An algebraic manipulation in Schutz's book on GR

[MathematicalPhysicist]

TL;DR Summary

Some algebraic manipulation, which I think there's a misprint in the book.
Please let me know what do you think?

Attached is a pic of the page in the book:
My problem is with equation (11.34) specifically with the term $\frac{6M^3}{L^2}y$ I get $L^4$ instead of $L^2$.
Here are my calculations (I also checked it with maple's expand command):
$$\frac{E^2-1}{L^2}+\frac{2M^2}{L^4}+\frac{2M}{L^2}y-[y^2+\frac{2yM}{L^2}+\frac{M^2}{L^4}]+$$
$$+2M[y^3+\frac{M^3}{L^6}+3y^2\frac{M}{L^2}+3y\frac{M^2}{L^4}]$$

so if we neglect the term $2My^3$, we should be getting as I wrote.
I don't see where did I make a mistake?
Can you spot it?

Thanks!

 

that's an alge-bruh moment. Yeah I agree, assuming I didn't f*ck it up too...
$$\begin{align*}
\left(\frac{dy}{d\phi}\right)^2 &= \frac{\tilde{E}^2}{\tilde{L}^2} - \left(1-2M\left(y + \frac{M}{\tilde{L}^2}\right)\right)\left(\frac{1}{\tilde{L}^2} + \left(y+\frac{M}{\tilde{L}^2}\right)^2\right) \\ \\
&= \frac{\tilde{E}^2 - 1}{\tilde{L}^2} + \frac{2M}{\tilde{L}^2} \left(y + \frac{M}{\tilde{L}^2}\right) - \left(y^2 + \frac{2My}{\tilde{L}^2} + \frac{M^2}{\tilde{L}^4}\right) + 2M\left(y^3 + \frac{3My^2}{\tilde{L}^2} + \frac{3M^2y}{\tilde{L}^4} + \frac{M^3}{\tilde{L}^6}\right) \\ \\
&= \frac{\tilde{E}^2 + M^2/\tilde{L}^2 - 1}{\tilde{L}^2} + \frac{2M^4}{\tilde{L}^6} + \frac{6M^3 y}{\tilde{L}^4} + \left(\frac{6M^2}{\tilde{L}^2} - 1\right)y^2 + \mathcal{O}\left(y^3\right)
\end{align*}$$

 

[MathematicalPhysicist]

It seems he carries this mistake in the definition of $y_0$ on the following page.

 

[MathematicalPhysicist]

I have a question to the experts, @vanhees71 @PeterDonis @Dale or others who know about GR.

Does this mistake appear also in the literature outside of Schutz's textbook?

 

[PeterDonis]

My problem is with equation (11.34) specifically with the term $\frac{6M^3}{L^2}y$ I get $L^4$ instead of $L^2$.

Just based on looking at units I think you are correct. The units of each term should be inverse length squared. The units of $y$ are inverse length; the units of $M$ and $L$ are both length; so for the units to be right you need $L^4$ in the denominator.

 

[Ibix]

My general impression of Schutz is that it needed a better proof reader.