Re: [Phys-L] treating force as a vector ... consistently

[Moses Fayngold <moshfarlan@yahoo.com>]

On Wednesday, September 7, 2016 11:50 PM, John Denker <jsd@av8n.com> wrote:


 
> That's not how it works.  Each location in a vector field has
> its own proprietary vector space.  Different point, different
> space.  Within each vector /space/ the vector has magnitude and
> direction but not location.  The space as a whole has a location
> within the field, but that's the answer to a different question.

  If thisis a definition, one can eitheraccept it or not. But one can always question its consistency. In my view, itis totally detached from physical reality. In reality, it is the field (withits sources!) that has a location within the space (and always distorts it !),not vice versa. The idea of innumerable set of different Euclidean spaces all packedwithin one physical space with the same dimension and common set of points anddiffering from each other only by their origins looks too far-fetched if notpreposterous. It might resemble, on the face of it, some known physical models withmany worlds, but it conceptually differs from them. Thus, in the model of "paralleluniverses" coexisting with our own, each universe forms a separate brane (4Dsubset) of multidimensional space with D>4, instead of being hashed withothers in one common (3+1)D space. Another analogy is the "many worlds"interpretation of QM, but it has a crucial distinction of being a hypothesis waitingfor experimental confirmations rather than decreed to be an unquestionable universallaw. 

  Actually, mathematical definition of a vectoradmits possibility of cases with a vector identified by its initial point togetherwith its norm and direction. Here is a quotation from https://en.wikipedia.org/wiki/Euclidean_vector: 

  "Being an arrow, a Euclideanvector possesses a definite initial point and terminal point. A vector with fixedinitial and terminal point is called a boundvector (knownalso as located vector – MF).[9]  When only the magnitude and direction ofthe vector matter, then the particular initial point is of no importance, andthe vector is called a freevector."  

  

 Thus, Math admits local vectors and free vectorsas two different legitimate cases. A free vector is realized, for e.g., as a vector eigenvalueof linear momentum operator in QM.  Weall know that it does not have a location. OTOH, a local vector is a crucial characteristicin CM, e.g., position vector of a classical particle, a force vector applied ata definite point, etc. Our discussion concerns precisely such situations, whichare vital in many physical phenomena. 


> The idea of a separate vector /space/ at each point in the
> vector /field/ is particularly important and obvious when
> you consider 2D vectors on the surface of a sphere.  Each
> vector space is /tangent/ to the sphere.  A tangent vector
> such as the velocity vector does not live in or on the
> surface of the sphere;  it lives in the tangent space.  See
> the second diagram (tangent space and tangent vector) here:
>  https://en.wikipedia.org/wiki/Tangent_space

  This isa beautiful illustration, but it calls for discretions. There are many subtlepoints here. 

  I) First, a continuous set of all planes tangentialto a sphere forms a 3D Euclidean space. So there was an implicit requirementthat the considered curved space be immersed in Euclidean space of higherdimensions. But what if this requirement is not met? There are no compellingexperimental evidences that our gravitationally curved 3D space is a subset ofsome embracing Euclidean space with N>3.Similarly, we could consider a 2D spherical surface in its own right without invokinganything embracing it. Would the concept of a vector as a directed straightsegment survive on such surface? This is a branching point for differentpossibilities. 

  1a)  A straight segment representing spatial displacement in Euclidean space cannot,by definition, exist on a sphere. It can be replaced with its directly relevantanalog – segment of the corresponding geodesic passing through the initial andfinal point. Would you insist that such directed segment must be ambivalent toits parallel transport on the sphere? If yes, it must remain the same after paralleltransport around a closed loop on the sphere. But we all know that this is notthe case. 

  1b)  Derivative vectors (velocity, momentum, acceleration,force etc.) can still be represented by a straightsegment since the formal "length" of such vectors is not spatial extension but merely auseful geometrical image of vector's norm, not requiring any embracing space. Butin this case the vector is still specified by its point of origin together withdirection and norm and it obeys the rules for parallel transport on the sphere.Ambivalence to parallel transport is intimately connected with openness and homogeneity of space; but curvedspaces are either closed, or non-homogeneous, or both ! 

  II) Second, if we allow for an embracingEuclidean space, there appears some ambiguity.

  2a) Any tangential vector automaticallybecomes a member of embracing space and therefore can be subjected to paralleltransport in all 3 dimensions. This, in turn, allowsany vector to travel parallel to itself from one point of the sphere to another.If you do this with a bound (local) vector, it will take us back to my originalobjections.     

 2b) If you do this by rules of paralleltransport on the sphere, you may wind up in conflict with results of paralleltransport by rules of the embracing space. A vector tangential to equator atpoint P (x=R, y=z=0), after paralleltransport along the equator in the direction of increasing polar angle phi to the antipodal point Q at (x=--R, y=z=0) will wind up opposing itsown self obtained by parallel transport preserving its Cartesian components.  


> I don't see any issues worth debating.  The exterior derivative
> ∇∧B would require a bit of work, but that's the answer to a
> different question.  The plain old wedge product between two
> plain old vectors A∧B is child's play.  Indeed the fact that
> A and B can be freely relocated makes it /easier/ to construct
> the parallelogram representing A∧B.
  Theplain old cross product AXB did not require any relocations. If A is position vector of a point in a solidbody with fixed center of mass, and I apply force B at A, then anyrelocation of B to another point A' is not only unnecessary but most ofthem are forbidden if we want to find the torque tau = AXB.The latest modifications towards outer products, bivectors etc., whilebeing useful, do not warrant universal cancellation of locality.  

Moses Fayngold,NJIT_______________________________________________
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