Inverted Time Theory: Can these maths work?

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Offline timey

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Inverted Time Theory: Can these maths work?
« on: 14/05/2016 22:53:03 »
Ok... I'm attempting here to juggle some established physics maths into a different format in order to express my idea of the existence of an additional inverted time dilation.  Inverted Time Theory (ITT).  The consequences of which result in a closed system, non expanding cyclic universe that finds its beginnings and ends of cycle within the black hole phenomenon.

Please note that I am making significant adjustments to the 'physics norm' here.  My idea does away with energy mass equivalence.  Mass remains set at rest mass and does not change with the addition or subtraction of energy... e=mc2 is energy inherent to mass, gravity potential is the energy of the gravitational force exerted by bodies of mass upon each other at distance,(f=ma) and the non zero energy of the vacuum is described as having energy that is proportional to the 'frequency' of gravity, and is inversely proportional to the energy inherent in mass.  ITT employs a different means of calculating the factor of kinetic energy, and includes an addition to the equivalence principal stating that as well as the speed of light being constant, the speed of light cannot exceed the local rate of time.

So...



In the first equation we see E=mc2 = KE+m0c2.
This is saying that to establish E, kinetic energy must be added to the equation: 'rest mass' times speed of light squared.

The equation is then presented in another form...
So, for a particle of zero rest mass: p=E/c, and p is traditionally calculated via mv=p, which for light is mc=p.  To calculate p for light we are saying that for light to have energy, it must have relativistic mass equal to e=mc2.

ITT states that particles of zero mass, such as the photon, are subject to the non zero energy of their location.  That this non zero energy affects the gravitational shift, which in itself is inherent with an inverted time dilation.
To explain what is meant by inverted time dilation we must look at gravitational time dilation and the caesium atom.

The caesium atomic clock keeps a standard second of time at a frequency of 9,192,631,770 Hz.   If you elevate the clock into a weaker gravitational field, the frequency of the clock increases and the length of a standard second decreases causing time to run faster.  Therefore the caesium atom gains energy proportionally to its increase in frequency within the scenario of a decreasing gravitational gradient, and it loses energy proportionally to its decrease in frequency within the scenario of an increasing gravitational gradient.
This is known as gravitational time dilation.  On the basis of this phenomenon of gravitational time dilation, it is thought that a gravitational field (body of mass) slows time, and that time will run faster in a weaker gravitation field. (space)

ITT states that this concept is wrong. A body of mass at distance from another body of mass will be experiencing an increase in energy due to:



My theory of an additional inverted time dilation is based on the concept that when we measure time via the caesium atom in a gravitational gradient, that we are only measuring what time is doing for the caesium atom, and we are not measuring what time is doing for the location of gravitational gradient that the caesium atom is elevated at.  ITT states that it is the frequency of massless light matching the frequency of the gravitational field that is indicative of what time is doing for the location of a gravitational gradient.

The caesium atom does however provide us with a standard second when operating at frequency  9,192,631,770 Hz.  (Because this frequency is subject to change within a subtle gravitational gradient of less than a meter, one assumes that a standard second must also come equipped with a standard elevation in relation to sea level. (?).)  The standard second is used to measure all time related considerations and all inverted time dilation considerations need to be related back to the caesium atom at frequency 9,192,631,770 Hz.

Going back to the De Broglie hypothesis:

So...for the photon:

We can see that Planck's constant h=6.62606957×10^−34 joules*seconds, has been employed to calculate energy:

E=hv=hc/lambda

Lambda was what Einstein retracted from relativity, and frequency-wavelength, ie: redshift, is what Hubble stated as proof of expansion of the universe and became the basis of the Big Bang theory.  These days lambda is associated with the cosmological constant and equations concerning frequency are denoted with the letter f.  I shall use f.

E=hv=hc/f ...so... p=hc/cf=h/f ... and... f=h/p being the De Broglie wavelength.

Frequency is a time integral.  The time measurement is a constant of 1 standard second, and 1 standard second is equal to 299 793 458 meters as per the constancy of the speed of light.
So...on the basis that vt=d, or d/t=v, or d/v=t, we can divide 299 792 458 meters (d) by frequency (f standing in for v, as in frequency being the velocity of time) which on the basis that frequency is a time integral holding both a standard second and the distance of 299 792 458 meters constant in relation to the constant speed of light, can be denoted as d/fc=t~ whereby t~ dependant on frequency, may be a longer or shorter second than our standard second.

ITT states that because light has no mass it will not be subject to the force energy of gravitational attraction and will only be affected by gravity via the non zero energy and frequency of the gravitational field itself, which ITT states as inverted time dilation.

I repeat, it is via the non zero energy and frequency (proven via Pound Rebka) of the gravitational gradient itself that is the cause of inverted time dilation...  Gravitational field energy denotes the gravitational shift in the energy of light, and it is the subsequent change in the frequency of the light that denotes the change in the length of a second, with any contraction or expansion of lights wave'length' being time related rather than distance related.

Light takes form in many different frequencies and associated wavelengths all of which will be affected by these changes of non zero energy within the gravitational gradient.

ITT is suggesting a direct relationship between energy and the phenomenon of time itself!



If the phenomenon of time is energy related and more energy denotes a faster rate of time, then an electrons frequency and subsequent wavelength can be calculated at the quantum level with the electrons exact position and momentum simultaneously.  Perturbation (this itself being a time integral) will not be necessary.

Now the factor of kinetic energy must be addressed.

ITT suggests that there are 4 types of energy that need to be considered.
1: inherent energy within mass
2: non zero energy of the gravitational field
3: the energy of the force of gravitational attraction between bodies of mass
4: kinetic energy due to motion
(In lights case only no2: the non zero energy of the gravitational field will apply)

This constitutes a matrix of 4 energy considerations that operate on a sliding scale of proportionality to other factors.  1, 2 and 3 are positive and must be added together and 4, kinetic energy is negative and must then be subtracted for a slowing of time due to motion.

This matrix may be used in conjunction with the space time matrix whereby there are 3 positives of space and one negative of time.  The resulting calculation of the energy matrix based on the co-ordinates of the 3 dimensions of space of the space time matrix, becomes the corresponding time aspect of the space time matrix for those co-ordinates.

Finally, the constancy of the speed of light in relation to the local rate of time:

We can cause man made light to shift frequency in a non changing gravitational field by adding or subtracting energy via temperature or charge. The rate of time is constant in the non changing gravitational field, but we still observe that the frequency wavelength relationship holds so how can a longer or shorter wave'length' be inverted time dilation related?
I counter this argument by asking how a photon can travel up and down wavelengths of differing frequencies and still get from a to b, this being a distance of 299 792 458 meters, in 1 second?
You may point out to me the 'boat analogy' whereby the photon is a boat and its wavelength the boats 'wake', to which I would reply that for there to be any change in the wake of a boat, a change in the boats speed must occur!

In any case, ITT states that light must travel at the speed of light, 299 792 458 meters per second, and that the speed of light cannot exceed the local rate of a second of time, and consequently the rate of the photons time will slow for a photon that is moving in line motion, (ie: Michelson Morley experiment).  And the rate of the photons time will speed up for a photon of low energy, and low frequency, in order that it travel at the speed of light in relation to the local rate of second. (Please remember that we 'play' with light in ways that would not necessarily be naturally occurring)

The speed of light is a constant.  ITT states distances and lengths are also constant, but the rate of time is highly variable.
It is variable between the scale of all masses, their energies and the non zero energies of the gravitational field, and runs at a faster rate where there is more energy, and at a slower rate where there is less.

The consequences of this concept result in a closed system, non expanding, cyclic universe that finds its beginnings and ends of cycle within the black hole phenomenon.
This concept relies solely upon the Standard Model and requires no unobserved entities in order to explain its mechanics.
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #1 on: 15/05/2016 01:28:34 »
I was just beginning to wonder where you were, and here you are!

My theory of an additional inverted time dilation is based on the concept that when we measure time via the caesium atom in a gravitational gradient, that we are only measuring what time is doing for the caesium atom,

If this were the case, then the measured time dilatation would not depend on the position of the observer.

Quote
E=hv=hc/f
Wrong. Confusion probably caused by the similarity of Roman v (vee), often used to denote velocity,  and Greek ν (nu) used by Planck and others to denote frequency. E = hν = hf = hc/λ where the lower case lambda signifies wavelength and has nothing to do with uppercase Λ, the cosmological constant.

Quote
Frequency is a time integral.
No. Frequency is the number of occurrences of an event per unit time. In the limit it is the time differential of a phenomenon.
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #2 on: 15/05/2016 01:35:17 »
We can cause man made light to shift frequency in a non changing gravitational field by adding or subtracting energy via temperature or charge.
Have I missed something here? You can't add temperature or charge to a photon, surely?
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Offline timey

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Re: Inverted Time Theory: Can these maths work?
« Reply #3 on: 15/05/2016 03:07:55 »
I was just beginning to wonder where you were, and here you are!

:)... Back in Birmingham again.

My theory of an additional inverted time dilation is based on the concept that when we measure time via the caesium atom in a gravitational gradient, that we are only measuring what time is doing for the caesium atom,

If this were the case, then the measured time dilatation would not depend on the position of the observer.[/quote]

The measure of time dilation does not rely on the position of the observer.  (See NIST 2010 ground level atomic clock tests of gravitational tie dilation)  It is true that the position of an observer within a gravitational field will affect the rate of the observers own time in relation to time dilation elsewhere in the gravitational field, but to measure any  differences in gravitational time dilation one simply relates the differences back to the standard second, which is measured via the caesium atom at frequency  9,192,631,770 Hz.

Quote
E=hv=hc/f
Wrong. Confusion probably caused by the similarity of Roman v (vee), often used to denote velocity,  and Greek ν (nu) used by Planck and others to denote frequency. E = hν = hf = hc/λ where the lower case lambda signifies wavelength and has nothing to do with uppercase Λ, the cosmological constant.[/quote]

That wasn't my confusion but you have cleared up that which was.  Thanks, I'll be back tomorrow to clean up those maths I copied from the link, even though it doesn't affect the maths I didn't copy from the link.

Quote
Frequency is a time integral.
No. Frequency is the number of occurrences of an event per unit time. In the limit it is the time differential of a phenomenon.
[/quote]

Because frequency is the number of occurrences per unit of time, this being a standard second, frequency is a time integral in that the maths used to calculate frequency are inherent with the constant of 1 standard second, which incidentally in turn gives us a constant distance of 299 792 458 meters as per the constant speed of light.

Yes, I am saying that frequency is the time differential (hope I've got terminology correct) of a standard second.  We use the frequency of a caesium atom operating at  9,192,631,770 Hz to measure a standard second, and the frequency of a caesium atom increases with elevation into a weaker gravitational field due to an increase in energy due to the force of gravitational attraction.  Light has no mass, does not experience any force of gravitational attraction and is indicative via its reduction in energy and frequency of the possibility of an inverted gravitational time dilation.

Of course light comes in the form of many energies and frequencies, however, I believe that Hubble used cepheid variables as his standard candle.

We can cause man made light to shift frequency in a non changing gravitational field by adding or subtracting energy via temperature or charge.
Have I missed something here? You can't add temperature or charge to a photon, surely?

Tut, tut Alan!  Lateral thinking?  Clearly the black body experiment adds temperature and produces photons of varying energy, frequency and wavelength.
Switch on a torch with enough battery 'charge' and the tungsten filament or LED glows when electricity flows through it, thus producing visible light.
Presumably if you wish to produce light of a lower energy and frequency you simply provide the suitable light conductor with a lesser charge of electricity (?).  The difference between non ionising radiation and gamma rays or X-rays?
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #4 on: 15/05/2016 12:04:09 »
Clearly the black body experiment adds temperature and produces photons of varying energy, frequency and wavelength.
Switch on a torch with enough battery 'charge' and the tungsten filament or LED glows when electricity flows through it, thus producing visible light.


But once the photons are emitted, temperature and charge have no effect on them.

More importantly, the frequency of a cesium clock is determined by the electric potential gradient in the cesium atom, or more precisely, the quantised difference in energy levels between the two hyperfine ground states of the atom's electrons. The frequency of a mossbauer photon is determined by the decay of a nucleon. The gravitational field at the surface of a nucleus is calculable in principle and is orders of magnitude greater than that at the surface of the earth, whereas at the mean electron radius, the earth's field dominates. But the mossbauer photon energy is exactly as predicted by E = mc^2, so you need to explain how the extra-atomic potential gradient affects both nuclear and electronic transitions equally. 

The conventional argument is that the emitted photons in both cases undergo frequency shift in a gravitational gradient, and the calculation gives us a very exact answer in both cases.

You want to argue that the initiating phenomenon produces a photon whose energy depends on the gravitational time compression at the point of generation. If your calculation gives the same answer, you need to show why it is not just a tautologous rearrangement of the standard equation, and why it applies to all photons regardless of source. If it doesn't give the same answer, it's wrong!
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Offline timey

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Re: Inverted Time Theory: Can these maths work?
« Reply #5 on: 15/05/2016 15:00:20 »
""But once the photons are emitted, temperature and charge have no effect on them.""

Clearly!  Where have I said otherwise?

""More importantly, the frequency of a cesium clock is determined by the electric potential gradient in the cesium atom, or more precisely, the quantised difference in energy levels between the two hyperfine ground states of the atom's electrons. The frequency of a mossbauer photon is determined by the decay of a nucleon. The gravitational field at the surface of a nucleus is calculable in principle and is orders of magnitude greater than that at the surface of the earth, whereas at the mean electron radius, the earth's field dominates. But the mossbauer photon energy is exactly as predicted by E = mc^2, so you need to explain how the extra-atomic potential gradient affects both nuclear and electronic transitions equally.""

I am simply stating that within the e=mc2 equation, when adding energy to mass, that the mass increase that is inherent to the equation itself in the scenario of adding energy, is indeed an increase in the rate of time, not an increase in the particles mass.  Both scenarios are equal to each other in that an increase in the rate of time for a particle can be viewed as an increase in gravitational attraction.

""You want to argue that the initiating phenomenon produces a photon whose energy depends on the gravitational time compression at the point of generation.""

Nope, not really.  I am arguing that photons that have already been produced at energies relevant to their source are energy shifted across the vacuum of space due to the gravitational energy of their location, and that the gravitational energy's corresponding frequency is an inverted time dilation phenomenon.  Thus the change in the wavelength of any frequency of light as it moves through a gravitational gradient is time related, not distance related and redshifted light is not indicative of light sources expanding away from us  in the way currently thought.
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #6 on: 15/05/2016 23:16:07 »
redshifted light is not indicative of light sources expanding away from us  in the way currently thought.
But the P-R experiment showed that it is, by matching Doppler redshift to a gravitational shift. If you want to explain the redshift of distant galaxies as being due to a gravitational contraction of time, you have to postulate that there is more mass just outside the universe than just inside it, which seems to me like a selfcontradiction.
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Re: Inverted Time Theory: Can these maths work?
« Reply #7 on: 15/05/2016 23:31:29 »
I'm sorry Alan, but you misunderstand (my fault no doubt).  I am saying that redshift of distant galaxies is due to a dilation of time, not a contraction of time.  That the extra length in wavelength is due to a longer second.

In my model there is no outside of the universe.

The Pound Rebka proved that a static gradient of gravitational shift has a frequency that can be matched by a man made Doppler shift created in the test signal.
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #8 on: 16/05/2016 18:42:36 »
But what makes the time dilate? All we know about distant galaxies is that they are distant. We infer that they are moving away from us because their spectra are Doppler redshifted. If you want to say that the redshift is actually due to a gravitaional effect, you need to postulate an external source of gravitation, or state that all distant galaxies are very dense (just possible) and their density increases with distance (most improbable).
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Offline jeffreyH

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Re: Inverted Time Theory: Can these maths work?
« Reply #9 on: 16/05/2016 21:19:53 »
Over time the central black holes of distant galaxies will consume some of the total mass of the galaxy over time. Since looking out to these galaxies means looking back in time then it is entirely possible that galaxy density does indeed increase with distance. Since as we view objects farther away less time has passed since the big bang.
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Re: Inverted Time Theory: Can these maths work?
« Reply #10 on: 16/05/2016 21:47:46 »
But what makes the time dilate? All we know about distant galaxies is that they are distant. We infer that they are moving away from us because their spectra are Doppler redshifted. If you want to say that the redshift is actually due to a gravitaional effect, you need to postulate an external source of gravitation, or state that all distant galaxies are very dense (just possible) and their density increases with distance (most improbable).

I am saying that it is the gravitational shift of gravity itself that dilates inverted time dilation into a decreasing gravitational gradient, or contracts inverted time dilation into an increasing gravitational gradient.

Light, at the speed of light takes a specific amount of time to cover a distance, and if you didn't know that the rate of time becomes progressively slower in a weaker gravity field, you would calculate the distance the light has travelled as being longer.

If distances are shorter than we think, then the magnitude of the luminosity of a light source recalculated at a closer distance would indeed constitute a larger star.
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Re: Inverted Time Theory: Can these maths work?
« Reply #11 on: 16/05/2016 22:50:10 »
Over time the central black holes of distant galaxies will consume some of the total mass of the galaxy over time. Since looking out to these galaxies means looking back in time then it is entirely possible that galaxy density does indeed increase with distance. Since as we view objects farther away less time has passed since the big bang.

I can appreciate where you are coming from...
However, in my model the inflation period results in a sea of particles that start clumping together, and distances between clumps or bodies of mass are formed by particles vacating their former position because they are clumping together.  After the initial inflation period, due to this clumping of mass, the actual spatial dimensions of the universe start very slowly reducing.  Black holes start forming as critical mass is achieved.  Eventually, after a scenario synonymous to a predator prey waveform, black holes become dominant until all the mass of the universe is contained in a galaxy of black holes.  These will merge until there is only one singular black hole, which without any equivalent gravitational force acting upon it, empties itself via its accretion disks until it own extinction, leaving a sea of particles.
This being the mechanics of Inverted Time Theory's closed system cyclic universe.
Having said this, your comment holds just as true, if not truer, in my model as it does in the Big Bang model, only that distances are not as far apart, so it would be of a significantly lesser effect.
Edit: Correction, (silly me...) ...even though the distance is shorter the effect would be 'the same' because it takes light exactly the same amount of time per 'standard' second to cover the shorter distance within the slower rates of time!
« Last Edit: 17/05/2016 00:27:46 by timey »
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Re: Inverted Time Theory: Can these maths work?
« Reply #12 on: 17/05/2016 19:35:33 »
So...

On the basis that the LIGO gravity wave has a frequency, presumably we can ascertain a frequency for any gradient of gravitational field.

On the basis that the gravity wave, and any changes within the gravity of a gravitational field travel at the speed of light, can these maths work to ascertain the longer or shorter second of my proposed inverted time dilation?

Frequency is, or has, a time integral.  This time integral being a constant of 1 standard second, and 1 standard second is equal to 299 793 458 meters as per the constancy of the speed of light.
So...on the basis that vt=d, or d/t=v, or d/v=t, we can divide 299 792 458 meters (d) by frequency (f standing in for v, as in frequency being the velocity of time. (Edit: or perhaps that should be the momentum of time)) which on the basis that frequency is, or has, a time integral holding both a standard second and the distance of 299 792 458 meters constant in relation to the constant speed of light, can be denoted as d/fc=t~ whereby t~ will be a longer second than our standard second.

When dealing with a gravitational field of a higher frequency than earth's gravitational field, which is what is used in relation to our measurement of a standard second, then: df/c=~t (?, or something like that anyway) whereby ~t is a shorter second...

Can it work?
« Last Edit: 17/05/2016 22:17:08 by timey »
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Re: Inverted Time Theory: Can these maths work?
« Reply #13 on: 17/05/2016 22:32:01 »
Too many pseudoscientific words in a meaningless jumble! Have another go, perhaps before the pubs open. 
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Re: Inverted Time Theory: Can these maths work?
« Reply #14 on: 17/05/2016 22:46:04 »
Too many pseudoscientific words in a meaningless jumble! Have another go, perhaps before the pubs open.

Oh dear me Alan, you seem to have suffered a rather detrimental character shift.  Personally, I very rarely visit pubs, only twice in the last 2 years.

A gravitational field has a frequency.  This frequency increases or decreases with changes in the gravitational field.

Can the frequency of gravity be used as I have used it in the 'very simple' maths that I have illustrated to describe a longer or shorter second relative to a standard second or not?

Don't be a dolt Alan.  It's unbecoming!

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Re: Inverted Time Theory: Can these maths work?
« Reply #15 on: 18/05/2016 12:17:02 »
No, a gravitational field does not have a frequency unless it is a repetitively time-varying gravitational field. You can generate such a field by, say, a pair of masses rotating about their center of gravity: the external field willl oscillate in the plane of rotation, but this has no connection with the mechanism of red shift, which occurs in a static field.
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Re: Inverted Time Theory: Can these maths work?
« Reply #16 on: 18/05/2016 13:46:06 »
And yet the Pound Rebka proved that the frequency of a man made Doppler shift can be matched within a static distance of gradient of a gravitational field.

https://en.m.wikipedia.org/wiki/Pound–Rebka_experiment

"if the emitting atom moves with just the right speed relative to the receiving atom the resulting Doppler shift cancels out the gravitational shift and the receiving atom can now absorb the photon. The "right" relative speed of the atoms is therefore a measure of the gravitational shift.""

So..  A static gradient of the gravitational shift itself has a 'speed' or a 'frequency'.
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #17 on: 18/05/2016 17:03:44 »
Rubbish. The gravitational gradient alters the frequency of the travelling photon - as you well know.

"Alters the frequency of" does not mean "has a frequency". Some people wear earrings, presumably to alter the length of their ear lobes. Earrings have weight, not length.

And don't bother with the phrase "static distance of gradient". The PR experiment measured gravitational shift by comparing it with a Doppler shift. That's all. If you keep it simple, you won't confuse yourself with unnecessary jargon. 
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Re: Inverted Time Theory: Can these maths work?
« Reply #18 on: 18/05/2016 18:38:19 »
GR states that... ""The gravitational gradient alters the frequency of the travelling photon - as you well know."" ... because it has a relativistic mass associated with its energy that is affected by gravity potential,ie: the strength of the gravitational field of its location.

Inverted time theory does not attribute light with having mass, only energy.

""Alters the frequency of" does not mean "has a frequency".""

A gravitational field has energy.  Simply view the energy of gravity as the non zero energy of space.  Where there is more gravity there is more energy.  Energy is inherent with a frequency.

""And don't bother with the phrase "static distance of gradient". The PR experiment measured gravitational shift by comparing it with a Doppler shift.""

And it was proved that there is the motion of a Doppler shift within the gravitational shift over a static distance.

""That's all. If you keep it simple, you won't confuse yourself with unnecessary jargon.""

I'm not being confused by any jargon.  I read respected physics books by respected physics authors, and I read articles and wiki links.  Then I simply repeat what I've read.  That's all!
Maths on the other hand do confuse me, and again and again, I stress that help with maths is my reason for being here.  Stated in my first post ever, and many times since.

This is a forum board of New Theories, and ITT is a new theory.  It stands to reason that any new idea is going to be taking a novel approach or novel view on aspects of accepted physics.

Unlike any other New Theory on this board Alan, this one does not seek to introduce or incorporate any unobserved inclusions to accepted physics, as indeed accepted physics does itself.  ITT only seeks to view any actual observed accepted physics from a slightly different perspective.

You have said ""The PR experiment measured gravitational shift by comparing it with a Doppler shift.""

The experiment shows us what is causing the motion of a Doppler shift in the test signal.  What is causing the motion of a Doppler shift in the gravitational shift measured?
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #19 on: 18/05/2016 19:01:16 »
GR states that... ""The gravitational gradient alters the frequency of the travelling photon - as you well know."" ... because it has a relativistic mass associated with its energy that is affected by gravity potential,ie: the strength of the gravitational field of its location. no, it's because a gravitational field is equivalent to an accelerating frame of reference

Inverted time theory does not attribute light with having mass, only energy.good, because that is true

""Alters the frequency of" does not mean "has a frequency".""

A gravitational field has energy.  Simply view the energy of gravity as the non zero energy of space.  Where there is more gravity there is more energy.  Energy is inherent with a frequency.nonsense. do your dimensinal analysis and you'll see why.

""And don't bother with the phrase "static distance of gradient". The PR experiment measured gravitational shift by comparing it with a Doppler shift.""

And it was proved that there is the motion of a Doppler shift within the gravitational shift over a static distance.A meaningless jumble of words. PR proved that you can measure G shift with D shift.

""That's all. If you keep it simple, you won't confuse yourself with unnecessary jargon.""

I'm not being confused by any jargon.  I read respected physics books by respected physics authors, and I read articles and wiki links.  Then I simply repeat what I've read. Obviously not. Please show me where you read  "there is the motion of a Doppler shift within the gravitational shift over a static distance", for instance. That's all!
Maths on the other hand do confuse me, and again and again, I stress that help with maths is my reason for being here.  Stated in my first post ever, and many times since.

This is a forum board of New Theories, and ITT is a new theory.  It stands to reason that any new idea is going to be taking a novel approach or novel view on aspects of accepted physics.Agreed, but it helps if you dopn't misquote the old physics, or reinvent it.

Unlike any other New Theory on this board Alan, this one does not seek to introduce or incorporate any unobserved inclusions to accepted physics, as indeed accepted physics does itself.  ITT only seeks to view any actual observed accepted physics from a slightly different perspective.

You have said ""The PR experiment measured gravitational shift by comparing it with a Doppler shift.""

The experiment shows us what is causing the motion of a Doppler shift in the test signal.  What is causing the motion of a Doppler shift in the gravitational shift measured? Doppler shift is not gravitational shift. Please mind your language!
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Re: Inverted Time Theory: Can these maths work?
« Reply #20 on: 18/05/2016 19:26:20 »
""no, it's because a gravitational field is equivalent to an accelerating frame of reference""

Yes, precisely... and an accelerating frame of reference (GR), and a gravitational field increasing the rate of time (ITT), can be viewed as being equivalent.  (the benefits of doing so result in a cyclic universe)

Alan, I really do not understand why you find it so difficult to grasp this simple concept:
If you are using a Doppler shift to measure a phenomenon, then the phenomenon you are measuring will be inherent with the properties of the measuring unit.
How can the gravitational shift match, ie: 'harmonise with' the created Doppler shift if it were otherwise?
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Re: Inverted Time Theory: Can these maths work?
« Reply #21 on: 18/05/2016 21:48:46 »
""no, it's because a gravitational field is equivalent to an accelerating frame of reference""

Yes, precisely... and an accelerating frame of reference (GR), and a gravitational field increasing the rate of time (ITT), can be viewed as being equivalent.  (the benefits of doing so result in a cyclic universe)

Alan, I really do not understand why you find it so difficult to grasp this simple concept:
If you are using a Doppler shift to measure a phenomenon, then the phenomenon you are measuring will be inherent with the properties of the measuring unit.
How can the gravitational shift match, ie: 'harmonise with' the created Doppler shift if it were otherwise?

Because they are both changes in photon frequency. "will be inherent with the properties of the measuring unit." is meaningless.

G shift alters the frequency of an emitted photon, D shift alters the absorption frequency of the receiver. Suppose we both have airband transceivers. I talk to you at 120.000 MHz and your receiver is set to 120.000MHz, so you can hear me. If I  change to 120.025MHz, you can't hear anything, unless you retune your receiver to 120.025 MHz. Now I transmit at X MHz, but I don't tell you what X is, so you sweep your receiver and read the frequency Y you have set on your dial. As soon as you can hear me, you know that Y = X. All that PR did was to use very precise transmitters and receivers, where G changed the transmit frequency and D was used to sweep the receiver band.
« Last Edit: 18/05/2016 21:58:14 by alancalverd »
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Re: Inverted Time Theory: Can these maths work?
« Reply #22 on: 18/05/2016 22:42:01 »
The man made Doppler shift in the Pound Rebka was 'not' a change in photon frequency at-all, Alan.  They created the man made Doppler shift of the emitted gamma ray in relation to the receiving atom, to cancel out the gravitational shift of the gamma ray in order that it be in the correct energy state to be absorbed by the receiving atom, as is the case in the equivalent horizontal experiment conducted without the additional man made Doppler shift.

So - What was it that the man made Doppler shift cancelled out?

The man made Doppler shift cancelled out the G shift, and the G shift that it cancelled out is gravity related, not photon related.
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Re: Inverted Time Theory: Can these maths work?
« Reply #23 on: 19/05/2016 00:27:15 »
The man made Doppler shift in the Pound Rebka was 'not' a change in photon frequency at-all, Alan.  They created the man made Doppler shift of the emitted gamma ray in relation to the receiving atom, to cancel out the gravitational shift of the gamma ray
What is Doppler shift? What is gravitational shift? A change in the frequency of a (gamma) photon! What do we observe in cosmology? G and D shifts of photon frequencies. That is why they are called red and blue shifts!

You can call it a change in photon energy if you like, but remember that you stated that E = hf = hc/λ, so a change in energy is a change in frequency, since h is a constant.

Thanks to the equivalence principle, it doesn't matter whether you move the source or the receiver in the PR experiment, as long as you move it at the velocity v at which D shift = G shift. In the horizontal experiment δg = 0 so v = 0.
« Last Edit: 19/05/2016 00:33:18 by alancalverd »
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Re: Inverted Time Theory: Can these maths work?
« Reply #24 on: 19/05/2016 01:05:02 »
The Doppler shift in the Pound Rebka is the man made speaker vibration of the emitted light source in relation to the motion of the receiving atom.

The gravitational shift is the change in the gravitational field gradient.

The Pound Rebka proved that the gravitational shift of the gravitational field has a frequency that can be harmonised with, and be cancelled out by a man made Doppler shift, (which was not a change in the frequency of a photon), created by speaker vibration and the motion of the receiving atom.

Why would someone call a change in the gravitational gradient, ie: gravitational shift, a change in photon energy?  The gravitational shift causes a change in photon energy, and in the case of any other particle with mass's energy, the gravitational shift causes a contrary (but not directly opposite) change in energy than it does with the photon, but you cannot describe these energy changes in the photon or any other particle as the gravitational shift itself.  The gravitational shift is the changes in the gravitational field, surely?

Edit: ...and in reply to your edit: where does this v exist in the gravitational field within a static distance?
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Re: Inverted Time Theory: Can these maths work?
« Reply #25 on: 19/05/2016 23:02:37 »
The Doppler shift in the Pound Rebka is No. It isn't the vibration. It is caused by the vibration which at some point gave the emitter or detector a velocity such that the Doppler shift matched the gravitational shift.the man made speaker vibration of the emitted light source in relation to the motion of the receiving atom.

The gravitational shift is the change in the gravitational field gradient.NJo, it is caused by the difference in gravitational potential between the emitter and the receiver.

The Pound Rebka proved that the gravitational shift of the gravitational field has a frequency that can be harmonised with, and be cancelled out by a man made Doppler shift, (which was not a change in the frequency of a photon), created by speaker vibration and the motion of the receiving atom. Poppycock

Why would someone call a change in the gravitational gradient, ie: gravitational shift, a change in photon energy? Nobody does. Just stick to the words on the card, and don't try to add more. The gravitational shift causes a change in photon energy,no, it is a change in photon energy, caused by gravitational potential difference and in the case of any other particle with mass's energy, the gravitational shift causes a contrary (but not directly opposite) change in energy than it does with the photon,If this means anything, can you prove or explain it? but you cannot describe these energy changes in the photon or any other particle as the gravitational shift itself.  The gravitational shift is the changes in the gravitational field, surely?No, it is the shift if photon frequency caused by the change in gravitational potential

Edit: ...and in reply to your edit: where does this v exist in the gravitational field within a static distance? It is the relative velocity of source and receiver - i.e. the determinant of Doppler shift. Nothing to do with gravitation.
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Re: Inverted Time Theory: Can these maths work?
« Reply #26 on: 20/05/2016 01:15:42 »
Ah... Ok, I see we are having a terminology problem.  You are referring to gravitational shift as being a photon related phenomenon. ie: a photons energy and frequency being affected by changes in the gravitational field.  I am referring to gravitational shift as being a phenomenon of gravity which causes all particles to shift energy and frequency with changes in the gravitational field.  And to explain: a photons energy and frequency reduces in a decreasing gravitational field, and increases in an increasing gravitational field.  And, any particle with mass's energy and frequency increases in a decreasing gravitational field, and decreases in an increasing gravitational field.  However, this is not a truly symmetrical contrary, because the changes in a photons energy and frequency reduces or increases far more drastically with changes in the gravitational field than a particle with mass experiences...  Although there will indeed be a mathematical proportionality between these contrary phenomenon. (if only I were a mathematician)

I said: 'Edit: ...and in reply to your edit: where does this v exist in the gravitational field within a static distance?'
You said: ""It is the relative velocity of source and receiver - i.e. the determinant of Doppler shift. Nothing to do with gravitation.""

What do you mean nothing to do with gravitation???  The relative velocity of source and receiver-ie: the determinant of Doppler shift was 'used' to measure a gravitational phenomenon.  This Doppler shift was """"matched"""" by whatever it is that gravity is doing when """"gravity"""" changes energy in a gravitational gradient.  And in the Pound Rebka this gravitational gradient was """"static""""!!!  Yet a velocity of Doppler shift was """"matched"""" by the gravity (not photon) related phenomenon being measured.  I repeat, it was the gravitational field, being measured.  The Doppler shift was created in the test signal in order that the emitted gamma rays """"did not"""" shift energy, and it achieved this by """"cancelling out"""" the energy shift effect of the gravitational gradient.

Therefore, logically speaking, this velocity of Doppler shift must exist within the energy changes in the gravitational field.
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Re: Inverted Time Theory: Can these maths work?
« Reply #27 on: 20/05/2016 18:01:44 »
velocity of Doppler shift
Try using the term "Doppler shift" instead of adding extraneous words, and it will all become clear.

Quote
I repeat, it was the gravitational field, being measured.
not at all. The gravitational potential difference was known:

Δg ≈ 0.31 x 10-6 h

where h is height in meters above the earth's surface.

The purpose of the experiment was to measure an unknown but theoretically calculable gravitational red shift in a known field by matching it with a known Doppler shift. Good experimental science is all about measuring one unknown at a time!
« Last Edit: 20/05/2016 18:13:36 by alancalverd »
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Re: Inverted Time Theory: Can these maths work?
« Reply #28 on: 20/05/2016 20:36:03 »
(...sigh). What was it that the 'determinant' of the Doppler shift matched, or harmonised with, and cancelled out?

Does the vibration of the speaker, or the relative motion of the receiving atom, affect the energy and frequency of the gamma ray?  No it doesn't.  The energy and frequency of the gamma ray when subjected to this determinant of Doppler shift remain unchanged by the 'gravitational potential', and are absorbed by the receiving atom because whatever it is about gravitational potential that causes light to shift in energy and frequency has been cancelled out by this determinant of Doppler shift.

So - what was it that was cancelled out by this determinant of Doppler shift in order for the gamma ray not to be gravitationally shifted in energy and frequency?

(May I politely hint to you at this juncture that I am repeating terminology that you have used yourself, which does rather render any complaint you have concerning linguistics as extraneous, and presumably you may now find yourself able to focus on the question posed within said annoying language without such distraction)
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Re: Inverted Time Theory: Can these maths work?
« Reply #29 on: 20/05/2016 21:46:49 »
whatever it is about gravitational potential that causes light to shift in energy and frequency has been cancelled out by this determinant of Doppler shift.

Nearly correct. Let's leave out a few words, then the gravitational shift equals the Doppler shift. Easy.

Gravitational shift is the difference between emitted and received frequency caused by the difference in gravitational potential between the source and the receiver. Doppler shift is the difference between emitted and received frequency caused by the relative motion of the source and receiver.

Quote
Does the vibration of the speaker, or the relative motion of the receiving atom, affect the energy and frequency of the gamma ray?  No it doesn't.
yes it does. It alters the received frequency!
 
To quote from your favorite Wikipedia entry

 
Quote
Pound and Rebka countered the gravitational blueshift by moving the emitter away from the receiver, thus generating a relativistic Doppler redshift

That's all there is to it.
« Last Edit: 20/05/2016 21:52:43 by alancalverd »
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Re: Inverted Time Theory: Can these maths work?
« Reply #30 on: 20/05/2016 22:05:37 »
And...also quoting from my favourite wiki link:

""This experiment is based on the principle that when an atom transitions from an excited state to a ground state, it emits a photon with a characteristic frequency and energy. Conversely, when an atom of the same species, in its ground state, encounters a photon with the same characteristic frequency and energy, it will absorb the photon and transition to the excited state. If the photon's frequency and energy is different by even a small amount, the atom cannot absorb it (this is the basis of quantum mechanics). When the photon travels through a gravitational field, its frequency, as well as its energy, will change due to the gravitational redshift. As a result, the receiving atom cannot absorb it. But if the emitting atom moves with just the right speed relative to the receiving atom the resulting Doppler shift cancels out the gravitational shift and the receiving atom can now absorb the photon. The "right" relative speed of the atoms is therefore a measure of the gravitational shift.""

Yes we can see that the determinate of the Doppler shift is a measure of what the gravitational shift of the gamma ray would have been, had it shifted energy and frequency.  But again I ask you, 'what' in the gravitational shift was 'cancelled out' in order for the gamma ray's energy and frequency 'not' to be shifted in the gravitational gradient?
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Re: Inverted Time Theory: Can these maths work?
« Reply #31 on: 20/05/2016 23:05:50 »
First, get rid of the unnecessary words you have added.

Yes we can see that the determinate of the Doppler shift is a measure of what the gravitational shift of the gamma ray would have been, had it shifted energy and frequency.
Now look at the physics. The frequency did indeed shift, exactly as predicted......

 
Quote
But again I ask you, 'what' in the gravitational shift was 'cancelled out' in order for the gamma ray's energy and frequency 'not' to be shifted in the gravitational gradient?

.....then they added a Doppler shift of equal and opposite magnitude, so that the received frequency was the same as the emitted frequency. What's the problem?

Think about a simple Doppler shift. A train is coming towards you. Its whistle emits an A at 440Hz. (How do we know the frequency? Because it's written on the whistle and I tuned it.) But because it is travelling towards you at 20 meters/second over the ground, you hear the note as

440 x (1 + v/c) = 440 x (1 + 20/340) = 466 Hz = A#  (note c here is the speed of sound, 340 m/s,  not light!)

So being a sensible person, you get in your car and drive away from the train at a ground speed of 50 m/s, so now your relative speed is -30 m/s and you hear

440  x (1 - 30/340) = 401 Hz (a bit sharp of G)

Having perfect pitch, you slow down until you hear an A. What speed are you travelling? It must be 20 m/s, to exactly cancel the Doppler shift from the train's movement. It's a dangerous way of measuring the speed of a train, but you gotta admit it works!

How does this fit with the PR experiment? Instead of a moving train, they used the gravitational gradient to produce the initial frequency shift. And of course c was the speed of light, not sound. That's all there was to it! 

Rule 1 of physics: don't go looking for complications.   
« Last Edit: 20/05/2016 23:42:50 by alancalverd »
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Re: Inverted Time Theory: Can these maths work?
« Reply #32 on: 20/05/2016 23:42:23 »
**sigh** That timey eh? If you have to explain once you have to explain a Million times! Brain not in gear girl?
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Re: Inverted Time Theory: Can these maths work?
« Reply #33 on: 20/05/2016 23:54:58 »
**sigh** That timey eh? If you have to explain once you have to explain a Million times! Brain not in gear girl?

Agreed wholeheartedly Jeff!  It is absolutely ridiculous how many times I am having to explain this incredibly simple concept.
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Re: Inverted Time Theory: Can these maths work?
« Reply #34 on: 20/05/2016 23:57:21 »
Quote from: timey on Today at 22:05:37
Yes we can see that the determinate of the Doppler shift is a measure of what the gravitational shift of the gamma ray would have been, had it shifted energy and frequency.
Now look at the physics. The frequency did indeed shift, exactly as predicted......

Yes the frequency shifted before they added the Doppler shift. Edit: and the gamma rays were not absorbed by the receiving atom.
« Last Edit: 21/05/2016 00:11:20 by timey »
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Re: Inverted Time Theory: Can these maths work?
« Reply #35 on: 20/05/2016 23:59:34 »
Quote
But again I ask you, 'what' in the gravitational shift was 'cancelled out' in order for the gamma ray's energy and frequency 'not' to be shifted in the gravitational gradient?

.....then they added a Doppler shift of equal and opposite magnitude, so that the received frequency was the same as the emitted frequency. What's the problem?

...and then the gamma ray's frequency did not shift.  Edit: and the gamma rays were absorbed by the receiving atom.
« Last Edit: 21/05/2016 00:12:55 by timey »
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Re: Inverted Time Theory: Can these maths work?
« Reply #36 on: 21/05/2016 00:00:24 »
You cannot change the energy or frequency of a gamma ray by shaking its emitter source via a speaker cone vibration, or by movement of the receiving atom, when conducting an equivalent horizontal experiment...  And you cannot change the energy or  frequency of an emitted photon by shaking its emitter source, or by the movement of the receiving atom, in the vertical experiment either!  Although there is indeed a case for the receiving atom having changed its energy as it moves through changes in the gravitational field towards the emitted photon.  However, as you yourself have previously pointed out to me somewhere else, the receiving atom will have travelled a minimal distance in relation to the distance travelled by the photon.

The only means of change in the energy and frequency of an emitted photon is via the changes in a gravitational field.  You yourself have previously pointed out to me elsewhere (to my agreement), that an already emitted photon cannot gain or lose energy via any means other than changes in the gravitational field.

So when the measure of the frequency that the gamma ray would have changed to is determined, ( ie: the 'measure of' (is that aloud?) the Doppler shift created by this shaking of the light source emitter in relation to the movement of the receiving atom), 'where' does the phenomenon/frequency that it 'matches', 'resonates' or 'harmonises with' exist?

This being my problem, because far as I can tell Alan, within the physical definition of 'match', 'resonate' or 'harmonise', at the very least 2 aspects are required for the function.  ie: to re-instate your radio frequency analogy, 2 transmitter/receivers would be necessary in order to tune into the same frequency.

The emitted gamma ray, no matter what you do with it, cannot match, resonate or harmonise with anything of its own self and change a phenomenon that is caused by changes in the gravitational field.  It has to be 'something' in the gravitational field that the Doppler shift created matches and cancels out.
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Re: Inverted Time Theory: Can these maths work?
« Reply #37 on: 21/05/2016 07:20:19 »
A teacher is someone who takes a subject he understands and phrases it in words a pupil can understand. An educationalist is someone who takes a subject he doesn't understand and rephrases it until nobody can.

Like when you add words to a perfectly simple statement such as

"Gravitational shift is the difference between emitted and received frequency caused by the difference in gravitational potential between the source and the receiver. Doppler shift is the difference between emitted and received frequency caused by the relative motion of the source and receiver."

until you can't understand it.

I suspect you are a covert educationalist. Or a New Age troll, injecting mystery where others seek simplicity. I enjoy matching, resonating and harmonising, but with girlfriends and musicians, not photons.

One last try. For chrissake (or at least your own) delete "energy and frequency". The speed of a photon is constant so the only thing that can change is its frequency as seen by a receiver. So let's talk about frequency.

If a Mossbauer photon is redshifted because it has come from a gravitational potential well, it will arrive at the detector with a lower frequency than expected, so it won't be absorbed. Now if we move the detector towards the source, we are adding a Doppler shift to the received frequency and at some critical velocity the D shift will be equal and opposite to the G shift, and the photon will be absorbed.

If, in the PR experiment, the source is at the top of the tower, the photon will be blueshifted by the time it reaches the bottom, so you have to move the detector away from the source, thus subtracting the D shift from the G shift.

In the horizontal case, there is no gravitational potential difference between source and receiver, so there must be absolutely no relative movement if the photon is to be absorbed.

The power of the Mossbauer effect is that the receiver bandwidth is very, very, very narrow, so you can measure small G shifts (as from a few meters vertical separation) with quite small D shifts (as from a loudspeaker movement).
« Last Edit: 21/05/2016 07:39:35 by alancalverd »
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Re: Inverted Time Theory: Can these maths work?
« Reply #38 on: 21/05/2016 11:32:25 »
Yes Alan - yawn.  And what is causing the photon to G-shift?  Is it the 'extra relative motion" that is causing the G shift?

Adding extra relative motion to a photons speed aye?  In a static distance of gravitational gradient aye?  Speed of light is 'constant' aye?

When the photon does suffer the gravitational shift in the Pound Rebka, before they added the Doppler shift that cancels the effect, either the photon is moving 'slower' for a 'reason' or the speed of light is 'not' constant, because the extra motion added by the Doppler shift that cancelled the effect has no extra 'distance' to exist in.

My theory says that the extra relative motion that the Pound Rebka measured as the G shift via the Doppler shift is caused by the photon moving through a slower rate of time in the weaker gravitational field.

On a more personal note, I'm becoming quite concerned Alan that you have picked up a touch of Simon Cowell from somewhere...  Far more dangerous than an STD and just as embarrassing, I've read that self diagnosis in these cases is nigh impossible.  So - as your sworn friend and internet buddy Alan, I feel it my social duty to inform you, that you may have your condition attended ASAP before it causes you any permanent damage.
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Re: Inverted Time Theory: Can these maths work?
« Reply #39 on: 21/05/2016 12:52:46 »
And what is causing the photon to G-shift?

I can do no better than quote Wikipedia

Quote
In astrophysics, gravitational redshift or Einstein shift is the process by which electromagnetic radiation originating from a source that is in a gravitational field is reduced in frequency, or redshifted, when observed in a region of a weaker gravitational field. This is a direct result of gravitational time dilation - as one moves away from a source of gravitational field, the rate at which time passes is increased relative to the case when one is near the source. As frequency is inverse of time (specifically, time required for completing one wave oscillation), frequency of the electromagnetic radiation is reduced in an area of a lower gravitational field.

which, if I understand your jargon, is exactly the opposite of

Quote
My theory says that the extra relative motion that the Pound Rebka measured as the G shift via the Doppler shift is caused by the photon moving through a slower rate of time in the weaker gravitational field.

though you might like to tell us what the words in purple mean to you.
« Last Edit: 21/05/2016 12:57:52 by alancalverd »
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Re: Inverted Time Theory: Can these maths work?
« Reply #40 on: 21/05/2016 14:17:06 »
As I understand it, the Doppler shift measurement is a plus for a redshift and a minus for a blueshift.  A plus will constitute a reduced frequency and a minus will constitute an increased frequency.

A frequency is ascertained via a time integral, ie: waves per 'second'.  The time integral used is the 'standard second' that also contains within its structure, the constants of the speed of light, and the distance of 299 792 458 meters.

By dividing the constants of the mathematical structure of the means of ascertaining frequency as though it were the time, distance, velocity formula, using the constant of 299 792 458 meters as d, the speed of light c standing in for t, and the measure of the frequency, (being the only variable), instead of v:

d/fc=t~ whereby t~ will be a longer second than our standard second.

And: Edit: because a standard second is a measurement of the frequency of cycles of a caesium atom when placed at ground level in earth's gravitational field, a gravitational field of greater strength than earth's gravitational field at ground level would have to be calculated as:

df/c=~t (?, or something like that anyway) whereby ~t is a shorter second...

However, for a correct interpretation of this 'inverted time dilation' length of second, the energy changes within the gravitational gradient need to be given frequencies associated with the energy of the field itself, and it is the frequency of the gravitational field that would constitute f in the maths I've illustrated, as light of 'any' frequency will gravitationally shift in a gravitational gradient and thus cannot be used.

Now Alan, I may have fluffed the maths here, but if you could tuck your 'physicist' character back in your top pocket and get into 'theoretical physicist' mode, perhaps you can appreciate what I'm attempting here - and even help?
« Last Edit: 21/05/2016 16:02:01 by timey »
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #41 on: 21/05/2016 16:55:13 »
As I understand it, the Doppler shift measurement is a plus for a redshift and a minus for a blueshift.  A plus will constitute a reduced frequency and a minus will constitute an increased frequency.

A frequency is ascertained via a time integral,I have explained previously that this is not true ie: waves per 'second'.  The time integral used is the 'standard second' that also contains within its structure, the constants of the speed of light,no it doesn't and the distance of 299 792 458 meters.not true either
Not relevant to the PR experiment. Gravitational frequency shift is measured as the amount of Doppler shift you need to apply in  order to compensate for it. Neither the emitter nor the detector cares or knows about the standard second, the speed of light, or anything else, except whether the mossbauer photon is emitted and absorbed.  If the emitter is above the detector, they have to be moving towards each other for absorption, and if the emitter is below the detector, they have to move apart. No numbers are required to demonstrate G shift, just "up" and "down".

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By dividing the constants of the mathematical structure of the means of ascertaining frequency as though it were the time, distance, velocity formula, using the constant of 299 792 458 meters as d, the speed of light c standing in for t, and the measure of the frequency, (being the only variable), instead of v:

d/fc=t~ whereby t~ will be a longer second than our standard second.
a moment's dimensional analysis will show you that this is nonsense.

The standard second is the elapsed time of 9,192,631,770 cycles of the electromagnetic radiation associated with the transition between the two hyperfine ground states of caesium. It is a fundamental unit, i.e. not dependent on any other measurement or assumption about the speed of light, the length of a meter, or anything else. In order to measure any time, all you need to do, in principle, is to count the cycles of your cesium source and divide by 9,192,631,770. If you want to measure the frequency of something else, you count the number of cycles it makes during 9,192,631,770 cycles of the cesium radiation. Not theoretical physics, but a neat bit of practical electronics.   

Now here's the fun bit. If we put a cesium clock in orbit, and count the number of cycles it transmits in one second, we find it is running faster than one on the ground. That is gravitational blue shift. But as far as the astronaut is concerned, his clock is running perfectly and yours is running slow - gravitational red shift. How does he know his clock is OK? Because everything else on his spaceship is keeping perfect time too. More to the point, he could claim that as his clock runs at the same speed as all  other clocks in a gravity-free area, i.e. practically the entire universe, his is right and yours is wrong, and clocks just misbehave in the vicinity of large masses. Indeed if your "ground" clock was on Mars, it would run faster, and on Jupiter, slower, but always slower than the gravity-free clock. . But as the clock is blissflly unaware of its surroundings, we must conclude that time runs slower in a gravitational field.
« Last Edit: 21/05/2016 16:57:55 by alancalverd »
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Offline timey

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Re: Inverted Time Theory: Can these maths work?
« Reply #42 on: 21/05/2016 17:51:30 »
Thank you Alan.  If you think for one minute that the 30 or so theoretical physics books that I have read over the last 8 years haven't covered all of that then you haven't read many theoretical physics books.

All I care about and comment on within the Pound Rebka is the plus and minus values of the Doppler shift in relation to the static distance and the 'causation' of a gravitational shift of energy in particles of zero rest mass.

Yes, you 'could' say that the caesium atom blue shifts when moving into a weaker gravitational field, and that every particle with mass will do so.

A photon does the opposite, it red-shifts into a weaker gravitational field.

I am looking at the possibility that because the photon has no mass that this reduction in frequency in a weaker gravitational field can be indicative of an additional phenomenon of an inverted time dilation, whereby the extra length in wavelength of shifted light is 'inverted time dilation' related, ie: it takes the light a longer 'time' to cover same distance, and the extra length of wavelength is then not an extra length in actual distance.

The standard second is what we use to measure everything in physics.  It is used to measure the speed of light, therefore the structure of any mathematics that measures anything in relation to a standard second, (edit: ie: per second), is, by default, also holding the speed of light and the distance of 299 792 458 meters constant, and if the initial equation is gravity or light related, (as both travel at speed of light) this sub-structure can be manipulated mathematically in relation to the result of the initial equation, surely?
« Last Edit: 21/05/2016 18:44:39 by timey »
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Offline alancalverd

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Re: Inverted Time Theory: Can these maths work?
« Reply #43 on: 21/05/2016 19:45:10 »
Yes, you 'could' say that the caesium atom blue shifts when moving into a weaker gravitational field, and that every particle with mass will do so.
but I didn't, because it isn't true and is wholly irrelevant. The cesium atom doesn't move.

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A photon does the opposite, it red-shifts into a weaker gravitational field.
And there's the source of your confusion.

Time slows down in a stronger gravitational field. Therefore a photon travelling towards a stronger field will appear to the observer to be blue-shifted, and a photon travelling towards a weaker field will appear to an observer in the weaker field to be red-shfted. 

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The standard second is what we use to measure everything in physics.  It is used to measure the speed of light
both irrelvant and untrue! Back to Wikipedia:

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The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its precise value is 299792458 metres per second (approximately 3.00×108 m/s), since the length of the metre is defined from this constant and the international standard for time.

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If you think for one minute that the 30 or so theoretical physics books that I have read over the last 8 years haven't covered all of that then you haven't read many theoretical physics books.
But why haven't you learned the most basic aspects of relativity from them? 


Quote
I am looking at the possibility that because the photon has no mass that this reduction in frequency in a weaker gravitational field can be indicative of an additional phenomenon of an inverted time dilation, whereby the extra length in wavelength of shifted light is 'inverted time dilation' related, ie: it takes the light a longer 'time' to cover same distance, and the extra length of wavelength is then not an extra length in actual distance.
Why invoke an additional phenomenon, which you have not defined and has no experimental or theoretical basis, to explain something that is  adequately explained without it? Occam would have a fit!


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Offline timey

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Re: Inverted Time Theory: Can these maths work?
« Reply #44 on: 21/05/2016 20:26:58 »
Yes, you 'could' say that the caesium atom blue shifts when moving into a weaker gravitational field, and that every particle with mass will do so.
but I didn't, because it isn't true and is wholly irrelevant. The cesium atom doesn't move.

We have moved a caesium atom into a decreasing gravitational field, and it's frequency increases.  No of course it's not a blueshift, but in that the frequency increases, blue shifted light also increases in frequency, but under the 'almost' opposite circumstances.

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A photon does the opposite, it red-shifts into a weaker gravitational field.
And there's the source of your confusion.[/qoute]

I have no confusion.

Time slows down in a stronger gravitational field. Therefore a photon travelling towards a stronger field will appear to the observer to be blue-shifted, and a photon travelling towards a weaker field will appear to an observer in the weaker field to be red-shfted.

Why is your light clock's frequency's direction of increase and decrease in a gravitational field the opposite of your caesium clocks direction of frequency when exposed to changes in the gravitational field?

Quote
The standard second is what we use to measure everything in physics.  It is used to measure the speed of light
both irrelvant and untrue! Back to Wikipedia:

Quote
The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its precise value is 299792458 metres per second (approximately 3.00×108 m/s), since the length of the metre is defined from this constant and the international standard for time.

What can I say?  I repeat, anything that measures per second, holds a second constant.  I can see the possibility of using other constants related to a second, ie: the speed of light, and via the speed of light, 299 792 458 meters of distance, as a substructure in relation to frequency, joules, ev, when calculating gravity and light.  I'm sorry you can't see it.

Quote
If you think for one minute that the 30 or so theoretical physics books that I have read over the last 8 years haven't covered all of that then you haven't read many theoretical physics books.
But why haven't you learned the most basic aspects of relativity from them?

But I have, the fact is I'm proposing something different.

Quote
I am looking at the possibility that because the photon has no mass that this reduction in frequency in a weaker gravitational field can be indicative of an additional phenomenon of an inverted time dilation, whereby the extra length in wavelength of shifted light is 'inverted time dilation' related, ie: it takes the light a longer 'time' to cover same distance, and the extra length of wavelength is then not an extra length in actual distance.
Why invoke an additional phenomenon, which you have not defined and has no experimental or theoretical basis, to explain something that is  adequately explained without it? Occam would have a fit!
[/quote]

What is adequately explained?

I am merely touching upon an aspect of the Pound Rebka to illuminate the possibility of an inverted time dilation that would mean that redshift is not indicative of expansion of this universe, to result in a theory of everything via a cyclic universe.
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Offline jeffreyH

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Re: Inverted Time Theory: Can these maths work?
« Reply #45 on: 22/05/2016 13:13:26 »
If the universe isn't expanding then the only feasible explanation would be that the force of gravity is diminishing over time. But that defeats the object since less gravity could not stop expansion from accelerating. This is the trap you have fallen into by trying to rewrite established theory.
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Offline timey

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Re: Inverted Time Theory: Can these maths work?
« Reply #46 on: 22/05/2016 14:46:54 »
You are looking at gravity from the point of view of 'everything' expanding outwards from a point.

I am looking at gravity from the point of view that everything is 'slowly' pulling together from a uniform 'sea' of particles, until all that is left is black holes that eventually merge into each other until there is only one left with everything of the universe in it.

This singular black hole, with no equivalent gravitational force acting upon it, ejects the matter of the entire universe (Big Bang) via its accretion disks (inflation period) until it's extinction, leaving a uniform sea of particles that start clumping together.  Distances of space between clumps of mass are created by particles vacating their former positions as mass is pulled together, but the actual spatial dimensions of the universe itself are slowly contracting as mass further clumps.

Viewing the mechanics of the universe under this alternative remit, the alternative perspective should indeed make a significant difference to your outlook on how mass, gravity, and time dilation may interact?
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Offline jeffreyH

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Re: Inverted Time Theory: Can these maths work?
« Reply #47 on: 22/05/2016 15:21:05 »
If you read up on what Beckenstein, Hawking and Susskind are saying then there is an upper limit on the amount of entropy in any region of spacetime. Since there is more spacetime in an expanding universe then entropy may be increasing, static or decreasing depending upon the rate of acceleration of the expansion. Also don't forget the important point that acceleration can be positive, zero or negative.
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Re: Inverted Time Theory: Can these maths work?
« Reply #48 on: 22/05/2016 17:13:12 »
I'm pretty certain that you are referring to the mass temperature conundrum concerning black holes: ie: contrary to usual physics the black holes temperature decreases inversely proportional to an increase in mass.  And the upper mass limit of a black hole being connected to Hawking's radiation and the jetting from accretion disks.  The concept of an upper mass limit for a black hole also finds its argument in that the temperature will drop too low, and that time runs slow for a black hole.  In fact within the GR equations for the space time of a black hole the time aspect of the space time matrix must be swapped with a space aspect to make the maths work.

My theory of inverted time means that the rate of time runs really fast for a black hole.  That the temperature of a black hole is in fact proportional to an increase in mass, but our observation of a black holes rate of time is time frame dependant and proportional to the rate of time we are viewing the black hole from.
Therefore the arguments for an upper  mass limit for black holes is negated.

Sure, black holes will jet and radiate particles, new stars will form from clouds of particles strew across space by black holes, but eventually the balance will tip, the black holes will become dominant, and a new cycle of the universe will be precipitated.

Entropy always increases in my model because when black holes 'scatter' particles into the vastly slower time in space, 'virtual particles' have 'the time' to become real particles and the 'size' of the universe increases.  The conservation of energy law is also upheld by these concepts.
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Re: Inverted Time Theory: Can these maths work?
« Reply #49 on: 22/05/2016 23:07:12 »
The mind is like a parachute. It works best when open.  -- A. Einstein
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