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The entropy of systems
16 years 10 months ago #18150
by GD
Replied by GD on topic Reply from
Hello Stoat,
The slowing down of the electron means: today there is slightly more available energy in a mass than tomorrow. This is what drives motion; a mass moves, spins, rotates in a direction which requires slightly less energy with time.
I think this is what gravity is: E (mass) tending towards E (energy).
Therefore E=mc^2 is an empirical equation which changes with time (or distance) -"E" in "m" varies continually-.
This concept is probably hard to grasp since the distances involved are astronomical and change is ever so slow at the scale of the universe.
What do you think Stoat? Is the available energy of matter in the universe changing with time? Is this what gravity is?
The slowing down of the electron means: today there is slightly more available energy in a mass than tomorrow. This is what drives motion; a mass moves, spins, rotates in a direction which requires slightly less energy with time.
I think this is what gravity is: E (mass) tending towards E (energy).
Therefore E=mc^2 is an empirical equation which changes with time (or distance) -"E" in "m" varies continually-.
This concept is probably hard to grasp since the distances involved are astronomical and change is ever so slow at the scale of the universe.
What do you think Stoat? Is the available energy of matter in the universe changing with time? Is this what gravity is?
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16 years 10 months ago #3061
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi GD, The graph for entropy is an exponential fall away graph. At some point x on the graph, we effectively have a straight line. At infinity we still have energy. It has fallen but by a ludicrously small amount. So, will we end up with a universe that's full of burnt out cinder stars? The so called heat death of the universe. Now, the fact that this is an appalling concept, a complete waste of space in effect, shouldn't stop us from looking at it.
I think that cinder stars will continue to radiate. They have a degenerate matter heart where the refractive index of space is negative. The stars will disappear like boiled sweets.
The energy available to the universe as ftl gravitational energy is vastly greater than that of electromagnetic energy. It can create matter from the vacuum. How it does so is anybody's guess but the principle doesn't fly in the face of what we think we know about the universe and its laws.
That's not very different from the argument of Fred Hoyle but I'm a lot nicer than he was. Fred came from Yorkshire and I don't live that far from there. Yorkshiremen would have everyone believe that they are the centre of the universe. Fred would have accepted the idea of a big bang (in fact he invented the term) if he could have proven that the centre was Leeds town hall[8D]
I think that cinder stars will continue to radiate. They have a degenerate matter heart where the refractive index of space is negative. The stars will disappear like boiled sweets.
The energy available to the universe as ftl gravitational energy is vastly greater than that of electromagnetic energy. It can create matter from the vacuum. How it does so is anybody's guess but the principle doesn't fly in the face of what we think we know about the universe and its laws.
That's not very different from the argument of Fred Hoyle but I'm a lot nicer than he was. Fred came from Yorkshire and I don't live that far from there. Yorkshiremen would have everyone believe that they are the centre of the universe. Fred would have accepted the idea of a big bang (in fact he invented the term) if he could have proven that the centre was Leeds town hall[8D]
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16 years 10 months ago #20573
by GD
Replied by GD on topic Reply from
Hi Stoat,
Do you think this theory breaks any physical laws as we know as of today?
Those "cinder stars" you are talking about, are they at the center of galaxies, center of galaxy clusters ?
These dead stars; where are they with respect to us (time & distance)? Are we moving towards (or away) from these locations?
-Stoat, thanks for your time for answering this ! We had 45 cm of snow a few weeks ago. Could you imagine all the shoveling we had to do ! I would have preferred a bit less physical work. I consider Gravity as a lazy force. So... it must be all over me!
I will be back only next week, So I wish you and everybody on the board a happy new year!
Do you think this theory breaks any physical laws as we know as of today?
Those "cinder stars" you are talking about, are they at the center of galaxies, center of galaxy clusters ?
These dead stars; where are they with respect to us (time & distance)? Are we moving towards (or away) from these locations?
-Stoat, thanks for your time for answering this ! We had 45 cm of snow a few weeks ago. Could you imagine all the shoveling we had to do ! I would have preferred a bit less physical work. I consider Gravity as a lazy force. So... it must be all over me!
I will be back only next week, So I wish you and everybody on the board a happy new year!
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16 years 9 months ago #19785
by GD
Replied by GD on topic Reply from
Hello Stoat,
Here is a portion of text which I posted on the 1st page of this thread (Aug 2004):
“Another technique used to test bonding theory is infrared spectroscopy. This technique measures the energy of bond vibrations. Double bonds require more energy to vibrate than single bonds (and less than triple bonds). This technique can then experimentally verify the predictions of Lewis and molecular orbital theory.”
Stoat: according to you, what does this mean with respect to the equation: hf=mc^2 ?
Here is a portion of text which I posted on the 1st page of this thread (Aug 2004):
“Another technique used to test bonding theory is infrared spectroscopy. This technique measures the energy of bond vibrations. Double bonds require more energy to vibrate than single bonds (and less than triple bonds). This technique can then experimentally verify the predictions of Lewis and molecular orbital theory.”
Stoat: according to you, what does this mean with respect to the equation: hf=mc^2 ?
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16 years 9 months ago #20593
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi GD, the frequency of an electron let's say is 1.23558E 20 but here we are talking about infrared which is about 10E 13 in the mid range. Infrared photons are going to be about ten million times less energetic than the soft gamma photons equivalent to the mass energy of an electron.
In short, the energy to make a particle is much higher than the energy needed to make a bond between atoms. Now I would argue, that when we talk about e = mc^2 we are only talking about the electromagnetic energy of a particle. The gravitational energy is going to have a frequency which is off the scale. In effect the gravitational mass radius of an electron is going to be about h. This would mean that the electromagnetic part of the electron can be big and "fuzzy" yet it "knows" where its gravitational mass centre is.
Now this has to relate to something like the stability of the benzene ring. Which has six carbon atoms arranged in a ring, with an alternate double and single bond. Normally compounds containing double or triple bonds are unstable i.e. reactive, they want to lose the extra bonds as quickly as they can.
If we "smear" the electromagnetic part of the electrons round the carbon atoms, to distribute the charge uniformly, then we can have the gravitational portion of the electrons arranged in that two one configuration round the ring. They can be right next to each other, because there's no electromagnetic repulsive force near them, it all having been smeared out.
I would argue that this follows from accepting an ftl speed of gravity but if you are after a mainstream explanation of what is happening with multiple bonds, then I think we might have to wait for a chemist to answer this one.
In short, the energy to make a particle is much higher than the energy needed to make a bond between atoms. Now I would argue, that when we talk about e = mc^2 we are only talking about the electromagnetic energy of a particle. The gravitational energy is going to have a frequency which is off the scale. In effect the gravitational mass radius of an electron is going to be about h. This would mean that the electromagnetic part of the electron can be big and "fuzzy" yet it "knows" where its gravitational mass centre is.
Now this has to relate to something like the stability of the benzene ring. Which has six carbon atoms arranged in a ring, with an alternate double and single bond. Normally compounds containing double or triple bonds are unstable i.e. reactive, they want to lose the extra bonds as quickly as they can.
If we "smear" the electromagnetic part of the electrons round the carbon atoms, to distribute the charge uniformly, then we can have the gravitational portion of the electrons arranged in that two one configuration round the ring. They can be right next to each other, because there's no electromagnetic repulsive force near them, it all having been smeared out.
I would argue that this follows from accepting an ftl speed of gravity but if you are after a mainstream explanation of what is happening with multiple bonds, then I think we might have to wait for a chemist to answer this one.
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16 years 9 months ago #20485
by GD
Replied by GD on topic Reply from
Stoat:
"Now I would argue, that when we talk about e = mc^2 we are only talking about the electromagnetic energy of a particle.
....
The gravitational energy is going to have a frequency which is off the scale. In effect the gravitational mass radius of an electron is going to be about h."
GD: Stoat, isn't gravitational energy included in the equation hf=mc^2?
What is missing according to you?
Let's say we take samples of matter (atoms) in the following regions :
1) a) 1 km above Earth's surface (atmosphere).
b) 500 km below Earth's surface.
c) 6000 km below Earth's surface.
2) a) Earth's surface.
b) Venus' surface.
c) Sun's surface.
3) a) Outer edge of our galaxy.
b) midway between edge and center of galaxy.
c) center of galaxy.
Does hf=mc^2 show how the electron behaves in these locations?
Is something constantly changing in this equation?
with distance? With time?
How tightly are electrons bound to protons in each case?
I would also appreciate if you could explain: "zero internal frequency" of an atom which you described earlier. How does an atom get to that stage?
Thanks!
"Now I would argue, that when we talk about e = mc^2 we are only talking about the electromagnetic energy of a particle.
....
The gravitational energy is going to have a frequency which is off the scale. In effect the gravitational mass radius of an electron is going to be about h."
GD: Stoat, isn't gravitational energy included in the equation hf=mc^2?
What is missing according to you?
Let's say we take samples of matter (atoms) in the following regions :
1) a) 1 km above Earth's surface (atmosphere).
b) 500 km below Earth's surface.
c) 6000 km below Earth's surface.
2) a) Earth's surface.
b) Venus' surface.
c) Sun's surface.
3) a) Outer edge of our galaxy.
b) midway between edge and center of galaxy.
c) center of galaxy.
Does hf=mc^2 show how the electron behaves in these locations?
Is something constantly changing in this equation?
with distance? With time?
How tightly are electrons bound to protons in each case?
I would also appreciate if you could explain: "zero internal frequency" of an atom which you described earlier. How does an atom get to that stage?
Thanks!
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