Newtonian force as a covariant or contravariant quantity


by bcrowell
Tags: contravariant, covariant, force, newtonian, quantity
WannabeNewton
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Jan28-13, 11:21 AM
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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.
WannabeNewton
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Jan28-13, 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
haushofer
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Jan28-13, 11:39 AM
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I think that's also what Stevendaryl means.
WannabeNewton
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Jan28-13, 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.
atyy
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Jan28-13, 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?
robphy
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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/Aeronautics-and-Astronautics/16-61Aerospace-DynamicsSpring2003/LectureNotes/index.htm
will probably be useful.
atyy
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Quote Quote by robphy View Post
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/Aeronautics-and-Astronautics/16-61Aerospace-DynamicsSpring2003/LectureNotes/index.htm
will probably be useful.
Perhaps more an aesthetic question - but is this natural or kludgey?

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?
stevendaryl
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Jan28-13, 09:01 PM
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Quote Quote by atyy View Post
Perhaps more an aesthetic question - but is this natural or kludgey?

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?
Lagrangians are never unique. You can always add terms to the Lagrangian that have no effect on the equations of motion.
TrickyDicky
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Jan29-13, 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 4-spacetime 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)
robphy
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Jan29-13, 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 metric-free 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 1-form... 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.)

Quote Quote by Burke ADG
p.xii
Now, to compare the volumes of two parallelopipeds does not require a metric structure. A linear structure is sufficient.

p.xiii
This emphasis on concrete applications and proper geometric structures helps us avoid the formal symbol manipulations that so often lead to nonsense or fallacious proofs of correct results. [Look at Figure 3.1 in Soper (1976) or the horrible calculus of variations manipulations and mistakes in Goldstein (1959).] Here we will be able to turn most of the
infinitesimals commonly seen in physics into the appropriate geometric objects, usually into either rates (tangent vectors) or gradients (differential forms). The distinction between these is lost in the metric-blinded symbol pushing of tensor calculus.

p.1
The mathematics of this book can be thought of as the proper generalization of vector calculus, div. grad. curl, and all that, to spaces of higher dimension. The generalization is not obvious. Ordinary vector calculus is misleading because the vector cross product has special properties in three dimensions. This happens because, for n = 3, n and (1/2)n(n -1) are equal. It is also important to divorce the formalism from its reliance on a Euclidean metric, or any metric for that matter.
...

Also, a metric allows some accidental identifications that obscure the natural properties of the geometric structures.


p.20
A blindness caused by the unnecessary use of metrics afflicts many physicists; so they do not distinguish between vectors and covectors. It does take practice to develop an eye that sees relations in a metric-free fashion.


p.271
Every classical field theory must have a force law. Electrodynamics is a very special classical field theory. Force is geometrically a 1-form. Think of it either as the rate of change of momentum or as the operator taking displacements into energy changes.

p.311
Not all forces come from the gradients of potentials. In general, we only have a force 1-form telling us how much it costs to "push the system" in different directions.
... [ abstract calculation involving differential forms]
Any force that is not derivable from a potential can be treated this way. If the force in configuration space is given by f dq, [then we lift it up to a force on the contact bundle of f alpha, and find the system motion from the preceding exterior differential system, equation 47.1.]

Quote Quote by Burke SGC
p.97
Some forces are derived from potentials. Such forces are clearly potentials gradients, that is, 1-forms. The amount of work done by a system in moving along a path in configuration space can be found by counting the net number of contour lines crossed. Any force field that can be derived from a potential does no work on a system which moves around any closed path. Not every force field can be written as the gradient of a potential. The force field sketched in Figure 18.3 cannot be derived from a potential. At any single point, however, any 1-form could have come from any number of functions.

How is it that we have been able to think of force as a tangent vector? If we are dealing with a configuration space which has a natural Euclidean space Euclidean geometry, then we can describe force as a tangent vector. The Euclidean metric lets us associate a tangent vector to every 1-form. This will be shown when we discuss metric tensors. We all learn mechanics by first studying the mechanics of a particle moving in Euclidean space. We must realize that this is a special situation.

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.

[from p xv. "Covariance": The changes in the rules and mathematical structures as one goes from one equivalent representation to another. The idea is that the representation is changing, whereas the physical situation is not.]
TrickyDicky
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Jan30-13, 07:27 AM
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Thanks for the relevant quotes.

Quote Quote by robphy View Post
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 metric-free formulation and regards that as more fundamental,
which then suggests the appropriate geometrical representations.
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.
Quote Quote by robphy View Post
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 1-form... and he'll never hide any use of the metric.
I'm all for not hiding the use of metrics, somthing that is sistematically done in physics textbooks. But as long as one uses a metric the assertion "force is always a 1-form" makes no sense. And in physics we need a metric, all measurements are done in the form of lengths or angles or can be reduced to them.



Quote Quote by Burke View Post
p.xiii
Here we will be able to turn most of the
infinitesimals commonly seen in physics into the appropriate geometric objects, usually into either rates (tangent vectors) or gradients (differential forms). The distinction between these is lost in the metric-blinded symbol pushing of tensor calculus.

p.20
A blindness caused by the unnecessary use of metrics afflicts many physicists; so they do not distinguish between vectors and covectors. It does take practice to develop an eye that sees relations in a metric-free fashion.
I think here Burke misses the target, I can't see how being aware of the metric blinds anyone, on the contrary, not being aware that the metric tensor is acting can lead to confusion as Burke himself admits.

Quote Quote by Burke View Post
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.
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.
stevendaryl
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Quote Quote by TrickyDicky View Post
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.
To see the problem with perpendicularity, let's take a look at a manifold having nothing to do with spatial distances. Imagine that your manifold represents thermodynamic states of some system (say, a certain quantity of a gas). We can label the states by a pair of numbers [itex](P,V)[/itex] representing the pressure and the volume.

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.
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Quote Quote by stevendaryl View Post
So for a general abstract manifold, you can always make sense of parallel lines, but you can't always make sense of perpendicular lines.
Again this requires some additional structure; your example is not a archetype of the norm. For an arbitrary smooth manifold M and [itex]p,q\in M[/itex] you can't just subtract p from q to get some vector in [itex]T_{p}(M)[/itex] like you could geometrically in euclidean space. Neither can you just simply subtract some [itex]v\in T_{p}(M)[/itex] from some [itex]u\in T_{q}(M)[/itex]. If you want to compare vectors that exist in different tangent spaces to the smooth manifold then you have to first define a connection [itex]\triangledown [/itex] to go with M (you also need this connection to even talk about "lines" i.e. geodesics). The sense of parallelism that makes sense without that extra structure, and is the kind I'm sure the author is talking about, is just plain old linear dependence / independence of some [itex]u,v\in T_{p}(M)[/itex] which we can make sense of simply by use of the vector space structure of the tangent space. If in addition we had a metric tensor defined on M, which is another additional structure that the author says need not be there a priori, then this metric tensor at each point is just an inner product on the tangent space at that point and of course we can then talk about orthogonality.
stevendaryl
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Quote Quote by WannabeNewton View Post
Again this requires some additional structure; your example is not a archetype of the norm. For an arbitrary smooth manifold M and [itex]p,q\in M[/itex] you can't just subtract p from q to get some vector in [itex]T_{p}(M)[/itex] 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.
micromass
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Jan30-13, 09:38 AM
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Quote Quote by stevendaryl View Post
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 Quote by stevendaryl View Post
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.
stevendaryl
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Quote Quote by micromass View Post
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 Quote by stevendaryl View Post
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 [itex]\mathbb{R}^n[/itex], 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 Quote by TrickyDicky View Post
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 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?


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