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

In summary, according to Dale Schutz on page 274 in the second edition of Schutz's textbook, equation (10.85) should be written as: ##M= (3k^3/(4\pi))^{1/2}##, but he gets only that ##M=k^{3/2}## without the numerical constant the includes ##\pi##. Dale Schutz also says that on page 268 in equation (10.57) the condition should be ##p_c/\rho_c >1/7##, but if this is true (perhaps there's a mistake in Eq. (10.57) and sign of inequality should be the other way around), then either ##12
  • #1
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Hopefully now my post won't be sucked to the BH... :oldbigger:
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?

MP.
 
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  • #2
Is there anybody in there, just nod if you can hear me?
Anyone knows how to answer my questions?
 
  • #3
I don't understand the question, because I don't have to book to look up the context. Sorry.
 
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  • #4
vanhees71 said:
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?
 
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Wait! This question was already asked and I commented on it. What happened?
 
  • #7
@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
martinbn said:
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
Nugatory said:
This reply?
Yes, but it is ok. I was worried that I am imagining things because of all the time I spend outside.
 
  • #10
martinbn said:
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
Nugatory said:
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.
 
Last edited:
  • #12
@martinbn or anyone else can answer my questions?

Thanks in advance.
 
  • #13
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...
 
  • #14
I am attaching a pic of the two pages where my second conundrum appears for those who don't have the book.
Schutz-274.png
 

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

1. What is the derivation in "A First Course in GR" by Schutz?

The derivation in "A First Course in GR" by Schutz is a mathematical proof that explains the foundations of general relativity, a theory of gravitation developed by Albert Einstein.

2. Why is the derivation in "A First Course in GR" by Schutz important?

The derivation in "A First Course in GR" by Schutz is important because it provides a rigorous and logical explanation of the principles and equations of general relativity, which is crucial in understanding the behavior of gravity and its effects on the universe.

3. Is the derivation in "A First Course in GR" by Schutz difficult to understand?

The level of difficulty in understanding the derivation in "A First Course in GR" by Schutz may vary depending on the reader's background and familiarity with mathematical concepts. However, Schutz presents the derivation in a clear and concise manner, making it accessible to those with a basic understanding of calculus and physics.

4. Can the derivation in "A First Course in GR" by Schutz be applied to real-world situations?

Yes, the derivation in "A First Course in GR" by Schutz is applicable to real-world situations. The principles and equations of general relativity derived by Schutz have been extensively tested and used in various fields such as astrophysics, cosmology, and space exploration.

5. Are there any alternative derivations of general relativity?

Yes, there are alternative derivations of general relativity proposed by other scientists and mathematicians. However, Schutz's derivation is widely accepted and considered as one of the most comprehensive and accurate explanations of general relativity.

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