Error in Goldstein?

Hello,

I have the third edition of Goldstein which I have been using to learn mechanics. I believe I have found an error in the book, however normally when I feel such things I tend to either be misreading the situation or misunderstanding the concept. I checked Professor Safko's site on Goldstein corrections but it made no mention of the error I believe I caught. Also, I am not sure what printing of the edition I have. The method for ascertaining the printing edition on the professor's site does not seem to apply to the book I have, but again it is possible that I am missing something obvious. However, since Professor Safko's site was last updated in 2010, and I very recently bought the book, I thought perhaps I have a more recent printing than those presented on Dr. Safko's site.

That said, here's the issue. Beginning on pg. 47 in my book, the first two paragraphs state:

As an example, consider a smooth solid hemisphere of radius a placed with its flat side down and fastened to the Earth whose gravitational acceleration is g. Place a small mass M at the top of the hemisphere with an infinitesimal displacement off center so the mass slides down without friction. Choose coordinates x, y, z centered on the base of the hemisphere with z vertical and the x-z plane containing the initial motion of the mass.

Let $\theta$ be the angle from the top of the sphere to the mass. The Lagrangian is $L = \frac{1}{2}M(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)-mgz$. The initial condition allows us to ignore the y coordinate, so the constraint is $a-\sqrt{x^2-z^2}=0$. Expressing the problem in terms of $r^2=x^2+z^2$ and $x/z = cos \theta$, Lagrange's equations are . . . . .

My objection to this is the book's definition of cos theta. It appears to me that based on what has been written cos theta should be z/a. Am I wrong in this? I don't think so, because the calculation directly following this seems to coincide with the way I'm thinking about it. Please let me know if I'm off base here.

Thanks.

mfb
Mentor
##cos(\theta)=\frac{z}{a}##, indeed.
Probably some weird typo, if the calculation afterwards is correct.

Andrew Mason
Not only is $\cos\theta \ne x/z$ but $a - \sqrt{x^2-z^2} \ne 0$. In fact $\sqrt{x^2-z^2}$ is not real for $\theta<\pi/4$