Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

TNB Frenet Frames

  1. Sep 11, 2008 #1
    I have some questions about the TNB frame.

    The T unit vector is defined this way:
    [tex]\hat{T} = \frac{dr(t)/dt}{ds/dt} = \frac{dr(s(t))}{ds}[/tex]
    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:
    [tex]\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|}[/tex]

    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:
    [tex]\frac{d\hat{T}}{ds} = \frac{d\hat{T}}{dt} = 0[/tex]
    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. jcsd
  3. Nov 25, 2009 #2
    I know this is digging up the archives but I still have these same questions. Anybody got any insight?
  4. Nov 25, 2009 #3
    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.
  5. Nov 26, 2009 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: TNB Frenet Frames
  1. Frenet formulas (Replies: 3)

  2. TNB Frame vectors (Replies: 1)

  3. Frenet-Serret Formulas (Replies: 1)