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Normal one-forms

  1. Jun 3, 2008 #1
    I'm reading through Schutz's first course in relativity book and am finding question 12 on page 83 a bit problematic.

    If I understand it correctly an normal one-form to a plane is a one-form that, when operating on a normal vector to the plane, will give the result 0. This seems fairly straight forward to me.
    The question is talking about the plane x=0.
    So all vectors normal to this must be of the form (a,0,0) (ie parallel to the x axis)
    In that case, the normal one form must have components (0,b,c) then
    [tex]\tilde{n}(\vec{V})= 0*a+b*0+c*0=0[/tex]

    Part (c) of the question says
    and the answer provided is:
    But my understanding is that (0,0,[tex]\beta[/tex]) is just a subset of all possible normal one forms to this plane, and I'd agree that of this subset any [tex]\tilde{n}[/tex] is a multiple of any other. But this isn't true for all [tex]\tilde{n}[/tex], as surley (0,[tex]\alpha, \beta[/tex]) is also valid.

    Obviously I'm missing something fairly fundamental here, and I just have to understand this before I move on... Please help :)
     
  2. jcsd
  3. Jun 3, 2008 #2
    Oh... I am being stupid.. Just realised that the one-form has to operate on a vector tangent to the surface, not the vector normal... I really should read more carefully....:blushing:
    now the one-form is actually perpendicular to the plane and so calling it a normal one-form to the plane makes much more sense!
     
    Last edited: Jun 3, 2008
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