I An algebraic manipulation in Schutz's book on GR

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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!
 

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that's an alge-bruh moment. Yeah I agree, assuming I didn't f*ck it up too...
$$\begin{align*}
\left(\frac{dy}{d\phi}\right)^2 &= \frac{\tilde{E}^2}{\tilde{L}^2} - \left(1-2M\left(y + \frac{M}{\tilde{L}^2}\right)\right)\left(\frac{1}{\tilde{L}^2} + \left(y+\frac{M}{\tilde{L}^2}\right)^2\right) \\ \\
&= \frac{\tilde{E}^2 - 1}{\tilde{L}^2} + \frac{2M}{\tilde{L}^2} \left(y + \frac{M}{\tilde{L}^2}\right) - \left(y^2 + \frac{2My}{\tilde{L}^2} + \frac{M^2}{\tilde{L}^4}\right) + 2M\left(y^3 + \frac{3My^2}{\tilde{L}^2} + \frac{3M^2y}{\tilde{L}^4} + \frac{M^3}{\tilde{L}^6}\right) \\ \\
&= \frac{\tilde{E}^2 + M^2/\tilde{L}^2 - 1}{\tilde{L}^2} + \frac{2M^4}{\tilde{L}^6} + \frac{6M^3 y}{\tilde{L}^4} + \left(\frac{6M^2}{\tilde{L}^2} - 1\right)y^2 + \mathcal{O}\left(y^3\right)
\end{align*}$$
 
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It seems he carries this mistake in the definition of ##y_0## on the following page.
 
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?
 
MathematicalPhysicist said:
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.
 
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My general impression of Schutz is that it needed a better proof reader.
 
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