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The Time Derivative of a Vector
in a Rotating Coordinate System

Note: In the following the typographical distinction between vectors and scalars is that a vector is shown in red. This distinction has a visual impact but the nature of a variable is usually readily apparent from the context in which it is used.

Consider two coordinate systems, one a fixed (inertial) rectangular system with unit vectors I, J and K, the other rotating on the surface of a sphere of radius R. The latter has unit vectors i, j, k. The sphere is rotating at a constant rate of Ω. In the inertial coordinate system the rotation vector is ΩK. In the rotating coordinate system, where the unit vector i is to the east, the unit vector j is to the north and the unit vector k is vertically normal to the surface, the rotation vector Ω has the representation

Ω = Ωcos(φ)j + Ωsin(φ)k

where φ is the latitude angle.

An arbitrary vector A has representations in both coordinate systems; i.e.,

A = Ax I + AyJ + AzK
A = A'x i + A'yj + A'zk

When considering the change in the vector A with time the crucial difference between the two coordinate systems is that in the inertial system the unit vectors are constant whereas in the rotating system they are not. Thus,

dA/dt = (dAx/dt)I + (dAy/dt)J + (dAz/dt)K
dA/dt = (dA'x/dt)i + (dA'y/dt)j + (dA'z/dt)k
+ A'xdi/dt + A'ydj/dt + A'zdk/dt

The terms

(dA'x/dt)i + (dA'y/dt)j + (dA'z/dt)k

represent the apparent time rate of change of A in the rotating coordinate system, which can be denoted as drA/dt. Thus

dA/dt = drAdt) + A'xdi/dt + A'ydj/dt + A'zdk/dt

The Time Derivatives of the Tangent Plane
Unit Vectors of a Rotating
Coordinate System

The point of origin of the tangent plane coordinate system can be expressed in terms of the inertial frame either in rectangular coordinates (x, y, z) or spherical coordinates (r, θ, φ), where r is the radius of the sphere, θ is the longitude and φ is the latitude. The relationship between the two representations is:

x = rcos(φ)cos(θ)
y = rcos(φ)sin(θ)
z = rcos(φ)

The local vertical unit vector k at (x, y, z) in terms of the inertial frame is:

k = (x/r)I + (y/r)J + (z/r)K
dk/dt = (1/r)((dx/dt)I + (dy/dt)J + (dz/dt)K)

Since r and φ are constant and θ = Ωt

dx/dt = -rcos(φ)sin(θ)(dθ/dt) = -rcos(φ)sin(θ)Ω
dy/dt = rcos(φ)cos(θ)(dθ/dt) = rcos(φ)cos(θ)Ω
dz/dt = 0.


dk/dt = Ω[-cos(φ)sin(θ)I + cos(φ)cos(θ)J]

On the other hand, Ωxk is equal to the determinant:


which reduces to

I((-y/r)Ω) - J((-x/r)Ω) + K(0)
= Ω((-y/r)I + (x/r)J)
y/r = cos(φ)sin(θ)
x/r = cos(φ)cos(θ)
Ωxk = dk/dt

The local east-pointing unit vector i expressed in the inertial frame is

i = -sin*θ)I + cos(θ)J
and therefore
di/dt = -cos(θ)ΩI - sin(θ)ΩJ
= -Ω(cos(θ)I + sin(θ)J

But Ωxi, given by the determinant


reduces to I(-Ωcos(θ) - J(sin(θ) + K(0) which is the same as -Ω(cos(θ)I + sin(θ)J which is Ωxi. Thus,

di/dt = Ωxi

And finally there is the local north-pointing unit vector j, which expressed in the inertial frame is

j = -sin(φ)cos(θ)I - sin(φ)sin(θ)J + cos(φ)K
and therefore
dj/dt = sin(φ)sin(θ)ΩI - sin(φ)cos(θ)ΩJ
Ω(sin(φ)sin(θ)I - sin(φ)cos(θ)J).

But Ωxj in determinant form is


which reduces to

I(sin(φ)sin(θ)Ω - J(sin(φ)cos(θΩ) + K(0)
= Ω(sin(φ)sin(θ)I - sin(φ)cos(θ)J).

But this the same as dj/dt so

dj/dt = Ωxj.

The General Form of the Time Derivative of a Vector

It was found in the previous section that for a general vector

dA/dt = drA/dt + A'xdi/dt + A'ydj/dt + A'zdk/dt

If the time derivatives of the local unit vectors, di/dt, dj/dt and dk/dt, are replaced by their values as Ωxi, Ωxj and Ωxk, and the rotation vector Ω factored from the cross products the result is:

dA/dt = drA/dt + Ωx(A'xi + A'yj + A'zk)

which is none other than

dA/dt = drA/dt + ΩxA


This says that the time derivative of a vector can be constructed from its apparent time derivative in the rotating frame plus the vector which is the vector cross product of the rotation vector for the frame and the vector itself. There are number of places in the literature where the time derivatives of the unit basis vectors are derived from the above formula on the basis of the argument that such unit vectors are just special cases of position vectors to which the formula applies. This is in valid because the formula has to be derived from the determination of the time derivatives of those basis vectors. The formula does of course apply to the basis vectors but it is logically invalid to derive its application to the basis vectors from the formula itself.

The derivation strictly holds for position vectors and its extension to axial vectors (vectors such as angular momentum and torque which are vector cross products of position vectors) requires additional analysis. For this extension see Axial Vectors.

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