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Differentials - is this valid or just sloppy but right?

  1. Apr 16, 2008 #1
    The proper time is defined by

    [tex]d\tau^2=g_{\mu\nu}dx^\mu dx^\nu[/tex]

    Suppose we have flat space time with one space dimension.


    Can this be rigorous?
  2. jcsd
  3. Apr 16, 2008 #2


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    Regard [itex]x[/itex] as [itex]x(t)[/itex], then [tex]d(\ x(t)\ )=\left(\frac{dx}{dt}\right)dt[/tex]
  4. Apr 16, 2008 #3
    Nothing wrong with juggling differentials --- just realise that they don't always obey the same algebraic rules as reals or complex numbers, which is sensible, since they're not. See non-standard analysis: http://en.wikipedia.org/wiki/Non-standard_analysis
  5. Apr 18, 2008 #4
    And something I only just noticed. From [tex]d\tau=\sqrt{dt^2-dx^2}[/tex] we see that one cannot express [tex]d\tau[/tex] in terms of first order changes in t and x. That is, there are no numbers A and B such that [tex]d\tau=Adt+Bdx[/tex]. The slope of the graph of [tex]\sqrt{x^2}[/tex] is singular at x=0. There is probably something significant for the proper time concept here.
  6. Apr 18, 2008 #5
    Let's just say that [tex]d\tau=\sqrt{dt^2-dx^2}[/tex] is short hand for the volume form on the worldline, i.e., it has no meaning until you integrate it. Let us write

    [tex]{\omega} = \sqrt{ \left( \frac{dt}{d\lambda}\right) ^2- \left(\frac{dx}{d\lambda}\right)^2} d\lambda[/tex]

    Here [tex]d\lambda[/tex] is a oneform. If you integrate this quantity over the parameter [tex]\lambda[/tex] you will get the volume of the wordline, or, more informally it's arclength.
  7. Apr 18, 2008 #6
    Do you mean to ask if it is rigorous? If so then yes, it is rigorous.

  8. Apr 18, 2008 #7
    Well... besides not being well defined =) I don't think you'll find any mathematician that would put their name to it =). It's just shorthand.
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