normal one-forms


by Mmmm
Tags: normal, oneforms
Mmmm
Mmmm is offline
#1
Jun3-08, 05:48 AM
P: 63
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
Show that any normal to S is a multiple of [tex]\tilde{n}[/tex]
and the answer provided is:
On the Cartesian basis, the components of [tex]\tilde{n}[/tex] are (0,0,[tex]\beta[/tex]) for some [tex]\beta[/tex]. Thus any [tex]\tilde{n}[/tex] is a multiple of any other.
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 :)
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Mmmm
Mmmm is offline
#2
Jun3-08, 05:58 AM
P: 63
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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