Mass is supposed to be dimensionless. It has nothing to do with volume. The job of density is to relate mass and volume to one another. It is not the responsibility of mass to do so on it's own.
F=ma is structured as it is because, experimentally, the acceleration of an object is uniquely determined by only the force being applied to it and it's mass, independent of volume.
F=ma is structured as it is because, experimentally, the acceleration of an object is uniquely determined by only the force being applied to it and it's mass, independent of volume.
If I have two spheres of differing density but identical mass, and applied an identical force, F=ma tells us that they'll both experience identical accelerations, and this is exactly what we see in experiment.
Your equation factors in volume. So F=da tells us that if one instead has twice the density of the other, it will experience half the acceleration. So according to F=da, if the two spheres weighed exactly the same, but one was twice as dense, meaning half the size, it would accelerate at half the rate. Or put another way, F=da means I can squish an object, without changing it's weight, and it will accelerate more slowly under equal load. Is this the behavior you're predicting? Have you been able to verify it?
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So a balloon filled with 1kg of helium and a balloon filled with 1kg of lead will have experience identical accelerations. That does not occur.
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Or do you have another definition of density other than d = m / v? For instance, you say: "Density is the vibration intensity within a volume in relation to the density of the surrounding medium." It's relative? So the density of an object changes when it's surrounding density changes?
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I do have an idea of what the complex density equation would look like but I lack the resources needed to provide the necessary data to substantiate the equation. I have created a general equation of baryonic density in the paper 'Variables involved in Baryonic Density'
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Anonymous said...
Regarding constants, there is no problem there. G or c are perfectly valid constructs, and don't point to any sort of illegitimacy of the theory.
If anything, they're simply the product of an unfortunate choice of units. We arbitrarily chose grams and seconds and miles and furlongs and hogsheads and bushels and meters, and those arbitrary decisions are reflected in the constants.
If our units aligned better with the universe, the silly constants would drop out of the equations.
For instance, everyone's favorite e = m c^2, under a particular choice of units for mass, would simply become e = m, which is really much more to the point, isn't it? Energy is mass is exactly what the equation says. c^2 was just a constant, and has no effect on the behavior of the equation.
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If our units aligned better with the universe, the silly constants would drop out of the equations.
For instance, everyone's favorite e = m c^2, under a particular choice of units for mass, would simply become e = m, which is really much more to the point, isn't it? Energy is mass is exactly what the equation says. c^2 was just a constant, and has no effect on the behavior of the equation.
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Constants are like perfect solids. They don't exist in the universe. It is not possible to have a natural sphere. I explain this in many postings throughout the blog.
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And the law of gravity, under a particular choice of units for mass, would become F = m1 m2 / r^2, which states exactly the idea intended, that each mass increases the effect linearly, and the distance decreases it quadratically.
And you might argue that I'm just hiding the constant in the unit, and yes! I am! But there's nothing wrong with that. The unit itself was arbitrary to begin with. The constant just scales the equation to fit the units of the universe, but the structure of the equation--it's behavior over the domain--stays the same.
If the equation is fundamentally wrong, there could be no constant that would result in correct behavior. The constant can't make an invalid equation valid, it can only scale the resultant units.
And you might argue that I'm just hiding the constant in the unit, and yes! I am! But there's nothing wrong with that. The unit itself was arbitrary to begin with. The constant just scales the equation to fit the units of the universe, but the structure of the equation--it's behavior over the domain--stays the same.
If the equation is fundamentally wrong, there could be no constant that would result in correct behavior. The constant can't make an invalid equation valid, it can only scale the resultant units.
Anonymous said...
Regarding multi-body... Gravity works just fine for N bodies. It's simply:
F-sub-i = G * m-sub-i * Sigma-sub-j [ ( m-sub-j / r-sub-ij ^ 2 ) * r-hat-sub-ij ]
where i and j range over 1..N, and r-hat-sub-ij is the normal vector pointing from i to j (I sincerely apologize for the phonetic syntax... not sure if your blog supports math markup). Or in other words, the force on each body is the superposition of the law of gravitation applied to each other body. There is no trickery here, and interestingly, to solve for a body, all I have to do is sum the mass/distance-scaled vectors to each body, then apply my mass.
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This is the best solution to the N-Body Problem I have seen.
The hat is an iterative process. It is a mathematical way of describing the mean after all the summations are completed iteratively. I agree that there is no trickery here. You would have one solution to a many iteration problem. This will help me in describing multiple bosons interacting in one ruleset. Thank you for that.
But that is not the real problem with gravity. Gravity does not allow for the interactions of magnetism, electricity, temperature or density. Each of these forces does affect the motion of an object. Each of these forces are (in this model) associated with a specific boson. This model does not need, and works well without; gravity or the higgs boson. In my opinion the higgs is an attempt to protect gravity by validating mass. Gravity describes a force that has not been found. Gravity is also 1 dimensional, it cannot be dimensionless. It is a scalar, it has a number.
We have found the planets are ordered by density.
Thanks for reading my work.
Aaron
Now, buoyancy.
ReplyDeleteBuoyancy does not fly in the face of gravity. Gravity, when acting on an object in isolation, holds. The theory of gravity says nothing more than this... that two masses, when considered in isolation, attract one another with some force. There might be other interactions adding additional forces, but somewhere in the big pile of forces acing on a mass is a contribution from gravity, and it conforms to a simple equation.
Buoyancy is a phenomenon that occurs in a large, complex system... A single solution to the simple 2-mass gravity law is not suffucicient in this situation, and nobody backing gravitational theory is saying that it is. But it does contribute in an important way. Allow me to attempt to explain...
If we consider the balloon and earth in complete isolation, it would accelerate downward at G*m-sub-earth/r-sub-earth-balloon^2, as expected.
But this isn't a two body system. There is a third player... air. And in this system, the three bodies are acting on each other. The gravitational force between air and the balloon can be neglected, since the balloon is so small in this instance and the air is so widely distributed (I hope we don't conflict on that assumption... but if we do, I'll happily defend it).
So let's start with the air. Being acted on by gravity, it is all trying to be pulled downward. This is creating pressure at all places within the air, at increasing values at lower altitudes, since there is more weight on top of the air at lower heights. This pressure is acting against gravity. It is the resistance of the air to being compressed into a singularity, primarily driven by kinetic forces of the matter being compressed. Everywhere there is a force pushing down (gravity) and everywhere there is a force pushing up resisting it. Thus we acheive equilibrium and a stable atmosphere.
Enter the balloon. Imagine, if you will, we identify the spherical surface, centered at the earth's origin, who's radius extends just to the bottom of the balloon. Thus, we can talk about the air beneath that altitude (which I'll call lower air), and the air above that altitude (higher air). So the lower air is all the air around the world beneath the altitude of the balloon, and higher air is the rest (which is where the balloon is). Everywhere over earth's surface, the lower air is pushing up at a certain force... the force required to hold the higher air bay... exactly the same force, but opposite in direction, as gravity is generating from the higher air. Without the balloon, it's a uniform equillibrium. But now, in one cylindrical column of higher air, there's a balloon... a balloon that is lighter than the surrounding air.
So now there's a particular place where the higher air is lighter. And so the lower air is no longer in equilibrium at that place. It's pushing up with the same force as all the rest of the lower air, but the air above it in that particular spot is pushing down with less force due to the lighter mass balloon. And so the net force, the combination of the upward pressure and the gravitationally-induced downward weight, is non-zero, and upward acceleration results.
This continues until a new equilibrium between air pressure and gravity is reached, which will occur when the balloon rises to a state of equi-density with it's surrounding.
And all of this was explained nicely with gravity as a key contributor.
Now, buoyancy.
ReplyDeleteBuoyancy does not fly in the face of gravity. Gravity, when acting on an object in isolation, holds. The theory of gravity says nothing more than this... that two masses, when considered in isolation, attract one another with some force. There might be other interactions adding additional forces, but somewhere in the big pile of forces acing on a mass is a contribution from gravity, and it conforms to a simple equation.
Buoyancy is a phenomenon that occurs in a large, complex system... A single solution to the simple 2-mass gravity law is not suffucicient in this situation, and nobody backing gravitational theory is saying that it is. But it does contribute in an important way. Allow me to attempt to explain...
If we consider the balloon and earth in complete isolation, it would accelerate downward at G*m-sub-earth/r-sub-earth-balloon^2, as expected.
But this isn't a two body system. There is a third player... air. And in this system, the three bodies are acting on each other. The gravitational force between air and the balloon can be neglected, since the balloon is so small in this instance and the air is so widely distributed (I hope we don't conflict on that assumption... but if we do, I'll happily defend it).
So let's start with the air. Being acted on by gravity, it is all trying to be pulled downward. This is creating pressure at all places within the air, at increasing values at lower altitudes, since there is more weight on top of the air at lower heights. This pressure is acting against gravity. It is the resistance of the air to being compressed into a singularity, primarily driven by kinetic forces of the matter being compressed. Everywhere there is a force pushing down (gravity) and everywhere there is a force pushing up resisting it. Thus we acheive equilibrium and a stable atmosphere.
Enter the balloon. Imagine, if you will, we identify the spherical surface, centered at the earth's origin, who's radius extends just to the bottom of the balloon. Thus, we can talk about the air beneath that altitude (which I'll call lower air), and the air above that altitude (higher air). So the lower air is all the air around the world beneath the altitude of the balloon, and higher air is the rest (which is where the balloon is). Everywhere over earth's surface, the lower air is pushing up at a certain force... the force required to hold the higher air bay... exactly the same force, but opposite in direction, as gravity is generating from the higher air. Without the balloon, it's a uniform equillibrium. But now, in one cylindrical column of higher air, there's a balloon... a balloon that is lighter than the surrounding air.
ReplyDeleteSo now there's a particular place where the higher air is lighter. And so the lower air is no longer in equilibrium at that place. It's pushing up with the same force as all the rest of the lower air, but the air above it in that particular spot is pushing down with less force due to the lighter mass balloon. And so the net force, the combination of the upward pressure and the gravitationally-induced downward weight, is non-zero, and upward acceleration results.
This continues until a new equilibrium between air pressure and gravity is reached, which will occur when the balloon rises to a state of equi-density with it's surrounding.
And all of this was explained nicely with gravity as a key contributor.
Okay, first up, F=da.
ReplyDeleteF=da would need to explain more than just buoyancy. Let's say you had two solid low-friction objects of different densities.. say a block of lead on wheels, and a block of balsa wood on wheels, where both had the same weight when placed on the same level surface at a fixed altitude. Clearly, the wood block is much larger due to it's much smaller density.
Now, F=ma tells us that if I push both of them with the same force, they'll accelerate at the same rate, because they weight the same. F=da tells us that if I push both of them with the same force, the much larger and less dense wood block will accelerate faster.
We can safely neglect friction from air, due to the obnoxiously slow speeds at which one can perform this experiment, but if you're concerned about drag, then it's entriely possible to simply do this in a vaccum chamber.
So this has nothing to do with buoyancy. If F=da is right, two solid objects of equal mass experiencing equal force in a vaccum will accelerate at different rates due soley to their differing volume.
This, from my experience with physics, has not been observed. And this is something one could easily test at home... ever made a rubber band powered wood car? Just use that, and load it with a huge block of something light, and again with a small block of something heavy, such that both blocks are equal weight. They will perform equivalently, despite the differing densities involved in each run.
This does not explain buoyancy yet. I'm simply attempting to confirm F=ma and refute F=da.
Regarding constants: I'm not saying that constants like G and c are part of nature. I'm saying they're part of the equation due to not using the primitive units of nature. They're a scaling factor. They don't alter the behavior of the function over its domain. The constants aren't there to describe nature, they're there to enable a function, that DOES describe nature independent of the constant, work at the correct scale.
ReplyDeleteOr to put it another way, if you experimentally observed what a system does and plotted it without units (no legend or axis tics), then calculated it from theory neglecting the constants and plotted it without tics, the curves shapes would match.
Thanks again for reading my work.
ReplyDeleteI don't really remember describing motion of baryons as f=da. That equation has the same problem as f=ma. The medium in which it travels.
The equation I propose for general baryonic motion does take into consideration the medium in which it travels. It is written in the variables involved in baryonic motion. That paper is on the top right side of the blog under published papers.
I hope you find that to be more useful the f=ma or f=da.
Aaron
The equation should be subtly tweaked... the i beneath the sigma should be "i=1" and the j above the sigma should be "N".
ReplyDeleteI wouldn't say "the hat is an iterative process". I'd say the "sigma is the summation of N vectors", which you can call iterative. The hat has nothing to do with it... it just signifies a normal vector.
And the end result isn't really a "mean"... nothing is being averaged. It's the vector sum.
And gravity isn't one-dimensional. It's a force, which is a vector.
And the planets being (almost) ordered by gravity is explainable in both models, so I don't think it helps one way or the other.
Regarding f=da, you took my comment out of context. in the original comment thread that I had replied to, you had stated "a 3 dimensional version of Newtons f=ma would be Force = Density* acceleration. This would be acceptable." This is what I was responding to.
ReplyDeleteHi,
ReplyDeleteThanks for the comments.
Funny story, I had a dream where the equation you posted was everywhere. I woke to a start, realizing that I had typed into LATEX, Sigma instead of Sum. So at 6am I was rewriting the equation. Yes I did get the i and j reversed. I will fix it tonight.
This model has nothing to do with Newtonian Mechanics. I know much has been done using Newton's work, but much is lacking using Newton's work.
In this model baryons express information. There is not a general law of attraction between bodies.
Thanks again
Aaron
Anonymous,
ReplyDeleteIf I did delete the comment was by accident and not intentional. Sorry, I do not know how it was deleted. I am having an issue with the comment moderation system. It tries to post each comment twice. I will look into it. For now here it is.
Anonymous has left a new comment on your post "The N-Body Problem":
Now, buoyancy.
Buoyancy does not fly in the face of gravity. Gravity, when acting on an object in isolation, holds. The theory of gravity says nothing more than this... that two masses, when considered in isolation, attract one another with some force. There might be other interactions adding additional forces, but somewhere in the big pile of forces acing on a mass is a contribution from gravity, and it conforms to a simple equation.
Buoyancy is a phenomenon that occurs in a large, complex system... A single solution to the simple 2-mass gravity law is not suffucicient in this situation, and nobody backing gravitational theory is saying that it is. But it does contribute in an important way. Allow me to attempt to explain...
If we consider the balloon and earth in complete isolation, it would accelerate downward at G*m-sub-earth/r-sub-earth-balloon^2, as expected.
But this isn't a two body system. There is a third player... air. And in this system, the three bodies are acting on each other. The gravitational force between air and the balloon can be neglected, since the balloon is so small in this instance and the air is so widely distributed (I hope we don't conflict on that assumption... but if we do, I'll happily defend it).
So let's start with the air. Being acted on by gravity, it is all trying to be pulled downward. This is creating pressure at all places within the air, at increasing values at lower altitudes, since there is more weight on top of the air at lower heights. This pressure is acting against gravity. It is the resistance of the air to being compressed into a singularity, primarily driven by kinetic forces of the matter being compressed. Everywhere there is a force pushing down (gravity) and everywhere there is a force pushing up resisting it. Thus we acheive equilibrium and a stable atmosphere.
Enter the balloon. Imagine, if you will, we identify the spherical surface, centered at the earth's origin, who's radius extends just to the bottom of the balloon. Thus, we can talk about the air beneath that altitude (which I'll call lower air), and the air above that altitude (higher air). So the lower air is all the air around the world beneath the altitude of the balloon, and higher air is the rest (which is where the balloon is). Everywhere over earth's surface, the lower air is pushing up at a certain force... the force required to hold the higher air bay... exactly the same force, but opposite in direction, as gravity is generating from the higher air. Without the balloon, it's a uniform equillibrium. But now, in one cylindrical column of higher air, there's a balloon... a balloon that is lighter than the surrounding air.
So now there's a particular place where the higher air is lighter. And so the lower air is no longer in equilibrium at that place. It's pushing up with the same force as all the rest of the lower air, but the air above it in that particular spot is pushing down with less force due to the lighter mass balloon. And so the net force, the combination of the upward pressure and the gravitationally-induced downward weight, is non-zero, and upward acceleration results.
This continues until a new equilibrium between air pressure and gravity is reached, which will occur when the balloon rises to a state of equi-density with it's surrounding.
And all of this was explained nicely with gravity as a key contributor.
I am going to take some time to read and think about your comment. Again I am sorry it was not posted correctly.
ReplyDeleteAaron
I agree with most everything you said in the Newtonian model. You paint an easy description of gravity in a medium.
ReplyDeleteBut,
There is never a uniform equilibrium and stable atmosphere.
And this is a model of force carrier that have been observed in supercolliders. The 4 Bosons.
The Gluon is the force carrier of the strong nuclear force, or density.
The Z boson transmits electricity.
The W+/- Boson transmits magnetism.
The Photon transmits temperature.
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I am going to move your whole comment to a new posting. The work you did here deserves a better atmosphere for discussion. I am going to post your comment tonight. Tomorrow I am going to add to it. Any comments you add I will add to the posting.
I hope to show what you are describing is density in motion. I will also show how the other bosons interact to show asymmetry and instability in the atmosphere. With those 4 bosons I will show everything you stated and more.
I just need time to formulate my solution.
Thanks for your patience.
Aaron
Agreed, there is never uniform equilibrium and stability in any actually observed natural systems. Large complex non-uniform systems are necessary complex, and involve a great deal of interactions beyond simple gravity. The Sun provides a ton of energy in a non-uniform manner. The earth itself is non-uniform in structure, emitting various energies into the atmosphere in differing amounts... deep oceans, shallow waters, volcano's, solar-induced thermal currents banging into abnormally shaped mountain ranges, organic matter emitting all manner of material, industrial byproducts... you name it. The net result is certainly not a "stable" atmosphere.
ReplyDeleteAnd no, I can't model all of that as part of the balloon example, but that alone doesn't invalidate gravity or buoyancy. But the theory has been shown to hold in controlled experiments, and has been observed to explain astronomical phenomena, where the complexity is a little more bounded than the chaos of an active planetary atmosphere. In other words, whenever we're able to simplify the environment to produce a targetted test that would result in gravitational force being the dominant observed quantity, we're able to do so successfully, and it conforms to Newton's law.
And more to the point: the above explanation, despite being explained in an "ideal" atmosphere, is readily observed in practice, which is actually saying a lot. It means that despite reality doing everything it can to complicate my "ideal" model, it still works as described.
Now, I'm perfectly happy with you proposing a theory that does not include gravity, provided it explains the same phenomena. Really... I'm not defending gravity "just because". But if you are going to attack gravity, for instance, by saying it's invalid due to zero-dimensional quantities, or due to the use of non-existent constants, or that it can't explain buoyancy, then I'll fight back by arguing gravity is in fact a sound and well verified theory. It might not be "right" in the sense of being an accurate explanation, but it is "right" in the sense of producing accurate results (same goes for Quantum Mechanics).
So thanks for being patient with me... I always love a good argument :) I'm quite interested to hear the explanation expressed in terms of your theory... it'll definitely help me form a better picture of your model.
Agreed, there is never uniform equilibrium and stability in any actually observed natural systems. Large complex non-uniform systems are necessary complex, and involve a great deal of interactions beyond simple gravity. The Sun provides a ton of energy in a non-uniform manner. The earth itself is non-uniform in structure, emitting various energies into the atmosphere in differing amounts... deep oceans, shallow waters, volcano's, solar-induced thermal currents banging into abnormally shaped mountain ranges, organic matter emitting all manner of material, industrial byproducts... you name it. The net result is certainly not a "stable" atmosphere.
ReplyDeleteAnd no, I can't model all of that as part of the balloon example, but that alone doesn't invalidate gravity or buoyancy. But the theory has been shown to hold in controlled experiments, and has been observed to explain astronomical phenomena, where the complexity is a little more bounded than the chaos of an active planetary atmosphere. In other words, whenever we're able to simplify the environment to produce a targetted test that would result in gravitational force being the dominant observed quantity, we're able to do so successfully, and it conforms to Newton's law.
And more to the point: the above explanation, despite being explained in an "ideal" atmosphere, is readily observed in practice, which is actually saying a lot. It means that despite reality doing everything it can to complicate my "ideal" model, it still works as described.
ReplyDeleteNow, I'm perfectly happy with you proposing a theory that does not include gravity, provided it explains the same phenomena. Really... I'm not defending gravity "just because". But if you are going to attack gravity, for instance, by saying it's invalid due to zero-dimensional quantities, or due to the use of non-existent constants, or that it can't explain buoyancy, then I'll fight back by arguing gravity is in fact a sound and well verified theory. It might not be "right" in the sense of being an accurate explanation, but it is "right" in the sense of producing accurate results (same goes for Quantum Mechanics).
So thanks for being patient with me... I always love a good argument :) I'm quite interested to hear the explanation expressed in terms of your theory... it'll definitely help me form a better picture of your model.
I found the missing comment. I am so sorry it took so long. I am going through everything to make it better. Thanks
ReplyDeleteAaron Guerami