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# Frenet-Serret formulas

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In vector calculus, the Frenet-Serret formulas describe the dynamic properties of a particle which moves along a continuous, differentiable curve in three-dimensional space [itex] \mathbb{R}^3 [itex]. 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.

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

The normal unit vector N is defined as

[itex] \mathbf{N} = {d\mathbf{T} / ds \over \| d\mathbf{T} / ds \|}, \qquad \qquad (2) [itex]

and the binormal unit vector B is defined as

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

and

Then the Frenet-Serret formulas are:

[itex] {d\mathbf{T} \over ds} = \kappa \mathbf{N}, [itex]
[itex] {d\mathbf{N} \over ds} = – \kappa \mathbf{T} + \tau \mathbf{B}, [itex]
[itex] {d\mathbf{B} \over ds} = -\tau \mathbf{N}. [itex]

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 that 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:

[itex] \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}.[itex]

## Proof

### Part one

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

[itex] \mathbf{B} \times \mathbf{T} = (\mathbf{T} \times \mathbf{N}) \times \mathbf{T} \qquad \qquad (6) [itex].

An identity from vector calculus states that

[itex] (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \mathbf{b} (\mathbf{a} \cdot \mathbf{c}) – \mathbf{a} (\mathbf{b} \cdot \mathbf{c}). \qquad \qquad (7) [itex]

Applying identity (7) to equation (6) yields

[itex] \mathbf{B} \times \mathbf{T} = \mathbf{N} (\mathbf{T} \cdot \mathbf{T}) – \mathbf{T} (\mathbf{N} \cdot \mathbf{T}) \qquad \qquad (8) [itex]

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

[itex] \mathbf{N} \times \mathbf{B} = (\mathbf{B} \times \mathbf{T}) \times \mathbf{B}. \qquad \qquad (9) [itex]

Applying identity (7) to equation (9) produces

[itex] \mathbf{N} \times \mathbf{B} = \mathbf{T} (\mathbf{B} \cdot \mathbf{B}) – \mathbf{B} (\mathbf{T} \cdot \mathbf{B}) \qquad \qquad (10) [itex]

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.

### 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 prove the third Frenet-Serret formula. From equation (3) it follows that

[itex] {d\mathbf{B} \over ds} = {d (\mathbf{T} \times \mathbf{N}) \over ds}. [itex]

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

[itex] {d\mathbf{B} \over ds} = {d \mathbf{T} \over ds} \times \mathbf{N} + \mathbf{T} \times {d \mathbf{N} \over ds}. [itex]

Applying the first Frenet-Serret formula to this last equation yields

[itex] {d\mathbf{B} \over ds} = \kappa \mathbf{N} \times \mathbf{N} + \mathbf{T} \times {d \mathbf{N} \over ds}, \qquad \qquad (11) [itex]

but the cross product of any vector with itself must be zero (due to an identity of vector calculus), so that equation (11) becomes

[itex] {d\mathbf{B} \over ds} = \mathbf{T} \times {d \mathbf{N} \over ds}. [itex]

The cross product of two vectors must necessarily be perpendicular to both vectors, so that dB/ds must be perpendicular to T, but it must also be perpendicular to B, since B is always a unit vector, so that dB/ds must be parallel to N. Let

[itex] {d\mathbf{B} \over ds} = -\tau \mathbf{N} [itex]

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

[itex] {d \mathbf{N} \over ds} = {d (\mathbf{B} \times \mathbf{T}) \over ds}. [itex]

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

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

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

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

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

## Alternative proof

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

[itex] \mathbf{T} := {\mathbf{r'} \over |\mathbf{r'}|} [itex]
[itex] \mathbf{N} := {\mathbf{T'} \over |\mathbf{T'}|} [itex]
[itex] \mathbf{B} := \mathbf{T} \times \mathbf{N} [itex]
[itex] \kappa := |\mathbf{T'}| [itex]
[itex] \tau := – \mathbf{B'} \cdot \mathbf{N} [itex]

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:

[itex] \mathbf{T'} = \kappa \mathbf{N}. [itex]

To prove the third Frenet-Serret formula, notice that

[itex] \mathbf{B'} = (\mathbf{B'} \cdot \mathbf{T}) \mathbf{T} + (\mathbf{B'} \cdot \mathbf{N}) \mathbf{N} + (\mathbf{B'} \cdot \mathbf{B}) \mathbf{B}. [itex]

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. [itex] \mathbf{B} \cdot \mathbf{B'} = 0 [itex]. The product [itex] \mathbf{B'} \cdot \mathbf{N}[itex] 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

[itex] (\mathbf{B} \cdot \mathbf{T})' = \mathbf{B'} \cdot \mathbf{T} + \mathbf{B} \cdot \mathbf{T'} [itex]

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

[itex] \mathbf{B'} = -\tau \mathbf{N}. [itex]

To prove the second Frenet-Serret formula, notice that

[itex] \mathbf{N'} = (\mathbf{N'} \cdot \mathbf{T}) \mathbf{T} + (\mathbf{N'} \cdot \mathbf{N}) \mathbf{N} + (\mathbf{N'} \cdot \mathbf{B}) \mathbf{B}. [itex]

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

[itex] (\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'} [itex]
[itex] (\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'} [itex]

therefore

[itex] \mathbf{N'} = -(\mathbf{T'} \cdot \mathbf{N}) \mathbf{T} – (\mathbf{B'} \cdot \mathbf{N}) \mathbf{B}. [itex]

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

[itex] \mathbf{N'} = -\kappa \mathbf{T} + \tau \mathbf{B} [itex]

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

## 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.