Newtonian force as a covariant or contravariant quantity

TrickyDicky

Quote by stevendaryl

Yes, any parametrized object, you can treat the parameter as another dimension. But there is a much more intimate connection between the dimensions in the case of time, which is that equations of motion explicitly relate what's happening on one time slice to what's happening on another time slice---that's what an "equation of motion" is.

No, it doesn't. A manifold is any object that is locally like [itex]R^n[/itex].

Sure, again, I'm not arguing here that the Galilean model is not a manifold, here I was talking about the parametrized 3-space versus the 4-manifold and wich looks more natural. I guess both views are ok, I just find it more natural to use the former, wich is the one we are used to from classical mechanics, the 4-manifold representation is more of a modern retelling in the light of what we know about relativity and Lorentzian manifolds.

WannabeNewton

Quote by stevendaryl

No, it doesn't. A manifold is any object that is locally like [itex]R^n[/itex].

There are more conditions than just that for a topological space to be a manifold. Anyways it seems like it is just turning out to be an issue of semantics regarding whether galiliean space - time is a "space - time". From what I'm seeing it seems to be a commonly used terminology and I don't see any reason to object it regardless. Also one can take a look at this: http://ls.poly.edu/~jbain/spacetime/....Spacetime.pdf

stevendaryl

Quote by TrickyDicky

But I'm not saying that Galilean "spacetime" is not a manifold, I was objecting to calling it spacetime if by spacetime we understand a manifold with Lorentzian metric.

Oh. Well, that's a matter of taste. Since the manifold includes both space and time, it seems to me that it should be called "spacetime".

stevendaryl

Quote by WannabeNewton

There are more conditions than just that for a topological space to be a manifold.

I don't think so. The definition of an n-dimensional manifold is a topological space such that every neighborhood is homeomorphic to the Euclidean space of dimension n.

WannabeNewton

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

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

I think that's also what Stevendaryl means.

WannabeNewton

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

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

robphy

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

Quote by robphy

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

Quote by atyy

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

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

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 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 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

Thanks for the relevant quotes.

Quote by robphy

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 by robphy

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 by Burke

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 by Burke

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

Quote by TrickyDicky

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.

WannabeNewton

Quote by stevendaryl

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.

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WannabeNewton Recognitions: Gold Member Science Advisor	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 Recognitions: Science Advisor	I think that's also what Stevendaryl means.

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

robphy Blog Entries: 47 Recognitions: Gold Member Homework Help Science Advisor	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.