|Jun3-08, 05:48 AM||#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
Part (c) of the question says
Obviously I'm missing something fairly fundamental here, and I just have to understand this before I move on... Please help :)
|Jun3-08, 05:58 AM||#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....
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!
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