What Should Cosmologists Look For?

Comments

January 19, 2009 at 9:25 am

(1) Burt Jordaan says:

I think the “rate of expansion of the universe” is not sufficient, but rather the evolution of the rate of expansion (or of the acceleration parameter). In a way it fixes the amount of dark energy in the universe, while the rate of expansion tells us nothing about that.

Burt

January 20, 2009 at 9:31 pm

(2) Bruce says:

I’d say some direct evidence of the universe expanding at all would be good. A Dopplerian approach to galactic red shift is only an assumption.

January 21, 2009 at 2:06 pm

(3) Birabhadra says:

Errors in the general theory of relativity. The whole scientific world is crazy about black holes. I am not at any rate convinced about this concept. Nature can’t violate her own laws. Black holes violate physics, mathematics as well as metaphyisics.

January 24, 2009 at 11:19 am

(4) saswot says:

it seems that universe is continuosly expanding from the red shift observed.now einstein did not believe it was expanding as a whole,rather, some parts of the universe were contracting,like during star formation. contradictly others like quasars and pulsars were moving away from their source which did prove the expansion.now considering that we are statically observing the universe from any part in space, the motion of the radiation of these quasars defined with respect to another observer in space,in inertial frame of referance,is the same for all, according to einstein. if einstein is right in his relativity thoery,then,the expansion of the universe itself seems contradictory,and actually is expanding or contracting with respect to another expansion or contraction!!!this was why einstein thought the indefinity of the expansion of the universe and related it to human stupidity,commenting that he was not sure of the latter one!!!!!!!

January 26, 2009 at 11:44 am

(5) joe nahhas says:

Kepler (demolish) Vs Einstein’s

Areal velocity is constant: r² θ’ =h Kepler’s Law
h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
r² θ’= h = S² w’
Replace r with S = r exp (ỉ wt); h = [r² Exp (2iwt)] w’
w’ = (h/r²) exp [-2(i wt)]
w’= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]
w’ = w’(x) + ỉ w’(y) ; w’(x) = (h/r²) [ 1- 2sine² (wt)]
w’(x) – (h/r²) = – 2(h/r²)sine²(wt) = – 2(h/r²)(v/c)² v/c=sine wt
(h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²

Δ w’ = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
Δ w’ = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians
Δ w’ = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π
Δ w’ = (-720×36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years
Δ w” = (-720×3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² seconds of arc by 3600

Δ w” = (-720×36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century
This Kepler’s Equation solves all the problems Einstein and all physicists could not solve

The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4) v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m

January 26, 2009 at 11:49 am

(6) joenahhas says:

Kepler (demolish) Vs Einstein’s
Areal velocity is constant: r² θ’ =h Kepler’s Law
h = 2π a b/T; b=a√ (1-ε²); a = mean distance value; ε = eccentricity
r² θ’= h = S² w’
Replace r with S = r exp (ỉ wt); h = [r² Exp (2iwt)] w’
w’ = (h/r²) exp [-2(i wt)]
w’= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]
w’ = w’(x) + ỉ w’(y) ; w’(x) = (h/r²) [ 1- 2sine² (wt)]
w’(x) – (h/r²) = – 2(h/r²)sine²(wt) = – 2(h/r²)(v/c)² v/c=sine wt
(h/ r²)(Perihelion/Periastron)= [2πa.a√ (1-ε²)]/Ta² (1-ε) ²= [2π√ (1-ε²)]/T (1-ε) ²
Δ w’ = [w'(x) – h/r²] = -4π {[√ (1-ε²)]/T (1-ε) ²} (v/c) ² radian per second
Δ w’ = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² radians
Δ w’ = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² degrees; Multiplication by 180/π
Δ w’ = (-720×36526/T) {[√ (1-ε²)]/(1-ε)²} (v/c)² degrees/100 years
Δ w” = (-720×3600/T) {[√ (1-ε²)]/ (1-ε) ²} (v/c) ² seconds of arc by 3600
Δ w” = (-720×36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century
This Kepler’s Equation solves all the problems Einstein and all physicists could not solve
The circumference of an ellipse: 2πa (1 – ε²/4 + 3/16(ε²)²- –.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4) v=√ [G m M / (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m

January 29, 2009 at 5:58 pm

(7) joe nahhas says:

Einstein is wrong and a Half
E=mc²/2
2009 is the end of Einstein’s space-jail of time and Fraud symbol E=mc²
Joenahhas1958@yahoo.com
Time is not a structure like space to allow space-to time-back to space jumping claimed by Physicists regardless of what physicists have to say about it because Physics is a business and not necessarily science or scientific and like every business it comes with fraud and fraud is Einstein’s space-time (x, y, z, it) continuum that led to fraud symbol E=mc² and yes I am saying that 109 years of Nobel prize winners physics and physicists are all wrong and space-time physics is based on scientific fraud. When “results” expected and “No” discovery, Physicists rigged Physics for grant money since the start of the industrial revolution. Physics today is at least 51 % fraud!
r ——————>>Exp (ì w t) ———->> S=r Exp (ì wt) Nahhas’ Equation
Orbit——–>> Orbit light sensing——>> Visual Orbit; Exp = Exponential
Particle —->> light sensing of moving objects———— >> Wave
Newton———>>light sensing———->> Quantum
Quantum = Newton x Visual Effects
Quantum – Newton = Relativistic = Optical Illusions
E (Energy by definition) = mv²/2 = mc²/2; if v = c
m = mass; v= speed; c= light speed; w= angular velocity; t= time
S = r Exp (ì w t) = r [cos (wt) + ì sin (wt)] Visual effects
P = visual velocity = change of visual location
P = d S/d t = v Exp (ì w t) + ì w r Exp (ì w t)
= (v + ì w r) Exp (ì w t) = v (1 + ì) Exp (ì w t) = visual speed; v = wr
E (visual energy= what you see in lab) = m p²/2; replace v by p in E = mv²/2
= m p²/2 = m v²/2 (1 + ì) ² Exp (2ì wt)
= mv²/2 (2ì) [cosine (2wt) + ì sine (2wt)]
=ì mv² [1 - 2 sine² (wt) + 2 ì sine (wt) cosine (wt)];v = speed; c = light speed
wt = π/2
E (visual) = ìmv² (1 – 2 + 0)
E (visual) = -ì mc² ≡ mc² (absolute value;-ì = negative complex unit) If v = c
w t = π/4
E (visual) = imv² [1-1 +ỉ] =-mc²; v = c
wt =-π/4+ỉln2/2; 2ỉ wt=-ỉπ/2 – ln2
Exp (2i wt) = Exp [-ỉπ/2] Exp [ln(1/2)]=[-ỉ (1/2)]
E (visual) = imv² (-ỉ/2) =1/2mc² v = c
Conclusion: E = mc² is the visual Illusion of E = mc²/2 joenahhas1958@yahoo.com. All rights reserved.
PS: In case of E=mc² claims to be rest energy claims then
E=1/2m (m v + m’ r) ² = (1/2m) (m’ r) ²; v = 0
E = (1/2m) (mc) ²; m’ r =mc
E=mc²/

February 8, 2009 at 12:44 am

(8) Alexander says:

Einstein’s Physics Dollar Store on Campus
MIT Harvard Cal-Tech
Sponsored by NASA
Why Relativity theory is not Physics and why Einstein’s “thought” = 0
Walking the walk and talking the talk taking on all space-time confusion of physics by
MIT Harvard and Cal-Tech and all other Physics dollar stores departments
And why LHC burned itself

Visual Effects and the confusions of “Modern” physics

r ——— Light sensing of moving objects ——- S
Actual object—– Light ——— Visual object
r – ——-cosine (wt) + i sine (wt) – S = r [cosine (wt) + i sine (wt)]
Newton– Kepler’s time visual effects — Time dependent Newton
Particle ————– Visual effects ——————– Wave

Line of Sight: r cosine wt

r ——————- r cosine (wt) line of sight light aberrations

A moving object with velocity v will be visualized by
light sensing through an angle (wt);w = constant and t= time
Also, sine wt = v/c; cosine wt = √ [1-sine² (wt) = √ [1-(v/c) ²]

A visual object moving with velocity v will be seen as S

S = r [cosine (wt) + i sine (wt)] = r Exp [i wt]; Exp = Exponential

S = r [√ [1-(v/c) ²] + ỉ (v/c)] = S x + i S y

S x = Visual along the line of sight = r [√ [1-(v/c) ²]

This Equation is special relativity length contraction formula
And it is just the visual effects caused by light aberrations of a
moving object along the line of sight.

In a right angled velocity triangle A B C: Angle A = wt; angle B = 90°; Angle C = 90° -wt
AB = hypotenuse = c; BC = opposite = v; CA= adjacent = √ [1-(v/c) ²]