Re: Problems of four-velocity

From: Tom Roberts (tjroberts_at_lucent.com)
Date: 02/20/05 

Date: Sun, 20 Feb 2005 19:07:48 GMT

ande452@attglobal.net wrote:
> Tom Roberts wrote:
>>That's a major cause of your confusions. In SR, spacetime is NOT a
>>vector space, it is a manifold. In particular, it does not make sense to
>>"add" two points to get a third point (addition of vectors is the
>>essence of a vector space). Geometry cannot be meaningfully modeled as a
>>vector space, which is why differential geometry on manifolds was invented.
>
> You can add displacements meaningfully in SR (and GR, at least for
> infinitesimal ones).

Sure. But those displacements are not points. And to make the addition
of displacements rigorous in an arbitrary manifold one must introduce
the tangent spaces....

>> Yes, the coordinates of a point wrt some origin consist of an
>> ordered set of 4 real numbers, and IN MINKOWSKI COORDINATES
>> those 4-element sets can be put in 1-to-1 correspondence with
>> the components of 4-vectors in some 4-d vector space. Do not
>> be confused by this -- it cannot possibly have physical
>> consequences as it is explicitly coordinate dependent. [...]

> The idea that the result is coordinate dependent is not correct.

For your addition of displacements specified in terms of DISTANCES (not
coordinate differences), you're right. But in my quote I was not
discussing that. Specifically: such addition does not work in spherical
coordinates, or most other coordinate systems (Minkowski and Cartesian
coordinates are special in that it does work for them)....

> Having observed your postings here for several years, I still don't
> understand why you're so hung up on this.

I'm not sure what you mean by "this". I don't recall ever discussing
this specific point before (a manifold not being a vector space).

What I am "hung up" on is the distinction between tensors and their
components -- when one confuses components with tensors one cannot
understand the beauty of the underlying invariance. And such invariance
has proven to be _essential_ in modern physics.

I am also "hung up" on precision in thought and writing. Especially
about the basics -- I think it's highly likely that progress in
theoretical physics will depend on the replacement of a manifold with a
new model of the world, and the details of what it means to be a
manifold are important here....

I apologize for neither of those hang-ups.

> You can define operations
> on a vector space without defining coordinates to "define" the vector
> components.

Sure. But I don't see how that relates to what I said above.

> On a manifold, you still define vector operations on the tangent
> space at an event which is a vector space.

Sure. But that tangent space is not the manifold.

Tom Roberts tjroberts@lucent.com

Relevant Pages

  • Re: Problems of four-velocity
    ... >> In physics spacetime is a space that satisfies certain ... treated as a vector space it leads to contradictions because such ... > | system to the manifold, ... applies a Minkowski coordinate system to the manifold. ...
    (sci.physics.relativity)
  • Re: Problems of four-velocity
    ... SPACETIME in SR is a static object of EVENTS, ... > vector space, it is a manifold. ... manifold is we can always find a tangent space Tp at each point; ...
    (sci.physics.relativity)
  • Re: Problems of four-velocity
    ... >>That's a major cause of your confusions. ... >>vector space, it is a manifold. ... spacetime IS a manifold; every manifold HAS a tangent space at ... I don't see where a vector space is needed to define a manifold. ...
    (sci.physics.relativity)
  • Re: Problems of four-velocity
    ... SPACETIME in SR is a static object of EVENTS, ... vector space, it is a manifold. ...
    (sci.physics.relativity)
  • Re: Problems of four-velocity
    ... definitions of various spaces (e.g. vector space, Hilbert space, etc.). ... geometry the relevant structure is a manifold. ... isomorphism between any pair of those tangent spaces. ...
    (sci.physics.relativity)