I How to write the Frenet equations using the vector gradient?

[HaniZaheer]

Hey. I am trying to self study from "Theoretical Physics" by Georg Joos and am stuck on this particular question. The question asks for the reader to write the equations $$\frac {dt} {ds} = \frac {\vec n} {\rho}$$ and $$\frac {db} {ds} = - \tau \vec n$$ using vector gradient. I don't even know where to begin. Any help will be appreciated.

 

[Charles Link]

I'm familiar with the first of these equations, which is $d \hat{T}/ds=\kappa \hat{N}$ in the notation I am familiar with where $\hat{T}$ is a unit tangent vector along the path and $\hat{N}$ is the unit vector perpendicular to it, with $\kappa$ being the curvature. I googled it and found $B$ in your second equation to be defined as a unit vector $\hat{B}=\hat{T} \times \hat{N}$, but I have not verified the accuracy of the second equation. In any case, I think I have something that might be useful. Along the curve x=x(t), y=y(t), and z=z(t), define $R=(x^2+y^2+z^2)^{1/2}$. If I computed it correctly, $R \nabla R=\vec{R}$ and $d \vec{R}/ds=\hat{T}$ so that $d(R \nabla R)/ds=\hat{T}$ and from this you can take another d/ds to generate your first equation containing a gradient operation. Perhaps this is what you are looking for. Hopefully it is helpful.

 

[HaniZaheer]

I'm familiar with the first of these equations, which is $d \hat{T}/ds=\kappa \hat{N}$ in the notation I am familiar with where $\hat{T}$ is a unit tangent vector along the path and $\hat{N}$ is the unit vector perpendicular to it, with $\kappa$ being the curvature. I googled it and found $B$ in your second equation to be defined as a unit vector $\hat{B}=\hat{T} \times \hat{N}$, but I have not verified the accuracy of the second equation. In any case, I think I have something that might be useful. Along the curve x=x(t), y=y(t), and z=z(t), define $R=(x^2+y^2+z^2)^{1/2}$. If I computed it correctly, $R \nabla R=\vec{R}$ and $d \vec{R}/ds=\hat{T}$ so that $d(R \nabla R)/ds=\hat{T}$ and from this you can take another d/ds to generate your first equation containing a gradient operation. Perhaps this is what you are looking for. Hopefully it is helpful.

Thanks for the help. But the form of the equations the question was looking for was to sort of not use derivative and instead use vector gradient. The solution to the problem is as follows: $$\vec t \cdot \nabla \vec t = \frac {\vec n} {\rho}$$ and $$\vec t \cdot \nabla \vec b = - \vec \tau \vec n$$ It's rather confusing.

 

[Charles Link]

For this one it helps to have the answer and then work to it. I have something that is close to showing the result for your first equation: $\hat{T}=\hat{T}(s)=\hat{T}(x,y,z)$ and $d \hat{T}=\frac{\partial \hat{T}}{\partial x}dx+\frac{\partial \hat{T}}{\partial y}dy+\frac{\partial \hat{T}}{\partial z}dz$ and $d \vec{s}=dx \hat{i} +dy \hat{j} +dz \hat{k} =\hat{T}ds$ Thereby $d \hat{T}=(\hat{T} \cdot \nabla) \hat{T}ds=(d \vec{s} \cdot \nabla) \hat{T}$. This gives the result $d \hat{T}/ds=(\hat{T} \cdot \nabla) \hat{T}$.

 

[Charles Link]

An item of interest: I saw the first of these equations in a calculus textbook by Purcell where they write velocity $\vec{v}=(ds/dt) \hat{T}$ and $\hat{T}=\cos{\phi} \hat{i}+\sin{\phi} \hat{j}$. They then find the acceleration $\vec{a}=d \vec{v} /dt$ and work in two dimensions. $d \hat{T} /dt=(d \hat{T} /ds)(ds/dt)$ and $d \hat{T} /ds=(d \hat{T} /d \phi)(d \phi /ds)$. The $d \hat{T} /d \phi =\hat{N}$ where $\hat{N}$ is perpendicular to $\hat{T}$. (Take the dot product of $\hat{T}$ and $d \hat{T}/ d \phi=-\sin{\phi} \hat{i} + \cos{\phi} \hat{j}$ to see this). Anyway the $\vec{a}=(d^2 s/dt^2) \hat{T}+(ds/dt)^2 \kappa \hat{N}$ where $\kappa=d \phi /ds= 1/r$ where $r$ is the instantaneous radius for the curvature $\kappa$. The equations in this form are quite useful and I even used them to solve a problem (one that I thought up) of having an applied force in the $\hat{N}$ direction (always perpendicular to the velocity) that increases linearly with time. The question is, does the object speed up and what path does it follow? The Frenet equations in this formalism yielded a very interesting solution. Notice also in the 2-D case that $d \vec{B} /ds=0$. For the 3-D case, the $d \vec{B} / ds$ is essentially telling you how quickly the instantaneous circle that occurs in two dimensions is changing its plane over the course of the motion.

 

[Charles Link]

Besides my post #5 (please read post #5), I could give a little more info on this method which may or may not be spelled out clearly in the descriptions of these equations. Assuming the trajectory is not a straight line, the two unit vectors $\hat{T}(t) = \hat{T}$ and $\hat{T}(t+dt)=\hat{T} +d \hat{T}$ determine the plane of the trajectory at a given time interval from $t$ to $t+dt$. Perpendicular to $\hat{T}$ and also in this plane is the unit vector $\hat{N}$. The unit vector $\hat{B}$ is normal to this plane and defined by $\hat{B}=\hat{T} \times \hat{N}$. Just maybe this helps clarify a couple of the concepts related to these equations. ...editing... $d \hat{B} /ds=d(\hat{T} \times \hat{N})/ds=(d \hat{T}/ds) \times \hat{N}+\hat{T} \times (d \hat{N}/ds)$. Now $d\hat{T}/ds$ points along $\hat{N}$ from the first Frenet equation so that only the second term is non-zero. Taking the derivative of the first equation, $d \hat{N}/ds=d((1/\kappa)(d \hat{T}/ds))/ds$ so that $d \hat{B}/ds$ depends on $d^2 \hat{T}/ds^2$. Also $d \hat{N}/ds$ can point along $\hat{T}$ or $\hat{B}$ so that $\hat{T} \times (d\hat{N}/ds)$ must point along $\hat{T} \times \hat{B}=\hat{N}$ and thereby the second Frenet equation $d\hat{B}/ds=-\tau \hat{N}$ with some proportionality constant which they call $-\tau$. There is a 3rd Frenet equation (that I see in a google) $d \hat{N}/ds=-\kappa \hat{T}+\tau \hat{B}$, and beginning with $\hat{N}=\hat{B} \times \hat{T}$ this one is readily derived from the previous expressions.

 

[Charles Link]

Besides posts #5 and #6, there is one item that arises in these unit vectors that puzzled me momentarily: If $\hat{T}$ undergoes an infinitesimal change, why is this change always perpendicular to $\hat{T}$ ? If a $d\hat{T}$ arises in the $\hat{N}$ direction, doesn't $\hat{T}$ need to change by some amount in the $\hat{T}$ direction, e.g. become somewhat shorter to keep the resultant vector a unit vector? And the answer is the amount of change(shrinkage) is second order in "dT" so that $d \hat{T} /ds$ will have no component in the $\hat{T}$ direction.

 

[Charles Link]

@HaniZaheer A very interesting topic. Still waiting on your feedback to my posts #5, #6, and #7. :-)

 

[swampwiz]

Besides posts #5 and #6, there is one item that arises in these unit vectors that puzzled me momentarily: If $\hat{T}$ undergoes an infinitesimal change, why is this change always perpendicular to $\hat{T}$ ? If a $d\hat{T}$ arises in the $\hat{N}$ direction, doesn't $\hat{T}$ need to change by some amount in the $\hat{T}$ direction, e.g. become somewhat shorter to keep the resultant vector a unit vector? And the answer is the amount of change(shrinkage) is second order in "dT" so that $d \hat{T} /ds$ will have no component in the $\hat{T}$ direction.

u → unit vector

u ⋅ u = | u |² = 1

d( u ⋅ u ) / ds = d( | u |² ) / ds = d( 1 ) / ds = 0 = u ⋅ ( du / ds ) + ( du / ds ) ⋅ u = 2 u ⋅ ( du / ds )

 

[Charles Link]

u → unit vector

u ⋅ u = | u |² = 1

d( u ⋅ u ) / ds = d( | u |² ) / ds = d( 1 ) / ds = 0 = u ⋅ ( du / ds ) + ( du / ds ) ⋅ u = 2 u ⋅ ( du / ds )

Yes, that proves it as well.