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Tuesday, December 30, 2014

Lingpoche, 凌波车

Consider this, every turn of time tg around the circular path, tg advances by tλc along the axis of rotation.

2πr=ktλ

k represents the resistance along the time axis, for every 2π around the circular path, it advances by one unit along the axis.  If k is measured in unit per radian, it is a constant.

In general, for a radial path of rv,

2πrv=kvtλc

kv=2πrvtλc

vt=ckv=ctλc2πrv

where

tgtt=vt

But when vc is normal time speed.

Tc=2πrcc=tλcvc --- (1)

ctλc=2πrcvc

Therefore,

vt=rcvcrv



When rv is varied in general, fv, the frequency varies, kv meausred in unit per perimeter changes such that for every cycle around the circular path, time advances by tλc still.  k when measured in unit per radian (per 2π) however, remains a constant.

Alternatively,

Tv=2πrvc=tλcvt --- (2)

We formulate (1) divided by (2),

TcTv=rcrv=vtvc

Since rc is small it seems that we can only go backwards in time with this.

rv>rc  

vt<vc

Tc<Tv

For every Tv, many Tc would have passed, vc rushes forward and we travel back in time.

Maybe vt can be reduced further by counter rotating, in which case,

Tv=2πrvv=tλcvt

And,

TcTv=rcrvvc=vtvc

When,

rv>rc

rvc>>rcv

such that,

vt<<vc

Tc<<Tv

If it is possible for,

rv<rc

then,

vt>vc

Normal time recedes, as we travel forward in time.

Time travels in a helix at constant speed.  Time advances after every cycle. k measured in unit per radian is a constant. When the radius of the time circle is increased, at constant speed around the circular path, time advances is slowed.  Normal time is in constant flux, as such normal time goes forward as we travel back in time.  If we rotate the sonic cone anti-clockwise, vt slows down further.

What exactly is "lingpo"?  A sonic cone at 7.489 Hz.  Two of such cones to form an enclosure.


And off you go, in time.

Time Speed Again, Two Pie Or Not Two Pie

Consider again,

Let,

r.Δθ=k.Δtc

ωc=θtc=kr

ω=ωctct=θtctct=krtct

(ωkr)tct=0

As, r,   k0

ωtct=0

since,

ω0

tct=0

This implies that as time stretches out, time speed is zero.

when r,

2πr=ktλ

2πrf=ktλf=ktct

tct=2πrfk=rωk=ck

What about,

tc=2πr

tct=2πrt=c

tct=2πrt=c

This is wrong because r for a given time speed is fixed.  Time tc, will always coil around x at a fixed r, gvien tct,

rt=0

r does not increase and the time coil does not expand as time progresses.

Time is in a helix with a phase associated with its position on its circular path along its axis of travel.  If time returns to its original phase at a distance tλ ahead, after time period T,

2πr=ktλ

2πrc=T=kctλ

tλ=ckT

so, time linear speed vt is,

vt=ck

Time speed vt, is not light speed, c.

Strained And Witchcraft

Consider the relation between time and space,

2π.tct=c

2π.tc=ct

2πtct=c

tct=c2π

In a similar way and considering the direction of these dimensions,

tTt=ic2π

tgt=ic2π

And

xt=xtctct+xtgtgt+xtTtTt --- (*)

xt=c2π{xtc+ixtgixtT}

If all time dimensions are equivalent,

xtc=xtg=xtT

xt=c2πxtc

We also have

(xt)2=(xtc)2+(xtg)2+(xtT)2

 xt=3xtc

This is possible only if,

 xt=xtc=0

At this point we can swap x for tc, since they are both dummy variables and conclude by symmetry that when xt=c2π,  tct=0

However, for the fun of it,

When x wraps around tc,

2π.xt=c

2π.x=ct

2πxt=c

and,

 2πxtc=cttc

So,

xt=xtctct --- (***)

Since,

xtc=xtgtgtc+xtTtTtc

As changes in x and tc does not effect tg and tT, and as far as x is concern tg and tT are equivalent, !?!

xtc=c2πttc=2xtgtgtc=2xtTtTtc

ie.

xtctct=c2π --- (**)

and

xtctctg=2xtg

From which we formulate,

xtctctgtgt=2xtgtgt --- (1)

Similarly,

xtctctT=2xtT

xtctctTtTt=2xtTtTt --- (2)

Therefore (1)+(2),

xtc{tctgtgt+tctTtTt}=2{xtgtgt+xtTtTt}

Substitute (**) into the above,

 c2π{tctgtgt+tctTtTt}=2tct{xtgtgt+xtTtTt}

But,

 xt=xtctct+xtgtgt+xtTtTt

So,

  c2π{tctgtgt+tctTtTt}=2tct{xtxtctct}

But from (***),

xt=xtctct

So,

c2π{tctgtgt+tctTtTt}=0

But,

tct=tctgtgt+tctTtTt

Thus,

c2πtct=0

tct=0

Blasphemy, I know.  The point is,


when xt=c2π, tct=0 and when tct=c2π, xt=0 are consistent.

The factor of 12π appearing before c is the result of circular motion around the orthogonal axis.  A particle traveling in circular motion in space about the time axis is not traveling in time, so time speed equals zero.  Similarly, the same particle traveling in circular motion in time about the space axis has zero space speed.

This relationship can be rewritten as,

v2+v2t=(c2π)2 --- (+)

where v is the particle velocity in space, vt is the particle velocity in time and c a constant.  Pythagoras' Theorem applies because v is perpendicular to vt.  Obviously expression (+) satisfies the boundary value conditions.

More importantly, the time speed limit occurs at c2π not at c, because the particle goes into circular motion.

More Portals

If two portals point directly at each other, then their forces should cancel BOTH outside and between the portal pair.

But Maxwell third and fourth equations suggest that we can just have the forces cancel outside the two face of the portals by letting one portal be passive.  The portal that we are traveling from is the active portal.



The active portal induces a corresponding force pair (Eψdλ) in the passive portal that in turn generate a force that cancels with the force from the active portal, beyond the passive portal as expected.

Furthermore,


this is also a possible portal configuration.  Two of such double pillars facing each other, with high gT between one of the double pillars can generate a conduit between the portals.

Monday, December 29, 2014

Sound Portal

My very own mandala.


How far did the Tibetan go with sound technology at 7.489 Hz?  Apparently, very far.  But can manipulating gravitational wave open up a portal?

And to where?  From the post "Where's The Charge?",  if

f=nc2πaψ,    n=1

So f confine to a smaller aψ will be trapped at a lower hn.

hf=2πaψmcf=2πaψmc7.489=mc2+x.F

x.F<0  (because aψ that produced 7.489 Hz is the size of Earth.)

where aψ is the radius of the confine into which sound waves of 7.489 Hz are directed as the diagram above shows.

Energy is being released.  Photons will be emitted.  EΔh<0  and EΔht<0, from previously (Post "Cold Jump"),

aψn=c2EΔh2π(EΔht)1

 the direction of travel is opposite to the right hand screw rule.  We point the mandala, front face, in the direction of the corresponding portal, face front and jump into the portal and travel backwards.


Time is always clockwise in the direction of x, even if time is slowed.

Note:  Not the mandala itself but the construct indicated by it.

Sunday, December 28, 2014

Spin Cool

Maybe ψ can be depressed by spinning the two particles.


Centrifugal forces on the two particles will space them further apart and cause a depression in ψ in the space between them.

Jump in.


But what is this particle pair?  These two particles are in orbit around each other, presenting a positive force density outwards.

Cold Jump

From the post "Time Travel Made Easy",

Δt=tgfttgit=1mc2(tgitEΔh+tgiEΔht)

From the post "Teleportation",

tgi=2πaψnc

and

tgit=c

We have,

tgftc=1mc2(cEΔh+2πaψncEΔht)

2πaψncEΔht=mc2(tgftc)cEΔh

aψn=mc32π(tgftc)(EΔht)1c2EΔh2π(EΔht)1 --- (*)

The plot below shows the ψ profile of two particles in pair, they are attracted to each other for x<aψ.  If aψ is increased, ψ in the space between the two particles is close to zero and negative.


When ψ is depressed,


The plateau value of ψ+n+ψn decreases further into the negative region.

From equation (*), we see that even without a time speed differential,

tgftc=0

aψn=c2EΔh2π(EΔht)1

And since ψ<0, any transitions from normal energy state, without the need to use high energy photons to first elevate the energy state, is negative,

 EΔh<0,

but,

EΔht<0

so the direction of travel is opposite to the right hand screw rule,


where tg around the spiral is slowed, ie tgft<0.  Such a scheme avoids the use of high energy photons to first slowly increase the particle's energy state and then allow the energy state to fall in a cascade.

Cool, very cool.  How then to depress ψ?

All Of Max

We can of course superimpose the results from "Two Wrongs Make Both Wrong, None Right", and obtain,

ddtEψdAl+1εnewJψdAl=2πaψFψdλ

Replacing εnew with εo and multiplying by μoεo,

μoεoddtEψdAl+μoJψdAl=μoεo2πaψFψdλ

μoεoddtEψdAl+μoJψdAl=1c22πaψFψdλ

which is the fourth Maxwell's Equations for a particle.

Since aψ is the radius of a circle, when we rotate Al by 2π along any axis of symmetry through Al, we see that Eψ cancels with Eψ when Al has been rotated by π. Similarly, Jψ cancels with Jψ at the rotation of π.  When so rotated, the integral 2πaψFψdλ describes a closed surface area containing the volume defined by Al after 2π rotation.  As such,

μoεoddtEψdV0+μoJψdV0=1c24πa2ψFψdAs=0

This is the second Maxwell's equation, Gauss's law for magnetism.

Consider the fourth Maxwell equation again, if we set Jψ=0 and integrate along aψ, the radius of the circle defined by the perimeter Al,

μoεoddtEψdAldx+μoJψ0dAldx=1c22πaψFψdλdx

The loop 2πaψ becomes an area Al=πa2ψ, therefore,

μoεoddtEψdAldx=1c2πa2ψFψdAl

It is totally arbitrary that we move at speed c, and since x is perpendicular to λ and Al.


Using Al are the reference direction

dxdt=ic

μoεoicddtEψdAldt=1c2πa2ψFψdAl

c=1μoεo

icEψdAl=πa2ψFψdAl

In a similar way, we consider Al as the integral of a loop 2πaψ outwards along dx.  This loop is obtained by integrating along dλ for 2πaψ.

ic2πaψEψdλdx=πa2ψFψdAl

Similarly, dxdt=ic

(ic)22πaψEψdλdt=πa2ψFψdAl

2πaψEψdλ=1c2ddt{πa2ψFψdAl}

This is the third Maxwell's Equation.  Consider again the fourth Maxwell equation, when we set Fψ=0

μoεoddtπa2ψEψdAl+μoπa2ψJψdAl=1c22πaψFψ0dλ

If we consider Al to be from the integration of 2πaψ along the radius x, using Al as reference,

iλ=Al  and dxdt=ic

εoddt2πaψEψdiλdx+πa2ψJψdAl=0

εo(i)2cddt2πaψEψdλdt+πa2ψJψdAl=0

εoc2πaψEψdλ=πa2ψJψdAl

Jψ=12πa2ψdqψdt

Substitute in Jψ and integrate over time, t.

εoc2πaψEψdλdt=πa2ψ12πa2ψdqψdtdtdAl

Eψ is constant,

εoct2πaψEψdλ=πa2ψq2πa2ψdAl

εo2πaψEψdλ=πa2ψq2πa2ψctdAl

where vol=2πa2ψct is the volume transcribed by J with a base of 2πa2ψ  over time t.

With t small,

ρ=qψ2πa2ψct

where ρ is the volume charge density.  So,

εo2πaψEψdλ=πa2ψρdAl

We then rotate 2πaψ into a surface containing the volume define by Al as Al rotates correspondingly,

4πa2ψEψdAl=1εoρdV

This is the first Maxwell's Equation, Gauss's Law.

Thanks to the fact that aψ describes circles and spheres, we have all four Maxwell's Equations for particles.  Eψ is the field around the particle, Fψ is a field analogous to the B field of charges, produce by the particle in motion.

Two Wrongs Make Both Wrong, None Right

From the post "Maxwell, Planck And Particles",

 1εo14πa2ψdqψdtdAl=2πaψd(2πmc)dtdλ=2πaψFψdλ

If, we formulate instead,

ddt{qψ4πεoa2ψ}dAl=2πaψd(2πmc)dtdλ=2πaψFψdλ

Eψ=qψ4πεoa2ψ

ddtEψdAl=2πaψFψdλ

This expression is without any appendages because if the 2π factor missing from the circular momentum term is the only reason

εoμo,

then,

εo=μo=1c  

after we made the correction, which would set,

Zo=1=μoεo

that space present equal resistance to B and E field.  With such a change to εo, the numerical value of q changes also.

Let, consider again,

 1εo14πa2ψdqψdtdAl=2πaψd(2πmc)dtdλ

14πεo12πa2ψdqψdtdAl=aψd(mc)dtdλ=2πaψFψdλ

without the correction to circular momentum.

Since,

Jψ=12πa2ψdqψdt

 14πεoJψdAl=2πaψFψdλ

And then we make the adjustment to εo, 4πεoεnew  to correct for the mssing 2π factor for circular momentum.

 1εnewJψdAl=2πaψFψdλ

 Since, μnew=εnew=1c

 1μnewJψdAl=2πaψFψdλ

Why not simply make the adjustment mc2πmc, and obtain,

12εoJψdAl=2πaψFψdλ

as in the post "Maxwell And Particles"???

This expression is wrong because it is still using the old definitions of μo and εo.

Saturday, December 27, 2014

Maxwell, Planck And Particles

From the post "Just Lots of Colors, Retro Disco" Jun2014, where the photon is modeled as a dipole, and the posts "de Broglie Per Unit Volume" and "de Broglie Per Person",

h=18πεoq=2πaψmc

where q is the dipole charge.

h=12πεoq4πa2ψπa2ψ=2πaψmcdλ

where the closed looped integral is around 2πaψ.

h=12πεoq4πa2ψdAl=2πaψmcdλ

where the area integral is over the area, Aaψ=πa2ψ surrounded by 2πaψ.

Differentiating with respect to time,

dhdt=12πεo14πa2ψdqdtdAl=2πaψdmcdtdλ --- (*)

Compare this with Ampere's Circuital Law, (The fourth Maxwell's Equation with time varying component set to zero),

μoJdAl=2πlBdl

In the first place, the factor of 12π in expression (*) suggests that for a body in circular motion at speed c, its momentum is not,

p=mc   but   p=2πmc

Consider the centripetal force, F that accounts for circular motion,

p=Fdt=mv2rdt

dpdt=F

Since,

θ=ωt   and v=rw

p=m(rω)2r1ωdθ=mrωdθ

Over one period θ=0θ=2π,

p=2πmv

And we got it all wrong.  This is important!  We have instead,

dhdt=1εo14πa2ψdqdtdAl=2πaψd(2πmc)dtdλ

Secondly, for a photon m=0, but a change in its ψ is momentum (from the posts "Bouncy Balls, Sticky Balls, Transfer Of Momentum" and "de Broglie Per Unit Volume"), so

d(2πmc)dt=Fψ

is a force that acts on ψ.  Depending on the nature of the photon it may be ψ along tT,tg or tc and one other space dimension, for which the photon is defined.  B is perpendicular to J in Maxwell's equation.  The direction of

dqdt is also perpendicular to Fψ

The nature of Fψ depends on the nature of q.  q is thus qualified as, qtT, qtg and qtc depending on the dimensions between which the photon is oscillating.  In all cases, these include a space dimension, the other being a time dimension, tT,tg or tc.  So a photon in motion exert a force around it, this force, Fψ is a force in space (ie.  F=ma) and also along one of the time axes tT,tg or tc.  Previously, in Maxwell's Equation, the effect of the force in the time dimension in which the charge is oscillating has been ignored.  So expression (*) in the case of a photon, ph(tc,tg) is,

 12εo12πa2ψdqtgdtdAl=2πaψd(2πmc)dtdλ=2πaψFψdλ

Fψ then acts on tg and a space dimension.

This equation for particles in motion through an area Al exert a force field, effecting similar particles, around a loop 2πaψ is just a restatement of Ampere's Law but extended to all particles.  It may be possible that all of Maxwell's equations can be similarly extended.

Let,

Jψ=12πa2ψdqtTdt

 12εoJψdAl=2πaψFψdλ

This however is using the old definition of μo and εo.

More importantly, the correct momentum in circular motion,

p=2πmc=m2πc

makes μo and εo equivalent but orthogonal.

Consider,

εnew=iμnew

σ2=(12εnew)2+(12μnew)2 --- (**)

σ2=ε2new=εnewμnew

When c is not in circular motion,

c=1μnewεnew=1σ

as such σ=εnew=1c is the resistance to c.

When c is in circular motion, where σ has been resolved into two components as in (**).  Along one of this component,

εo=12εnew

μo wrongfully incoperated a factor of 12π when momentum was taken to be mv and not 2πmv.  So,

2πμo=12μnew

And so,

μoεo=14πμnewεnew

This means in the expression for c,

c=1μnewεnew=14πμoεo

μoεo has to be adjusted by a factor of 4π because the 2π factor needed when momentum is circular was not accounted for. (2εoεo and 2πμoμo)

μo is defined as 4π×107 NA-2.  This is wrong for failing to account for the fact that in circular motion, the momentum is 2πmv and not mv.

It is correct and consistent to set

μo=εo=1c

at the same time, all momentum p in circular motion is replaced with 2πp, ie.

pcir=2πp

where pcir is the momentum in circular motion. Equivalently, all velocity v in circular motion is replaced with 2πv.

vcir=2πv

Equivalently,

pcir=2πp=2πmv=m2πv=mvcir

Correcta, Errata...

Note: This derivation for Ampere's Law applicable to all particles is not rigorous, but indicative none the less.

Squeezing Photons

From the post "Just Lots of Colors, Retro Disco" on Jun 2014, photon modeled as dipole that can resonate at

freso=15.088kHz

From the post "Increase Photon Frequency" (Sep 2014), we see that photons can be squeeze to high frequency with the use of a tapered coil.  The coil will be more effective operating at a frequency of 15.088 kHz.

This allows photons of higher aψ to be squeezed to lower aψ.


This might allow photons needed to energize for teleportation be produced without reaching for ultra ultra high frequencies (1042Hz).

Squeeze and telepop.

More ScFi, Resonator And Wrap Speed

Consider the expression for EΔh,

EΔh=FΔhidaψi=mc21aψi(aψfaψi)

Differentiating with respect to aψi,

EΔhaψi=FΔhi=mc2aψfa2ψi

And consider,

EΔh=FΔhfdaψf=mc21aψi(aψfaψi)

Differentiating with respect to aψf,

EΔhaψf=FΔhf=mc21aψi

So, when both aψi and aψf are changing,

EΔhaψ=EΔhaψf+EΔhaψi=FΔhT=mc21aψi(1aψfaψi)

And we consider the second derivative,

2EΔha2ψ=aψ{EΔhaψf+EΔhaψi}=FΔhT

FΔhT=mc21a2ψi+2mc2aψfa3ψimc21a2ψi

FΔhT=2mc21a2ψi(aψfaψi1)=k

And an approximation,

FΔhT=FΔh.Δaψ

If such a system is prompted to oscillate,

ωn=km=2caψfa3ψi

where ωn is the natural frequency of the system.  Naturally, such a system is damped as it emits photons, the loss due to damping is,

Ploss=EΔht=mν(daψidt)2

where ν (rads-1) is the damping factor using the model f(x,˙x)=kxmν˙x and the rate at which f(x,˙x) does work is,

P=f.˙x=kx˙xmν˙x2=Posc+Ploss

And consider, aψi making a round trip to aψf and back in time Td=2π1ωd,

daψidt=2(aψiaψf)Td=1π(aψiaψf)ωd

where ωd is the damped resonance frequency.  We have,

EΔht=mν(daψidt)2=mν1π2(aψiaψf)2ω2d

As,

EΔht=EΔhaψfdaψfdt+EΔhaψidaψidt

Assuming daψfdt=daψidt, that the external driving force affects both equally,

mc21aψi(1aψfaψi)=mν1π(aψiaψf)ωd

Since, aψ at aψf loses almost all its energy, the amplitude of the wave at aψf, Af is small,

ν1 rads-1

then,

wd=πc21a2ψi

which is a very high frequency, (1034Hz).

At this frequency, the particle will be "resonating" with high output of photons emitted as a result of the energy state transition aψiaψf.

If this high frequency, ωd can be achieved, we have a photon resonator that might drive that particular particle to wrap speed.

Note:  What? Wrong to replace ˙x with average speed?

Friday, December 26, 2014

645 Hard Science Fiction

645 pages of hard science... fiction, already.  Ah, Ah, Ah.  Maybe we should attempt at science not already in science fiction.

Until Next time.

The Lights Is On And Off You Go

In the post "Charge Photons Creation", it was suggested that charges on relaxation from an excited state emit photons of the nature, ph(tc,tg) and ph(tc,tT) where tc is the time axis on with the photons exist and tg, or tT is the time axis on which they oscillates.

If the process is reversible, then a photon of the correct nature will impart energy onto charges of the corresponding nature ch(tg) or ch(tT).

In the post "Amplitude, An", it was suggested that ψ at aψn, makes a state transition to the next level, aψn+1 when its amplitude has gain enough magnitude.

It might be possible to bathe a particle in photons of the correct nature such that its aψn increases slowly, and then induce a sudden collapse such that we have,

EΔht<0

and very large.

As long as the photons shines on the particle,

Ept=EΔht>0

where Ep is the photon energy,

this itself maybe enough to stop the relaxation of energy states, as aψn is prompted to the next higher level continuously.  Conversely if the photons are suddenly switched off, the absence of a driving force that increases aψn continuously maybe enough to trigger a collapse of energy states, as the particle is now at a very high aψn.  Otherwise, an increase in ψ from a magnetic/electric field around the particle as the photons are switched off, might prompt a continuous energy state relaxation.  Such continuous relaxations are accompanied by emissions of photons at light speed, and are expected to produce a large negative EΔht.

Teleportation/Warp speed could be as simple as turning a switch on and off.  The presence of an external field that would effect EΔh of the particle is important because its rate of change with time must follow the profile,

EΔht=At    A<0

over the cascade of transitions to lower aψn.

How then to generate massive amount of photons of high energy, of a particular nature?

Imagine such a massive photon generator at the center of a spaceship.  When the generator is on, photons from it energize the whole craft to high aψn.  When the generator is suddenly switched off, all the particles that constitute the craft relax to a lower energy state under a force field, B or E, that is turned on as the generator is switched off.  The craft is teleported.

But to which direction?  In the case of a portal, the nature of tc around x dictates that tc slows down in a circular plane perpendicular to x.  Photons are inject from a circular perimeter clockwise, toward the center to slow time, tc within the circle.  The direction of travel is then into the circle, perpendicular to the plane of the circle.

In the case of a photon generator, the photons need be radiated outwards in a clockwise circular manner from stern to bow, then the spaceship will be driven forward.


The geometry of the photons, around the photon generator at the center, dictates the direction of travel.

Jump!

Undefined At What? High Five!

Fortunately, we are able to formulate,

EΔht=At+t

such that t0 as t0 and we have the following illustrative plot.


where,

EΔht=At,  and A<0

In practice, EΔht on the throw of the switch, but it can be a high value that,

aψno=12πmc4(|EΔht|)1

is sufficiently large.

The question remains, how to induce EΔht in a particle.

Stargate

If we stare down one of the space dimension with one time dimension curled around it,


where r is very small.  tg is at light speed, v=c.  If it is possible to slow down time,

tgft<c   (in ticks per second)

such that,

tgft=rω

r=1ωtgft<cω

where both ω and r are physically manageable number, consider the equation from the post "Time Travel Made Easy" again,

tgft=1mc2(tgitEΔh+tgiEΔht)+tgit ---(*)

tgit=c

tgft=1mc2(cEΔh+tgiEΔht)+c

tgiEΔht=mc2tgftcEΔhmc3

tgi=2πaψnc

aψnEΔht=12π{mc3tgftc2EΔhmc4}

aψn=12π{mc4(EΔht)1+mc3tgft(EΔht)1c2EΔh(EΔht)1}

Since,

EΔht<0

aψn=12π{mc4(|EΔht|)1mc3tgft(|EΔht|)1+c2EΔh(|EΔht|)1}

replacing aψno=12πmc4(|EΔht|)1

aψn=aψno{11ctgft+1mc2EΔhtt}

From (*)

tgft=1mc2(tgitEΔhtt+tgittEΔht)+tgit

tgft={2tmc2EΔht+1}tgit --- (**)

So,

aψn=aψno{1{2tmc2EΔht+1}+1mc2EΔhtt}

aψn=aψno{tmc2EΔht}

aψn=12πc2.t

The particle travels at a forward velocity of,

vtele=12πc2

Is this speed possible?  Yes, time speed has been slowed down, that implies greater than light speed space travel.  However we are NOT traveling through space this way.

At time t=0, just before t=0,

aψn,t=0=aψno{11ctgft+1mc2EΔhtt0}

aψn,t=0=aψno{11ctgft}

in this case, tgft0, and the teleported distance is aψnt=0.  As with the case considered in the post "Teleportation", with

tgft=constant

 the particle return velocity is,

aψno1mc2EΔhtt=12πc2.t=vtele.t

vtele=12πc2

At t>0 however,

vtele=12πc2

It seems paradoxical that the particle can be at different locations and have different velocities just before and after t=0.  The trick here is,

tgft={2tmc2EΔht+1}tgit = constant

The expression EΔht.t is held constant such that 

tgft={2tmc2EΔht+1}tgit

from (**) is a constant.  This means

 EΔht=At

where A is a constant.

This way, only the situation for t=0 applies and as will be discussed below t0.

The problem is, if EΔht is induced in a particle, how to turn it off when it is at a distance aψn,t=0 away?

Simple, we beam the particle through another portal that induces EΔht.  After passing through the second portal, EΔht=0.  Equation (*) collapses to,

tgft=tgit

as at t=0,  EΔh=0.

And the particle does not return from x=aψn,t=0,  t0.



Since time, tg is curled around x where itg=x as required in the post "Time Travel Made Easy".  x is perpendicular to the circular plane containing tg.  The portal is directional.  This sort of teleportation unfortunately is restricted to line of sight.  Unless we can draw a straight line between the two points, teleportation is not possible.

And the billion dollar question is, how to induce EΔht=At in a particle?

Teleportation

Why did the Philadelphia Experiment traveled into the future time?  From the post "Rotating Al Again", if they have just rotated the (tc-tT) plane to the (tg-tT) plane by rotating the projection of tnow on the (tc-tT) plane by 60o onto the (tg-tT) plane, and the space dimensions are coiled up along the time axes such that any space dimension is perpendicular the time dimension and at the same time orthogonal to all other dimensions, space and time cannot cross nor superimpose no matter what rotation or transformations.


In our 3 space dimension world time flows freely, continuously.  If we are able to rotate/transform a space dimension on to a time dimension then space along that dimension will flow freely.  The time dimension that has been swapped with a space dimension will stand still unless kinetic energy is applied.

From the post "Eternal Embrace",  time and space loop around each other.  As we shorten time, space elongates.  It might be possible that using the equation from the post "Time Travel Made Easy",

tgft=1mc2(tgitEΔh+tgiEΔht)+tgit

and setting

tgft=0

1mc2(tgitEΔh+tgiEΔht)+tgit=0

tgiEΔht=(EΔh+mc2)tgit

but,

tgit=c

So,

tgiEΔht=c(EΔh+mc2)

since,

tgi>0,   EΔht<0

Furthermore,

tgi=2πaψnc=c(EΔh+mc2)(EΔht)1

aψn=12πc2(EΔh+mc2)(EΔht)1

Initially when, EΔh=0

aψn=12πmc4(EΔht)1

As (EΔht) is negative,

aψno=12πmc4(|EΔht|)1 --- (*)

So, depending on the value of (|EΔht|)1,  the particle is transported to a distance defined by equation(*).  Afterwards,

EΔh=EΔhtt

aψn=12πc2{t+mc2(EΔht)1}

aψn=12πmc4(EΔht)112πc2t

aψn=aψno12πc2t

After the initial jump, if the particle is still subjected to EΔht<0, as time passes, the particle travels back to its origin with a velocity vtele,

 vtele=12πc2

If EΔht is set to zero immediately at aψno then the particle remains at aψno.

Teleportation!  Warp Speed!  When the particle is allowed to return, on a smaller scale, Brownian motion.

Thursday, December 25, 2014

Invisible, Right Here!

What?  Expecting invisibility in the previous post "Now You See Me, Now You Don't"?  Here is the Time Flow Differential Equation again, but this time along tc,

Δt=1mc2(tcitEΔh+tciEΔht)

What if tc is sufficiently slowed?  If optics are charge phenomenon, due to photons that exist along the tc time dimension, then slowing tc slow down the photons.  Our eyes and equipments may not be able to detect such slowed photons, and we effectively turn invisible.

Remember, you are in my dream.

What would changing tTt do?

Δt=1mc2(tTitEΔh+tTiEΔht)

Maybe it will stop aging.

Wouldn't tg effect gravity and gravity only,

Δt=1mc2(tgitEΔh+tgiEΔht)

Floating around invisible, totally ignored and aging slowly.  I'm a see through plastic banana balloon.

Wednesday, December 24, 2014

Now You See Me, Now You Don't

From the previous post "Time Travel Made Easy",

Δt=1mc2(tgitEΔh+tgiEΔht)

EΔh occurs naturally as state transitions of a particle, sometime with the emission of a photon.  Time travel is thus a natural process and happens all the time.

Such phasing in and out of time might be the root of Brownian motions.  The key is the frequency of the random motion and the frequency of energy state transitions.  If these phenomena are related, these frequencies are equal if not of the same nature, assuming that all EΔh results in observable discrepancies in time and position.

Note:  If there is a flash of light on time travel, photons are emitted and EΔht<0, the time travel direction is backward in time.  If there is a flash of light on arrival from time travel, EΔht is still negative, the time traveler is from the future.

EΔh is the change in energy experienced by the particle, an external agent effecting such energy changes on the particle will have to do the opposite; absorb energy, the external agent itself gaining energy for the particle to have a negative EΔht.

Time Travel Made Easy

From the post "What Does It Mean to Be Excited",

EΔh=mc2(aψfaψi1)

and from the previous post "Charge Photons Creation",

iT=2πaψnc=tg

aψfaψi=tgftgi

We have,

EΔh=mc2(tgftgi1)

tgf=tgi(EΔhmc2+1)

But what is tgf and tgi?  However,

tgft=1mc2(tgitEΔh+tgiEΔht)+tgit

This suggests that the passage of time can be affected by an absolute change in energy, EΔh or an high rate of change in  EΔh,  EΔht.  In both cases, we are restrained by mc2.

Δt=tgfttgit=1mc2(tgitEΔh+tgiEΔht)

It is possible to increase or decrease time flow by changing the sign of EΔht.  When Δt>0, the rest of the world flows backward and we travel forward in time; when Δt<0, the rest of the world flows forward and we travel back in time.

Changing EΔht is the same as manipulating ψ as discussed in the post "Time Travel By Manipulating ψ".  The relationship presented here is much more clear.

Charge Photons Creation

From the post "Why Should Amplitude Remain Constant?", it was argued that a change in parameter ninf changes λn and aψn but does not change An.  λn is orthogonal to aψn and both are orthogonal to An.

Since,

c=fnλn=λnT

T=λnc

where T is along the tc time dimension.

2πaψn=iλn

λn=i2πaψn

So,

T=i2πaψnc

iT=2πaψnc

iT is a time dimension perpendicular to T.  It is either tT or tg.  iT is not any of the two space dimensions as aψn does not affect An.

iT is separately,

iT=tT

 iT=tg

for the two type of charges.

A change in aψn, therefore changes tT or tg, and a change in λn changes tc.


As long as aψn and λn are valid solutions to the wave equation,  the particle is still a wave in 2 space dimensions and one time dimension tg or tT after the changes (ie the particle is not destroyed).  That means v=c along both tc and, tT or tg.

If a change forces v>c along tg or tT, the energy is released along that dimension instead.  v remains constant at c.  If a changes results in v<c, energy is absorbed along the same time dimension, such that v increases to c again.  In both changes, other orthogonal dimensions are not affected.  The particle has no access to other dimensions in which it is not oscillating, and not existing (The charge exist along tc).  So the charge has access to tc, two space dimensions and tT or tg.

A slow down in time tc is associated with greater then light speed, vs>c in space.  When time returns to normal, vs=c, the particle velocity is at light speed.  (Time speed changes after the particle has reach light speed in space).  As such a change along tc, the time dimension in which the particle exist, is coupled to the space dimension, (s) along which it travels (c2=v2s+v2tc).  As the particle returns to light speed in tc, it is in light speed in s.

Taken altogether,  changes in aψn and λn of a particle creates another wave in the same set of dimensions as the particle, as long as the final aψf and λf are still valid solutions to the wave equation.  If the particle is destroyed, there is then no light speed constrain on the dimensions involved and no emission/radiation of energy would be required.  That the original particle is not destroy is itself a constrain leading to the creation of the new particle.

The diagram above shows that changes in aψn and λn of charges lead to the creations of two types of photons oscillating along tg or tT, existing along tc.  These are charge photons associated with each type of charge.

Makan Time!  If you are Swedish it means something else.

Geometric Mean And Trickery

Consider a transition from ninf,

EΔh1=mc21aψi(aψfaψi)

Consider another transition from aψf+aψiaψf,  (This is an extrapolation, it is not proven that aψf+aψi is a solution.)

EΔh2=mc2aψf+aψi(aψf(aψf+aψi))=mc2aψf+aψi(aψi)

such that, aψi is in between the paths of EΔh1 and EΔh2.

If we take the geometric mean of the two,

ˉEΔh=EΔh1.EΔh2=mc2aψiaψfaψf+aψi

and let

Eo=ˉEΔhaψi=mc21aψiaψiaψfaψf+aψi  Jm-1

We have an expression for Eo.  A negative sign is necessary because a change from a higher value of aψ to a lower value of aψ (ie. a positive value of (aψiaψf) ) results in a loss of energy.

Note:  If,

EΔhf=2πaψfmc.f

EΔhi=2πaψimc.f

And,

EΔhi+EΔhf=2πaψimc.f+2πaψfmc.f

2π(aψi+aψf)mc.f=EΔhi+f

And,

EΔhi+f=mc2aψf+aψi(aψf(aψf+aψi)) --- (*)

This does not proof that aψf+aψi is a solution to x for ψ but it shows that (*) is a valid expression.

Simply

Consider,

EΔh=mc2(aψfaψi1)

In the case of a photon, aψ is the radius of its helical path.  This shows that a change in radius is accompanied by an exchange of energy.

and

2πaψn=λn

So,

EΔh=mc2(λfλi1)

EΔh=mc2(fiff1)

Which suggests a photon can change frequency, given an appropriate energy input, EΔh.  If EΔh resulted in an emission of a photon,

hffp=EΔh=mc2(1fiff)

λfmcfp=1ffmc2fp=mc2(1fiff)

fp=ff(1fiff)=fffi

Surprise! Surprise!  This suggests that an emission of a photon occurs only if the lower energy states have been folded up, where a lower energy state actually have higher frequency.  This is illustrated on the plot below,

Frequency Intersections
The curve marked n=1 in red, is folded up and has a higher intersection for frequency with the line y=x.  Transition from all other states with lower frequencies will result in an emission of a photon. Transition from states n=10 however, will not result in a photon emission because the associated frequency is higher than that for state n=1.

And,

1λp=1λf1λi

for photon emission.

Since,

hifi=mc2

hfff=mc2

EΔh=mc2(hfhi1)

which relates the change in energy as a result of a change in hn directly.

We still don't have justifications for Eo

Touch And Go

From the previous post "Amplitude An", it is also possible that a state transition, nfni, occurs when,

r=aψi

then,

a2ψi=a2ψf+A2f

Af=a2ψia2ψf

In which case we still have,

EΔnAf,  because ψ is already energy density.

EΔn=Eoa2ψia2ψf

And we define Eo as,

Eo=mc21aψi(aψiaψf)(aψi+aψf)

In this case a state transition occurs when a lower ψ wave touches the path of an outer wave.  This would occur before the perimeters of the paths are equal.

Tuesday, December 23, 2014

Amplitude, An

Obviously,

|ψn|=An

The following diagram shows two energy states, at aψi and aψf.


As the particle gain energy, Af increases, and

r2=a2ψf+A2f --- (*)
r increases.

Could it be that when the perimeters are equal,

2π(a2ψf+r2)2=2πaψi

a transit of energy states nfni occurs;  ie when

r2=2a2ψia2ψf

By substituting r into (*)

Af=2a2ψia2ψf

Ai=0

Just on transition, Ai is small.  This can explain a lower aψ to a higher aψ transition, but what about the reverse?

The reverse occurs when aψi collapses back to aψf but the amplitude, AiAf  remains small.

The excess energy as a result of a reduced Af at aψf is emitted as a photon.  In the limit when Ai0, this excess energy is just the rms value of Af (not A2f as ψ is already energy),

EΔn=12Af=a2ψia2ψf

Af is measured in meters (m), so more correctly,

EΔn=Eoa2ψia2ψf

where Eo (Jm-1) maps meters (m) to Joules (J) in the diagram above.

Is this the same result as before?

EΔn=Eoa2ψia2ψf

EΔn=Eo(aψiaψf)(aψi+aψf)

If Eo is,

Eo=mc21aψi(aψiaψf)(aψi+aψf)

then,

EΔn=mc21aψi(aψiaψf)(aψi+aψf)(aψiaψf)(aψi+aψf)

EΔn=mc2(aψfaψi1)

Since,

aψn=eAaψn+Cn2

EΔn=mc2{eAaψf+Cn2f+Aaψi+Cn2i1}

which is the same expression for EΔn in the post "What Does It Mean to Be Excited".

But why is,

Eo=mc21aψi(aψiaψf)(aψi+aψf)  (Jm-1)???

Unless Eo is justified, this scenario is just probable.

Note:  The perimeter of a ellipse is used here,

P=2πa2+b22

where a and b are the radii of the ellipse.


But...


There is a problem, from the post "Schrödinger's Equation", ψ is actually energy density, in which case, the collapsing atom paradox is unresolved.  Unless the proposed radiated energy from an orbiting electron is actually ψ that remains around the particle, and does not actually radiated away. ψ is the energy field around the particle, the negative gradient of which is the force field.  In the case of an electron, this force field is the E field.

The particles are still not orbiting about any nucleus, although from the post "Boundary Between Wave And Particle Interaction", they can be superimposed onto a common center, provided that they interact as particles and not affect each others' ψ.

It is still unanswered why the particle make energy transitions in the first place.  What happen to ψ when a particle gain energy?

Why Should Amplitude Remain Constant?

The amplitude of the wave, An is perpendicular to both λn and aψn.  If there are forces changing λn and aψn, these forces will have no effect on An.

So, because there is no force acting on An, An remains unchanged and the system loses energy as the particle transit from a higher aψ to a lower aψ.

λn and aψn changes when we set ninf, but this by itself does not change An.

What Does It Mean to Be Excited

From the post "de Broglie Per Person"

2πaψ=nλ

hn=2πaψmc=mc.nλ

with n=1.

2πaψ=λ1

h1=2πaψmc=mc.λ1

At naψ,

hn=2πaψn.mc=mc.λn

When a particle is excited we have,

hfhi

and

λfλi

And when the particle returns from an excited state, λi is applied to hf and we formulate,

Eif=hffi=2πaψf.mc.fi=λfmc.cλi

which is the energy of the particle at hf.

Since,

2πaψn=2πeAaψn+Cn2=λn

λfλi=eAaψf+Cn2f+Aaψi+Cn2i  --- (*)

We have,

Eif=mc2.eAaψf+Cn2f+Aaψi+Cn2i=ψf+x.F

The energy required at nf is Eif, the energy in excess is x.F, the work function.

EΔn=EifEf=mc2.eAaψf+Cn2f+Aaψi+Cn2imc2=x.F

EΔn=mc2{eAaψf+Cn2f+Aaψi+Cn2i1}=x.F --- (1)

where EΔn is the energy gained by the particle.  In this case the, the particle loses energy and EΔn is negative.

In this derivation, the Planck constant h, is formulated for each solution of aψ.  The resulting hi is applied at the appropriate level to obtain ψi+x.F, the work function.  x.F is emitted as a photon.

Alternatively,

ΔE=EfEi=mc2mc2=0

What happened?

When we apply hf to λi, we are suggesting that a wave at λi is caught at energy level hf.  λi adjusts itself to λf and there are energy changes in the process.  When λi is due to a larger aψn, its amplitude is also smaller  (cf. post "H Bar And No Bar"; this amplitude need to be quantified).  When λi constrict to λf at energy state n=f it does so at smaller amplitude, it loses energy.


This lost in energy appears as the work function x.F and is emitted as a photon.  The material cools.  The wave will regain its amplitude as heat is applied.  In the diagram above we considered only the fundamental frequency, when 2πaψn=λn.

In the post "Particle Spectrum", we took reference at n=1, where ho=h1.  The expressions for EΔn and λp are valid when the final energy state is n=1.

Intuitively, ψ has a longer path length at ni, the difference in energy EΔh due to a change in hn as the wave move from ni to nf is,

EΔh=hffihifi=cλi(hfhi)=c2πaψi(hfhi)

EΔh=1aψimc2(aψfaψi)

EΔh=mc2(aψfaψi1)

EΔh=mc2(eAaψf+Cn2f+Aaψi+Cn2i1)

Which is the same expression as (1).  Both derivations are equivalent.

ψ of a particle is excited to ni energy state, aψ=aψi.  On its return to a lower nf energy state, aψ=aψf where aψf<aψi, it loses energy in the form of a photon.

This is consistent with the observation of spectra lines, but this particle is not orbiting around any nucleus.

To be excited is then aψiaψf where aψf>aψi; to have a higher aψ.

Note: EΔh and EΔn are not the difference in energy as the particle move from ni to nf energy state.  But are energy changes as a result of hn changes between energy states.

In all cases, for,

ni>nf,    fi>ff

which is consistent with the post "Discreetly, Discrete λ And Discrete Frequency, f", unless the frequency profile is folded upwards.