Newtonian force as a covariant or contravariant quantityby bcrowell Tags: contravariant, covariant, force, newtonian, quantity 

#1
Jan2513, 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,
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 Einsteinstyle 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 energymomentum tensor, which means it changes the massenergy.) 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 stressenergy tensor...? 



#2
Jan2513, 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.




#3
Jan2513, 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, 



#4
Jan2513, 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.




#5
Jan2513, 03:46 PM

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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. 



#6
Jan2513, 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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Jan2513, 03:51 PM

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Jan2513, 03:55 PM

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#9
Jan2513, 04:04 PM

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#10
Jan2513, 04:06 PM

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#11
Jan2513, 04:21 PM

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The entire reason for defining 1forms is to be able to take line integrals. A line integral should only ever be defined on 1forms (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 1form 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 1form. Since taking line integrals of force fields is something that is very common, this is an explanation of why force should be a 1form and not a vector. 



#12
Jan2513, 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 4acceleration as contravariant vectors. 4momentum was treated inconsistently, sometimes contravariant, sometimes covariant.




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Jan2513, 04:51 PM

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#14
Jan2513, 05:07 PM

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#15
Jan2513, 05:29 PM

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This point of view about force as a 1form 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. 



#16
Jan2513, 05:35 PM

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#17
Jan2513, 06:09 PM

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to what fundamental structures (e.g. metric, volumeform, connection, etc...) are being used in a construction or calculation. So, we have this blurring of whether something is naturally a vector, covector (oneform), or maybe even a bivector or twoform. 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 metricindependent formulations of electromagnetism... where, e.g., the magnetic field is fundamentally a twoform. Then, only with additional structure, can it be thought of as an axialvector [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. 



#18
Jan2513, 10:41 PM

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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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