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A vector space as defined by a mathematician lacks the invariant scalar product that lies at the heart of what I would define as a physicist's definition of a vector space that models the physical world. What type of mathematical structure is a physicist's definition of a vector space?

So far, I've managed to find the inner product space which lacks the additional invariant property I'm looking for.

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    $\begingroup$ What's your question? $\endgroup$ – garyp Jun 3 '14 at 1:12
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    $\begingroup$ @garyp it's in the title. $\endgroup$ – Larry Harson Jun 3 '14 at 1:15
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    $\begingroup$ What invariant property does the inner product space not have? Your question still isn't clear to me. Maybe I'm dense. $\endgroup$ – garyp Jun 3 '14 at 1:30
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    $\begingroup$ @garyp He may be thinking of the invariance of distances under Galilean boosts and Lorentz scalars under Lorentz boosts. Certainly those are the only examples that occur to me, but I think that a plain old inner-product space has those. $\endgroup$ – dmckee --- ex-moderator kitten Jun 3 '14 at 1:49
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    $\begingroup$ This seems to be asking more or less the same thing as this question I posted on Mathematics. $\endgroup$ – David Z Jun 3 '14 at 6:10
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Physicists use three vector spaces: Euclidean space, Minkowski space, and a separable Hilbert space (with some exceptions). All of these have a quadratic form (in the first and third, it is an inner product, but it is not positive definite for Minkowski space) allowing one to define angles. (Okay, for the Hilbert space, it is twisted by the complex conjugation, but you get the idea.)

But what does "invariant" mean? Really, it doesn't mean anything because given the quadratic form, then one defines the group $O$ (or, in the complex case, $U$) as the linear transformations of the space that are invertible and continuous and preserve the quadratic (or Hermitian, as the case may be) form. So "invariant" means nothing, it comes for free. Given any form, one can find a group that preserves it, and then it is invariant under that group.

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Much of theoretical physics is linear. Some is not. Only when it is linear can one use nothing but vector spaces. Vector spaces are inherently linear. When the phenomena are non-linear, then linear forms (and a quadratic form is really a kind of bi-linear form) become less useful, and have, at any rate, to be supplemented with other, more difficult, constructs. Like flows and foliations etc.

GR is non-linear, and uses curved manifolds, which are not vector spaces. Even then they are locally like vector spaces, and so linear analysis plays an important part of it, and the vector spaces appear as the tangent spaces to a curved manifold. But that was not the original question.

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  • $\begingroup$ What about tangent spaces on curved manifolds or the space of $p$-forms (as in: we use more than the three vector spaces you refer to)? $\endgroup$ – Hunter Jun 3 '14 at 1:58
  • $\begingroup$ Those are Euclidean spaces, or, in the case of relativity, Minkowski spaces (or if one generalises the dimension, Lorentz spaces). Someone else's comment also seemed to neglect the fact that the curved manifolds usually dealt with are all locally Euclidean (or Lorentzian, as the case may be) and so the tangent spaces all fall into one of those three categories (with some exceptions). Um...is the problem that you are assuming "Euclidean" means three dimensions? $\endgroup$ – joseph f. johnson Jun 3 '14 at 2:00
  • $\begingroup$ ok yeah, good point (I didn't think about the fact that in general relativity a curved manifold looks locally like a Euclidean or Lorentzian space). Thanks $\endgroup$ – Hunter Jun 3 '14 at 2:05
  • $\begingroup$ I'd personally feel like your answer would be a lot stronger if you wrote "Physicists primarily use..." even given your parenthetical "with some exceptions" $\endgroup$ – joshphysics Jun 3 '14 at 2:18
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    $\begingroup$ Well, in Stat Mech non-separable Hilbert spaces arise, but I honestly think that that is a mistake. Other than that, which other vector spaces do you think are important? $\endgroup$ – joseph f. johnson Jun 3 '14 at 2:22

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