Exercise 20 absolute values

Gary

It's a good question, and if we do take the absolute value of the result of part (iii) then the result for part (iv) comes out automatically with the absolute value in the numerator as required by the exercise.

As the treatment of curvature in the book regards it as a signed quantity then I would say that when doing algebraic manipulations it is better not to remove the sign until the end of any calculation, as in certain situations it could cause an error. For example when using (23), as in this example, if we look at (30), which is equivalent to (23), we can see that this could be thought of as $\widetilde{\kappa}=\kappa_1+\kappa_2$, and if $\kappa_1$ and $\kappa_2$ were opposite in sign, then if we made them both positive the value of $\widetilde{\kappa}$ would be calculated incorrectly. In this particular case $\kappa$ is zero and so there is no problem. In the exercise the result in (iii) is an intermediate result on our way to the result in part (iv). If we were happy with the result in part (iii) and wanted to go no further, then we could write it as $|Im(\bar{v}\dot{v})|/|v|^3$. But since it is an intermediate result in our case I think it is better to take the absolute value at the end.

Vasco

Gary

The quantity ${\textbf{v}\times\dot{\textbf{v}}}$ is a real quantity here because in chapter 1 Needham has redefined the cross product - it does not have its usual meaning, it is not a vector quantity, and so $\widetilde{\kappa}=\frac{\textbf{v}\times\dot{\textbf{v}}}{|\textbf{v}|^3}$ is a real number and therefore can be positive or negative. All the final result does is make sure $\widetilde{\kappa}$ is positive.
I think that to be mathematically consistent the result of part (iv) should be written as $|\widetilde{\kappa}|=\frac{|\textbf{v}\times\dot{\textbf{v}}|}{|\textbf{v}|^3}$.

I suppose I'm really saying that the exercise contains a small error.

Vasco