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Newtonian force as a covariant or contravariant quantity 
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#73
Jan2813, 11:21 AM

C. Spirit
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There are various definitions of topological manifolds and they all take conditions on top of locally Euclidean. One very common definition is a topological manifold must also be Hausdorff and second countable. Other definitions pertaining to separability but not Hausdorff are also present. In GR we take manifolds that are Hausdorff and second countable as well as locally Euclidean. The condition you stated is not enough.



#74
Jan2813, 11:32 AM

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Also it is not every neighborhood. There exists A neighborhood for every point on the manifold that is homeomorphic to an open subset of R^n



#75
Jan2813, 11:39 AM

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I think that's also what Stevendaryl means.



#76
Jan2813, 11:58 AM

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Then it should be clarified because there is a huge difference between saying each point has a neighborhood homeomorphic to R^n and saying every non empty element of the topology is homeomorphic to R^n.



#77
Jan2813, 02:11 PM

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So what's the consensus on force as a covector if one considers things like friction? Is energy still a useful concept?



#78
Jan2813, 06:54 PM

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It's probably fine.
Not all covectors are the "d" of some potential function... so, presumably friction will be of that type. This lecture note http://bazzim.mit.edu/NR/rdonlyres/A.../lecture10.pdf from bazzim.mit.edu/oeit/OcwWeb/AeronauticsandAstronautics/1661AerospaceDynamicsSpring2003/LectureNotes/index.htm will probably be useful. 


#79
Jan2813, 07:33 PM

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Could one try to argue that in classical GR one should take the Einstein equations as primary, rather than the Hilbert action, because the differential equations are unique, but the Lagrangian formulations of GR are not? 


#80
Jan2813, 09:01 PM

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#81
Jan2913, 05:27 AM

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I certainly can't see why force and momentum would be "naturally" covectors, I just see that in certain physical situations they act as covectors, but just as naturally they act as as tangent vectors in other situations, basically by definition in F=ma, and p=mv, they are both proportional to clearly tangent velocity and acceleration vectors. Now of course just by using a metric tensor one can turn even velocity to a covector, but that's the point that has been made by several people here already, as long as one has a metric tensor it makes no sense to tell apart quantities as vectors or covectors in general. I don't think that is what Burke or the WP is trying to say at all. Burke was simply a very visual physicist that tried to see physics as geometrically as possible and from that POV he tried to make clear how ech physical quantity was acting geometrically in each situation.. It makes perfect sense that if one considers how a field of force acts on something to obtain a scalar (work), that force is acting as a covector field.
To introduce the Galilean 4spacetime here can only bring confusion (though it might be useful in other instances like introducing relativity spacetime from classical mechanics as have been commented earlier) in this context due to its apparently degenerate metric (but a well defined affine connection , that as commented by stevendaryl gives an easy account of the fictitious forces of Newtonian mechanics) 


#82
Jan2913, 07:34 AM

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I think Burke is trying to find the correct geometrical model of a physical quantity.
To do so, he asks what is the minimum needed structure to perform a certain calculation of physical quantities. In particular, he asks if a metric is needed. If not, then he seeks the metricfree formulation and regards that as more fundamental, which then suggests the appropriate geometrical representations. Then, if there happens to be a metric is available, he certainly does not want to blur the distinctions because of that... that is to say, for forces, force is always a 1form... and he'll never hide any use of the metric. Here are some quotes from Burke's Applied Differential Geometry that support my statement above http://books.google.com/books?id=58S...metric&f=false and quotes from Burke's Spacetime, Geometry, Cosmology http://books.google.com/books?id=nDGuQgAACAAJ (highlighting mine) (apologies for the excessive quoting... but these might clarify [my interpretation of] Burke's approach.) 


#83
Jan3013, 07:27 AM

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Thanks for the relevant quotes.



#84
Jan3013, 08:55 AM

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Suppose I have four states: [itex]S_1 = (P_1,V_1)[/itex], [itex]S_2 = (P_2,V_2)[/itex], [itex]S_3 = (P_3,V_3)[/itex] and [itex]S_4 = (P_4,V_4)[/itex] Define: [itex]\delta(P)_{12} = P_2  P_1[/itex] [itex]\delta(V)_{12} = V_2  V_1[/itex] [itex]\delta(P)_{34} = P_4  P_3[/itex] [itex]\delta(V)_{34} = V_4  V_3[/itex] To say that the line from [itex]S_1[/itex] to [itex]S_2[/itex] is parallel to the line from [itex]S_3[/itex] to [itex]S_4[/itex] is to say that there is some nonzero real number [itex]\lambda[/itex] such that: [itex]\delta(P)_{34} = \lambda \delta(P)_{12}[/itex] [itex]\delta(V)_{34} = \lambda \delta(V)_{12}[/itex] So parallel displacement vectors are defined for this space. On the other hand, how would you define perpendicularity for displacement vectors? What does it mean to say that the line from [itex]S_1[/itex] to [itex]S_2[/itex] is perpendicular to the line from [itex]S_3[/itex] to [itex]S_4[/itex]? If the points were points in Euclidean space, and the coordinates were Cartesian, then we could say that the displacements are perpendicular if [itex]\delta(P)_{12} \delta(P)_{34} + \delta(V)_{12}\delta(V)_{34} = 0[/itex] But that equation doesn't even make any sense for pressures and volumes. You can't add a square pressure to a square volume. In order to make sense of adding squared pressures and squared volumes, you need a conversion factor that relates pressure to volume. So for a general abstract manifold, you can always make sense of parallel lines, but you can't always make sense of perpendicular lines. 


#85
Jan3013, 09:14 AM

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#86
Jan3013, 09:33 AM

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#87
Jan3013, 09:38 AM

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#88
Jan3013, 09:44 AM

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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. 


#89
Jan3013, 11:44 AM

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#90
Jan3013, 11:47 AM

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"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. 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 groupsay, 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? 


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