3. The Material Derivative in Spherical Coordinates

Spherical coordinates are of course the most intimidating for the untrained eye. For engineers and fluid dynamicists, the farthest we go is usually cylindrical coordinates with rare pop-ups of the spherical problem. Here, I want to derive the material derivative of the velocity field in spherical coordinates. First, let us do that for a scalar.

Assume that at point r and time t a fluid particle has a property Q. As this particle moves about, this property will change with time (and space). Again, in the Lagrangian description, Q is only a function of time, i.e.

(Eq. 1)

However, from the Eulerian point of view, any property of the fluid is a function of time and space, which is also a function of time implicitly. Then

(Eq. 2)

Then, the time rate of change of any scalar fluid property is given by the following

(Eq. 3)

where we have used the chain rule to account for the spatial dependence on time. Remembering some of the formulas from dynamics, we have

(Eq. 4)

upon substitution of Eq. 4 into Eq. 3, we finally obtain the material derivative for a scalar

(Eq. 5)

To obtain the material derivative for a vector field, we follow a similar procedure keeping in mind the directional nature of a vector. We illustrate this using the velocity field - keep in mind that this works for any kind of vector field. Again, in a Lagrangian reference, the velocity is only a function of time. In the Eulerian view, the velocity has the following form

(Eq. 6)

Using the chain rule, the material derivative of the velocity field is written as

(Eq. 7)

Again, noting that the partial derivative with respect to time in Eq. 7 (first term) is evaluated at a fixed position in space, the unit vectors associated with the fluid particle at that point are fixed as viewed from an Eulerian reference, therefore,

(Eq. 8)

To evaluate the remaining terms in Eq. 7, we have to first remember some equations from dynamics or vector calculus about differential changes in unit vectors in spherical coordinates. These are given by

(Eq. 9)

Now we can evaluate the spatial terms in Eq. 7. The radial derivative is

(Eq. 10)

while the tangential derivative takes the following form

(Eq. 11)

Finally, the azimuthal derivative is as follows

(Eq. 12)

Voila!!!

Once Eqs. 8 through 12 are put together, one obtains the full expression for the material derivative of a vector field in spherical coordinates.

In the next post, I will present an invariant vector form for the material derivative so that we don't have to go through all the hassle of using chain rule differentiation to evaluate the material derivative. But it was worth it to see how it works using good old calculus.