by robphy Here are some quotes from Burke's Applied Differential Geometry that support my statement above
The quotes are great, robphy -- thanks for going to the trouble of posting them. I found the one about forces of constraint to be particularly helpful.

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 Quote by stevendaryl I don't mean to be condescending, it's just that in an informal setting, as opposed to writing a textbook, it seems to me that pitching at the right level is more important than being perfectly general. Yes, there's the danger that someone will take something more literally than it was intended, and could be led astray, but I think that people generally can recognize the difference between a textbook style formal definition and an intuitive argument, and should know that the latter is only meant to steer someone in the right direction, rather than to be the definitive last word on the subject.
If you don't want to be completely literal and precise, then that's perfectly ok. But you should say that you're being imprecise. Nobody benefits from people getting misconceptions.

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 Quote by William Burke A further payoff for having a clear geometric picture of force comes in situations involving constraints. A slippery constraint is one having no frictional forces. One usually says that the constraint force is perpendicular to the constraint surface. Now if you are truly learning to think covariantly with respect to linear transformations, you will see that this is a nonsense statement. Perpendicularity is a meaningless concept in any configuration space that does not accidentally happen to have a metric. The correct geometric view of a constraint is sketched in Figure 18.4. As a 1-form, the constraint force is parallel to the constraint surface, and parallelism is a properly covariant notion. Again, the non-covariant language comes from excessive attention to the peculiar features of particle mechanics in Euclidean space, features which do not generalize.
I wanted to see if I could work out an example of this that was as simple as possible. I think my example makes sense, but maybe others here could tell me if this makes sense and help me smooth out the stuff I'm unsure of.

I actually made two examples. Example #1 is a billiard ball of unit mass in two dimensions, constrained by a diagonal wall to have y<x. The Lagrangian formalism just leads to the expected Newtonian expressions, $p_x=\dot{x}$, $p_y=\dot{y}$. The force of constraint is $F_a=dp_a/dt$. Let $w^a$ be a vector parallel to the wall. Since we do happen to have a metric in this example,we can say that $F_aw^a=0$; most people would say that the force was perpendicular to the wall.

Example #2 is a human arm with a heavy mass gripped in the hand. The upper arm is raised at an angle $\theta$ and the lower arm raised at an angle $\phi$ (both measured relative to the vertical). The arm's weight is negligible compared to the unit mass of the gripped weight, and both the upper and lower arm have unit length. Because of the construction of the elbow joint, we have a constraint $\theta \le \phi$. The conjugate momenta (which are actually angular momenta) turn out to be $p_\theta=\dot{\theta}+\cos(\phi-\theta)\dot{\phi}$ and a similar expression for $p_\phi$. The force of constraint is $F_a=dp_a/dt$. The surface of constraint can be represented by a vector $w^a$, which is on a diagonal line in the $(\phi,\theta)$ plane. Since there is no metric, it doesn't make sense to say that $F_a$ is perpendicular to $w^a$.

Geometrically, Burke has a nice representation of a 1-form as a pair of parallel lines, with one of the two lines marked with an arrowhead (see http://www.scribd.com/doc/37538938/Burke-DivGradCurl ). In this representation, the parallel lines representing the force of constraint are clearly parallel to the surface of constraint.

But I'm not clear on how to notate this idea that in both examples, the force 1-form is parallel to the surface. Do you represent the surface as, say, a 1-form created by taking the (infinite) gradient of a step function across the wall?

Apart from my notational confusion, does the rest of this seem right?
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When I say there's no metric, I mean that there's no metric on the two-dimensional space of $(\phi,\theta)$.