# TNB Frenet Frames

1. Sep 11, 2008

### Prologue

I have some questions about the TNB frame.

The T unit vector is defined this way:
$$\hat{T} = \frac{dr(t)/dt}{ds/dt} = \frac{dr(s(t))}{ds}$$
So, it is parametrized by arc length. Why can't t be left as the parameter? Is this just for definition-of-curvature-sake? If so is there any reason why the TNB frame couldn't be parametrized all in time, not by arc length?

Ok, next question. I see that the N unit vector is defined this way:
$$\hat{N} = \frac{d\hat{T}/ds}{|d\hat{T}/ds|} = \frac{(d\hat{T}/dt)/(dt/ds)}{|d\hat{T}/dt|/|dt/ds|} = \frac{d\hat{T}/dt}{|d\hat{T}/dt|}$$

Ok, confusion. The T unit vector is defined by parameter s, but then let's compute the N unit vector and for fun make it parametrized by t?! What is going on?

Another thing about the N unit vector. It seems to me that if this TNB frame is so great it should be able to deal with a straight line.

However if the path is a straight line:
$$\frac{d\hat{T}}{ds} = \frac{d\hat{T}}{dt} = 0$$
So, there would be no way to find an N unit vector.

For this reason I am assuming that it must be a curve in order to be defined. Is this a reasonable assumption?

Last edited: Sep 11, 2008
2. Nov 25, 2009

### Prologue

I know this is digging up the archives but I still have these same questions. Anybody got any insight?

3. Nov 25, 2009

### chingkui

Hi Prologue,

In general, if you think of t as time as s as arc length, then dr/dt is the velocity which might not be a unit vector. But T=dr/ds is the unit tangent vector. Other than that, there is really no difference between using s or t since they are 1-1. The choice of whether to use s or t or some other parametrization makes no different at all except maybe one choice is easier to compute than another.
dT/ds measures the change of the unit tangent. If the curve is a straight line, there is of course no change to the tangent and N doesn't exist.

4. Nov 26, 2009

### HallsofIvy

Staff Emeritus
If the parameter, t, is not arclength, then dr/dt is not a unit vector. The unit tangent vector would be T= (dr/dt)/||dr/dt||. You can leave t as the parameter but it makes the formulas much more complicated.

The point is that they use s to do the calculation but then convert back to the more general t so that you don't have to convert to areclength just to use the formula.

Yes, a straight line does not HAVE a normal vector. For a straight line, the tangent vector is a constant. It's derivative is the 0 vector so you can't divide by ||dT/ds||. The geometric problem is that because the curvature is 0, all vectors normal to the line have equal right to be called "the" normal vector.