Tuesday, September 16, 2008

2. The Material Derivative in Cylindrical Coordinates

This one a little bit more involved than the Cartesian derivation. The reason for this is that the unit vectors in cylindrical coordinates change direction when the particle is moving.

In the Lagrangian reference, the velocity is only a function of time. When we switch to the Eulerian reference, the velocity becomes a function of position, which, implicitly, is a function of time as well as viewed from the Eulerian reference. Then


and the material derivative is written as (with the capital D symbol to distinguish it from the total and partial derivatives)

Special attention must be made in evaluating the time derivative in Eq. 2. In dynamics, when differentiating the velocity vector in cylindrical coordinates, the unit vectors must also be differentiated with respect to time. In this case, the partial derivative is computed at a fixed position and therefore, the unit vectors are "fixed" in time and their time derivatives are identically zero. Then, we have

we can now evaluate the remaining terms in Eq. 2 as follows


finally

(Eq. 6)

When these are put together, the material derivative in cylindrical coordinates becomes

(Eq. 7)

This was a rather tedious way of deriving the material derivative as one could have used vector technology to obtain an invariant form that works for all coordinates. Nonetheless, it is interesting to see the intricacies of the derivation using chain rule differentiation. Note that if were computing the material derivative for a scalar, the extra terms in Eq. 7 (in the radial and tangential components) would disappear. These are purely reminicsent of the vectorial nature of the velocity field (or any other vector field for that matter).

It is very interesting to note the intimate link between the physical nature of the velocity and its mathematical description through vectors. One would pose the following argument: why don't we treat the material derivative of the velocity as that of three scalars, namely, u_r, u_theta, and u_z? Doing this will obviously remove the hassles of dealing with derivatives of unit vectors, but will eventually lead to inconsistent results. So what's the issue here?

The problem with that treatment is that in essense, the velocity is one quantity that we describe using vectors: a magnitude and a direction. If we are to use three scalars to describe the velocity we lose an essential ingredient which is the direction. In the end, the material derivative of the velocity can be decomposed into the material derivatives of three scalars (u_r, u_theta, and u_z) plus some correction. This correction stems from the directional nature of the velocity field. In other words, this correction can be thoguht of as being the material derivative of the direction of the velocity field.

[Next Article: The Material Derivative in Spherical Coordinates]

Cite as:
Saad, T. "2. The Material Derivative in Cylindrical Coordinates". Weblog entry from Please Make A Note. http://pleasemakeanote.blogspot.com/2008/08/derivation-of-navier-stokes-equations_17.html

6 comments:

  1. hey can you explain why some of the derivatives of the unit vectors e are cancelled, left behind, and equal to some other unit vector

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  2. you can find the answer to the question above here: http://www.csupomona.edu/~ajm/materials/delcyl.pdf

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  3. hello. the link doesnt work. i do not understand how the unit vector in theta direction when partiallly derivated with theta changes to -er . Really appreciate if you could give a reply. ty.

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  4. in equation 6 should the last partial derivative be d(uz)/dz and not d(uz)/d(theta)?

    ReplyDelete