Hi
@PeterDonis:
Thank you very much for your post. I have learned quite a bit from it. I apologize for my previous careless reading of your article. I realized it will take me several readings to grasp all the concepts in it. For the present, I did a quick scan to look for particular topics I am currently trying to learn about. At my age I use the excuse that I fail to see things that are there, and I see things that are not.
I bolded your text from your previous post. I did this because this was more convenient for me than using the Quote feature.
As you have no doubt guessed, the function m(r) can be physically interpreted as the mass inside radial coordinate r.
Does this not mean that m(r) is related to ρ(r)?
ρ(r) = m(r)/V(r)
where V(r) is the volume of a sphere of radius r. I understand that
V(r) ≠ (4/3)πr3
Let A be the area of a sphere of radius r in a space with curvature.
A(r) = 4πR sin(r/R)2
V(r) = ∫[x= 0 to r] A(x) dx
= 4πR2 ∫[x= 0 to r] sin2(x/R) dx
NOTE: ∫sin
2(x)dx = x/2 - sin(2x)/4 [ CRC pg 291 Eq 230 ]
Let w = x/R
V(r) = 4πR2 ∫[w=0 to r/R] sin2(w) R dw
= 4πR3 { (1/2) (r/R) - (1/4) sin(r/R) }
I did the following just as a check on my math. My first try had an error, so the effort turned out to be useful.
V(πR) = 4πR3 π/2
= 2π2 R3
The spacetime discussed in the Insights article is static. Nothing is a function of t. There is no scale factor a.
This is a surprise. I did not think that the tensor equations could be specialized for such a static model. Although I probably could not understand an explanation of the GR tensor equations, does the article include a simplification of the tensor equations based on the assumption of a static spacetime? I would like to take a look at that even if I cannot understsand the math. I can still understand the tensor notation.
If you are referring to the Einstein static universe,...
I am. However, I am assuming that the equations
[1] dm/dr = 4πr2 ρ(r)
[2] (1/2J) dJ/dr = (m+4πr3p)/(r(r-2m))
[3] dp/dr = -(ρ+p)[(1/2J) dJ/dr) - (2(p-s)/r) ]
are intended for a universe which may or may not be static, and that you specialize them to be static Is this correct? If not, where in the article was the step where the simplification for static spacetime was explained? I am now getting the message that deriving equations for a static universe from more general equations by assuming
dr/dt = d2r/dt2 = 0
may not reach the same result as the approach in your article. Is this correct?
Also, I assume for the Einstein 1917 model, pressure p=s=0, so the 2nd and 3rd equations become
[2' -> 4] (1/2J) dJ/dr = m/(r (r-2m))
[3' -> 5] 0 = -rho (1/2J) dJ/dr
[4&5 -> 6] 0 = m/(r (r-2m))
But, this implies
ρ=0 or dJ/dr = 0
and
m = 0.
These results (implied by p=s=0) makes no sense to me, so I must have made an error (which I cannot find) in my analysis. Can you help me?
The function J(r) (or more precisely its square root) has the obvious interpretation of being the “redshift factor” for an observer who is static at radial coordinate r.
This "redshift" factor is what suggested to me that it implied a radial speed associated with distance. That is what lead me to think there was a relationship between J and H.
The spacetime discussed in the Insights article is static. Nothing is a function of t. There is no scale factor a.
If there is no function of time, the redshift must be only an effect of seeing an object with a redshift corresponding to it kinetic energy. Is that correct? If so, why is redshift a function of r?
ρ is mass density.
More precisely, energy density as measured by a static observer.
Does this mean that the static observer sees the energy of something as including the energy equivalent of its mass together with its kinetic energy?
p is pressure
More precisely, radial pressure as measured by a static observer. And s is the tangential pressure (or stress).
I confess I do not understand why in a homogeneous and isotropic model the tangential and radial pressure should be different.
Regards,
Buzz