- #1

MathematicalPhysicist

Gold Member

- 4,699

- 369

- TL;DR Summary
- 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?

Best, MP.

##(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.