*Originally posted by joe shmo*

**[b]r** is the vector

**u_r** is the unit vector of **r**

**u_r'** is the unit vector for **r'** describing the position of a partical after it moves through some angle δTheta in time δt.

What im saying is that the second term in the inner product cannot actually be zero becuase **u_r'** = **u_r** are both unit vectors with the sam ...[text shortened]... mation in the derivation. What type of equation can be be developed without any approximations?[/b]

You are treating what you call theta as a large angle and it is not only small, but we are going to take the limit that it goes to zero. Let's choose our axes so that the vector

**u_r** points along the x-axis so its components in Cartesian coordinates are (1, 0). Then the vector

**u_r'** has components (cos(δw), sin(δw)), where δw is the small angle (intended to be read as delta omega, but actually the roman letter w) between the vectors (you call this theta, but it should really be δw). It is now easy to calculate

**δu_r'**:

**δu_r'** = (cos(δw), sin(δw)) - (1, 0) = (cos(δw) - 1, sin(δw)) ~ (δw^2/2 + O(δw^4), δw)

now let's take the dot product with

**u_r**:

**δu_r'**.

**u_r** = (δw^2/2 + O(δw^4), sin(δw)) . (1, 0) = (δw^2/2 + O(δw^4), 0)

So, to first order in δw the dot product is zero. Yes, you are right we are relying on an approximation - but you are trying to do calculus and the "error term" goes to zero as δw goes to zero if it is anything more than first order. Let w' be the angular velocity so that w' = limit (δt -> 0) δw/δt. We have:

d

**u_r**/dt = limit (δt -> 0)

**δu_r'**/δt = limit (δt -> 0) (δw^2/2 + O(δw^4), 0)/δt = limit (δt -> 0) ((δw/δt)δw/2, 0) = (limit(δt -> 0) δt w'^2/2, 0) = (0, 0)

So the rate of change of the unit vector is zero, because the change in the vector is second order in the angle and only first order terms survive.

A more rigorous argument can be made by actually using polar coordinates (what you present above has the components of the vectors in Cartesian coordinates parameterised by the radial and angular coordinates). In a properly constructed coordinate system the unit vectors in the radial direction and in the direction of increasing angle

*are automatically orthogonal*.

The components of the vector in plane polar coordinates is (r, w). The components of the

*same* vector in Cartesian coordinates is (r cos(w), r sin(w)). The rate of change of the vectors can be worked out just by differentiating (using r' and w' to mean the derivatives):

cartesian_components = (dx/dt, dy/dt) = (d(r cos(w))/dt, d(rsin(w))/dt) = (r'cos(w) + r sin(w)w', r'sin(w) - r cos(w)w' )

polar_components = (dr/dt, dw/dt) = (r', w' )

If we set r' = 0 then:

cartesian_components = (y, x) w'

polar_components = (0, w' )

I'm guessing you're looking at a book on elementary vector calculus. The problem with the treatment in those books is that they do not cover changes in coordinate systems properly (their vector spaces have unit vectors that correspond to Cartesian axes even when they are working in polar or cylindrical coordinates.). More advanced books which call themselves things like "Introduction to Differential Geometry." explain this aspect far better.