# Schutz, page 224

1. Oct 13, 2005

### Jimmy Snyder

This question is just for those who have a copy of Schutz, A First Course in GR. I have tried to plug equations 9.43 and 9.44 into equation 9.42 in order to verify equations 9.45 and 9.46. So far, I have not been successful. However, I have come to the conclusion that probably 9.45 is incorrect. The book has:
$$R = \frac{1}{2}l_{0}\Omega^{2}A/[(\omega_{0} - \Omega)^2 + 4\Omega^{2}\gamma^{2}]^{1/2}$$
But I believe it should be the following:
$$R = \frac{1}{2}l_{0}\Omega^{2}A/[(\omega_{0}{}^{2} - \Omega^{2})^2 + 4\Omega^{2}\gamma^{2}]^{1/2}$$
Unfortunately, I haven't been able to successfully justify either equation. The reason I think that my version may be the correct one is by looking at equation 9.46
$$tan \phi = 2\gamma \Omega / (\omega_{0}{}^{2} - \Omega^{2})$$
which implies:
$$cos \phi = (\omega_{0}{}^{2} - \Omega^{2}) / [(\omega_{0}{}^{2} - \Omega^{2})^2 + 4\Omega^{2}\gamma^{2}]^{1/2}$$

2. Oct 16, 2005

### mitchellmckain

Yes I get it to work out with your correction
$$R = \frac{1}{2}l_{0}\Omega^{2}A/[(\omega_{0}{}^{2} - \Omega^{2})^2 + 4\Omega^{2}\gamma^{2}]^{1/2}$$
$$tan \phi = -2\gamma \Omega / (\omega_{0}{}^{2} - \Omega^{2})$$

3. Oct 16, 2005

### Jimmy Snyder

Thank you, thank you, thank you, thank you. I have spent hours on this one equation trying many different tricks, but somehow this one escaped me.