Newtonian force as a covariant or contravariant quantity


by bcrowell
Tags: contravariant, covariant, force, newtonian, quantity
bcrowell
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Jan25-13, 03:02 PM
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I recently came across a very cool book called Div, Grad, and Curl are Dead by Burke. This is apparently a bit of a cult classic among mathematicians, not to be confused with Div, Grad, Curl, and All That. Burke was killed in a car accident before he could put the book in final, publishable form, but it can be found in various places online, e.g., on scribd.com. At the beginning of section 12, in the context of Newtonian mechanics, he says,

We now come to the first surprise. Force is not a vector, but a 1-form. The most direct way to see this is to think of the work done by a force. Force is the operator that takes in a displacement, a vector, and tells you how much work was done. This makes forces dual to vectors, i.e., 1-forms.
In the language of Einstein-style index gymnastics, applied in a nonrelativistic context, this amounts to a statement that energy is a scalar, and displacement is a contravariant (upper-index) vector, so force should naturally be considered as a covariant (lower-index) vector.

The first thing I'm unsure about here is whether energy is really a scalar in the context of nonrelativistic mechanics, if "scalar" is taken to have its full Einstein-style interpretation of "invariant under any change of coordinates." Doesn't nonrelativistic energy change when you rescale your coordinates, implying that it's a scalar density rather than a true scalar? (After all, in relativity, rescaling coordinates changes all the components of the energy-momentum tensor, which means it changes the mass-energy.)

The second thing that bugs me is that if you were to reason from Newton's second law, it seems like you would "naturally" conclude the opposite, that force is a contravariant vector.

If we follow the usual but arbitrary convention of saying that upper indices are used for distances measured on a ruler, then this breaks the otherwise perfect symmetry between vectors and their duals. It then seems clear that things like velocity, which can be obtained by differentiation with respect to a scalar, should also take upper indices (be contravariant). But I'm less convinced that this then breaks the duality symmetry in the case of Newtonian force, or a relativistic quantity like the stress-energy tensor...?
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WannabeNewton
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Jan25-13, 03:25 PM
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I think here when we talk about [itex]W({x}) = \int_{{x_{0}}}^{{x}}F_{i}dx^{i}[/itex] being a scalar, it just means it is a real valued function defined on the image of the curve that the particle is traveling on and not necessarily a coordinate invariant quantity. Note, regarding your second point if I understood the question correctly, that when we are dealing with [itex]\mathbb{R}^{n}[/itex], the dual vectors and vectors are naturally identified.
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Jan25-13, 03:31 PM
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A similar but slightly different example...

This WP article http://en.wikipedia.org/wiki/Covaria...nce_of_vectors says,

Examples of vectors with contravariant components include the position of an object relative to an observer, or any derivative of position with respect to time, including velocity, acceleration, and jerk.
But if I follow Burke's philosophy that work is obviously a scalar, then power should be a scalar as well. If there is only one force acting on a particle of mass m, then the power is given by [itex]P=m a\cdot v[/itex]. This implies that a and v can't both naturally be contravariant as claimed by WP.

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Jan25-13, 03:44 PM
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Newtonian force as a covariant or contravariant quantity


Yes if you write it in that form i.e. [itex]P = ma_{i}v^{i}[/itex] then the acceleration shows up as a co - vector but if we are talking about particles moving in [itex]\mathbb{R}^{3}[/itex], the dual space of [itex]\mathbb{R}^{3}[/itex] IS [itex]\mathbb{R}^{3}[/itex] so it isn't wrong to also say it is a vector if we are just talking about euclidean space with the canonical basis.
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Jan25-13, 03:46 PM
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Quote Quote by WannabeNewton View Post
I think here when we talk about [itex]W({x}) = \int_{{x_{0}}}^{{x}}F_{i}dx^{i}[/itex] being a scalar, it just means it is a real valued function defined on the image of the curve that the particle is traveling on and not necessarily a coordinate invariant quantity. Note, regarding your second point if I understood the question correctly, that when we are dealing with [itex]\mathbb{R}^{n}[/itex], the dual vectors and vectors are naturally identified.
What you're saying seems sensible, but it would seem to completely contradict the statements by Burke and in the WP article to the effect that this quantity or that quantity "is" covariant or contravariant. That is, they seem to be claiming that the natural identification provided by duality is not so natural.

As a side issue, I'm not sure what you mean by "when we are dealing with [itex]\mathbb{R}^{n}[/itex]." As far as I can see, we're always dealing with a tangent space on a manifold, which is always isomorphic to [itex]\mathbb{R}^{n}[/itex]. Whenever you have a vector space, you can make duals of vectors.
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Jan25-13, 03:51 PM
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I think power is [tex]F_a v^a[/tex].
I think that to write Newton's Second Law with acceleration,
one needs a metric: [tex]F_a =mg_{ab}a^{b}[/tex].

Using momentum (thought of as a covector), the Second Law is
[tex]F_a= \frac{d}{dt} p_a [/tex]
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Jan25-13, 03:51 PM
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Quote Quote by WannabeNewton View Post
the dual space of [itex]\mathbb{R}^{3}[/itex] IS [itex]\mathbb{R}^{3}[/itex]
I don't think this is true. They're isomorphic, but not the same. An element of [itex]\mathbb{R}^3[/itex] is a 3-tuple of real numbers. An element of its dual space is a linear function. Physically, elements of the two spaces have different transformation properties.
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Jan25-13, 03:55 PM
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Quote Quote by robphy View Post
I think power is [tex]F_a v^a[/tex].
I think that to write Newton's Second Law with acceleration,
one needs a metric: [tex]F_a =mg_{ab}a^{b}[/tex].
Sure. If we assert complete symmetry between vectors and their duals, then the whole issue becomes a non-issue. But both Burke and the WP article seem to be asserting that the symmetry is not complete, and that there is some mode of reasoning that tells you which form is preferred, covariant or contravariant, for a given physical quantity.
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Jan25-13, 04:04 PM
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Quote Quote by bcrowell View Post
I don't think this is true. They're isomorphic, but not the same. An element of [itex]\mathbb{R}^3[/itex] is a 3-tuple of real numbers. An element of its dual space is a linear function.
Hi ben yes strictly they are not equal but their vector space structures are identical so for all practical purposes they are the same. For vector spaces over the reals, of with which we are concerned with, we can readily go from a vector to a corresponding dual vector via the Riesz representation so why should there be some a priori preference for saying say force must be co - vector as opposed to a vector: is that what is being asked?
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Jan25-13, 04:06 PM
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I recently came across a very cool book called Div, Grad, and Curl are Dead by Burke.
Bill was a remarkable person, and one of my best friends. His biography is here.
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Jan25-13, 04:21 PM
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Quote Quote by WannabeNewton View Post
Hi ben yes strictly they are not equal but their vector space structures are identical so for all practical purposes they are the same. For vector spaces over the reals, of with which we are concerned with, we can readily go from a vector to a corresponding dual vector via the Riesz representation so why should there be some a priori preference for saying say force must be co - vector as opposed to a vector: is that what is being asked?
If that is the question, then the answer is easy: because you want to define line integrals.

The entire reason for defining 1-forms is to be able to take line integrals. A line integral should only ever be defined on 1-forms (or differential forms). Taking line integrals of vector fields is done in practice, but is not a good standard.

So there is a reasonable process to decide whether something should be a 1-form or a vector. The question you should ask: do you intend to take line integrals of your object? If the answer is yes, then you are dealing with a 1-form.

Since taking line integrals of force fields is something that is very common, this is an explanation of why force should be a 1-form and not a vector.
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Jan25-13, 04:44 PM
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Just to add to this, Bergmann, in his 1942 book on SR+GR, treated force as a covariant vector, and velocity and 4-acceleration as contravariant vectors. 4-momentum was treated inconsistently, sometimes contravariant, sometimes covariant.
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Quote Quote by PAllen View Post
Just to add to this, Bergmann, in his 1942 book on SR+GR, treated force as a covariant vector, and velocity and 4-acceleration as contravariant vectors. 4-momentum was treated inconsistently, sometimes contravariant, sometimes covariant.
Just to further add on to this PAllen if you don't mind =D, Soper in his text on classical field theory also defines force as a one form and you can see it in the tensorial nature of the euler lagrange equations which are one - form equations.
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Quote Quote by Bill_K View Post
Bill was a remarkable person, and one of my best friends. His biography is here.
I really like the book, and it's a shame that it seems to have been condemned to this sort of dusty junk-shop existence on the internet. Do you have any idea whether his heirs (Violet John?) would be interested in seeing it spiffed up and distributed for free under a creative commons license? I'd been thinking of making inquiries, but didn't know who to contact initially, was thinking maybe his colleagues at UCSC.
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Jan25-13, 05:29 PM
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This point of view about force as a 1-form appears in Burke's
Applied Differential Geometry and in Spacetime, Geometry, Cosmology.
A related article is a J.Math Phys article on Twisted Differential Forms for Electromagnetism ( http://dx.doi.org/10.1063/1.525603 )

see also
http://www.ucolick.org/~burke/home.html

Earlier sources include Schouten's Tensor Analysis for Physicists.

Momentum as a covector seems natural from a Hamiltonian Mechanics viewpoint.


...all of this talk is nudging me to get back to a project of mine to visualize tensors and differential forms.
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Quote Quote by robphy View Post
Momentum as a covector seems natural from a Hamiltonian Mechanics viewpoint.
Here I think it is more strictly required than seen as one possible viewpoint in order for the phase space to be the cotangent bundle to the configuration space manifold. Unlike this case that you mentioned, in GR I find it hard to keep track of quantities that are naturally taken to be one - forms or vectors in the context of index based computations because one freely uses the metric tensor to go form one to the other.
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Quote Quote by WannabeNewton View Post
Here I think it is more strictly required than seen as one possible viewpoint in order for the phase space to be the cotangent bundle to the configuration space manifold. Unlike this case that you mentioned, in GR I find it hard to keep track of quantities that are naturally taken to be one - forms or vectors in the context of index based computations because one freely uses the metric tensor to go form one to the other.
In typical presentations of GR or other subjects, there is little attention paid
to what fundamental structures (e.g. metric, volume-form, connection, etc...)
are being used in a construction or calculation.

So, we have this blurring of whether something is naturally a vector, covector (one-form), or maybe even a bivector or two-form. This is especially true when one deals with a space (e.g. Minkowski space or Euclidean space) with lots of symmetries.

The references by Burke and Schouten (and others) try to tease out what could be the fundamental structures... For example, there is work on metric-independent formulations of electromagnetism... where, e.g., the magnetic field is fundamentally a two-form. Then, only with additional structure, can it be thought of as an axial-vector [in three spatial dimensions].


I actually like the [abstract-]index computations because they help me keep track of
what kinds of objects I am dealing with. Ideally, it would be nice to a have a geometrical picture
to go along with the computations to suggest what is physically going on.
Ben Niehoff
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Jan25-13, 10:41 PM
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Quote Quote by robphy View Post
Using momentum (thought of as a covector), the Second Law is
[tex]F_a= \frac{d}{dt} p_a [/tex]
I think this is the most sensible answer, and eliminates the discrepancy. Force is a covector in every place it appears. Momentum is naturally a covector, being canonically conjugate to position.

What is missing from this picture is how you relate velocity and momentum. Incidentally, this is exactly the same question as relating Lagrangian to Hamiltonian dynamics. At some point you are forced to identify the tangent and cotangent bundles. Luckily there is a canonical way to do so.


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