# Normal one-forms

1. Jun 3, 2008

### Mmmm

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
$$\tilde{n}(\vec{V})= 0*a+b*0+c*0=0$$

Part (c) of the question says
But my understanding is that (0,0,$$\beta$$) is just a subset of all possible normal one forms to this plane, and I'd agree that of this subset any $$\tilde{n}$$ is a multiple of any other. But this isn't true for all $$\tilde{n}$$, as surley (0,$$\alpha, \beta$$) 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. Jun 3, 2008

### Mmmm

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!

Last edited: Jun 3, 2008