Is the Hubble Constant Constant?

[kmarinas86]

-> = implies

Dark Energy -> Repulsion

Inflation -> Repulsion

Gravitation -> Attraction

(Repulsion = Attraction -> Steady Region) & Euclidean Universe -> Light can travel on parallel lines that remain the same distance from each other :: matter, light, and energy maintaining present density
(Repulsion > Attraction -> Opening Region) & Euclidean Universe -> Light can travel on parallel lines that increase in distance from each other :: matter, light, and energy are being segregated
(Attraction > Repulsion -> Closing Region) & Euclidean Universe -> Light can travel on parallel lines that decrease the same distance from each other :: matter, light, and energy are returning

According to the Big Bang, "Since the beginning of the universe, it has been opening."

Opening Regions > Closing Regions -> Opening Universe
Closing Regions > Opening Regions -> Closing Universe

Gravitation + Inflation -> ~%15 of Omega=1 :: Opening Universe
Gravitation + Inflation + Dark Energy -> ~%100 of Omega=1 :: Accelerating Universe
exactly %100 of Omega=1 -> Hubble Constant is smaller that if Omega=0.15 instead :: Universe "is" Euclidean, not Hyperbolic
Accelerating Universe -> Opening Regions > Closing Regions -> Increasing Hubble Constant -> Universe "is" not asymptotically decelerating, and it is not reversing direction.

Spherical Balloon Analogy -> Closed Universe (a sense of "closed" independent from the size, expanding, or contracting of the balloon/universe) -> Non-Euclidean Universe

Euclidean Universe -> Light does not orbit the universe -> Spherical Balloon Analogy Uncogent

Periods of deceleration (early universe) and acceleration (late universe) -> Acceleration of the Universe is variable -> Attraction/Repulsion = not constant -> Hubble Constant is not constant -> Omega changes with time -> Omega will stop changing or Omega will continue to change

((Omega~1 (late universe) && periods of deceleration (early universe) and acceleration (late universe)) -> Omega will change) && This change is continuous not discrete -> Omega is changing right now -> If Omega is exactly 1 right now, it will cease equaling 1 in short order

Decelerating Universe -> Hubble Constant is decreasing and positive -> Omega is increasing

Accelerating Universe -> Hubble Constant is increasing and positive -> Omega is decreasing

3rd Year WMAP Observations -> Universe is perhaps slightly closed -> Omega "perhaps >" 1

Supernova Observations -> Accelerating Universe -> Hubble Constant is Increasing -> Omega is decreasing

(3rd Year WMAP Observations & Supernova Observations) -> Omega is "perhaps approaching 1 from the plus side" -> Omega >1 by a little && Hubble constant is "perhaps small" but is "evidently getting larger" -> Opening Regions > Closing Regions -> Light travels in straight lines parallel to each other which themselves are being separated from each other via the Hubble Velocity (which is increasing)

Result: No contradictions in terminology found. Mainstream cosmology "OK".

 

[George Jones]

Accelerating Universe -> Hubble Constant is increasing and positive

This isn't true. It is possible that a universe has accelerating expansion and a decreasing Hubble constant, and, apparently, this is true for our universe.

The proper distance D between 2 galaxies is given by D = R \chi, where R is the scale factor, and \chi is the comoving coordinate distance. Thus,

\dot{D} = \dot{R} \chi = \frac{\dot{R}}{R} R \chi = H D,

with
H = \frac{\dot{R}}{R}.
Taking the time derivative of this last equation shows that \dot{H} can be negative when the acceleration \ddot{R} is positive.

Regards,
George

 

[kmarinas86]

This isn't true. It is possible that a universe has accelerating expansion and a decreasing Hubble constant, and, apparently, this is true for our universe.

The proper distance D between 2 galaxies is given by D = R \chi, where R is the scale factor, and \chi is the comoving coordinate distance. Thus,

\dot{D} = \dot{R} \chi = \frac{\dot{R}}{R} R \chi = H D,

with

H = \frac{\dot{R}}{R}.

Taking the time derivative of this last equation shows that \dot{H} can be negative when the acceleration \ddot{R} is positive.

Regards,
George

I recently took higher level math courses, I want to see if I can do this derivative right:

\dot{H} = -\frac{\dot{R}}{R^2}

Correct?

Can \dot{H} be positive in our universe?

Also, can Omega increase in an accelerating universe?

 

[George Jones]

\dot{H} = -\frac{\dot{R}}{R^2}

This isn't quite right - my notation wasn't the best. Now, I'm going to denote differentiation with respect to t by prime. Also, R is a function of time.

H \left( t \right) = \frac{R' \left( t \right)}{R \left( t \right)} = R' \left( t \right) R \left( t \right)^{-1}

Now use the product rule and the chain rule to find

\frac{d H}{d t} \left( t \right) = H' \left( t \right).

Can \dot{H} be positive in our universe?

This seems unlikely for our universe,

Also, can Omega increase in an accelerating universe?

If \Omega < 1, then \Omega will increase towards one in an acclerating universe.

Regards,
George

 

[kmarinas86]

This isn't quite right - my notation wasn't the best. Now, I'm going to denote differentiation with respect to t by prime. Also, R is a function of time.

H \left( t \right) = \frac{R' \left( t \right)}{R \left( t \right)} = R' \left( t \right) R \left( t \right)^{-1}

Now use the product rule and the chain rule to find

\frac{d H}{d t} \left( t \right) = H' \left( t \right).
This seems unlikely for our universe,
If \Omega < 1, then \Omega will increase towards one in an acclerating universe.

Regards,
George

I find this strange, because if the universe was accelerating, I would have expected (km/s)/Mpc to increase, not decrease.

When I do the product rule (I hope correctly), I get:

\frac{d H}{d t} \left( t \right) = -R' \left( t \right)R \left( t \right)^{-2}dt+R'' \left( t \right)R \left( t \right)^{-1}dt

The first term is what I had originally.

dt is positive, so it doesn't have an effect on the sign of the answer - I think.

Is R'' \left( t \right) dt the acceleration of the expansion (m/s^2)? If it was 0, \frac{d H}{d t} \left( t \right) would be negative, unless if R' \left( t \right) was negative (that is, if the universe is contracting). Since the universe is expanding, R' \left( t \right) is positive. R \left( t \right)^{-2} has to be positive since it is an inverse square. Thus, the first term must be negative in an expanding universe. In an expanding universe, if R'' \left( t \right), the acceleration of the expansion (m/s^2), is "large enough", that would mean \frac{d H}{d t} \left( t \right) would positive. I think what you're saying is that is that you think R'' \left( t \right) is not large enough - correct?

Now I get it. The universe is said to be expanding, say at velocity v. While this is happening, the radius of the universe has been increasing. Since the radius of the universe, in this case, in increasing faster than the velocity is increasing, (dist/time)/dist is getting smaller, even if velocity is increasing in the case of the accelerating universe. If the change of the velocity over time is faster than the change of the distance during that same time, then the (dist/time)/dist would increase.

 

[hellfire]

Accelerated expansion means \ddot a > 0 (strictly speaking a deceleration parameter q < 0). You can check that an increasing Hubble parameter \dot H > 0 is equivalent to a deceleration parameter q < -1. It can be shown that this implies a violation of the dominant energy condition (and one talks about "phantom energy").

 

[George Jones]

I find this strange, because if the universe was accelerating, I would have expected (km/s)/Mpc to increase, not decrease.

When I do the product rule (I hope correctly), I get:

\frac{d H}{d t} \left( t \right) = -R' \left( t \right)R \left( t \right)^{-2}dt+R'' \left( t \right)R \left( t \right)^{-1}dt

To be strictly correct, either the dt should be removed from the left side, or all the dt's should be removed from the right.

dt is positive, so it doesn't have an effect on the sign of the answer - I think.

Is R'' \left( t \right) dt the acceleration of the expansion (m/s^2)?

Yes, (without the dt) it is.

If it was 0, \frac{d H}{d t} \left( t \right) would be negative, unless if R' \left( t \right) was negative (that is, if the universe is contracting). Since the universe is expanding, R' \left( t \right) is positive. R \left( t \right)^{-2} has to be positive since it is an inverse square. Thus, the first term must be negative in an expanding universe.

Yes.

In an expanding universe, if R'' \left( t \right), the acceleration of the expansion (m/s^2), is "large enough", that would mean \frac{d H}{d t} \left( t \right) would positive. I think what you're saying is that is that you think R'' \left( t \right) is not large enough - correct?

Yes, and, as hellfire has noted, there are physical reasons for believing this.

Now I get it. The universe is said to be expanding, say at velocity v. While this is happening, the radius of the universe has been increasing. Since the radius of the universe, in this case, in increasing faster than the velocity is increasing, (dist/time)/dist is getting smaller, even if velocity is increasing in the case of the accelerating universe. If the change of the velocity over time is faster than the change of the distance during that same time, then the (dist/time)/dist would increase.

I think this is roughly correct.

Regards,
George