The Unsinkable Standard Model

The big physics news of the week last week came while I was in transit on Wednesday: The MiniBooNE (the odd capitalization is because it’s sort of an acronym) neutrino experiment released their first results on the neutrino oscillation studies they’ve been doing, and found, well, nothing new. In contrast to a previous experiment that hinted at the possible existence of a fourth type of neutrinos, the MiniBooNE results were entirely consistent with having only the three previously known types. There’s a news article here, and one of the MiniBooNE experimenters did a excellent guest post explaining the results at Cosmic Variance.

I don’t have much to say about the results themselves because, well, I’m not actually a neutrino physicist. One thing about this that’s really interesting, though, is that it’s yet another “success” for the Standard Model of particle physics.

It occurs to me that the Standard Model is in kind of a unique position among scinetific theories. I know of lots of examples of theories that everybody thought were right that turned out to be wrong, and there are plenty of examples of theories that at least some people think are right but that they can’t prove right. The Standard Model is the only theory I can think of that everybody knows is wrong, but nobody can prove is wrong.

OK, “wrong” may be a little too strong– “incomplete” is probably a better word. The Standard Model consists of a set of twelve material particles: six quarks (up, down, strange, charm, top, and bottom) and six leptons (electron, muon, tau, and electron, muon and tau neutrinos) with their associated antiparticles. It also includes four forces: gravity, electromagnetism, the strong nuclear force, and the weak nuclear force, plus their associated force carriers. Taken together, these particles and forces describe everything about the structure and organization of ordinary matter.

The problem is, they don’t explain everything. Most of the universe is made up of “dark matter” that we see only indirectly through its gravitational interactions with stars and galaxies. For various reasons, we know that this matter, whatever it is, can’t be made up of quarks, but beyond that, we have no idea what it is. There are lots of proposals of different sorts of particles not included in the Standard Model that could account for this extra mass, but nobody has ever conclusively seen one.

There’s also the question of mass: The Standard Model enumerates the particles and their masses, but doesn’t say why they have those masses. There’s a proposed mechanism by which fundamental particles could acquire their masses from interactions with another sort of particle– the interaction is called the “Higgs mechanism” and the particles are “Higgs bosons,” and there ought to be one for every type of material particle. Nobody has ever seen conclusive evidence of a Higgs boson, though, despite many active searches for them.

This is really a strange and awkward position to be in. The Standard Model works extremely well for those things that it describes, but we know it can’t be the whole story. And yet, every attempt to find physics beyond the Standard Model has come up empty. Nobody has yet found a particle or force that isn’t accounted for in the theory, despite a couple of decades’ worth of searching. In a certain sense, it’s a theory that works too well. We’ve got excellent indirect evidence that says it can’t be the whole story, but we can’t find any direct evidence of anything that doesn’t fit the theory.

It’s sort of like being in the early stages of one of those old Infocom text adventure games. We’ve explored all the obvious rooms, and picked up all the obvious items, but we haven’t really gotten anywhere. There’s got to be more to the game, because it takes up a lot of disk space, but we can’t find any way to get into any of the other rooms even though we know they have to be there…

I’m not sure what the particle physics analogue of being eaten by a grue is, though– loss of funding, maybe?

Comments

  1. #1 Moshe
    April 16, 2007

    Great post, it is puzzling, and maybe a little worrisome, how well the SM works. Hopefully news will come at the LHC.

    Small correction, the standard model has a single Higgs boson (some extensions have a couple), and all particles presumably get their masses through their interactions to that boson. This was a little ambiguous the way phrased.

  2. #2 Uncle Al
    April 16, 2007

    Mass is physics’ bad boy. Mass’ is a lump in Sevres, France. Gravitation cannot be usefully quantized. The Standard Model’s masses are explicitly inserted and Higgs rationalized.

    Perhaps all mass theories are heuristics for faulty founding postulates. Credible failure requires consistency with all prior success. Violate the Equivalence Principle with opposite parity mass distributions. Both metric gravitation (EP) and quantum field theory (conservation of angular momentum in isotropic space) would be falsified. It’s a two-day experiment in existing apparatus.

    EP parity violation might fail (420+ years of that with mass composition) or it might succeed – spacetime geometry falling to test mass geometry. Isn’t anybody curious to know the empirical outcome? Theory has no defense against observation.

  3. #3 Jeff
    April 16, 2007

    Am I the only one who was disappointed to see that Chad didn’t link to an Infocom emulator?

  4. #4 Neil
    April 16, 2007

    To use an Infocom analogy, the Standard Model seems to me rather like the thing your aunt gave you but you don’t know what it is from the Hitch-hiker’s Guide game – it takes whatever you can throw at it, and you can’t get rid of it.

    PS. Just for Jeff, some Infocom links:

    Flash version of H2G2 from the BBC, with illustrations by, among others, Rod Lord (who did the hand-animated TV series graphics).

    If you prefer old-skool, the Java version is here.

    Frotz, a full Z-machine emulator (for any Infocom games) is available for:
    Windows and Unix/Linux (for Debian/Ubuntu systems you should just be able to apt-get install frotz). Share and Enjoy!

  5. #5 andy
    April 16, 2007

    Eaten by a grue = making a strangelet/mini-black-hole that destroys the Earth?

  6. #6 Rob Knop
    April 16, 2007

    I think loss of funding is right.

    Here’s the grue:

    We build LEP in CERN. It’s hugely expensive, but that’s the cost of doing high-energy physics nowadays. (The energies have gotten really high.) The successor to LEP is the ILC (International Linear Collider), which will be that much more expensive.

    But LEP doesn’t find the Higgs boson, nor does it find supersymmetry. It does find all sorts of interesting physics, much as the colliders now are, but it’s the sort of stuff only really of interest to other particle physicists, and not the sort of stuff of interest to somebody watching a Nova special.

    No Higgs boson, no supersymmetry, and it will become an EXTREMELY hard sell to tell people that we need spend even more money to go to the next energy frontier.

    Nuclear and particle physics had a watershed century in the 20th century. Every time we went to a new energy, we found new stuff, and cool new stuff. Things were expanding all the time. There were dark periods, sure, such as when we had this zoo of particles without a great way to unify them, but the eightfold way and the quark model did nice things to all of that. But the experiments were always finding new stuff.

    If we don’t find something big and new — something like supersymmetry and “dark matter” LSP, or the Higgs boson, or, even better, something unexpected — at LEP, it’s going to be a long time before we have the stomach to look harder. Maybe the next energy threshold is what’s needed, but without some payoff at each step, it’s hard to make the next step.

    That’s the grue.

    Oh, and text adventures are a lot of fun!

    sometimes I wonder why I make long comments like this in others’ blogs instead of just making it a post on my own blog.

    -Rob

  7. #7 KeithB
    April 16, 2007

    I think that string theory might provide the answer to dark matter. I think that gravity is just as strong as the electrical forces, but some is projected out of our Brane and either into someone elses brane, (I am the dark matter in someone elses universe!) or back into our own.

  8. #8 onymous
    April 16, 2007

    Rob Knop wrote:

    “No Higgs boson, no supersymmetry, and it will become an EXTREMELY hard sell to tell people that we need spend even more money to go to the next energy frontier.”

    No Higgs boson and no other new physics would mean our understanding of quantum mechanics is fundamentally wrong (unitarity would be violated). I don’t have any idea what that would mean for funding for future experiments. Of course, it won’t happen, because unitarity breakdown is crazy. So there will be (at least) a Higgs.

  9. #9 Aaron Bergman
    April 16, 2007

    I think you mean LHC, not LEP, Rob.

    Everyone’s nightmare is a single scalar Higgs and nothing else. Anything that’s not that is good.

  10. #10 Chad Orzel
    April 17, 2007

    Am I the only one who was disappointed to see that Chad didn’t link to an Infocom emulator?

    I was going to, but once I found one, I spent the rest of the afternoon playing it…

  11. #11 dileffante
    April 18, 2007

    Same point to Rob as onymous, though I can’t be sure since I’m not a physicist. This exchange near the end of The Newtonian Legacy made me hope that the LHC will find something interesting, whatever it is:

    “”No, actually, this experiment is the Holy Grail of science in that it is a ‘’no lose’’ machine! If you take our theories of particles and omit the Higgs, funny stuff happens. In particular if you scatter W particles, the ones responsible for the weak force, then the chance of them interacting grows with energy. Eventually the probability of interaction becomes bigger than one!””
    ““What does that mean?””
    ““It means the theory is rubbish! Something has to happen that’’s new. The LHC will be able to probe this behaviour for sure. So we’’re guaranteed to find another part of the puzzle.””

  12. #12 Alexander Nahhas
    February 14, 2009

    It is sinkable because the experimental proofs readings are fraud
    Einstein’s Relativity Theory derived from Kepler’s Light Visual Deceptions Equation: S = r Exp ỉ ω t; sin ω t= v/c; v=speed; c=light speed
    By Joe Nahhas
    Abstract: Relativity theory can be derived from Kepler’s light visual deception equation S = r Exp ỉ ω t; sin ω t = v/c; v = speed and c = light speed. And all the experimental data used to support “proofs” of relativity theory fits deceptions formulas better than all of published papers of Einstein and all other physicists and astrophysicists combined.
    A- Special theory of relativity: Length contraction and Time dilations and Δ E = mc² and
    B- General theory of relativity: Advance of perihelion light bending gravitational red shifts and Shapiro’s time delay

    Object at r ——-Light sensing of moving objects ———– (seen as) S
    r —— Cosine (wt) + i sine (wt) ——– S = r [cosine (wt) + i sine (wt)]
    Particle ————————- Light —————————— Wave
    Newton ——– Kepler’s Time dependent ——– Newton’s Time dependent

    A-Special theory of relativity

    1-Lenght contraction

    Line of Sight: r cosine wt: light aberrations
    A moving object with velocity v will have when visualized through light sensing a light aberration angle (wt); w = constant and t= time

    Also, sine wt = v/c; cosine wt = √ [1-sine² (wt)] = √ [1-(v/c) ²]
    Where v = velocity; c = light velocity
    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 location 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 and 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 = c √ [1-(v/c) ²]

    2- Time dilatations
    Along the line of sight; S x = r cosine wt
    Hypotenuse = S x = [c t x] = c t √ [1-(v/c) ²];
    Where t = self time; t x = time by others

    t x = t √ [1-(v/c) ²]; and
    t = {1/√ [1-(v/c) ²]} t x

    These are time dilatation equations given by Einstein’s special relativity theory.

    3- Δ E= mc²

    S = r Exp (ỉ ω t); sin ω t =v/c; v = c sin ω t; r = -(c/ω) cosine ω t;
    And r. v = (-c²/ω) sin ω t cosine ω t

    P = d S/d t = (v + ỉ ω r) Exp (ỉ ω t); v² = c² sin² ω t; ω² r² = c² cosine² ω t
    P² = (v + ỉ ω r). (v + ỉ ω r) Exp [2(ỉ ω t)] = [v² -ω² r² +2ỉ ω (r. v)] Exp [2(ỉ ω t)]

    P² = [c² sin² ω t - c² cosine² ω t - 2c²ỉ sin ω t cosine ω t] Exp [2(ỉ ω t)]
    P² = – c² [cosine² ω t - sin² ω t + ỉ sin 2ω t] Exp [2(ỉ ω t)]

    P² = -c² Exp [4(ỉ ω t)]
    E = mP²/2 = – mc²/2 [cosine² 2ω t - sin² 2ω t + 2ỉ sin 2ω t cosine 2ω t]

    E = (-mc²/2) {1-2sin² 2ω t + 2ỉ [1- 2sin² ω t] 2[sin ω t cosine ω t]}
    E = (-mc²/2) {1- 2(v/c) ² + 4ỉ [1- 2(v/c) ²] (v/c) √ [1- (v/c) ²]}

    If v = 0 then E (1) = (-mc²/2); and
    If v = c then E (2) = (mc²/2) then

    Δ E = E (2) – E (1) = (mc²/2) – (-mc²/2)
    Δ E = mc²

    B- General Theory of relativity

    What is the visual effect for angular velocity along the line of sight? At Perihelion It is called the Advance of perihelion. Let us derive that

    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
    S = r Exp (ỉ wt); r² θ’= h = S² w’

    h = S²w’= [r² Exp (2iwt)] w’=r²θ’; w’ = (θ’) exp [-2(ỉ wt)]
    And w’= (h/r²) [cosine 2(wt) - ỉ sine 2(wt)] = (h/r²) [1- 2sine² (wt) - ỉ sin 2(wt)]

    With w’ = w’ (x) + ỉ w’(y); w’(x) = (h/r²) [1- 2sine² (wt)]
    Δ w’= w’(x) – (h/r²) = – 2(h/r²) sine² (wt) = – 2(h/r²) (v/c) ² v/c=sine wt

    Angular velocity (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
    [180/π; degrees][100years=36526days; century] x [3600; seconds in degree]

    Δ w” = (-720x36526x3600/T) {[√ (1-ε²]/(1-ε)²} (v/c)² seconds of arc per century

    This equation gives the rate of advance of perihelion of Mercury with better results than all of Albert Einstein’s publications and better than all of published physics.

    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<

    1- Advance of Perihelion of mercury.

    G=6.673x10^-11; M=2x10^30kg; m=.32x10^24kg
    ε = 0.206; T=88days; c = 299792.458 km/sec; a = 58.2km/sec
    Calculations yields:
    v =48.14km/sec; [√ (1- ε²)] (1-ε) ² = 1.552
    Δ w”= (-720x36526x3600/88) x (1.552) (48.14/299792)²=43.0”/century

    2- DI Herculis Apsidal motion solution: derived from S= r exp [ỉ ω t]
    (See other articles by Joe Nahhas)

    W° (ob) = (-720x36526/T) x {[√ (1-ε²)]/ (1-ε) ²} [(v*/c) + (v°/c)] ² degrees/ century

    Where v* = v (center of mass) = 106.38km/sec; v° (spin difference) = 0
    T = orbital period; ε = eccentricity; c =light speed

    Application 3: Gravitational red shift: Pound Rebka Experiment

    S = r Exp [î ω t]

    1/S = 1/r Exp [-ỉ ω t]
    And λ (S) = λ (r) Exp [-ỉ ω t]; λ = wavelength

    Then υ(s) = υ(r) Exp [ỉ ω t]; υ = frequency
    And υ(S) = υ (r, t) = υ(r, 0) υ (0, t) = υ(r) υ (0, t)
    With sin ω(r) t = v/c; cosine ω(r) t = √ [1-(v/c) ²]

    Then υ (r, t) = υ(r, 0) {√ [1-(v/c) ²] + ỉ (v/c)} = Real {υ(r, t)} + Imaginary {υ(r, t)}
    Real {υ (r, t)} = υ (r, 0) √ [1-(v/c) ²] ≈ υ (r, 0) [1 - 1/2(v/c) ²]

    Δ υ (r, t) = real {υ (r, t)} - υ (0, t)
    Δ υ (r, t) = -υ (r, 0)/2 [(v/c) ²]
    Δ υ(r, t)/υ(r, 0) = -1/2(v/c)²[up]-{1/2(v/c)²[down]} = - (v/c) ²
    v² = 2gh; g = 9.81km/s² gravitational acceleration; h = height

    Δ υ/υ [Total] =-[2gh/c²]
    4- Light bending: Lord Edenton experiment
    S = r Exp [ỉ ω t]; From Kepler's Equation: r² θ' = h = 2A/T
    h = S²(r, t) θ'(r, t) = r² (θ, t) θ' (θ, t) = r² (θ, 0) Exp [2ỉ ω t] θ' (θ, t) = 2A/t
    And θ' (θ, t) = θ' (θ, 0) θ'(0, t) = [h/ r² (θ, 0)] Exp [-2ỉ ω(r) t]
    Then θ '(θ, t) = [2A/t r² (θ, 0)] {1 - 2sin²ω(r) t - 2ỉ sin ω(r) t cosine ω(r) t}
    Now [t θ'(θ, t)] = [2A/r² (θ' 0)] [1 - 2sin²ω(r) t] -2ỉ [2A/r² (θ, 0)] [sin ω(r) t cosine ω(r) t]
    = Δ x + i Δ y
    Δ θ = Δ x - [A/r² (θ, 0)] = - [A/r² (θ, 0)][4sin²ω(r)t]; sin ω(r)t = v/c
    Δ θ = - [A/r² (θ, 0)](v/c) ²
    (v/c) ² ≈ 1.75"; v² = GM/R; G = Gravitational constant; M = Sun mass; R = sun radius
    Δ θ = [A/r² (θ, 0)] [1.75"]; A = area
    The values depend on near by stars and the measured values fit this equation.
    Russians in 1936; Δ θ = 2.74
    [A/r² (θ, 0)] = π/2
    Δ θ = π/2(1.75") = 2.74"

    Application 5: Shapiro time delay (Vikings 6, 7; 1977)
    Mars --------------------------- Middle---- Sun ------------- Earth
    The center of mass is the sun. The sun produces a velocity field given by
    v = √ [GM/a (1- ε²/4)]
    From above t =2 arc length/c=2d Δ w/c = (8π r/c) (v/c) ²; Δ w=4π (v/c) ²; r = 2a=d
    t = 16πGM/c³ (1-ε²/4); ε = [a (1) -a (2)]/ [a (1) + a (2)] = .2075
    t = (8πd/c) (v/c) ²= 8π (377,536,987.5/299792.458) (26.6575872/299792.458)²=250μs
    If d = 2a (1-ε²/4), then t = 247.597μs value theorized actual measured value is 250μs
    All this is not due to space-time but due to light aberration caused by moving planets.
    θ'(0,0) = h(0,0)/r²(0,0) = 2π/T
    θ' (0,t) = θ'(0,0)Exp(-2ỉwt)={2π/T} Exp (-2iwt)
    θ'(0,t) = θ'(0,0) [cosine 2(wt) - ỉ sine 2(wt)] = θ'(0,0) [1- 2sine² (wt) - ỉ sin 2(wt)]
    θ'(0,t) = θ'(0,t)(x) + θ'(0,t)(y); θ'(0,t)(x) = θ'(0,0)[ 1- 2sine² (wt)]
    θ'(0,t)(x) – θ'(0,0) = - 2θ'(0,0)sine²(wt) = - 2θ'(0,0)(v/c)² v/c=sine wt; c=light speed
    T [θ'(0, t) - θ'(0, 0)] = -4π (v/c) ²
    Δ θ = -4π (v/c) ² Earth-Mars
    Sun-Photon:
    The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²---) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
    v=√ [Gm M/ (m + M) a (1-ε²/4)] ≈ √ [GM/a (1-ε²/4)]; m< ΔΓ = 2 arc length/c = 2[Δ θ] 2d/c = 2[- 4π (v/c) ²] 2d/c; ΔΓ = -8πd/c (v/c) ²;
    ΔΓ = 8πd/c³ [GM/a (1-ε²/4)] =16πGM/c³ (1-ε²/4) = Γ0 (1 - ε²/4)
    ε = [a (planet 1) - a (planet 2)]/ [a (planet 1) + a (planet 2)] =0.2075 Mars-Earth
    Γ0 = 16 πGM/c³= 247.5974607μs=universal constant; ΔΓ = 250μs Mars-Earth.
    Joe nahhas1958@yahoo.com All right reserved

  13. #13 joe nahhas
    February 26, 2009

    Einstein’s Nemesis: DI Herculis Apsidal motion solution
    It is not about just throwing relativity theory but throwing relativity theory with NASA Astrophysicists Attached to it.
    Abstract: This is the Solution to the “quarter of a century” motion puzzle that Einstein and all other 100,000 space-time physicists and Astrophysicists could not solve by space-time physics or any other said or published physics including 109 years of Noble Prize winner physics and 400 years of Astro-physics and dedicated to Dr Edward Guinan and Dr Frank Maloney of Villanova University who posted this motion puzzle in 1985 as not solvable by space-time physics and in “Apparent” inconsistency with General relativity theory and started the binary stars collections with motion that can not be explained by space-time physics or any said or published physics. This motion puzzle is posted on Smithsonian-NASA website SAO/NASA.

    Introduction Time is not a structure like space- to scientifically accept space- to imaginary time – back to space jumping continuum told by Einstein and taught by MIT Harvard Cal-Tech Stanford NASA and all other space-time physics departments regardless what the 100,000 Space-time Physicists have/had said about it because space-time physics is not good enough to solve the simplest problem in physics introduced here and solved.

    Universal Mechanics Solution:

    For 350 years Physicists Astronomers and Mathematicians missed Kepler’s time dependent equation introduced here and transformed Newton’s equation into a time dependent Newton’s equation and together these two equations explain Quantum – Relativistic effects; it combines classical mechanics and quantum mechanics into one mechanics and explains “relativistic” effects as the difference between time dependent measurements and time independent measurements of moving objects and in practice it amounts to “Visual” effects.

    All there is in the Universe is objects of mass m moving in space (x, y, z) at a location
    r = r (x, y, z). The state of any object in the Universe can be expressed as the product

    S = m r; State = mass x location:

    P = d S/d t = m (d r/d t) + (dm/d t) r = Total moment
    = change of location + change of mass
    = m v + m’ r; v = velocity = d r/d t; m’ = mass change rate

    F = d P/d t = d²S/dt² = Total force
    = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r
    = m γ + 2m’v +m” r; γ = acceleration; m” = mass acceleration rate
    In polar coordinates system

    r = r r (1) ;v = r’ r(1) + r θ’ θ(1) ; γ = (r” – rθ’²)r(1) + (2r’θ’ + r θ”)θ(1)

    F = m [(r"-rθ'²) r (1) + (2r'θ' + r θ") θ (1)] + 2m’[r' r (1) + r θ' θ (1)] + (m” r) r (1)

    = [d²(m r)/dt² - (m r)θ'²]r(1) + (1/mr)[d(m²r²θ')/d t]θ(1) = [-GmM/r²]r(1)

    Proof:

    r = r [cosine θ î + sine θ Ĵ] = r r (1)
    r (1) = cosine θ î + sine θ Ĵ
    v = d r/d t = r’ r (1) + r d[r (1)]/d t = r’ r (1) + r θ’[- sine θ î + cosine θĴ]
    = r’ r (1) + r θ’ θ (1)
    θ (1) = -sine θ î +cosine θ Ĵ; r(1) = cosine θ î + sine θ Ĵ

    d [θ (1)]/d t= θ’ [- cosine θ î - sine θ Ĵ= - θ' r (1)
    d [r (1)]/d t = θ’ [-sine θ î + cosine θ Ĵ] = θ’ θ(1)

    γ = d [r' r(1) + r θ' θ (1)] /d t = r” r(1) + r’ d[r(1)]/d t + r’ θ’ r(1) + r θ” r(1) +r θ’ d[θ(1)]/d t
    γ = (r” – rθ’²) r(1) + (2r’θ’ + r θ”) θ(1)

    d² (m r)/dt² – (m r) θ’² = -GmM/r² Newton’s Gravitational Equation (1)
    And d (m²r²θ’)/d t = 0 Central force law (2)

    (2) : d(m²r²θ’)/d t = 0 < ==> m²r²θ’ = [m²(θ,0)φ²(0,t)][ r²(θ,0)ψ²(0,t)][θ'(θ, t)]
    = [m² (θ, t)] [r² (θ, t)] [θ' (θ, t)]
    = [m² (θ, 0)] [r² (θ, 0)] [θ'(θ, 0)]
    = [m² (θ, 0)] h (θ, 0); h (θ, 0) = [r² (θ, 0)] [θ'(θ, 0)]
    = H (0, 0) = m² (0, 0) h (0, 0)
    = m² (0, 0) r² (0, 0) θ’(0, 0)
    m = m (θ, 0) φ (0, t) = m (θ, 0) Exp [λ (m) + ì ω (m)] t; Exp = Exponential
    φ (0, t) = Exp [ λ (m) + ỉ ω (m)]t

    r = r(θ,0) ψ(0, t) = r(θ,0) Exp [λ(r) + ì ω(r)]t
    ψ(0, t) = Exp [λ(r) + ỉ ω (r)]t

    θ’(θ, t) = {H(0, 0)/[m²(θ,0) r(θ,0)]}Exp{-2{[λ(m) + λ(r)]t + ì [ω(m) + ω(r)]t}} ——I

    Kepler’s time dependent equation that Physicists Astrophysicists and Mathematicians missed for 350 years that is going to demolish Einstein’s space-jail of time

    θ’(0,t) = θ’(0,0) Exp{-2{[λ(m) + λ(r)]t + ỉ[ω(m) + ω(r)]t}}

    (1): d² (m r)/dt² – (m r) θ’² = -GmM/r² = -Gm³M/m²r²

    d² (m r)/dt² – (m r) θ’² = -Gm³ (θ, 0) φ³ (0, t) M/ (m²r²)

    Let m r =1/u

    d (m r)/d t = -u’/u² = -(1/u²)(θ’)d u/d θ = (- θ’/u²)d u/d θ = -H d u/d θ
    d²(m r)/dt² = -Hθ’d²u/dθ² = – Hu²[d²u/dθ²]

    -Hu² [d²u/dθ²] -(1/u)(Hu²)² = -Gm³(θ,0)φ³(0,t)Mu²
    [d²u/ dθ²] + u = Gm³(θ,0)φ³(0,t)M/H²

    t = 0; φ³ (0, 0) = 1
    u = Gm³(θ,0)M/H² + A cosine θ =Gm(θ,0)M(θ,0)/h²(θ,0)

    And m r = 1/u = 1/[Gm(θ,0)M(θ,0)/h(θ,0) + A cosine θ]
    = [h²/Gm(θ,0)M(θ,0)]/{1 + [Ah²/Gm(θ,0)M(θ,0)][cosine θ]}

    = [h²/Gm (θ, 0) M (θ, 0)]/ (1 + ε cosine θ)
    Then m r = [a (1-ε²)/ (1+εcosθ)] m (θ, 0)

    Also, r (θ, 0) = [a (1-ε²)/ (1+εcosθ)] m r = m (θ, t) r (θ, t)
    = m(θ,0)φ(0,t)r(θ,0)ψ(0,t)

    Then, r (θ, t) = [a (1-ε²)/ (1+εcosθ)] {Exp [λ(r) + ω(r)] t} —————— II

    This is Newton’s time dependent Equation

    If λ (m) ≈ 0 fixed mass and λ(r) ≈ 0 fixed orbit; then

    θ’(0,t) = θ’(0,0) Exp{-2ì[ω(m) + ω(r)]t}

    r(θ, t) = r(θ,0) r(0,t) = [a(1-ε²)/(1+εcosθ)] Exp[i ω (r)t]

    m = m(θ,0) Exp[i ω(m)t] = m(0,0) Exp [ỉ ω(m) t] ; m(0,0)

    θ’(0,t) = θ’(0, 0) Exp {-2ì[ω(m) + ω(r)]t}

    θ’(0,0)=h(0,0)/r²(0,0)=2πab/Ta²(1-ε)²

    = 2πa² [√ (1-ε²)]/T a² (1-ε) ²; θ’(0, 0) = 2π [√ (1-ε²)]/T (1-ε) ²

    θ’(0,t) = {2π[√(1-ε²)]/T(1-ε)²}Exp{-2[ω(m) + ω(r)]t

    θ’(0,t) = {2π[√(1-ε²)]/(1-ε)²}{cosine 2[ω(m) + ω(r)]t – ỉ sin 2[ω(m) + ω(r)]t}

    θ’(0,t) = θ’(0,0) {1- 2sin² [ω(m) + ω(r)]t – ỉ 2isin [ω(m) + ω(r)]t cosine [ω(m) + ω(r)]t}

    θ’(0,t) = θ’(0,0){1 – 2[sin ω(m)t cosine ω(r)t + cosine ω(m) sin ω(r) t]²}

    – 2ỉ θ’(0, 0) sin [ω (m) + ω(r)] t cosine [ω (m) + ω(r)] t

    Δ θ (0, t) = Real Δ θ (0, t) + Imaginary Δ θ (0.t)

    Real Δ θ (0, t) = θ’(0, 0) {1 – 2[sin ω (m) t cosine ω(r) t + cosine ω (m) t sin ω(r) t]²}

    W (observed) = Real Δ θ (0, t) – θ’(0, 0)
    = – 2 θ’(0, 0) {(v°/c) √ [1-(v*/c) ²] + (v*/c) √ [1- (v°/c) ²]} ²

    With, v ° = spin velocity; v* = orbital velocity; v°/c = sin ω (m)t; v*/c = cosine ω (r) t

    And, v°/c < < 1; (v°/c)² ≈ 0; v*/c << 1; (v*/c)² ≈ 0

    W (ob) = - 2[2π √ (1-ε²)/T (1-ε) ²] [(v° + v*)/c] ²

    W (ob) = (- 4π /T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² radians
    W (ob) = (-720/T) {[√ (1-ε²)]/ (1-ε) ²} [(v° + v*)/c] ² degrees; Multiplication by 180/π

    W° (ob) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} [(v°+ v*)/c] ² degrees/100 years
    The circumference of an ellipse: 2πa (1 - ε²/4 + 3/16(ε²)²- --.) ≈ 2πa (1-ε²/4); R =a (1-ε²/4)
    Where v (m) = √ [GM²/ (m + M) a (1-ε²/4)]
    And v (M) = √ [Gm² / (m + M)a(1-ε²/4)]
    DI Her Apsidal motion solution:

    Data: T=10.55days r(m) = 0.0621 m=5.15M(0) R(m)=2.68R(0) [v°(m),v°(M)]=[45,45]
    ε = 0.4882; r(M) = 0.0574 M=4.52M(0) R(M) =2.48 m + M=9.67M(0)

    1-ε = 0.5118; (1-ε²/4) = 0.94; [√ (1-ε²)] / (1-ε) ² = 3.33181
    G=6.673x10^-11; M (0) = 1.98892x19^30kg; R(0) = 0.696x10^9m
    Calculations
    V (m) = √ [GM²/ (m + M) a (1-ε²/4)] = 99.88 km/sec
    V(M) = √ [Gm²/ (m + M) a (1-ε²/4)] = 113.9km/sec
    Apsidal motion is given by this formula:
    W° (observed) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²]}[(v*/c) + (v°/c)]²
    And, v° (spin) = v° (m) - v° (M)
    = 45 km/s - 45 km/s = 0 km/s
    With, v*(Orbit) = v (cm)/c
    And, v* = v (cm) = [m v (m) + M v (M)]/ (m + M)
    Then, v* = v (cm) = ∑ m v/∑m = [(5.15x99.88) + (4.52x113.8)]/9.67 = 106.38km/s
    W° (observed) = (-720x36526/T) {[√ (1-ε²)]/ (1-ε) ²} {[v* + v°]/c] ²
    = (-720x36526/10.55) (3.33181) (106.38/300,000)²
    W° (observed) = 1.04°
    References: Go to Smithsonian/NASA website SAO/NASA and type:
    1- Apsidal motion of DI Her: Dr Edward Guinan and Dr Frank Maloney; 1985.
    2- New Apsidal Motion of DI Her: Dr Edward Guinan and Dr Frank Maloney; 1994.

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