Redshift (2)

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My new article on Electro-Weak redshift. 1

Transcript of Redshift (2)

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My new article on Electro-Weak redshift.

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Can Gravitational Redshift and Cosmic Expansion Redshift

Be Separated?

Douglas Leadenham

March 30, 2016

1 Introduction

In a 2005 seminar for high school teachers at SLAC National Accelerator Laboratory ateacher asked the moderator how much of observed cosmic microwave background redshiftis gravitational and how much results from recessional velocity from us, the observers. Sheresponded that the answer depends on whether the early universe was ultra relativistic ornot, and that question can’t be answered at present.

1.1 Motivation

Years later in 2014 an astrophysics researcher also claimed that the two factors can’tbe separated. Now that the Higgs field and the field strength of electro-weak symmetrybreaking are known the two factors may be separable.

2 Higgs Field

Section 7.4, Eq.7.14, of 21st Century Physics1 provides the Higgs field:

EH =

√ln 2

4πkFEP (1)

In this simple expression, kF is the Fermi level of the degenerate primordial fireball andEP is the Planck energy.

1Douglas Leadenham, Topics in 21st Century Physics - The Universe as Presently Understood, DJLe-Books, 2016

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2.1 Field Curvature

From basic physics one knows the energy density of Earth’s gravitational field.

ug =g2

8πG(2)

Section 8.7.1, Eq. 8.25, of the same book gives the corresponding energy density of thefireball at electro-weak (E-W) symmetry breaking:

uw =3 (ln 2)2 k2F c

2

(4π)3G~2(3)

From this the functional equivalent of the E-W field strength is

g =

√6 ln 2

kF c

~(4)

Simple dimensional analysis confirms that this g represents an acceleration like Earth’s g,but a kinematical formula is not defined before de Sitter space came into existence. Divide(4) by c2 to convert it to an analog of more general curvature. This can form a ratio withthe effective curvature of the whole universe at its logical beginning.

g

c2=

√6 ln 2

kF~c

= 59 (5)

This curvature is measured in meters−1. For comparison and perspective the curvature ofg in (2) is 9.8

c2meters−1.

2.2 Universal Curvature

Start with a spherical universe with a horizon:

2Gµ

c2=

2GMu

Ruc2= 1 (6)

Here Mu is the effective mass of the universe and Ru is the Hubble length. In the inflation-ary model, now confirmed by modulation of the cosmic microwave background, althoughsome skeptics attach more credence to the overall signal, there must be more of the universebeyond the horizon, because inflation is superluminal, assuming that the speed of light wasconstant at the beginning. The speed of light may have been much greater then, but wecan’t be sure of that. Section 8.5, Eq. 8.19, gave the Hubble constant:

H0 =mec

π exp 1

e2c

~(7)

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The result of Chapter 3, obtained by a linear optimization on galaxy rotation, gave thetension at GUT symmetry breaking:

c2= 4πµ0α =

µ20e2c

~(8)

So that the curvature factor we seek for the universe is

H0

c=

exp 1

me

µ0(9)

However, this is for the observable universe. There is an unknown part that must be ac-counted for. We will use an adjustable factor n, as follows below. Use the analogy ofNewton’s universal law of gravitation and divide by c2 to convert the kinematical formu-lation to one of curvature.

2GMu

c2R2u

=1

Ru

2Gµ

c2(10)

Substitute for the Hubble length and obtain

2GMu

c2R2u

=4α

exp 1

me

µ0

2Gµ

c2(11)

The last factor on the right, 2Gµc2

, represents the whole universe, including the unobservablepart beyond the horizon. The left hand side represents the curvature of this mostly unseenuniverse, so the right hand side must include the adjustable factor n. Express (10) thisway:

2GMu

c2R2u

=4α

exp 1

me

µ0

2Gµ

c2=

exp 1

me

µ0

2G

c2mP

lP(12)

The ratio of Planck mass to Planck length is by definition c2

G . That leaves 2Gc2

, which is 1µ ,

the inverse of a curvature, less than an unknown, higher-dimensional universal curvatureµu by a dimensionless adjustable factor n. If we try n = 6, (11) evaluates to 63 meters−1,for comparison to (5). Rewrite (11) as:

2GMu

c2R2u

=4α

exp 1

me

µ0

1

µ

c2

G=

exp 1

me

µ0

1µun

c2

G(13)

The dimensionless ratio of curvatures becomes:√6 ln 24π

kF~c

4αexp 1

meµ0

nc2

G

=exp 1 ln 2

√6

8πα

kFµ0G

n~mec3(14)

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3 Redshift

At the very beginning the general relativistic gamma-factor is

γ =1√

1− GEplpc4

. 108 (15)

With knowledge that redshift results from frequency reduction due to time dilation, replacethe hypothetical

GEplpc4

with (14). Vary n to see how redshift z responds.

zg + 1 =1√

1− exp 1 ln 2√6

8παkFµ0Gn~mec3

(16)

Observed redshift z is the product of the gravitational field redshift zg with the cosmicexpansion redshift zc as

z + 1 = (zg + 1)(zc + 1) (17)

The Planck Law of black body radiation is

I(λ, T ) =2πhc2

λ5(ehc

λkBT − 1)(18)

The black body radiation intensity has a maximum at hcλkBT

= 2.82144. This λ is thewavelength of the cosmic background radiation that is red shifted by a factor of 1091,according to recent measurement. To preserve symmetry we will put back into (14) thefactor 2 that goes with G.

exp 1 ln 2√

6

8πα

kFµ02G

n~mec3< 1 (19)

Now the parameter n can be given integer values from 11 to 14 to show how (16) variesin response. Taking n = 11 makes γ imaginary. Take n = 12 and zg = 3.02; n = 13 giveszg = 2, and n = 14 gives zg = 1. The last one is unreasonably small, but zg = 3 is certainlyunderstandable, as is zg = 2.

At this point one could reason that the integers just chosen may have something to do withthe number of dimensions of the universe of three mutually orthogonal sectors. It is knownthat there must be at least 11 dimensions, so the integer 12 possibly is significant.

If we take the composite redshift to be 1091, then the cosmic expansion redshift factorthen is zc = 272 if zg = 3, and zc = 363 if zg = 2. These are reasonable results. However,there is a problem.

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4 The Problem

Here is the situation: the field strength of (5) is that of the primordial universe at E-Wsymmetry breaking, and it is ratioed to the field at the very beginning 13.8 billion yearsago, while the observed CMB radiation originated at the epoch of last scattering whenneutral hydrogen could form. The redshift measured now is relative to the emitted CMBradiation. These two pairs of events occurred at very different times and circumstances,most notably the ambient field strength. The analysis presented here may be correct inprinciple, but while the fields of the analysis can be right, the CMB radiation did notoriginate with them. While the present day universal average field strength is known, therelevant field at last scattering is not. The challenge is to find it.

5 Resolution

Here we will redo the redshift calculation of §2.2 for the relevant fields. Energy densityuCMB at the surface of last scattering is the equivalent energy of the observable universedivided by the volume of the (Euclidean flat) universe at the age of the universe then. Thistime is 375,000 years, because the surface of last scattering is more like a dense ball ofwater than a transparent balloon. We will call that time tCMB. We have

uCMB =3Ruc

8πGt3CMB

=1

8πGg2CMB (20)

gCMB

c2=

√3Ru

c3t3CMB

(21)

Ru =exp 1µ04αme

(22)

The corresponding energy density of the present-day universe is the same universal equiv-alent energy divided by the equivalent volume of the Euclidean universe.

uNOW =3c4

8πGR2u

(23)

gNOWc2

=

√3

Ru(24)

The dimensionless ratio desired to substitute into (15) is√c3t3CMB

R3u

(25)

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This evaluates to 1.48 × 10−7. This looks close to the pattern, 1.16 × 10−7, found in §3.6of the cited e-book, see §8 below. Just for curiosity, put in an age of the CMB that givesthat figure. It is 323,000 years. This age falls into the inflationary period prior to thesurface of last scattering where the CMB radiation originates. The CMB corresponds to atemperature . 3000 K for which the radiation does not ionize hydrogen. The earlier timecorresponds to the time when stable electrons and quarks condensed from a radiation fieldof temperature 250 million K, left over after anti-particles annihilated or aggregated intonascent black holes. This modeling does not conflict with our limited knowledge of theearly universe, so we can be satisfied with our estimates of both components of redshiftz = 1091.

6 Conclusion

The redshift factor zg = 3 of §3 above is correct in principle, but it is unobservable. Whatit tells us anyway is that redshift due to field strength is quite small compared to redshiftresulting from recessional velocity of cosmic inflation. Redshift of field strength resultsfrom a large field strength difference between two points of measurement. In theory we seethat this difference between E-W symmetry breaking and the very beginning is modest.That means that the universe was modestly relativistic at early times, not ultra-relativistic.That the redshift of the CMB is mostly due to cosmic inflation squares with recent CMBpolarization results that the universe began with exponential inflation showing gravitationalwave effects from an early epoch. The 1091 CMB redshift factor gives a β = 0.9999996factor for the recessional velocity of the neutral hydrogen of the young universe of 375,000years.

The most informative information to be gleaned from this analysis could be the ratio ofcurvature at E-W symmetry breaking to curvature calculated for the CMB. This ratio is59/1.16× 10−7, 500 million times greater. If curvature of the nascent universe at the timeof last scattering is modest, then at the time of E-W symmetry breaking it would have beenmodestly relativistic with β = 0.205. Recent measurement of B-mode polarization effectsof gravitational radiation show a strong signal in the CMB background where a weak onewas expected.2 Maybe we know more about the early universe than we thought.

2BICEP2 Collaboration, Bicep 2 I: Detection of B-mode polarization at degree angular scales. 18 Mar2014. Phys. Rev. Lett. 112, 241101 (2014) arXiv:1403.3985v2 [astro-ph.CO]

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7 Implications

Here is the widely publicized graphic portrayal of the growth of the universe from NASA:

ImageCredit : NASA/WMAPScienceTeam

This graphic shows the universe as it evolved from the Big Bang to now, the 13.77-billion-year-long history of our universe.

The present-day gentle Dark Energy Accelerated Expansion shows as the flaring horn atthe right end of the graphic. The exponential inflation resulting from GUT symmetrybreaking in the instant of Quantum Fluctuations is at the extreme left end. The slowlyexpanding part between the ends is the matter dominated phase that slows expansionafter the inflaton field shuts off. Present-day acceleration would shut off also, if all chargedparticles paired up and fell into the ground state, and the Higgs field shuts off.

Exponential inflation is described in Chapter 3 of the cited book, and dyons, the firstparticles of quantum fluctuations, pair up to make quadrons. In the fleeting moment thatthey are free, they exist in an enormously strong field analogous to the Higgs field of our era.We call this field the inflaton, in that it expands the universe in the era labeled Inflation in

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the graphic. Formation of quadrons neutralizes the dyons except for their electric charges,as described in Chapter 5, and that shuts down the inflaton field when quadrons fall intostable states. An acceleration is seen now as an effect of the Higgs field of luminous matter,described in Chapter 8, that gives the observed universal acceleration that will continueindefinitely. Eons from now, if the universe cooled off completely and all matter fell intothe ground state, the impetus of expansion would shut off, and the remaining matter woulddecelerate gravitationally until it heated up again, restoring the Higgs field.

Food for thought: the Higgs field is known to have a higher energy component, and hintsof it have been observed in experiments at the Large Hadron Collider. Is this higher energyfield the remnant of the inflaton?

8 §3.6 of the Book; Dimensionless Magnetic Constant µ0

This constant first appeared in 19th century laboratory work with the field of a current.In textbooks it shows up as below:

dB =µ04π

ids sin θ

r2

−→B is the magnetic field in Teslas at a point P , i is the current in Amperes, s is the arclength in meters, θ is the angle between d−→s and the radius vector −→r from ds to P , withmagnitude r also in meters. In this definition µ0 ≡ 4π × 10−7T-m/A. In current lists ofconstants it is given as µ0 ≡ 4π × 10−7N/A2, akin to a pressure. These classical physicsdefinitions are very far from the quantum mechanical general relativistic conditions at thefounding of the universe being discussed here. In general relativity it is customary togive physical observables in units of an equivalent length. For example the solar massM� = 1.988435Ö1030kg is usually expressed as a length RS = 2.96 km, the radius of ablack hole enclosing a solar mass. Likewise, a single kilogram of mass is expressible as2Gc2

= 1.49×10−27m. Now return to the regression result 4πµ0α = 1.152×10−7N/A2. This

SI unit is equivalent to kg−m/s2

(Cs )2 = kg−m

C2 because µ0 has units of N/A2. The mass part can

be converted to a length equivalent by multiplying by 2Gc2

. Now observe that

µ0 = 4π × 10−7kg −mC2

× 2G

c2m

kg=

8πG

c2× 10−7

m2

C2=

2G

c2µ0m2

C2(26)

If we express a single charge q expressed in Coulombs C as

q ⇒ q

√2G

c2µ0 = 4.32× 10−17

m

C× q(C) (27)

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then µ0 is truly a dimensionless constant. This is a new general relativity expression forelectric charge that we should call the Coulomb length to go with Planck length and Planckelectric charge set out in §3.2. For example, the fundamental electric charge e in generalrelativity can be expressed as

e = 4.32× 10−17m

C×(−1.6021766× 10−19C

)= −6.92× 10−36m (28)

This carries a profound implication, namely that the founding of the universe began withelectromagnetism. There is one force only: electromagnetism, that warps space-time givinggravity and the electromagnetic field. Every observable in physics results from these fieldeffects. This now begs the question: What is mass?

Related this question is the often-quoted electron charge-to-mass ratio: 1.758820024×1011

C/kg, called a ratio in the sense of “amount of this per unit of that”. Now that it ispossible to express both charge and mass in consistent units of measurement, the ratio isbetter understood as a true dimensionless quotient.

e

me= (1.7588×1011C/kg)×

√2G

c2µ0m/C/(

2G

c2m/kg) = (1.7588×1011C/kg)

õ0c2

2G= 5.116×1021

(29)Note that this is over 10 billion times larger that the traditional numerical value. It is nowonder that J.J. Thomson and many physics students after him were able to move theelectrons so easily with electric and magnetic fields in their lab experiments.[?]

There is more. If in fact there is only the electromagnetic force in nature, then the gravi-tational constant G is not as fundamental as thought and could be expressible in terms ofelectromagnetic constants. Invert the last equation, take the cube root and multiply by 2.This may be contrived, but notice the result:

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√√√√me

e

√2G

µ0c2= 1.16× 10−7. (30)

An obvious pattern has emerged. It is thought that electric charge and mass involvedifferent dimensions of 11-dimensional space-time, so this may suggest that mass involves3 times as many of them as charge. We use a value of G obtained by laboratory experimentor observational astrophysics that is true today. We applied it in the case of the electroncharge-to-mass ratio to get the measurement dimensions right. If the observed pattern hasany validity, we can obtain a value of G that was true at the founding of the universe withGUT symmetry breaking. That value of G need not equal the modern observed value, and

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there have been many speculative articles published that suggest that G has changed overthe 13.8 billion years of the universe’s existence. Let us see what we get from this.

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√√√√me

e

√2G

µ0c2= 4πµ0α (31)

Isolating our hypothetical old value of G on the lhs, we get

GGUT =(µ0c

2)( e2

m2e

)(32π6µ60α

6). (32)

This gravitational constant is proportional to the 7th power of µ0 and the 6th power ofthe fine structure constant α. Even minute changes in either of them in 13.8 billion yearswould affect the gravitational constant profoundly, but as follows, the situation is muchsimpler.

Anyway we getGGUT = 6.39138× 10−11 (33)

as compared to today’s value, G = 6.67408×10−11m3/kg−s2, as reported by NIST. This is4.2% smaller than the presently accepted value, not unreasonable given the long-speculatedpossible change in G from the founding of the universe.[?]

Why would G be different then? Look at the denominator of eq. 3.26. You see theelectron’s rest mass, but at symmetry breaking it is certainly not at rest. We must replacethat mass-energy by its special relativity value

√p2c2 +m2

ec4, at 2.1% larger than the rest

value. This momentum −→p is directed radially and is not random as if thermal, so wewould use 1

2kBT instead of 32kBT to calculate a temperature for the kinetic energy. We

get 250 million Kelvin, also reasonable. Modify eq. 3.26 with a weak correction factor asfollows:

GGUT =G

(1 + f (u))2=(µ0c

2)( e2

m2e (1 + f (u))2

)(32π6µ60α

6). (34)

In the calculation (1 + f (u))2 = 1.042, and f (u) = 0.021. This looks like special relativitywith (1 + f (u))2 = 1.042 = 1

1− v2c2

= γ2 and vc = 0.205, reasonable once more. The weak

correction runs from 1 to 1.021 as electron energy runs from laboratory temperature to250 million K. By this reasoning, any experiment to measure G must take into account themotion of the particles in the apparatus, if they are at all relativistic.

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See what this principle means for the gravitational field of a moving electron ei towardanother, stationary electron ej at an inter-electron distance rij :

Vij = −Gγ2i

γimemerijγi

= −Gm2e

rij(35)

There is no change from Newton’s Law even though the gravitational constant is affectednoticeably. A possible test of such a field effect as this could be made near the beam ofthe Large Hadron Collider. Higher energy of newly formed electrons at symmetry breakingreduces the measured G, which could suggest that any or all ambient energy affects themeasurement. Could precision measurements of G be made at varying distances fromthe LHC beam, both parallel to and perpendicular to the beam in order to estimate thishypothesized weak effect?

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