Inquiry about a derivation in "A First Course in GR" by Schutz

  • #1
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Summary:

Hopefully now my post won't be sucked to the BH... :oldbigger:

Main Question or Discussion Point

on page 269 it's written in the second edition of Schutz's textbook that
##(10.69)p_c/\rho_c=\beta (2-5\beta)^{-1}##.
Demanding that this be less than ##1/7## gives:
##(10.70) 0<\beta < 1/6##

Now, if I am not mistaken on page 268 in equation (10.57) the condition should be ##p_c/\rho_c >1/7## (since ##\rho < 7p_*## ).
If this is true (perhaps there's a mistake in
in Eq. (10.57) and sign of inequality should be the other way around).

Anyway, if Eq. (10.57) is the one that should be used on (10.69) then I get the following:
$$b/(2-5b)>1/7 \Leftrightarrow (7b+5b-2)/(2-5b)>0 \Leftrightarrow (12b-2)/(2-5b)>0$$
which means either ##12b-2, 2-5b >0## or ##12b-2, 2-5b<0## the first condition means that (by the fact that ##b\le1##) ##1/6<\beta<2/5## the second condition isn't met.

I have another question regarding his textbook in its second edition:
on page 274 it's written down equation (10.85) as: ##M= (3k^3/(4\pi))^{1/2}##, but I get only that ##M=k^{3/2}## without the numerical constant the includes ##\pi##.

Here are my calculations:
$$M^{1/3} = R\bar{\rho}^{1/3}$$
thus, ##R= M^{1/3}/(\rho^{1/3})##, plug the last equation to (10.84) to get:
##M/(M^{1/3}/(\bar{\rho}^{1/3}))=k\bar{\rho}^{1/3}##
i.e ##M^{2/3}=k##.

How did they get the numerical factor ##(3/(4\pi))^{1/2}##?

Sorry @Dale if you got pissed off of me, I believe it's better to forgive than to be pissed off of someone, is it now OK?

Best, MP.
 
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Answers and Replies

  • #2
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Is there anybody in there, just nod if you can hear me?
Anyone knows how to answer my questions?
 
  • #3
vanhees71
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I don't understand the question, because I don't have to book to look up the context. Sorry.
 
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  • #4
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I don't understand the question, because I don't have to book to look up the context. Sorry.
Ok, thanks anyway @vanhees71 appreciate you finding the time to look at this thread of mine.

In the past there was a user called Jimmy Snider who went through this book, is he still on PF?
 
  • #6
martinbn
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Wait! This question was already asked and I commented on it. What happened?
 
  • #7
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@martinbn it got sucked into the BH due to me breaking the rules once more.

Anyway now I edited it to the will of the moderators.
Can you answer these questions again?

Thanks!
 
  • #8
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Wait! This question was already asked and I commented on it. What happened?
This reply?
About your first question, you have that ##p<p_*## and ##\rho<7p_*##, which is the same as ##\frac{p}{\rho}<\frac{p_*}{\rho}## and ##\frac17<\frac{p_*}{\rho}##. From this you cannot conclude anything about ##\frac{p}{\rho}## and ##\frac17##. But if you choose that ##\frac{p}{\rho}<\frac17<\frac{p_*}{\rho}##, then the above inequalities will be satisfied.
 
  • #9
martinbn
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This reply?
Yes, but it is ok. I was worried that I am imagining things because of all the time I spend outside.
 
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Yes, but it is ok. I was worried that I am imagining things because of all the time I spend outside.
Can you help me with my second question?
BTW, I'll read your answer to my first question when I'll find the time, quite hectic schedule I have.
 
  • #11
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This reply?
I don't understand how are ##p_c ,\rho_c## defined here with respect to ##p^*## and ##\rho##?
It's not clear to me from the book.
 
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  • #13
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No one bothers about an incorrect numerical factor?

Pity.
I can upload screenshots of the book, but it won't be from my hardcover copy...
 
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