Frenet-Serret formulas

From Wikipedia, the free encyclopedia

In vector calculus, the Frenet-Serret formulas describe the kinematic properties of a particle which moves along a continuous, differentiable curve in three-dimensional space $\mathbb{R}^3$ . More specifically, the formulas describe the derivatives of the tangent, normal, and binormal unit vectors in terms of each other. The formulas are named after their (independent) discoverers: the Frenchmen Jean Frédéric Frenet and Joseph Alfred Serret. The formulas were first described in the mid 19th century; however, the use of vector notation and linear algebra in writing them came much later.

1 Frenet-Serret formulas
2 Proof
- 2.1 Part one
- 2.2 Part two
3 Alternative proof
4 References
5 External links

[edit] Frenet-Serret formulas

Let s(t) represent the distance which the particle has moved along the curve. Let r(t) represent the position vector of the particle. Then the tangent unit vector T is defined as

$\mathbf{T} = {d\mathbf{r} \over ds}. \qquad \qquad (1)$

The normal unit vector N is defined as

$\mathbf{N} = {\frac{d\mathbf{T}}{ds} \over \| \frac{d\mathbf{T}}{ds} \|}, \qquad \qquad (2)$

and the binormal unit vector B is defined as

$\mathbf{B} = \mathbf{T} \times \mathbf{N}. \qquad \qquad (3)$

From equation (2) it follows, since T always has unit magnitude, that N is always perpendicular to T. From equation (3) it follows that B is always perpendicular to both T and N. Thus, the three unit vectors T, N, and B are all perpendicular to each other.

It can be proven, from the above, that

$\mathbf{B} \times \mathbf{T} = \mathbf{N} \qquad \qquad (4)$

and

$\mathbf{N} \times \mathbf{B} = \mathbf{T} \qquad \qquad (5)$ .

Then the Frenet-Serret formulas are:

${d\mathbf{T} \over ds} = \kappa \mathbf{N},$

${d\mathbf{N} \over ds} = - \kappa \mathbf{T} + \tau \mathbf{B},$

${d\mathbf{B} \over ds} = -\tau \mathbf{N},$

where $κ$ is the curvature and $τ$ is the torsion.

The three vectors: tangent, normal, and binormal, are collectively called the Frenet vectors. Together they form a basis for 3-space, which in turn defines a reference frame with its own coordinate system. The reference frame is called a Frenet frame, and it is neither static nor inertial, since the Frenet frame moves tangentially to the (non-straight) curve, so it is constantly accelerating. (The curve can be assumed to be parametrically dependent on time: i.e. the Frenet frame can be visualized kinematically).

The Frenet-Serret formulas are also known as Frenet-Serret theorem, and can be stated more concisely using matrix notation:

$\begin{bmatrix} \mathbf{T'} \\ \mathbf{N'} \\ \mathbf{B'} \end{bmatrix} = \begin{bmatrix} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0 \end{bmatrix} \begin{bmatrix} \mathbf{T} \\ \mathbf{N} \\ \mathbf{B} \end{bmatrix}.$

This matrix is skew-symmetric.

[edit] Proof

[edit] Part one

First prove equations (4) and (5). From equation (3) it follows that

$\mathbf{B} \times \mathbf{T} = (\mathbf{T} \times \mathbf{N}) \times \mathbf{T} \qquad \qquad (6)$ .

An identity for the vector triple product gives

$(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \mathbf{b} (\mathbf{a} \cdot \mathbf{c}) - \mathbf{c} (\mathbf{a} \cdot \mathbf{b}). \qquad \qquad (7)$

Applying identity (7) to equation (6) yields

$\mathbf{B} \times \mathbf{T} = \mathbf{N} (\mathbf{T} \cdot \mathbf{T}) - \mathbf{T} (\mathbf{T} \cdot \mathbf{N}) \qquad \qquad (8)$

but N and T are perpendicular so that their dot product is zero. Also, T is always a unit vector, so the dot product of T with itself is always one. Therefore equation (8) simplifies to equation (4).

Now prove equation (5). From equation (4) it follows that

$\mathbf{N} \times \mathbf{B} = (\mathbf{B} \times \mathbf{T}) \times \mathbf{B}. \qquad \qquad (9)$

Applying identity (7) to equation (9) produces

$\mathbf{N} \times \mathbf{B} = \mathbf{T} (\mathbf{B} \cdot \mathbf{B}) - \mathbf{B} (\mathbf{B} \cdot \mathbf{T}) \qquad \qquad (10)$

but T and B are perpendicular so that their dot product is zero, and B is always a unit vector so that the dot product of B with itself is always unity. Therefore equation (10) simplifies to equation (5).

Equations (4) and (5) should be intuitively rather obvious.

[edit] Part two

Let κ equal the magnitude of dT / ds. Then the first Frenet-Serret formula follows from equation (2), where κ is the curvature of the particle's path.

Now to prove the third Frenet-Serret formula. We show that dB/ds is parallel to the normal by showing that it is perpendicular to both the tangent and binormal. Two vectors are perpendicular if their dot product is zero.

$\mathbf{B} \bullet \mathbf{B} = 1$

${d\mathbf{B} \over ds} \bullet \mathbf{B} + \mathbf{B} \bullet {d\mathbf{B} \over ds} = 0$

${d\mathbf{B} \over ds} \bullet \mathbf{B} = 0$

Therefore dB / ds is perpendicular to B. Applying the same method to the dot product of the binormal and tangent yields

$\mathbf{B} \bullet \mathbf{T} = 0$

${d\mathbf{B} \over ds} \bullet \mathbf{T} + \mathbf{B} \bullet {d\mathbf{T} \over ds} = 0$

${d\mathbf{B} \over ds} \bullet \mathbf{T} = 0$

Hence dB / ds is also perpendicular to T. It then follows that dB / ds is parallel to N and can be written as a scalar multiple. Let

${d\mathbf{B} \over ds} = -\tau \mathbf{N}$

where τ is torsion, so that the third Frenet-Serret formula has been proven.

Lastly, prove the second Frenet-Serret formula. Equation (4) implies that

${d \mathbf{N} \over ds} = {d (\mathbf{B} \times \mathbf{T}) \over ds}.$

Applying the product rule for the cross product to this last equation yields

${d \mathbf{N} \over ds} = {d \mathbf{B} \over ds} \times \mathbf{T} + \mathbf{B} \times {d \mathbf{T} \over ds} \qquad \qquad (12)$

Next, substitute the third and first Frenet-Serret formulas into equation (12), producing

${d \mathbf{N} \over ds} = -\tau \mathbf{N} \times \mathbf{T} + \mathbf{B} \times \kappa \mathbf{N}.$

Then, applying equations (3) and (5) to the last equation yields the second Frenet-Serret formula. Q.E.D.

[edit] Alternative proof

A different proof applies the chain rule to dot products instead of to cross products. Let us restate the definitions:

$\mathbf{T} := {\mathbf{r'} \over |\mathbf{r'}|}$

$\mathbf{N} := {\mathbf{T'} \over |\mathbf{T'}|}$

$\mathbf{B} := \mathbf{T} \times \mathbf{N}$

$\kappa := |\mathbf{T'}|$

$\tau := - \mathbf{B'} \cdot \mathbf{N}$

These five quantities: the (1) tangent, (2) normal and (3) binormal vector fields and the (4) curvature and (5) torsion scalar fields are collectively known as the Frenet-Serret apparatus of the curve. By these definitions, T and N must be unit vectors, and then B must be a unit vector as well since it is the cross product of unit vectors.

From the definitions of N and κ the first Frenet-Serret formula follows immediately:

$\mathbf{T'} = \kappa \mathbf{N}.$

To prove the third Frenet-Serret formula, notice that

$\mathbf{B'} = (\mathbf{B'} \cdot \mathbf{T}) \mathbf{T} + (\mathbf{B'} \cdot \mathbf{N}) \mathbf{N} + (\mathbf{B'} \cdot \mathbf{B}) \mathbf{B}.$

The last term on the right side is equal to zero and can be dropped: B is a unit vector so its derivative B′ must be perpendicular to B, i.e. $\mathbf{B} \cdot \mathbf{B'} = 0$ . The product $\mathbf{B'} \cdot \mathbf{N}$ in the second term on the right is equal to −τ by the definition of torsion. Now for the dot product in the first term. Due to the chain rule for the dot product, it is known that

$(\mathbf{B} \cdot \mathbf{T})' = \mathbf{B'} \cdot \mathbf{T} + \mathbf{B} \cdot \mathbf{T'}$

but $\mathbf{B} \cdot \mathbf{T} = \mathbf{T} \times \mathbf{N} \cdot \mathbf{T} = 0$ so $\mathbf{B'} \cdot \mathbf{T} = - \mathbf{B} \cdot \mathbf{T'}$ . Applying the first Frenet-Serret formula, $\mathbf{B'} \cdot \mathbf{T} = -\mathbf{B} \cdot \kappa \mathbf{N} = -\kappa \mathbf{T} \times \mathbf{N} \cdot \mathbf{N} = 0$ , and the first term also drops out. We are left with the third Frenet-Serret formula:

$\mathbf{B'} = -\tau \mathbf{N}.$

To prove the second Frenet-Serret formula, notice that

$\mathbf{N'} = (\mathbf{N'} \cdot \mathbf{T}) \mathbf{T} + (\mathbf{N'} \cdot \mathbf{N}) \mathbf{N} + (\mathbf{N'} \cdot \mathbf{B}) \mathbf{B}.$

The second term on the right drops out: N is a unit vector so its derivative N′ must be perpendicular to N, thus $\mathbf{N} \cdot \mathbf{N'} = 0$ . From application of the chain rule we know that

$(\mathbf{N} \cdot \mathbf{T})' = \mathbf{N'} \cdot \mathbf{T} + \mathbf{N} \cdot \mathbf{T'}, \quad \mathbf{N} \cdot \mathbf{T} = 0, \quad \mathbf{N'} \cdot \mathbf{T} = -\mathbf{N} \cdot \mathbf{T'}$

$(\mathbf{N} \cdot \mathbf{B})' = \mathbf{N'} \cdot \mathbf{B} + \mathbf{N} \cdot \mathbf{B'}, \quad \mathbf{N} \cdot \mathbf{B} = 0, \quad \mathbf{N'} \cdot \mathbf{B} = -\mathbf{N} \cdot \mathbf{B'}$

therefore

$\mathbf{N'} = -(\mathbf{T'} \cdot \mathbf{N}) \mathbf{T} - (\mathbf{B'} \cdot \mathbf{N}) \mathbf{B}.$

Applying the first and third Frenet-Serret formulas, this last equation reduces to

$\mathbf{N'} = -\kappa \mathbf{T} + \tau \mathbf{B}$

which is the second Frenet-Serret formula, Q.E.D.

See also: Frenet frame.

[edit] References

Salas and Hille's Calculus -- One and Several Variables. Seventh Edition. Revised by Garret J. Etgen. John Wiley & Sons, 1995. p. 896.
Elements of Differential Geometry, by Richard S. Millman and George D. Parker.