Chinese Scholar Yang Jian liang Putting Wrongs to Rights in Astrophysics

[heyuhua]

Where is the value of $$q_0$$ coming from? It's being set without justification, which pretty much looks like it was set after the fact in order to fit the data.


Except not really. That equation is derived for a static universe, where all forces cancel. So a sign-flip wouldn't show up in a equation of motion, because there literally is no motion.

q_0 is obtaind from calculation according to observational cosmic density and Hubble parameter value of today. And (1.15) is geodesic equation which is the dynamical equation of particle in gravitational field and equivalent to the standard geodesic equation, don't means to be applied to static universe only, about its derivation you may see the following Yang's article page 142,or arbitrary standard textbook of post Newton's mechanics
Yang's crticle:
http://www.journalrepository.org/media/journals/BJMCS_6/2011/May/1306478...

 

[heyuhua]

Let's focus on one step that's missing from the article. How does:
$$h_{00}=-\frac{\gamma}{4\pi}\int\frac{\rho+3p}{\xi}dx'dy'dz'$$
where $$\xi=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$$

Get you to:
$$h_{00}=\frac{2GM}{r}$$
with $$\gamma=4\pi G$$?

Edit: (The given equation for $$g_{jj}$$ is obviously wrong in this article; a typo, I suspect.)

the calculations in Yang's article are correct and you need hard read and check with an open mind

 

[NotEinstein]

q_0 is obtaind from calculation according to observational cosmic density and Hubble parameter value of today.

You are misreading Yang's article. $$q_o$$ is explicitly introduced as being set to $$0.14$$, for no justifiable reason.

(1.15) is geodesic equation which is equivalent to the standard geodesic equation, don't means to be applied to static universe, about its derivation you may see the following Yang's article page 142,or arbitrary standard textbook of post Newton's mechanics
Yang's crticle:
http://www.journalrepository.org/media/journals/BJMCS_6/2011/May/1306478...

Right, so it's a standard GR equation, and doesn't prove anything related to the sign of the force of gravity, which was what we were talking about. Why did you dodge the question?

the calculations in Yang's article are correct

I just pointed out a location where all GR-experts in the world seem to disagree with what Yang has written. Prove your assertion that Yang is correct.

and you need hard read and check with an open mind

No, you need to prove your assertion real hard.​

 

[heyuhua]

You are misreading Yang's article. $$q_o$$ is explicitly introduced as being set to $$0.14$$, for no justifiable reason.


Right, so it's a standard GR equation, and doesn't prove anything related to the sign of the force of gravity, which was what we were talking about. Why did you dodge the question?

I just pointed out a location where all GR-experts in the world seem to disagree with what Yang has written. Prove your assertion that Yang is correct.


No, you need to prove your assertion real hard.​

all your question had already been answered in Yang's article, now I indeed doubt you don't read seriously or you don't understand those equations at all, I said many times that, but you seem no attention, the sign of coupling coefficient doesn't express attraction or repulsive, the attraction or replusive is reflected from the direction of acceleration of moving particle, substituting the new metric components worked out in Yang's article into (1.15) the acceleration of radial component is obviously negative, which means that a particle of free motion in the gravitational field is attracted by central body. Note that the space-space components of new metrics are different from the previous,

 

[NotEinstein]

all your question had already been answered in Yang's article,

Which you cannot point to, because...?

now I indeed doubt you don't read seriously or you don't understand those equations at all,

Well, if my apparently-not-understanding matches that of all GR-experts in the world, I'm fine with that.

I said many times that, but you seem no attention, the sign of coupling coefficient doesn't express attraction or repulsive,

Yes, you've said that many times. And I've qualified my statement already. Why are you so stuck on that, and refuse to respond to the other (more serious) things I've brought up?

the attraction or replusive is reflected from the direction of acceleration of moving particle,

True.

substituting the new metric components worked out in Yang's article into (1.15) the acceleration of radial component is obviously negative,

There is no acceleration in a static universe, so no, that doesn't work out.

which means that a particle of free motion in the gravitational field is attracted by central body

There are no particles is all you have is a perfect fluid.

It seems you don't understand Yang's articles at all. Perhaps you should be reading them more carefully.

 

[heyuhua]

your foundation is too poor, unexpectedly you don't know that geodesic equation is to describe free particle's motion in gravitational field, for example the motion the earth's revolution round the sun. Of course, so-called gravitational fields include the static. to describe perfect fluid is the equations that 4-dimension diverence of energy-momentum tensor is zero but not geodesic equation, they aren't the same thing

 

[NotEinstein]

your foundation is too poor,

Says the person that can't explain where a specific minus-sign is coming from.

unexpectedly you don't know that geodesic equation is to describe free particle's motion in gravitational field,

I do know that and I've never claimed otherwise, so I don't know where you got that idea from?

But it's also irrelevant in this situation: there is no moving (because: static) and there is no particle (because: fluid).

of course, so-called gravitational fields include the static.

That's misrepresenting Yang's article, and my position. Of course in a static universe the gravitational fields will be static, but that's not the problem. The problem is that there in a static universe, all equations of motion will be zero, because there is (obviously) no motion. You can't read off the sign of a force's contribution from the equations of motion if it all cancels out anyway.

to describe perfect fluid is the equations that 4-dimension diverence of energy-momentum tensor is zero but not geodesic equation, they aren't the same thing

That's not my argument, so I don't feel the need to comment on the validity of that statement.

 

[heyuhua]

Says the person that can't explain where a specific minus-sign is coming from.


I do know that and I've never claimed otherwise, so I don't know where you got that idea from?

But it's also irrelevant in this situation: there is no moving (because: static) and there is no particle (because: fluid).


That's misrepresenting Yang's article, and my position. Of course in a static universe the gravitational fields will be static, but that's not the problem. The problem is that there in a static universe, all equations of motion will be zero, because there is (obviously) no motion. You can't read off the sign of a force's contribution from the equations of motion if it all cancels out anyway.


That's not my argument, so I don't feel the need to comment on the validity of that statement.

In Yang's article http://www.journalrepository.org/media/journals/BJMCS_6/2011/May/1306478...
the chapter 2 and the chapter 3 is to treat the static gravitational field of spherical symmetry and the motion of free particle in the gravitational field, and don't refer to any cosmological question at all, why do you always pull to the static universe?

 

[NotEinstein]

In Yang's article http://www.journalrepository.org/media/journals/BJMCS_6/2011/May/1306478...
the chapter 2 and the chapter 3 is to treat the static gravitational field of spherical symmetry and the motion of free particle in the gravitational field, and don't refer to any cosmological question at all, why do you always pull to the static universe?

You need to re-read chapter 3 of that article; it's not talking explicitly about a static gravitational field of spherical symmetry. Re-check what the static is referring to, and what that means.

 

[heyuhua]

You need to re-read chapter 3 of that article; it's not talking explicitly about a static gravitational field of spherical symmetry. Re-check what the static is referring to, and what that means.

you are too absurd, it is me who translated Yang's articles into English, don't I know the content?

 

[NotEinstein]

you are too absurd, it is me who translated Yang's articles into English, don't I know the content?

Well apparently, yes. You've clearly not understood the perfect fluid part, the static universe part, the equation of motion, you can't explain where the minus-sign in the partially contracted Ricci tensor is coming from, and I've probably forgotten a few things. And this is just looking at the first part of the article!

But don't feel bad. GR is complicated, and not many people understand it well. I for one wouldn't dare to claim to be an expert on it, but I do know a bit of the basics. And it seems that's enough to point at problems in Yang's articles. Is there a way you can contact Yang, to have him/her respond to these issues?

 

[heyuhua]

n Yang's article http://www.journalrepository.org/media/journals/BJMCS_6/2011/May/1306478...
the chapter 2 and the chapter 3 is to treat the static gravitational field of spherical symmetry and the motion of free particle in the gravitational field, and don't refer to any cosmological question at all, why do you always pull to the static universe?

the static gravitational field of spherical symmetry is just the gravitational field Newton's law describes, now use general relativity describe, through comparing the results decide the coupling ceoffierent

 

[NotEinstein]

the static gravitational field of spherical symmetry is just the gravitational field Newton's law describes, now use general relativity describe, through comparing the results decide the coupling ceoffierent

That sentence is garbled. What do you mean?

 

[heyuhua]

Well apparently, yes. You've clearly not understood the perfect fluid part, the static universe part, the equation of motion, you can't explain where the minus-sign in the partially contracted Ricci tensor is coming from, and I've probably forgotten a few things. And this is just looking at the first part of the article!

But don't feel bad. GR is complicated, and not many people understand it well. I for one wouldn't dare to claim to be an expert on it, but I do know a bit of the basics. And it seems that's enough to point at problems in Yang's articles. Is there a way you can contact Yang, to have him/her respond to these issues?

the minus-sign isn't a question at all, you may calculate yourself, it will take on, note Minkowski's metric according to the convention in Yang's article, besides, note the defintion of 4-dimension speed

 

[heyuhua]

That sentence is garbled. What do you mean?

this is the content of chapter 2 and chapter 3

 

[NotEinstein]

the minus-sign isn't a question at all, you may calculate yourself,

No, I can't, because my answer always turns out to match that of all GR-experts on the planet, and not Yang's.

Can't you calculate it? Why do you keep refusing to post it here?

it will take on, note Minkowski's metric according to the convention in Yang's article

Yes, that convention matches what (for example) Carroll uses, and Carroll comes to the same equation Wikipedia has. In other words, Yang's article is in conflict with Carroll, who uses the same convention. So no, it's not a convention thing that's changing the minus signs.

So now we have an additional question: why didn't you know that? Why can't you answer any questions? Do you even know any GR?

 

[NotEinstein]

this is the content of chapter 2 and chapter 3

In that case, you've done a really terrible job translating to English, because that's word-garbage.

 

[heyuhua]

No, I can't, because my answer always turns out to match that of all GR-experts on the planet, and not Yang's.

Can't you calculate it? Why do you keep refusing to post it here?


Yes, that convention matches what (for example) Carroll uses, and Carroll comes to the same equation Wikipedia has. In other words, Yang's article is in conflict with Carroll, who uses the same convention. So no, it's not a convention thing that's changing the minus signs.

So now we have an additional question: why didn't you know that? Why can't you answer any questions? Do you even know any GR?

please send your email addree to me I will use mathematics equation renew derivate to you

 

[NotEinstein]

please send your email addree to me I will use mathematics equation renew derivate to you

No, post them here. Use the tex-tag, like so: [ tex]y=x^2[ /tex] (without the spaces after the [ ).

Edit: Or post them as images. Or post them elsewhere, and post a link here.

 

[heyuhua]

No, post them here. Use the tex-tag, like so: [ tex]y=x^2[ /tex] (without the spaces after the [ ).

Edit: Or post them as images. Or post them elsewhere, and post a link here.

so writing is very bad, especially up or down index cann't be expressed explictly in the same time