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Differential Forms' Point of View |
Differential geometry is like a fabulous city standing at the confluence of two rivers, drawing upon the riches of both river systems. Only differential geometry stands at the confluence of three streams of mathematics; analysis, geometry and algebra. Its beauty comes from the combining of elements of these three streams.
The stage for differential geometry is Euclidean three space. A point in three space is specified by its three componenets, (x, y, z) or alternatively (x_{1}, x_{2}, x_{3}). Functions may be defined over regions of this three space. The range of a function could be real numbers. That is to say, at every point P=(x,y,z) of a region a real number is given; i.e., f(P). The range of a function could be a vector, an element with three components and a member of a vector space V; i.e., v(P). This type of function is called a vector field. An example of a vector field is the wind direction and magnitude. A field might have two or three vectors specified at every point of a region. An interesting case of this type is where three basic vectors of a vector space are defined at every point of the region. This is called a frame field.
Although one might think that the way to define curves in three space is simply as a set of points this is not the approach that has proven fruitful. The preferred way to define curves is in terms of parametrization. A function is defined over some interval for a parameter, say t, that gives the coordinates of a point; i.e., a function f is specified such that I:t → P or (x(t), y(t), z(t)) is given for t belonging to some interval I. As t varies over its interval a path in three space is traced which is a curve.
If the parametrized curve is interpreted as the path of a particle as a function of time then some of the aspects of the curve have an interesting physical interpretation. The vector given by:
is the velocity of the particle. The speed of the particle is the magnitude of the vector v. A unit speed vector u(t) can be defined by dividing the vector v by its magnitude; i.e., u(t) = v(t)/|v(t)|. This vector u(t) then has the property that u(t)·u(t) = 1. If this equation is differentiated with respect to t the result is
which says that the derivative of u(t), du(t)/dt, is always perpendicular to u(t).
At this point it is desirable to switch to a special parametrization of the curve. The natural parameter for the curve is arc length. If s is arc length, distance along the curve, from the start of the curve then
If this equation is solved to obtain arc length as a function of t, say s = F(t) and F has an inverse then t could be replaced in the parameterization by F^{-1}(s) and the curve would be given by (x(s), y(s), z(s)). Now we find that the vector in the direction of movement as the parameter s changes is a unit vector; i.e.,
T(s) is the unit vector which is tangent to the curve. It then follows that dT(s)/ds·T(s)= 0; i.e., dT(s)/ds is perpendicular (or orthogonal or normal) to T(s). The magnitude of the vector dT(s)/ds is called the curvature of the curve and is denoted κ. That is to say, κ(s) = |dT(s)/ds|. The unit vector in the direction of dT(s)/ds is defined as the normal to the curve and denoted as N(s) where N(s) = (dT(s)/ds)/κ. Note that T(s) and N(s) are orthogonal; i.e., T(s)·N(s) = 0.
The unit vector that is perpendicular to both T(s) and N(s) is called the binormal B(s) of the curve and is given by the cross product of T(s) and N(s); i.e., B(s) = T(s)xN(s).
The set of vectors {T(s), N(s), B(s)} at any point P of the curve is an
orthonormal basis and consequently any vector V can be represented in terms
of this set; i.e.,
V = v_{1}T(s)+v_{2}N(s)+v_{3}B(s).
Furthermore the coefficients in this representation can be easily expressed
in terms of the dot product of the vector V with the orthonormal basis
vectors; i.e., v_{1} = V·T(s),
v_{2} = V·N(s),
and v_{3} = V·B(s).
There are six independent conditions which are satisfied by an orthonormal frame field on a curve. The differentiation of each of these conditions gives us information revelevant for finding the representation of vectors. Below is a table of those conditions and the implication resulting from differentiation.
Condition | Differentiation | Implication |
---|---|---|
T(s)·T(s) = 1 | 2T(s)·dT/ds = 0 | T(s)·dT/ds = 0 |
T(s)·N(s) = 0 | T(s)·dN/ds + dT/ds·N(s) = 0 | dT/ds·N(s) = -T(s)·dN(s)/ds |
T(s)·B(s) = 0 | T(s)·dB/ds + dT/ds·B(s) = 0 | dT/ds·B(s) = -T(s)·dB(s)/ds |
N(s)·N(s) = 1 | 2N(s)·dN/ds = 0 | N(s)·dN/ds = 0 |
N(s)·B(s) = 0 | N(s)·dB/ds + dN/ds·B(s) = 0 | dN/ds·B(s) = -N(s)·dB(s)/ds |
B(s)·B(s) = 1 | 2B(s)·dB/ds = 0 | B(s)·dB/ds = 0 |
Now let us determine the representations of dT/ds, dN/ds and dB/ds in terms of T(s), N(s) and B(s). We know be definition that dT/ds = κN(s) so this one is easy. But as a preliminary exercise we can verify its representation.
Thus we verify that dT(s)/ds = 0(T(s))+κ(N(s))+0(B(s)) = κN(s).
For dN(s)/ds we need T(s)·dN(s)/ds, N(s)·dN(s)/ds and B(s)·dN(s)/ds. We know that T(s)·dN(s)/ds is equal to -N(s)·dT(s)/ds = -κN(s)·N(s) = -κ. We know that N(s)·dN(s)/ds = 0. Finally, B(s)·dN(s)/ds is a term we cannot further evaluate, but we can give it a more familiar designation. It will be called τ. Thus the representation of dN(s)/ds is -κT(s) + 0(N(s)) τB(s).
For dB(s)/ds we need T(s)·dB(s)/ds, N(s)·dB(s)/ds and B(s)·dB(s)/ds. We know that T(s)·dB(s)/ds is equal to -B(s)·dT(s)/ds = -κB(s)·N(s) = 0. N(s)·dB(s)/ds is equal to -B(s)·dN(s)/ds which has been given the designation τ. Thus the representation of dB(s)/ds is -κT(s)-τN(s)+0(B(s). When theseee representations are expressed in matrix form the result is:
| dT(s)/ds | | | 0 | κ | 0 | | | T(s) | | ||
| dN(s)/ds | | = | | -κ | 0 | τ | | | N(s) | | |
| dB(s)/ds | | | 0 | -τ | 0 | | | B(s) | |
These are the Frenet formulas, first discovered by Frenet (documented in his doctoral dissertation in 1847 but not published in a mathematics journal until 1852) and independently discovered again by Serret and published in a mathematics journal in 1851. The matrix which gives the rate of changes of the frame field in terms of the frame field itself is a skew-symmetric matrix; i.e., element (j,i) is the negative of element (i,j) and thus the elements on the diagonal must be zero.
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