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A different take on gravity
- Larry Burford
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15 years 5 months ago #22891
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
<b>[neilderosa] "I argued ... that we don't see a delay for the effects of gravity, as we do for light propagation, because the gravitons are impinging on the two attracting bodies from opposite directions, and they do so simultaneously. There is no need for them to (hypothetically) travel from one object to the other before the effect can be felt, as is the case with light or EM radiation."</b>
It is a good question.
The similarities and differences between a pulling model of gravitational force and a pushing model of gravitational force are subtle, and profound.
<ul>
<li>These two physical models are as different as night and day.
<li></li>They can both be wrong, but they cannot both be right.
<li></li>And yet - <u>both produce the same math model</u> for describing and predicting, in excrutiating numerical detail, the gravititional force between two masses.
<li></li>Physics and math are NOT synonyms. Please think about this.
</li></ul>
It is not the propagation time of the gravitons from the direction of the source mass that hit the target mass that count. It is the propagation time of the the gravitons from the direction of the source mass that do <u>not</u> hit the target mass that count.
===
Gravitons that are traveling on a trajectory that does not pass through both source mass and target mass cannot carry any information about the source mass to the target mass. The following applies only to gravitons that are on an "interesting" trajectory, a trajectory that does pass through both masses.
<ul>
<li>Most gravitons pass thought both the source mass and the target mass without interaction. They remain on their inital trajectory until they have passed through both masses.
They carry no information about the source mass to the target mass.
</li><li>A few gravitons pass through the source mass without interaction, but do interact with the target mass. They remain on their initial trajectory until they have passed through both masses.
They also carry no information to the target mass about the source mass.
</li><li>A few other gravitons do interact with the source mass, and as a result they do not remain on their initial trajectory and do not interact with the target mass.
These are the gravitons that carry information to the target mass about the source mass. They do so by their absense rather than their pressence. It is the propagation delay of this "gap" in the stream of gravitons that causes the delay in the push between the masses.
And this delay is very much speed dependent.</li></ul>
LB
It is a good question.
The similarities and differences between a pulling model of gravitational force and a pushing model of gravitational force are subtle, and profound.
<ul>
<li>These two physical models are as different as night and day.
<li></li>They can both be wrong, but they cannot both be right.
<li></li>And yet - <u>both produce the same math model</u> for describing and predicting, in excrutiating numerical detail, the gravititional force between two masses.
<li></li>Physics and math are NOT synonyms. Please think about this.
</li></ul>
It is not the propagation time of the gravitons from the direction of the source mass that hit the target mass that count. It is the propagation time of the the gravitons from the direction of the source mass that do <u>not</u> hit the target mass that count.
===
Gravitons that are traveling on a trajectory that does not pass through both source mass and target mass cannot carry any information about the source mass to the target mass. The following applies only to gravitons that are on an "interesting" trajectory, a trajectory that does pass through both masses.
<ul>
<li>Most gravitons pass thought both the source mass and the target mass without interaction. They remain on their inital trajectory until they have passed through both masses.
They carry no information about the source mass to the target mass.
</li><li>A few gravitons pass through the source mass without interaction, but do interact with the target mass. They remain on their initial trajectory until they have passed through both masses.
They also carry no information to the target mass about the source mass.
</li><li>A few other gravitons do interact with the source mass, and as a result they do not remain on their initial trajectory and do not interact with the target mass.
These are the gravitons that carry information to the target mass about the source mass. They do so by their absense rather than their pressence. It is the propagation delay of this "gap" in the stream of gravitons that causes the delay in the push between the masses.
And this delay is very much speed dependent.</li></ul>
LB
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- neilderosa
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15 years 5 months ago #23609
by neilderosa
Replied by neilderosa on topic Reply from Neil DeRosa
Thanks Phil J and Larry for your answers. They are similar to the answer Tom gave me which I reproduce here:
"The Sun and Earth are a finite distance apart. So if either one
causes an effect, such as a graviton shadow, it takes a finite time for that
shadow to reach the other. The two bodies also have a relative motion. So
when the propagation-delayed shadow reaches the other body, it will appear
to come from the retarded position of the first body, not its instantaneous
position.
You can think of a single unbalanced graviton instead of a whole
shadow of missing gravitons. Compare the two cases, one where the Earth is
bathed in a uniform, continuous rain of gravitons from all directions and
feels no net force, and the other where a distant, moving body blocks a
single graviton. That missing graviton will cause the Earth to feel a small
net force toward where the moving body was when it blocked the graviton, but
only after the graviton would have reached Earth if it had not been blocked.
So the propagation time to Earth is indeed relevant." [Tom]
My disagreement with all of you is only tentative since this is a dynamic process and Im not that sure of my facts or opinion, so I'll leave it there. [Neil]
"The Sun and Earth are a finite distance apart. So if either one
causes an effect, such as a graviton shadow, it takes a finite time for that
shadow to reach the other. The two bodies also have a relative motion. So
when the propagation-delayed shadow reaches the other body, it will appear
to come from the retarded position of the first body, not its instantaneous
position.
You can think of a single unbalanced graviton instead of a whole
shadow of missing gravitons. Compare the two cases, one where the Earth is
bathed in a uniform, continuous rain of gravitons from all directions and
feels no net force, and the other where a distant, moving body blocks a
single graviton. That missing graviton will cause the Earth to feel a small
net force toward where the moving body was when it blocked the graviton, but
only after the graviton would have reached Earth if it had not been blocked.
So the propagation time to Earth is indeed relevant." [Tom]
My disagreement with all of you is only tentative since this is a dynamic process and Im not that sure of my facts or opinion, so I'll leave it there. [Neil]
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- JAaronNicholson
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15 years 5 months ago #23795
by JAaronNicholson
Replied by JAaronNicholson on topic Reply from James Nicholson
What if the solar system like all other star systems and galactic systems are closed frames of references? Like in the same Elevator enclosure. Or like inside a car moving along the highway with its windows up?
Then there would be no delay from where the sun "was" having anything to do with the speed of gravity getting to that planet. Like a fly in the car with you will take a certain amount of time to fly across the width of the enclosure, (similar to Light traveling to Earth from the Sun e.g.) but there is no sixty mile an hour wind blowing it (the fly / or the "graviton") specifically back against the back window relative to the other objects in the car?
With a cosmic geometry model of gravity, with gravity just being all the particles coming from all distant star "winds" being balanced by and averaging each other, isn't this what you would get?
Planets would have to be taken as part and parcel of a particular star's <u>major</u> gravity "well" or gravity "sink" as minor or secondary reflections of that star, where the close influence of the star's unique solar wind/s meets in balanced numbers the cosmic particles crossing or coming into the major focal point of the Star.
If the star is "moving", its movement is predicated by the movement of the distant conglomerate of stars and is linked to how they are moving moment to moment as part of that greater Frame of reference.
Thought? Aaron
Then there would be no delay from where the sun "was" having anything to do with the speed of gravity getting to that planet. Like a fly in the car with you will take a certain amount of time to fly across the width of the enclosure, (similar to Light traveling to Earth from the Sun e.g.) but there is no sixty mile an hour wind blowing it (the fly / or the "graviton") specifically back against the back window relative to the other objects in the car?
With a cosmic geometry model of gravity, with gravity just being all the particles coming from all distant star "winds" being balanced by and averaging each other, isn't this what you would get?
Planets would have to be taken as part and parcel of a particular star's <u>major</u> gravity "well" or gravity "sink" as minor or secondary reflections of that star, where the close influence of the star's unique solar wind/s meets in balanced numbers the cosmic particles crossing or coming into the major focal point of the Star.
If the star is "moving", its movement is predicated by the movement of the distant conglomerate of stars and is linked to how they are moving moment to moment as part of that greater Frame of reference.
Thought? Aaron
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15 years 5 months ago #22901
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Sorry for the delay in getting back to this topic. Did I check out the dimensionality of those stated pure numbers? The answer's no, It's a bog standard cosmological argument, if it was good enough for Dirac and Hoyle, then it's good enough for me. Obviously I have some arguments with the idea of the age of the universe in atomic units but then so must Hoyle have had. A for instance would be, do we get a pure dimensionless number if we replace the Hubble constant with a Creil effect equation?
Now if I say that the area of a circle is pi sq metres, I simply divide that by pi to get the radius, no problems there, it's hardly a trick. Now i believe that the speed of gravity is a lot faster than the low figure of twenty billion times c. Using that figure in the famous E =mc^2 equation, the gravitational energy has exactly the same numerical value as the electromagnetic frequency. but hf = E. Now I happen to believe that Planck's constant is a measure of angular momentum and is not in Js. Let's look at the electron for a moment, it's e.m. frequency is about 1.2E 20. So, hf = 1.2E 20 its gravitational energy. Now I want to say that a particle, the electron in this case, in e.m. space, has an angular momentum of h. The mass times the velocity, which is c, times the Compton wavelength, it's radius. However, in gravitational space, it's angular momentum goes up to one. It's spinning faster than light, or it would e if it were in e.m. space. Likewise its mass would be about 1.3 tonnes if it were in e.m. space. However in gravitational space it's mass is simply the electron mass.
We can consider grav space to e informationally smaller than e.m. space, or we can think of it as being much larger. The e.m radius of an electron is going to e the Compton wavelength but the grav radius is going to be much much larger, and that 1.3 tonnes is spread out in a low density ball. Though we can also say that it's inside the Swartzchild radius of the electron. It doesn't matter which way you want to look at it the two spaces are in one to one correspondence, they are the same "size".
The idea of phase change. We write the Lorentzian in terms of a refractive index, then we allow ourselves negative refractive indices, this changes the sign in the Lorentzian to positive at the speed of light. It also lets us have a wavicle that's an f.m wave.
Well, as I don't like the concept of negative mass, a particle entering a region of negative refractive space will will stay as it is, it's frequency will stay the same but its wavelength will change. i also don't like the concept of negative time, so I propose a "chassis" energy level, rather like we use when we show the waveform of a transistor. Stick a nine volt battery on and then say that the wave goes negative but it's not really, it's negative in respect to positive to the chassis voltage of 4.5 So there's no reverse time (I don't think we're anywhere near a theory for a metric of time yet)
I'll eave it there for no but add that I think there's something in Tom's last article in the bulletin. Where he talks about passive mass and acceleration. A particle has a grav mass and an e.m mass, the e.m mass is really going it some at the Compton wavelength!
Now if I say that the area of a circle is pi sq metres, I simply divide that by pi to get the radius, no problems there, it's hardly a trick. Now i believe that the speed of gravity is a lot faster than the low figure of twenty billion times c. Using that figure in the famous E =mc^2 equation, the gravitational energy has exactly the same numerical value as the electromagnetic frequency. but hf = E. Now I happen to believe that Planck's constant is a measure of angular momentum and is not in Js. Let's look at the electron for a moment, it's e.m. frequency is about 1.2E 20. So, hf = 1.2E 20 its gravitational energy. Now I want to say that a particle, the electron in this case, in e.m. space, has an angular momentum of h. The mass times the velocity, which is c, times the Compton wavelength, it's radius. However, in gravitational space, it's angular momentum goes up to one. It's spinning faster than light, or it would e if it were in e.m. space. Likewise its mass would be about 1.3 tonnes if it were in e.m. space. However in gravitational space it's mass is simply the electron mass.
We can consider grav space to e informationally smaller than e.m. space, or we can think of it as being much larger. The e.m radius of an electron is going to e the Compton wavelength but the grav radius is going to be much much larger, and that 1.3 tonnes is spread out in a low density ball. Though we can also say that it's inside the Swartzchild radius of the electron. It doesn't matter which way you want to look at it the two spaces are in one to one correspondence, they are the same "size".
The idea of phase change. We write the Lorentzian in terms of a refractive index, then we allow ourselves negative refractive indices, this changes the sign in the Lorentzian to positive at the speed of light. It also lets us have a wavicle that's an f.m wave.
Well, as I don't like the concept of negative mass, a particle entering a region of negative refractive space will will stay as it is, it's frequency will stay the same but its wavelength will change. i also don't like the concept of negative time, so I propose a "chassis" energy level, rather like we use when we show the waveform of a transistor. Stick a nine volt battery on and then say that the wave goes negative but it's not really, it's negative in respect to positive to the chassis voltage of 4.5 So there's no reverse time (I don't think we're anywhere near a theory for a metric of time yet)
I'll eave it there for no but add that I think there's something in Tom's last article in the bulletin. Where he talks about passive mass and acceleration. A particle has a grav mass and an e.m mass, the e.m mass is really going it some at the Compton wavelength!
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15 years 5 months ago #22902
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
You have me at a disadvantage; I never heard of atomic units until you came along. Apparently,
Hartree atomic units
were developed for their convenience in calculating the structure of atoms. But what does atomic structure have to do with cosmology? Do you like them because Hartree was a Brit? If I decide to specialize in atomic structure, I'll bone up on Hartree atomic units.
Stoat: 05 Jul 2009 : 06:13:42 <blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Obviously I have some arguments with the idea of the age of the universe in atomic units but then so must Hoyle have had. A for instance would be, do we get a pure dimensionless number if we replace the Hubble constant with a Creil effect equation?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">A dimensionless number to represent age??? Age is time, and time is a dimension; how can it be dimensionless?
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Now if I say that the area of a circle is pi sq metres, I simply divide that by pi to get the radius, no problems there, it's hardly a trick.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"> NO! NO! NO! You take the square root of an area to get a length. R = sqrt(A/pi).
Dimensional analysis cannot prove you right, but it can easily prove you wrong.
This is me ignoring you, again, until you start making sense.
Edited to depersonalize. Sorry I lost patience.
Stoat: 05 Jul 2009 : 06:13:42 <blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Obviously I have some arguments with the idea of the age of the universe in atomic units but then so must Hoyle have had. A for instance would be, do we get a pure dimensionless number if we replace the Hubble constant with a Creil effect equation?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">A dimensionless number to represent age??? Age is time, and time is a dimension; how can it be dimensionless?
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Now if I say that the area of a circle is pi sq metres, I simply divide that by pi to get the radius, no problems there, it's hardly a trick.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"> NO! NO! NO! You take the square root of an area to get a length. R = sqrt(A/pi).
Dimensional analysis cannot prove you right, but it can easily prove you wrong.
This is me ignoring you, again, until you start making sense.
Edited to depersonalize. Sorry I lost patience.
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- Larry Burford
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15 years 5 months ago #23612
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
PhilJ,
Looks like you had the right answer all along.
Sorry about that.
Looks like you had the right answer all along.
Sorry about that.
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