# The Naked Scientists Forum

### Author Topic: Inverted Time Theory: Can these maths work?  (Read 2783 times)

#### alancalverd

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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.

#### timey

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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.

#### alancalverd

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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 »

#### timey

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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)

#### alancalverd

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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.

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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 »

#### timey

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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?

#### alancalverd

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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 »

#### jeffreyH

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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?

#### timey

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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.

#### timey

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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 »

#### 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 »

#### 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.

#### alancalverd

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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 »

#### timey

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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.

#### alancalverd

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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 »

#### timey

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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 »

#### 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".

Quote
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 »

#### 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 »

#### 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.

Quote
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:

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.

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?

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!

#### 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.

Quote
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]

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.

#### 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.

#### 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?

#### 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.

#### timey

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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.

#### timey

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##### Re: Inverted Time Theory: Can these maths work?
« Reply #49 on: 22/05/2016 23:07:12 »
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##### Re: Inverted Time Theory: Can these maths work?
« Reply #49 on: 22/05/2016 23:07:12 »