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Old 22-October-2006, 05:25 PM
Uclock Uclock is offline
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Default A quantum theory of gravity

Please forgive they way I have presented these equations. I seem to be having trouble entering equations onto the site so I have used a fortran style.

As this part of the website is here to air ATM theories I want to put forward a purely geometric quantum theory of gravity whose value matches that Newtonian gravity within the confines of the solar system where we know Newtonian gravity works well (except for the pioneer anomaly).

To do this I will only use simple geometry but a paradigm shift in thinking is required. I will contend that the presence of matter (therefore energy) generates a super-luminal quantum field around itself and a mass-less particle considered at rest, will generate the largest quantum field.

Mass will have a retarding effect on the generation of an objects quantum field and the base of any measurement will be taken to be that of the quantum field of a mass-less particle considered at rest.

I assert that the radius of a quantum field generated by a mass-less particle in one second will be equal to ‘c’ or 299 792 458 m giving a volume of V = 4/3*pi*c^3 and here V = Phi = 1.129 x 10^26 m^3 s^-1.

To make this ATM a quantum theory I need a time and space quanta which will be 1/Phi = 8.86 x 10^-27 m^-3 s^-1.

Space will be measured by the radius of the quantum field and time will be measured by the volume of the quantum field. This simple contention will separate time and space during measurement but space-time will remain one entity.

As I said earlier mass will retard or distort the quantum field from its true value. I will use the volume of the distortion of the Earth’s quantum field caused by the Earth’s mass to gain a space-time constant ‘STC’. I will use it to calculate the quantum field distortion caused by any mass and give an example of the Sun’s quantum gravity field.

At the Earth surface and according to Newton, gravity has a distortion of g = 9.81 m s-2 so the volume of this distortion caused by the Earth will be Vde = (4/3*pi*(Re + g)^3) – (4/3*pi*Re^3) = 5.01 x 10^15 m^3 where Re = the radius of the Earth.

Now I will simply divide the volume of Earth’s distortion by the mass of the Earth to gain the space-time constant, so STC = Vde/Me = 8.386 x 10^-10 where ME is the mass of the Earth.

Now let’s see if this geometric quantum method can match Newtonian gravity for the gravity field of the Sun.

As stated earlier the base of any measurement will be the flow of a quantum field generated by a mass-less particle considered at rest. Here space is ST = ((3*t*Phi)/(4*pi))^1/3 where ‘t’ is the time input in seconds.

The Sun’s quantum field will be retarded by a small amount because of its mass. The volume of the distortion of the Sun’s quantum field will be Vds = Ms x STC = 1.67 x 10^21 m3 where Ms = Mass of the Sun. Now I need to convert this value to time using the space-time quanta.

The time distortion for the Sun will be tds = (1/Phi)*Vds = 1.48 x 10^-5 s. This means the Sun’s quantum field is retarded by tdS and the flow of the Sun’s quantum field itself will be ts = t – tds.

Using this method the measurement of space and time are completely interchangeable. It will take a mass-less particle considered at rest t = ((4*pi*Rs^3)/(3*Phi)) = 12.49 s, where Rs = radius of the Sun, for the quantum field to reach the surface of the Sun from its centre of mass so ST = (3*t*Phi/4*pi)^1/3 = 6.956 x 10^8 m.

The retardation of the Sun’s own quantum field will be STS = ((3*ts*Phi)/(4*pi)) = 6.955 997 256 79 x 10^8 m giving a quantum distortion value for the Sun’s gravity of qgs = ST – STs = 274.321 m s^-2 if we check this with Newtonian gravity Ngs = G*Ms / Rs^2 = 274.321 m s^-2 and as you can see they are in agreement.

Combining the geometric equations and checking the value of the Sun’s gravity at 1 AU we have ST = (3*(((4/(3*phi))*pi*AU^3)*phi)/(4*pi))^(1/3) = 1.495 978 706 91 x 10^11 and
STs = ((3*(((4/(3*phi))*pi*AU^3)-TdisSun)*phi)/(4*pi))^(1/3) = 1.495 978 706 909 94 x 10^11

Giving a value for the distortion of quantum gravity at 1 AU from the Sun of qgs = ST -STs =5.93 x 10^-3 m s^-2

The value for Newtonian gravity will be Ngs = GMs/AU^2 = 5.93 x 10^-3 m s^-2.

Newtonian gravity and quantum gravity differ by only 9.1 x 10^-9 m s^-2 over this distance.

This method of calculating quantum gravity remains very close to Newtonian gravity for any planet or moon over any distance.

If anyone would like to test this for their self, I will be happy to provide the equations in a form that can be directly entered into the free Console calculator which is accurate to 80sf and can be downloaded at http://ccalc.shanebweb.com/

This paradigm shift in thinking about time and space will take some people by surprise but the mathematics work and work well. I hope you will try it yourself and that it gives everyone food for thought.

Last edited by Uclock; 23-October-2006 at 06:07 PM.. Reason: Typing error
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Old 22-October-2006, 11:33 PM
Nereid Nereid is offline
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How does this ATM idea account for:

a) the advance of the perihelion of Mercury (and Venus, and Eros, and ...)?

b) the deflection of 'light' as it passes by the Sun (and Jupiter)?

c) the data from binary pulsars, esp the double pulsar, and the observations which earned Hulse and Taylor their Nobel prize?

What do you predict the results of Gravity Probe B will be?
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Old 23-October-2006, 04:25 AM
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I assert that the radius of a quantum field generated by a mass-less particle in one second will be equal to ‘c’ or 299 792 458 m giving a volume of V = 4/3*pi*c^3 and here V = Phi = 1.129 x 10^26 m^3 s^-1.

To make this ATM a quantum theory I need a time and space quanta which will be 1/Phi = 1.86 x 10^-27 m^-3 s^-1.
I don't have a lot of time, but this one was glaring. Leaving aside the dimensional values, when I invert the numerical portion of your V (and Phi) value, I get 8.85 X 10-27, not 1.86. Getting back to the dimensional values, how exactly do your dimensional quantities go from m3s-1 to m-3s-1 by inverting them?
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Old 23-October-2006, 01:09 PM
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How does this ATM idea account for:

a) the advance of the perihelion of Mercury (and Venus, and Eros, and ...)?

b) the deflection of 'light' as it passes by the Sun (and Jupiter)?

c) the data from binary pulsars, esp the double pulsar, and the observations which earned Hulse and Taylor their Nobel prize?

What do you predict the results of Gravity Probe B will be?
Hi Nereid

Before I attempt to answer these questions I must reiterate that I am an amateur not a professional. The long climb from conception of the idea to a working mathematical model has been hard and I do not posses all the answers to everything about physics because this concept is so new it will take years of study to find all the answers.
I have great chunks of my education missing so in each area of study I have to start from scratch but the point here is the concept has given me a working model of quantum gravity that matches Newtonian physics within the confines of the solar system.
My main area of study with this concept has been acceleration and inertia and this has been very productive from my point of view.
As far as your questions above are concerned, as you can appreciate, if a super-luminal quantum field is generated by the Sun, or other stars and planets, then the space around them will also be rotating and twisting. Mercury happens to be close to the Sun so this should account for the advance of the perihelion. Gravity is still a distortion of space as it is in GR but space becomes local to the object.
If you take a little time to think about the consequences of the concept you might find it fits the observational evidence very well.
I have been waiting in anticipation for the results of Gravity Probe B since launch. It is a wonderful technical achievement by all those concerned. If I am right and each and every object in this Universe generates a quantum field then I would expect the frame dragging value to be above what is expected for GR.
I wish I had the mathematical skills to give an exact predicted value but I haven’t. If the results for frame dragging are above what is expected by GR then perhaps trained physicists might engage with the concept and study the consequences for their self.
As an amateur with very limited mathematical skills I can only take this concept so far.
I would love a professional to see how far they can take this but up until now, and as far as I am aware, no other person has come up with a theory of quantum gravity that matches Newton gravity within the confines of the solar system so I will continue to study this concept on my own with the limited tools I have available, which includes my incomplete education.
Thank you for at least looking at it.
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Old 23-October-2006, 01:29 PM
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Hi Tensor

I could describe this as being a bit picky but unfortunately I have used = instead of  because I seem to difficulty in entering symbols on this site.
One day I might find out how to do it right! As far as your assumption that the math is wrong I can assure you that phi = 4/3*pi*c^3 = 1.128 627 734 036 37^26 so 1/phi = 8.860 317 444 297 14^-27. Try it yourself.

You have a point with the dimensional values and this only comes down to my lack of mathematical skills. But the concept is original and it has provided a quantum theory of gravity that matches or very closely matches Newtonian gravity and I defy anyone who says different. You may be a professional, I don’t know, but I would urge you to take a time to run the math properly.

Thank you also for taking a quick look.
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Old 23-October-2006, 04:05 PM
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To nitpick a little further, you define the volume V as that which a photon can "create" in one second, and therefore your volume has to be dimension m3 and not m3s-1.

secondly
Quote:
At the Earth surface and according to Newton, gravity has a distortion of g = 9.81 m s-2 so the volume of this distortion caused by the Earth will be Vde = (4/3*pi*(Re + g)^3) – (4/3*pi*Re^3) = 5.01 x 10^15 m^3 where Re = the radius of the Earth.
Well, here you are adding the Earths radius Re (unit m) to the Earth's gravitational accelerations g (unit m s-2) which you cannot do.
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Old 23-October-2006, 04:36 PM
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I seem to be having trouble entering equations onto the site.
The remedy for that would be Tex or MathML (preferably) support.
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Old 23-October-2006, 04:39 PM
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The remedy for that would be Tex or MathML (preferably) support.
I am Nereid and I support this message*.

*If you have not been in the USA around the time of an election, this joke will go over your head (I laughed myself silly the first time I encountered one of these, on a trip to the US, years ago).
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Old 23-October-2006, 04:44 PM
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Hi Tensor

I could describe this as being a bit picky but unfortunately I have used = instead of  because I seem to difficulty in entering symbols on this site.
One day I might find out how to do it right! As far as your assumption that the math is wrong I can assure you that phi = 4/3*pi*c^3 = 1.128 627 734 036 37^26 so 1/phi = 8.860 317 444 297 14^-27. Try it yourself.
Hey Uclock.

I did, that's why I pointed it out. In your post, you gave the numerical value of the iverted phi as 1.129. I pointed out it should be 8.85. Now you are saying it's 8386 (which is close enough to the value I provided not to matter). So which one is it? Your original value, or the one you presented above?

Quote:
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You have a point with the dimensional values and this only comes down to my lack of mathematical skills. But the concept is original and it has provided a quantum theory of gravity that matches or very closely matches Newtonian gravity and I defy anyone who says different. You may be a professional, I don’t know, but I would urge you to take a time to run the math properly.

Thank you also for taking a quick look.
As tusenfem points out, there are other problems with dimensions. If you had done a dimensional analysis on the idea, you would have realize the the units are completely off and lead to operations that are not allowed. Unless, of course, you are redefining the units, but then you should mention that and define all the units before using them in the equations. As it stands now, the idea does nothing more than provide some numbers that come close, arrived at through forbidden operations. Nothing more than a coincidence.


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Thank you also for taking a quick look.
Not a problem. Unfortunately, that is all I have time for at this time.
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Old 23-October-2006, 04:44 PM
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The remedy for that would be Tex or MathML (preferably) support.
I have always found it ironic that the World Wide Web, which owes its origin to CERN, a scientific establishment, has never had adequate support for display and typesetting of scientific equations. I guess the original goal was to have a text-only hypertext front-end for serving up TeX files of the papers where the math really was!
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Old 23-October-2006, 05:50 PM
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To nitpick a little further, you define the volume V as that which a photon can "create" in one second, and therefore your volume has to be dimension m3 and not m3s-1.

secondly

Quote:
At the Earth surface and according to Newton, gravity has a distortion of g = 9.81 m s-2 so the volume of this distortion caused by the Earth will be Vde = (4/3*pi*(Re + g)^3) – (4/3*pi*Re^3) = 5.01 x 10^15 m^3 where Re = the radius of the Earth.

Well, here you are adding the Earths radius Re (unit m) to the Earth's gravitational accelerations g (unit m s-2) which you cannot do.
It is the understanding of time and space according to known physics that is the problem, not the way I have presented it. Time is measured by seconds and space is measured by metres. In this concept the volume of time generated creates space. Space is not a fixed three dimensional metric as it is in GR, here it is a multitude of flowing dynamic fields.
An amount of space is created by every object in a given period of time that is why both m and seconds are used. In the case a mass-less particle considered at rest, it generates a given volume (m^3) in one second.
As stated in the concept, time is measured by the fixed flow of the volume and space is measured by the radius of that volume. Time and space are the same entity so this idea that they are completely separate is nonsense in my opinion.
By all means stick with what you know but no other theory has ever explained the flow of time, unless you can point me to one.
It is true that three dimensional space is measured in m^3 but according to this concept, space in created in a given amount of time that is why I have used m^3 s^-1.
I have divided space by a given amount of time, in this case one second.
If space and time are a dynamic field as the concept suggest and not fixed then logically you can add subtract, divide and multiply both in m and s. You need to grasp the concept and then maybe you will see where I’m coming from.
I agree in known physics you cannot interchange these dimensions but this theory is not within known physics yet still it can match Newtonian gravity.
Acceleration is just a distortion of space-time in this concept and I have shown it is possible to measure this distortion by volume and use it to retard the quantum field. I will ask you to try to understand the concept before criticising the way I have used the units, if that is ok with you.
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Old 23-October-2006, 06:26 PM
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I
Quote:
did, that's why I pointed it out. In your post, you gave the numerical value of the iverted phi as 1.129. I pointed out it should be 8.85. Now you are saying it's 8386 (which is close enough to the value I provided not to matter). So which one is it? Your original value, or the one you presented above?
No I did not give the inverted value of Phi as 1.129^26 nor have I said the value is 8386 (whatever that is).

I gave the inverted value of Phi as 1/Phi = 1.86^-27 when it should read 8.86^-27 (2dp) it is just a typing error, just as you have missed the ‘n’ out of ‘inverted’. Please don’t nitpick, it does not get us anywhere.
In my original post I rounded up the inverted Phi value to 2dp. To clarify, again, in this concept Phi = 4/3*pi*c^3 = 1.12862773403637e26, rounded to 1.129e26 (3dp) so the inverse of this is 1/Phi = 8.86031744429714e-27 when rounded up, 8.86e-27 (2dp). I hope that clears that up.

Quote:
As tusenfem points out, there are other problems with dimensions. If you had done a dimensional analysis on the idea, you would have realize the the units are completely off and lead to operations that are not allowed. Unless, of course, you are redefining the units, but then you should mention that and define all the units before using them in the equations.
When dealing with time and space we are dealing with the same entity, unless you are saying time and space are not the same entity.
There is no need to redefine the units. I just use them in an unconventional manor because this concept is unconventional.

Quote:
As it stands now, the idea does nothing more than provide some numbers that come close, arrived at through forbidden operations. Nothing more than a coincidence.
Is that right, so all I have done is provide some numbers that come close, a pure coincidence, yet it can match Newtonian gravity. Because I have used an unconventional approach you decide it can be dismissed without further investigation. Very scientific!

I don’t want this to turn into a slanging match because that’s pointless.
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Old 23-October-2006, 07:06 PM
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No I did not give the inverted value of Phi as 1.129^26 nor have I said the value is 8386 (whatever that is).

I gave the inverted value of Phi as 1/Phi = 1.86^-27 when it should read 8.86^-27 (2dp) it is just a typing error, just as you have missed the ‘n’ out of ‘inverted’. Please don’t nitpick, it does not get us anywhere.
In my original post I rounded up the inverted Phi value to 2dp. To clarify, again, in this concept Phi = 4/3*pi*c^3 = 1.12862773403637e26, rounded to 1.129e26 (3dp) so the inverse of this is 1/Phi = 8.86031744429714e-27 when rounded up, 8.86e-27 (2dp). I hope that clears that up.
Yes it does, I compounded the error by misquoting your orginal 1.86 in my second post. My apologies for my misquote.

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When dealing with time and space we are dealing with the same entity, unless you are saying time and space are not the same entity.
They are a combined entity, that are measured with different dimensions. When doing the the different calculations, the dimensions have to come out correctly.

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There is no need to redefine the units. I just use them in an unconventional manor because this concept is unconventional.
This is just flat our wrong. If you are going to use the dimensions unconventionally, you have redefinded them. Which you have to explain when you are presenting your idea. And even then, your dimensions have to come out correctly when you do your calculations.

Quote:
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Is that right, so all I have done is provide some numbers that come close, a pure coincidence, yet it can match Newtonian gravity. Because I have used an unconventional approach you decide it can be dismissed without further investigation. Very scientific!
I suggest your reread my first few words:

Quote:
Originally Posted by Tensor
As it stands now....
So again, as it stands now, with your "unconventional" usage, the dimensions in your calculations do not work. As a result, your calculations are wrong. With the wrong calculations all you have a coincidence. I've looked at one part of your idea, tusenfem look at another. We both have found dimensional errors in your calculations. This shows your calculations to be wrong. As a result, your idea isn't viable as it is. I did not dismiss it out of hand. There were mathematical and logical reasons for dismissing it. And, yes, I consider that scientific. Much more so than throwing equations together, with wrong dimensions and then getting upset when those calculations are show to be wrong.

Quote:
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I don’t want this to turn into a slanging match because that’s pointless.
There is nothing to slang. Either your equations are correct, or they are wrong. If they are wrong, there is nothing to your idea, no matter how close you get to Newtonian gravity. Your equations, as they stand right now, are wrong.
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Old 23-October-2006, 11:25 PM
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The equations are not wrong. A given volume of quantum field (time) m^3 is generated each second, (per second s^-1). What other way can you put it?

Time and space are the same entity therefore all calculations should be completely interchangeable and with this concept they are completely interchangeable.

The concept is not wrong and the equations are not wrong. Gravity, acceleration and inertia are all the same in this concept there is no difference between them, they are all a distortion of space-time.

Perhaps it is best if I use the example of acceleration in free space, well away from any gravitational field.

When you accelerate at 1g in free space you mimic the gravitational field of the Earth at the surface of the Earth, you still feel a force, the same force as you would feel here on the surface of the Earth. According to known physics the illusive Higgs field is supposed to be responsible but this is not the case.

If an astronaut was in deep space well away from the influence of gravity and his space craft began to accelerate at 1g he has velocity within his own quantum field, he could walk around on the rear bulkhead just as he walks on the surface of the Earth there is no difference.

In this concept his quantum field has collapsed by 9.81 m and the centre of his quantum field has moved 9.81 metres below him or in this case 9.81 m from his centre of mass and behind the spacecraft. When his spacecraft ceases acceleration his quantum field is realigned with his centre of mass and his field re-expands to its natural size and because he has no velocity within his own quantum field he will float around weightless.

We all generate our own quantum field, you, me, the cup sitting on your desk. Every object or particle in this Universe generates a quantum field.
The way acceleration is explained in our text books leaves a lot to be desired and here is a typical example of acceleration due to gravity taken from such a text book.

In the absence of air resistance, an object falling freely under the influence of the Earth’s gravity, close to the surface of the Earth, experiences an acceleration of about 9.81 m s-2 in the downward direction.’

This description is really a description of the rate of increasing velocity towards the Earth (as viewed from an observer’s inertial frame of reference), second by second and is described as acceleration towards the Earth. But we are held still by the surface of the Earth and have no observable velocity. However, according to the concept we do have velocity within our own quantum space-time field.

The retarded quantum field of the Earth or the Earth’s distorted quantum field, which ever way you like, causes our own quantum field to be pulled off alignment from our centre of mass and collapses by 9.81 m for the first second of its generation. It is exactly the same as acceleration in free space. In both cases we have velocity within our own quantum field. Time and space are the same entity and by keeping these dimensions separate you will never understand space-time.

I still maintain the equations are correct and are valid for any object with mass.
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Old 24-October-2006, 04:21 AM
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I still maintain the equations are correct and are valid for any object with mass.
You can maintain this all you want, but until the dimensionallity is fixed, by definitions or correcting the equations so the dimnesionallity works out, you're going to have a hard time convincing anyone who knows even the most elementary dimensional analysis, that the equations are correct. The fact that you also can't show how your idea can lead to the other predictions Nereid asked about will continue to be a weak point. I'm puzzled on how you expect us to simply accept assertations, about the equations, when you admitted that you don't have the math skills to show how the dimensionallity is correct.

If your only purpose in presenting it was to say here it is, it's not wrong, deal with it. No matter what anyone shows, you wouldn't acknowledge any problems anyway. So, there really isn't any reason to discuss it. There are quite a few threads here with that kind of presented idea, some of them with much more involved math (which didn't make them any more correct).

If you were actually looking for any input to correct possible problems, simply declaring that they are correct leads to nothing more than the first case, so again, there is no reason to discuss.

If your honestly looking for input to help you correct problems, people here will be more than happy to help. But, you have to acknowledge the errors found, provide possible fixes and only then is there really anything to discuss.

With your last post, simply stating that the equations are correct, without acknowledging threre are problems, while expecting us to accept your unconventional usage without definitions, your posts have entered the first or second case. Since, in those cases, there really isn't anything left to discuss, I will simply bow out here.
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Old 24-October-2006, 07:28 AM
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It is the understanding of time and space according to known physics that is the problem, not the way I have presented it. Time is measured by seconds and space is measured by metres. In this concept the volume of time generated creates space. Space is not a fixed three dimensional metric as it is in GR, here it is a multitude of flowing dynamic fields.
An amount of space is created by every object in a given period of time that is why both m and seconds are used. In the case a mass-less particle considered at rest, it generates a given volume (m^3) in one second.
As stated in the concept, time is measured by the fixed flow of the volume and space is measured by the radius of that volume. Time and space are the same entity so this idea that they are completely separate is nonsense in my opinion.
Sorry to not agree with you here, but if you use a reference volume here for your calculations then you have specifically used a 1 second time interval which means that your volume has a dimension m3 and the per-second disappears. Otherwise you will be trying to measure with a ever changing reference volume measured in m3s-1, which naturally can be done, but then you will have to tell us how you define the radius of the Earth in something per second squared, if you want to add it to the gravitational acceleration.

Otherwise, you may try to start off clearly with your paper, and define some of the ways in which "normal" units transform into your "STC" units, otherwise there is nothing to discuss, as you can come up with any explanation. So, reconcider your premises and explain them clearly to us, then a useful discussion can start.
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Old 24-October-2006, 08:22 PM
Uclock Uclock is offline
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Otherwise you will be trying to measure with a ever changing reference volume measured in m3s-1, which naturally can be done, but then you will have to tell us how you define the radius of the Earth in something per second squared, if you want to add it to the gravitational acceleration.
Ok, let’s start with the basic premiss. My contention is that a super-luminal quantum field radiates from every object or particle in this Universe and has done so since the big bang. The field has a constant volume per second but it is retarded according to its mass so it has a constant volume flowing each and every second which is why I use per second s^-1. This volume flows each and every second, just as a when you move at a fixed velocity you do so each and every second, (per second s^-1).

Quote:
Otherwise, you may try to start off clearly with your paper, and define some of the ways in which "normal" units transform into your "STC" units, otherwise there is nothing to discuss, as you can come up with any explanation. So, reconcider your premises and explain them clearly to us, then a useful discussion can start.
I have used the volume of quantum field radiated by the Earth to calculate the space-time constant (STC). I also maintain that acceleration and gravity are exactly the same entity therefore at the surface of the Earth the volume of quantum field is retarded by a given amount, in the case of the earth this retardation has a radius of g = 9.81 m. This retardation or distortion has a fixed amount of volume which can be calculated.

Perhaps a simple analogy such as the flow of water from a pipe might help here. Imagine two separate water pipes ‘A’ and ‘B’ with a balloon stuck over the end of each pipe and both have a valve which can be either completely open or completely closed. For the start of this analogy both are turned off. You turn on valve ‘A’ and a fraction of a second later you turn on valve ‘B’ and the balloons begin to fill up and expand rapidly. Just for the purpose of measuring the difference in volume you turn both off together after a given amount of time. Now you need to find the difference in volume of each balloon because balloon ‘A’ started to fill before balloon ‘B’.

In the case of the analogy you measure the circumference of each balloon to find the radius of balloons ‘A’ and ‘B’. Once we know the radius of each balloon we need to find the difference in the volume between ‘A’ and ‘B’. To find this difference you subtract the volume of ‘B’ from the volume of ‘A’ and this will give you the difference in the volume between balloon ‘A’ and balloon ‘B’.

In the case of the space-time quantum fields ‘A’ would be the volume of quantum field for a mass-less particle considered at rest and ‘B’ is the volume of quantum field generated by the Earth, g is just the difference in the radius of the quantum field between the Earth and a mass-less particle considered at rest so the volume of the distortion (difference in volumes) Vdiff = 4/3*pi*Ra^3 – 4/3*pi*Rb^3 will give the difference in the volume between the two balloons and in the case of the quantum fields Vdis = 4/3*pi*(RE + g)^3 – 4/3*pi*RE^3 gives the volume of the distortion. I hope that’s not too long winded and you understand it.

A mass-less particle considered at rest always generates the maximum volume of quantum space-time field possible, in this Universe, each and every second which is Phi = 4/3*pi*c^3. The Earth has its volume of quantum space-time field retarded because of its mass.

In the case of the analogy, if you work out the difference in the volumes you can work backwards and calculate the difference between the times when valve ‘A’ and valve ‘B’ were turned on.

In this concept time and space are completely interchangeable, the volume of field will tell you how much time has passed and the radius of the field will tell you how much space has been generated.

If space was just a three dimensional metric then I can understand why you cannot mix the dimensions but in this concept space is being generated all the time and it is made up of billions and billions of quantum fields and the size of these fields are constantly increasing.

One thing about this concept that everyone seems to of overlooked is; if I am right and each and every object generates a fixed volume of space-time quantum field each and every second, then gravity is not a force that works over an infinite distance. If the age of the Universe is about 13.75 billion years then the largest size of a quantum field (or gravitational field) can be is around 1500 AU. Now think about the consequence of such a possibility.
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Old 25-October-2006, 10:18 AM
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you still do not explain how you can add Re and g, which in normal live have units m and m/s2. Now you suddenly say that g has unit m, how, why? Did you devide "normal" g through your volume-per-second? Nope, then you would have as unit for g m-2.

I do understand that you can define your volume-per-second with unit m3s-1. Sure, you have an ever expanding "ball", but if you then "scale" everything with this volume-per-second, you will find that the Earth's volume is shrinking with time.

Also, should the volume of your quantum-wave-field not have a volume:
VQWF = 4/3 * pi * (c * t)3?
This does not mean that every second VQWF increases by 4/3 * pi * c3 m3s-1 but 4/3 * pi * c3 m3s-3, unless I do not understand your physics and mathematics.
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Old 25-October-2006, 07:30 PM
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Hi Tusenfem thank for the reply.

Quote:
you still do not explain how you can add Re and g, which in normal live have units m and m/s2. Now you suddenly say that g has unit m, how, why? Did you devide "normal" g through your volume-per-second? Nope, then you would have as unit for g m-2.
The quantum space-time field is dynamic and constantly flowing. To work out the volume of the distorted quantum field I have taken a snapshot, a freeze-frame if you like, which will mean the measurement is in m only. If you could see the quantum field of a mass-less particle considered at rest and take a photo graph of it at about 9.65^-6 sec as it expands from the particle, then take a picture of the quantum field of the Earth (without the Earth mucking up the picture) when you overlay the pictures you will find the difference in the radius between the two photos would be 9.81 m. It is a still snapshot of dynamic quantum fields which is being measured in m.


Quote:
I do understand that you can define your volume-per-second with unit m3s-1. Sure, you have an ever expanding "ball", but if you then "scale" everything with this volume-per-second, you will find that the Earth's volume is shrinking with time.
It is not the volume of the Earth that I am measuring, it is the volume of the Earths quantum field which is generated at the Earth’s centre of mass.
The field moves through the Earth and about 9.56^-6 seconds later it reaches the surface of the Earth, I hope that clears it up.

Quote:
Also, should the volume of your quantum-wave-field not have a volume:
VQWF = 4/3 * pi * (c * t)3?
This does not mean that every second VQWF increases by 4/3 * pi * c3 m3s-1 but 4/3 * pi * c3 m3s-3, unless I do not understand your physics and mathematics.
Most definitely not but it can be written as V = (4/3*pi*c^3)*t but in 1 second the natural volume of a quantum field generated by a mass-less particle considered at rest is Phi = 4/3*pi*c^3.It is the base measurement from where all distortions of time and space can be measured also the field is not a wave or anything akin to wave.
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Old 25-October-2006, 07:59 PM
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One thing about this concept that everyone seems to of overlooked is; if I am right and each and every object generates a fixed volume of space-time quantum field each and every second, then gravity is not a force that works over an infinite distance. If the age of the Universe is about 13.75 billion years then the largest size of a quantum field (or gravitational field) can be is around 1500 AU. Now think about the consequence of such a possibility.
What about the galaxy? If the largest size that a gravitational field can be is 1500AU, how does the galaxy hold itself together? How do you account for our home galaxies globular clusters, which are gravitationally bound to the galaxy, even though the furthest is almost 50 kpc (about 10 billion AU) away?
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Old 25-October-2006, 11:20 PM
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What about the galaxy? If the largest size that a gravitational field can be is 1500AU, how does the galaxy hold itself together? How do you account for our home galaxies globular clusters, which are gravitationally bound to the galaxy, even though the furthest is almost 50 kpc (about 10 billion AU) away?
Hi Tensor

A very fair question and I hope I can answer it to your satisfaction.

The reason I asked you to think about the consequences of the gravitational field not being infinite is because it explains the almost fixed rotation of stars inside spiral galaxies.

Remember each and every object or particle in this Universe generates a quantum field according to this concept. Each object can only be gravitationally influenced where these fields overlap, a bit like watching a small cluster of bubbles rotating in a sink full of water. The bubbles rotate as one because each bubble is connected to the one next to it. A bubble on one side of the cluster is not directly connected to a bubble on the other side but if you try to slowly pull at one of the bubbles, all will move so all are connected indirectly.

It is the same inside our galaxy and every other galaxy. Each star or object is directly connected to the quantum field of the objects next to it and indirectly connected to every other object in the galaxy by a chain of connections. The beauty of this is that dark matter is not required to hold the galaxy together. Even the IGM is full of primordial gas so this links the galaxy clusters together.

In the case of our own globular clusters the effect is the same. Each star is linked to every other star by the overlapping quantum fields of dust, gas and larger objects although they are not directly connected, what happens on one side of the cluster will have an effect (albeit a small one) on the other side of the cluster, the same goes for all objects in this Universe.

Earlier this year I was reading an article in the Astronomy Now magazine about Sagittarius A*, the black hole believed to be at the centre of our galaxy. Very near to the black hole there is a small cluster of stars that are rotating around the black hole and astronomers are at a loss to explain why the cluster is not being ripped apart by the gravitational forces of Sagittarius A*. The reason is because the black hole only has an indirect affect on the cluster, that’s why it is rotating around Sagittarius A* as a cluster.

There is no real direct evidence for dark matter and I believe dark matter does not exist, although you may not agree with me.

I hope that explains it for you.

Last edited by Uclock; 25-October-2006 at 11:21 PM.. Reason: typing error
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Old 26-October-2006, 12:04 AM
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[snip]

There is no real direct evidence for dark matter and I believe dark matter does not exist, although you may not agree with me.

I hope that explains it for you.
Please provide, in a quantitative, consistent way, an explanation* of the good astronomical observations that are used as the basis for concluding that there is (lots of) dark matter in the universe.

Specifically, please account for such data for rich clusters. You may, as you wish, start with Zwicky's 1929 paper.

*or, if you prefer, a quantitative account of ...
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Old 26-October-2006, 03:41 AM
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It is the same inside our galaxy and every other galaxy. Each star or object is directly connected to the quantum field of the objects next to it and indirectly connected to every other object in the galaxy by a chain of connections.
OK, lets stay in our galaxy. The closest star to our sun is Proxima Centauri. It is 4.22 light years away. That equals ~266,700 AU.

To quote you:

Quote:
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then the largest size of a quantum field (or gravitational field) can be is around 1500 AU.
That means, even if you allow the largest gravitational field your idea allows between the sun and Proxima, it can only reach 3000 AU (1500 away from the Sun and 1500 AU away from Proxima), which is ~263,000 AU short of the two having any kind of gravitational effect on each other. And that is for the closest star. Heck, by the reasoning of your idea, Proxima is even too far away from Alpha and Beta Centauri to have any kind of gravitational effect with them(about 13,000 AU and I'm not even talking about an orbit, just any kind of gravitational interaction)

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There is no real direct evidence for dark matter and I believe dark matter does not exist, although you may not agree with me.
I don't agree with you, but what I'm asking has nothing to do with dark matter.

Quote:
Originally Posted by Uclock View Post
I hope that explains it for you.
Well, as you can see, not really. If even the closest star to the Sun, by your idea, can't have a gravitational interaction with the Sun, how is the galaxy holding on to the sun? Much less, hold on to stars even further away than the sun from the galaxy's center.
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Old 26-October-2006, 01:35 PM
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Thanks for your input Nereid

In a quote from the Van Den Bergh paper you cited as proof of dark matter; “These observations clearly showed that the rotation curve of M31 did not exhibit a Keplerian drop-off. In fact, its rotational velocity remained constant over radial distance of 16-30 kpc.These observations indicated that the mass in the outer regions of the Andromeda galaxy increased with galactocentric distances, even though the optical luminosity of M31 did not.

The rotational velocity will remain constant if the gravitational field is localised to each and every object as I stated in the reply to Tensor.

In the paper Zwicky is quoted as saying “Present estimates of the the masses of nebulae are based on observations of the luminosities and internal rotations of nebulae. It is shown that both these methods are unreliable; that from the observed luminosities of extragalactic systems only lower limits for the values of their masses can be obtained.”

Well they would be wouldn’t they if you view gravity as working over an infinite distance. The problem appears to be that the mass to light ratio does not work unless you include dark matter but all of this is based around the force of gravity working over an infinite distance. If the gravitation field is localised it will explain all these observations and for this reason I cite my answer above in my reply to Tensor.

Last edited by Uclock; 26-October-2006 at 01:35 PM.. Reason: Typing error
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Old 26-October-2006, 01:47 PM
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Hi Tensor

Quote:
That means, even if you allow the largest gravitational field your idea allows between the sun and Proxima, it can only reach 3000 AU (1500 away from the Sun and 1500 AU away from Proxima), which is ~263,000 AU short of the two having any kind of gravitational effect on each other. And that is for the closest star. Heck, by the reasoning of your idea, Proxima is even too far away from Alpha and Beta Centauri to have any kind of gravitational effect with them(about 13,000 AU and I'm not even talking about an orbit, just any kind of gravitational interaction)
If it was only each star that generated a quantum field then I would agree with you but space is a thick soup of quantum fields all overlapping. Each and every particle of dust and gas is generating a quantum field. This soup of fields will move and rotate as one so our Sun has an indirect affect on Proxima via the quantum fields of all the dust, gas, comets meteors and every other object in between our Sun and Proxima.
The galaxy has a thick soup of quantum fields and it is not just stars that generate quantum fields, all objects do. Now do you understand?
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Old 26-October-2006, 01:56 PM
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Taking these last two posts (#24 and #25) together: as there are huge numbers of high quality observations showing that there is ordinary matter (ions, atoms, electrons, molecules, dust, stars, ....) between galaxies (and even between clusters of galaxies), where does gravity 'end', in the Uclock ATM idea?
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Old 26-October-2006, 02:55 PM
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Taking these last two posts (#24 and #25) together: as there are huge numbers of high quality observations showing that there is ordinary matter (ions, atoms, electrons, molecules, dust, stars, ....) between galaxies (and even between clusters of galaxies), where does gravity 'end', in the Uclock ATM idea?

In this concept gravity is localised so the gravitational field of each and every object extends out to about 1500 AU if the Universe is around 13.75 billion years old.

This ordinary matter, (ions, atoms, electrons, molecules, dust, stars, ....) between galaxies makes the IGM like a soup of fields which extends all over the Universe. In fact you could describe our Universe as a ‘bubbleverse’ but I think that destroys the beauty of the Universe in which we are privileged to exist in for a finite amount of time.

Each object, whether that is a molecule or a star, will only feel the gravitational influence of the objects next to it (wthin 1500 AU) outside of this there is no ‘direct’ gravitational effect on the object but each and every object is linked by a chain of quantum gravity fields. So although quantum gravity fields are localised in effect gravity will work over cosmological distances which is why we see clusters of galaxies reacting as they do.
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Old 26-October-2006, 03:17 PM
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In this concept gravity is localised so the gravitational field of each and every object extends out to about 1500 AU if the Universe is around 13.75 billion years old.

This ordinary matter, (ions, atoms, electrons, molecules, dust, stars, ....) between galaxies makes the IGM like a soup of fields which extends all over the Universe. In fact you could describe our Universe as a ‘bubbleverse’ but I think that destroys the beauty of the Universe in which we are privileged to exist in for a finite amount of time.

Each object, whether that is a molecule or a star, will only feel the gravitational influence of the objects next to it (wthin 1500 AU) outside of this there is no ‘direct’ gravitational effect on the object but each and every object is linked by a chain of quantum gravity fields. So although quantum gravity fields are localised in effect gravity will work over cosmological distances which is why we see clusters of galaxies reacting as they do.
Which brings us back to the question(s) re dark matter: how does this Uclock ATM account for the observations ... quantitatively?

Earlier you wrote:
Quote:
If the gravitation field is localised it will explain all these observations and for this reason I cite my answer above in my reply to Tensor.
However, "this reason" (cited) is not quantitative.

Please provide a quantitative explanation.
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Old 26-October-2006, 03:50 PM
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I am working towards a mathematical solution but at the moment, with my current mathematical skills, it will be some time before I can give you a quantitative explanation. It is only a matter of time and a good deal of work, and hopefully with some help, I will eventually be able to model the movement of galaxy clusters using this concept. That is if someone doesn’t beat me to it.

I know the concept is radical but it will pay dividends in the end.

This concept has modelled quantum gravity in a way that matches Newtonian gravity within the confines of the solar system, which is a start. I do not at present have the facilities I need to model all of the physical world but give it time and you will see that I am onto something with this concept.
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Old 26-October-2006, 05:05 PM
Nereid Nereid is offline
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I am working towards a mathematical solution but at the moment, with my current mathematical skills, it will be some time before I can give you a quantitative explanation. It is only a matter of time and a good deal of work, and hopefully with some help, I will eventually be able to model the movement of galaxy clusters using this concept. That is if someone doesn’t beat me to it.

I know the concept is radical but it will pay dividends in the end.

This concept has modelled quantum gravity in a way that matches Newtonian gravity within the confines of the solar system, which is a start. I do not at present have the facilities I need to model all of the physical world but give it time and you will see that I am onto something with this concept.
So, from an earlier post of yours:
Quote:
I believe dark matter does not exist
what is the objective basis for your belief? Or, perhaps I should ask: do you have an objective basis for this belief?

To put this into context: without an objective way to select between statements such as the one I quoted, why should anyone choose any one of them? To turn up the contrast, why choose yours over mine (that it's invisible pink fairies monkeying with motes on the transparent sphere that's just a few thousand au away)?
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