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Newtonian force as a covariant or contravariant quantity 
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#145
Feb113, 02:19 AM

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PF Gold
P: 1,961

f(φ, θ) = θ  φ = 0 A constraint force which is parallel to the constraint surface can be written as a multiple of df, since df is parallel to the constraint surface. F = λdf = λ(dθ  dφ) 


#146
Feb113, 05:55 AM

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In differential geometry, a vector [itex]V[/itex] is defined via its effects on scalar fields. Specifically, if we choose a particular coordinate basis, we can write: [itex]V(\Phi) = V^\mu \partial_\mu \Phi[/itex] with the implicit summation convention. Now, if [itex]V[/itex] happens to be a basis vector [itex]e_\sigma[/itex], then that means that [itex]V_\sigma = 1[/itex], and all the other components are equal to 0. So we have, in this case: [itex]e_\sigma(\Phi) = \partial_\sigma \Phi[/itex] Since this is true for any [itex]\Phi[/itex], it's an operator equation: [itex]e_\sigma = \partial_\sigma[/itex] 


#147
Feb113, 06:10 AM

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To complete the confusion, even though [itex]dx^\mu[/itex] is a 1form, the prototypical vector is a tangent vector to a parametrized path [itex]P(\lambda)[/itex], which has components [itex]\dfrac{dx^\mu}{d\lambda}[/itex]. 


#148
Feb113, 07:34 AM

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So I'm still waiting for someone to explain why the rate of change of momentum is not naturally a tangent vector. 


#149
Feb113, 09:07 AM

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Another way to see that momentum should be considered contravariant might be: In curvilinear coordinates, the equations of motion for a free particle is [itex]\dfrac{d}{dt} U_\mu = 0[/itex] not [itex]\dfrac{d}{dt} U^\mu = 0[/itex] 


#150
Feb113, 09:36 AM

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PF Gold
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I think the basic cause of the ugly inconsistency in Einstein notation is not that there's anything wrong with Einstein notation but that it's being mixed inappropriately with Sylvester's notation. Sylvester was the one who originally introduced terms like "contravariant" and "covariant." The pure Sylvester notation is described in this WP article: http://en.wikipedia.org/wiki/Contravariant_vector Unfortunately Sylvester's notation and terminology are much more cumbersome than Einstein notation, and it doesn't correspond to the way physicists customarily describe the objects themselves as having transformation properties, rather than the objects staying invariant while their representations transform.
In Einstein notation, a sum over updown indices represents a tensor contraction that reduces the rank of an expression by 2. In Sylvester notation, a vector is expressed in terms of a basis as a sum over updown indices, which is not a tensor product. Mixing Einstein and Sylvester notation produces these sillylooking things where it looks like we're expressing a covector as a sum of vectors or a vector as a sum of covectors. The abstract index form of Einstein notation is carefully designed so that you *can't* inadvertently refer to a coordinate system; anything written in abstract index notation is automatically coordinateindependent. This means that in abstract index notation, you can't express a vector as a sum over basis vectors. You don't want to and you don't need to. There is also an issue because in Einstein notation we refer to vectors as covariant and contravariant, whereas in Sylvester notation a given vector or 1form is written in terms of components and basis vectors that are covariant and contravariant. Neither is right or wrong, but mixing them leads to stuff that looks like nonsense. 


#151
Feb113, 09:50 AM

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Ok, I actually can't understand the Lewis notes I linked to in #127.
These guys http://www.google.com/url?sa=t&rct=j...41867550,d.b2I do it differently. In proposition 7.8.1 they do define a force via contraction with a vector field. However they contract with a symplectic 2 form, not a metric. I think the relevant definition of the 2 form is on p161. Incidentally the Lewis notes say ad hoc "motivation" is that the EulerLagrange equations transform as the components of a one form. He say this is handwaving because the EL equations are not a oneform. 


#152
Feb113, 06:23 PM

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I guess the Lewis notes http://www.mast.queensu.ca/~andrew/t...f/439notes.pdf are fine after all. Force is a oneform. What I was confused by is their notation F: R X TQ > T*Q. All they mean is that a force is an assignment of a one form to every velocity at every position. Just as they say a vector field is V: Q > TQ by which they mean a vector field is an assignment of a vector at every position.



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