Andre Lefebvre Posted July 16, 2015 Author Posted July 16, 2015 Weight is the measure of force on an object due to gravity. Plus Change this to oppose my free fall acceleration. So the weight is the difference of my accelerating velocity (blocked) toward the center of gravity and the velocity of the ground toward the same center. Weight is then a difference of speed between the velocity of two objects.
Mordred Posted July 16, 2015 Posted July 16, 2015 (edited) So the weight is the difference of my accelerating velocity (blocked) toward the center of gravity and the velocity of the ground toward the same center. Weight is then a difference of speed between the velocity of two objects. When you accelerate an object you change its velocity. Gravity is measured in the same units as acceleration m/s^2. Velocity is just m/s. Inertia is the tendency of an object to maintain the same velocity. Even if that velocity is zero. Any change in velocity (including direction) requires an influence( force, space time curvature can be described as a type of force). Weight can be described as the relation between an objects resistance to inertia (mass)*the acceleration due to gravity. (At least for simplicity sake, the Earths rotation also influences weight) were not ready to go into this detail. We would need to cover net acceleration due to location on Earth. W=ma and W=mg are in this case equivalent. Speed is just the magnitude scalar portion of velocity. Units and definitions are important. As they describe specific relations Edited July 16, 2015 by Mordred
Andre Lefebvre Posted July 16, 2015 Author Posted July 16, 2015 I shouldn't have tried to be "physic vocabulary correct". I'm not ready yet. So I should have said: "So the weight is the difference of my speed (blocked) toward the center of gravity and the speed of the ground toward the same center. Weight is then a difference of the speed of two objects". In flat universe (nothing falls there) two objects, moving at different speed, that make contact, the less massive object will show weight in regard to the more massive object.
Mordred Posted July 16, 2015 Posted July 16, 2015 Sorry the last post makes even less sense. It sounds like your trying to find a way to describe shell theorem. https://en.m.wikipedia.org/wiki/Shell_theorem
Andre Lefebvre Posted July 17, 2015 Author Posted July 17, 2015 (edited) That link was the one I went to, when I asked Strange what was the shell theorem. That's why I added : « Oh. Sorry. I'm only saying that as long as a center of gravity doesn't join (unify with) another center of gravity to become one, it doesn't add mass to the greater deformation containing it…”. Once again you’re both right in recognising part of the shell theorem’s implication. But I “work” with the geometry of the inside of the shell instead of equations. And it shows a slight difference in the understanding of the event. So if you want to talk about the shell theorem, let’s do it (from your link) : “Isaac Newton proved the shell theorem and said that: 1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. 2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.” What does it mean exactly? I don't know about you; but to me it means that if you consider a galaxy as a “shell” (hollow ball), its mass (that can be considered at the center of the galaxy, doesn’t exert any gravitational force whatsoever on any object inside the shell, regardless of the object’s location. So explain to me why we add all stars inside a galaxy to find it’s mass and, most important of all, how can the mass of the galaxy affect the speed of those inside situated stars. It’s directly against what the shell theorem says. You’ll need mathematical logic here; because reasoning doesn’t “fit”. Edited July 17, 2015 by Andre Lefebvre
Mordred Posted July 18, 2015 Posted July 18, 2015 (edited) https://www.google.ca/url?sa=t&source=web&cd=16&ved=0CDYQFjAFOApqFQoTCJ_GjL-848YCFQSKDQod3PcGnA&url=http%3A%2F%2Fwww.saddleback.edu%2Ffaculty%2Fbhubbard%2Fdocuments%2FSHELLTHEOREMPROOF1_000.pdf&rct=j&q=shell%20theorem&ei=V6SpVd--CoSUNtzvm-AJ&usg=AFQjCNEdTSOlRrQpfMyVdnkyOeTtahSKMA&sig2=z93zf7__9bC8TxZw3s_mIg Think of shell theorem as not just one shell but a series of shells. Like layers of an onion. At each layer you apply the net force. At the center the force from each point on the shell pointing toward the center cancel each vector out. Remember the vector direction of force in the shell theorem always points towards the center of mass. So say you have x force due to mass pointing to the COM at 90 degrees and the same amount of force going 270 degrees. The net sum is zero. In the case of a galaxy you don't just calculate the outer shell but every shell as a net sum of force as you approach the center. So imagine your flying a ship towards the center. At every point during your flight the net sum of force varies. Once the ship reaches the center of mass of a perfectly spherical galaxy the net sum of forces from all directions equals zero. Edited July 18, 2015 by Mordred
Andre Lefebvre Posted July 18, 2015 Author Posted July 18, 2015 (edited) Think of shell theorem as not just one shell but a series of shells. Like layers of an onion. I agree to think that way if you want me to; but an onion is a funny kind of "hollow ball", don't you think? I suddenly wonder why they used the word "shell" instead of "onion"? Once the ship reaches the center of mass of a perfectly spherical galaxy the net sum of forces from all directions equals zero. But just before that you become spaghetti because of all those zero sums of each layer (of onion) you went through; if I remember right what you told me before regarding the black hole at the center of a galaxy. Edited July 18, 2015 by Andre Lefebvre
Mordred Posted July 18, 2015 Posted July 18, 2015 I agree to think that way if you want me to; but an onion is a funny kind of "hollow ball", don't you think? I suddenly wonder why they used the word "shell" instead of "onion"? But just before that you become spaghetti because of all those zero sums of each layer (of onion) you went through; if I remember right what you told me before regarding the black hole at the center of a galaxy. Bh space time curvature is an extreme case. Were dealing with far reduced gravitational bodies. Remember the strength of gravity from any object falls off at a rate of 2/r^2. [latex]f=\frac{GM_1m_2}{R^2}[/latex] The other detail you missed is in the case of the Earth or galaxies the gravitational strength curve is different. Let's use the Earth. Remember the Earth is made up of particles, each particle with its own point mass. Also remember the term net sum of vector forces. I hope you know how to add and subtract vectors? http://www.physicsclassroom.com/class/vectors/Lesson-1/Vector-Addition Now as you go from the surface inside the Earth to its center the net sum force of gravity decreases. Not increases. This is because there is now mass on both sides of you. Mass ahead and mass behind. This also occurs in the galaxy as you approach it's center of gravity the force of gravity acting at each point decreases in terms of net sum of gravitational force. Mass ahead and mass behind. In the case of a BH, your always outside the shell as all its mass is at the singularity.
Andre Lefebvre Posted July 18, 2015 Author Posted July 18, 2015 Now as you go from the surface inside the Earth to its center the net sum force of gravity decreases. Not increases. This is because there is now mass on both sides of you. Mass ahead and mass behind. I can't understand that, since there's no "attraction" between mass. This also occurs in the galaxy as you approach it's center of gravity the force of gravity acting at each point decreases in terms of net sum of gravitational force. Mass ahead and mass behind.In the case of a BH, your always outside the shell as all its mass is at the singularity. So I could install myself right on the event horizon and be stabilised by the mass behind me that would prevent me from getting inside the horizon. I have some difficulties with that idea.
Strange Posted July 18, 2015 Posted July 18, 2015 (edited) 1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. 2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.” I don't know about you; but to me it means that if you consider a galaxy as a “shell” (hollow ball), its mass (that can be considered at the center of the galaxy, doesn’t exert any gravitational force whatsoever on any object inside the shell, regardless of the object’s location. The trouble is, the galaxy isn't a hollow ball. So, obviously, treating it as such will obviously give you the wrong results. The only bit of the shell theorem relevant to the galaxy as a whole is that from outside the shell (or any spherically symmetrical distribution of mass) you can treat it as iff all the mass were at a central point. However, the shell theorem can be used at any point in the galaxy to show that only the mass inside a sphere of that radius matters; mass outside the sphere doesn't affect the orbits of stars at that distance, for example. (This is only an approximation, because only the central bulge is spherically symmetrical.) A more detailed analysis can be used to show the need for more mass in the galaxy than we can see. I can't understand that, since there's no "attraction" between mass. Eppure, si muove. So I could install myself right on the event horizon and be stabilised by the mass behind me that would prevent me from getting inside the horizon. I have some difficulties with that idea. I think everyone would have some difficulties with that idea. As it is wrong. Which bit of "you are always outside the shell as all its mass is at the singularity" did you not understand? Edited July 18, 2015 by Strange
Mordred Posted July 18, 2015 Posted July 18, 2015 (edited) I can't understand that, since there's no "attraction" between mass. I'm not going to bother trying to type my answers to suit your viewpoint on what you think is involved. Both GR and Newtonian physics use the term mass. When you study Newtonian based models you use force. This helps to understand the formulas involved. Perhaps if you had done this in the first place. You would have had a better comprehensive understanding of basic physics. Use the terminology of the model under study. I have enough difficulty getting you to understand basic physics math relations. Your math skills is not near enough to sit down and explain how geodesics cause acceleration change without applying a force. Quite frankly we've just scratched the surface on the field of kinematics. In point of detail we've only covered the basic terminology involved. Of which you still need to improve your understanding of including units, unit conversions, and vector calculus. Without these firmly in place you won't understand a single geodesic formula. For example inertia as opposed to moment of inertia. On basic vector addition. I have 20 Newton of force in direction 90 degrees and 120 Newtons of force in direction 180 degrees. What is the sum of force (net force) and direction ? Edited July 18, 2015 by Mordred
Andre Lefebvre Posted July 18, 2015 Author Posted July 18, 2015 The trouble is, the galaxy isn't a hollow ball. So, obviously, treating it as such will obviously give you the wrong results. I guess you're right; but if a galaxy is not a "hollow" ball with "object inside,", what use is that theorem that says "no net gravitational force is exertedby the shell on any object inside, regardless of the object's location within the shell.” I'm not going to bother trying to type my answers to suit your viewpoint on what you think is involved. Both GR and Newtonian physics use the term mass. That's not the point. The point is that the "gravitational force" doesn't exist because gravitation is a consequence of space-time deformation. If you keep on thinking of forces, you end up saying that at mid-point inside the Eartrh, the gravitation "pull" is equal in front and in back of you so you feel no gravitation. When, in fact, there's no "pull" whatsoever anywhere in or around a massive object whatever the equation that says otherwise. So I agree that if I keep on basing my opinion in this fact, I will not get the same results that if I based it on the "attraction" of mass. But it sure doesn't mean that I'm wrong until GR is proven wrong and gravity is not a consequence of the deformation of the geometry of space-time.
Mordred Posted July 18, 2015 Posted July 18, 2015 (edited) The other difficulty will come in Cartesian to Polar coordinate transformations. Needed for SR and GR. I guess you're right; but if a galaxy is not a "hollow" ball with "object inside,", what use is that theorem that says "no net gravitational force is exertedby the shell on any object inside, regardless of the object's location within the shell. That's not the point. The point is that the "gravitational force" doesn't exist because gravitation is a consequence of space-time deformation. If you keep on thinking of forces, you end up saying that at mid-point inside the Eartrh, the gravitation "pull" is equal in front and in back of you so you feel no gravitation. When, in fact, there's no "pull" whatsoever anywhere in or around a massive object whatever the equation that says otherwise. So I agree that if I keep on basing my opinion in this fact, I will not get the same results that if I based it on the "attraction" of mass. But it sure doesn't mean that I'm wrong until GR is proven wrong and gravity is not a consequence of the deformation of the geometry of space-time. Tell you what prove you understand basic physics and terminology first. Otherwise no explanation I can offer will make any sense. Considering I posted the Einstein field equations stress energy tensor for Cartesian coordinates to Polar coordinate transformation in the stress energy tensor. ( Which defines how space time curves) several pages ago. You didn't understand it then, still doubtful your any closer to understanding it The issue your missing is you need a clear understanding of Basic physics before handling the complexities in GR and SR. If you don't know the difference between velocity and acceleration and how mass is defined in its relationship to inertia. How do you expect to understand GR? Newtonian physics teaches those relations. ( it would be like building an automobile with just a hammer) Edited July 18, 2015 by Mordred
Andre Lefebvre Posted July 18, 2015 Author Posted July 18, 2015 (edited) Tell you what prove you understand basic physics and terminology first. The first of all basic physics is that gravity is a consequence of the deformation of the geometry of space-time. If we start from there and protect this concept, what ever tool we use that doesn't change it, we shouldn't have any problem. This is exactly what I understand. But bringing equations that cancels gravity anywhere in the Earth because those equations are based on "attraction" of masses, won't do, that's for sure. I'm sorry. If we use "tidal effect" equations not based on "attraction" of masses, we'll have no problem either. If we consider flat space-time without involving to and fro "attractions" to accept it "flat", but consider its flatness as an observed fact and accept that it was "flat" from the start because of the nature of Big bang itself, when mass or pressure could not be involve in a non-pressurized universe because the "radiating" movement was free of whatever opposition, producing space-time as it needed, we shouldn't have any problems; and so on. I cannot base my opinions on what people thought two or three hundred years ago or even opinions debated 70 years ago; I want to study the "facts» that our advance technology supplied us with, these last few years. Those should be the base of our physics; not things that were based on observations at a time when the quality of observation was a million times less effective than today. New results shouldn't be subjected to old interpretations. That's what my opinion is. I’m sorry. Considering I posted the Einstein field equations stress energy tensor for Cartesian coordinates to Polar coordinate transformation in the stress energy tensor. ( Which defines how space time curves) several pages ago. See what I mean? "...the Einstein field equations stress energy tensor for Cartesian coordinates to Polar coordinate transformation in the stress energy tensor." Is only the "name" to indicate what we are talking about. And this is only to explain how space-time is curved. What will look like the explication of when it is "flat"? Geez! If expansion is the increase of space-time metric and gravitation is the decrease of space-time metric, don't you think that a simpler title to name it couldn't be possible? Don't you think that an equation explaining the pressure on the center of a deformation of space-time couldn't not be simpler than what you’re working with only because you keep that "attracting force" notion in all the equations? Don't you think that a topology or geodesic (even none of those two words really mean the sense of direction of a movement) inside a gluon would explain "naturally" what is observed of the characteristic of that gluon, instead of attributing to it a "force" that comes from nowhere to hold the quarks, the neutrons and proton together and even extends indirectly to the electronic portion of an atom? Don't you agree that there's something wrong on the notion of gravity "force" diminishing at the square or the distance when you have stars distributed on great distances from the center of a galaxy that orbit at all the same speed? That’s what bothers me in the actual description of physics. I think that physics is fascinating because it studies the universe not because the people doing it are more intelligent than average. That's not the case anyway. They are more informed in a way to seeing things and have access to all the new results of research. What is fascinating is the universe; nothing else. Trying to understand it is a noble enterprise. It has to be protected to be kept that way. That is why I give so much importance to the notion of gravity being only a passive "consequence" and not a "force". In fact the whole universe was made by a series of consequences where not one single "force" was involved. This is exactly what I think. And again, I'm very sorry if it doesn't fit mainstream physics or can't be adapted to it. Edited July 18, 2015 by Andre Lefebvre
Mordred Posted July 18, 2015 Posted July 18, 2015 That's the gist of the problem. You won't accept any as you put it older and far simpler concepts. Instead you want to comprehend complex partial differential geometry equations. When you don't even understand the basic terminology. Quite frankly that's impossible.
Andre Lefebvre Posted July 18, 2015 Author Posted July 18, 2015 I have nothing against terminology we can use. I'm just asking not to deviate from the basic fact of GR. It's possible to explain things without using "forces" coming out of nowhere. I did it here myself in explaining "tidal effect" with two deformations of space-time and I didn't use any magical force to do it. If I could do it with simple geometrical figures, there should be a mathematic equation that could describe these figures without using "forces" that aren't present at all in the explanation. I can’t believe it impossible. Sorry.
Mordred Posted July 18, 2015 Posted July 18, 2015 Oh it's possible but you wouldn't understand the equations. Not how the units themselves are defined. Why do you think the Feyman textbook is organized in the sequence its written in? Do you believe the author isn't aware that GR involves space time curvature as opposed to force? When you study a textbook you start at chapter 1 not skip to the portion that agrees with you. Here look at the description on this page in regards to force in GR. " By the early 20th century, Einstein developed a theory of relativity that correctly predicted the action of forces on objects with increasing momenta near the speed of light, and also provided insight into the forces produced by gravitation and inertia." Then look at the definition of force on the same page. "In physics, a force is any interaction that tends to change the motion of an object.[1] In other words, a force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate. " Then read the section including the formulas under Special relativity. https://en.m.wikipedia.org/wiki/Force Your fighting a battle based upon personal feelings instead of actual knowledge. The term force can be used anywhere there is any form of interaction that influences motion. 1
Andre Lefebvre Posted July 18, 2015 Author Posted July 18, 2015 The term force can be used anywhere there is any form of interaction that influences motion. And it is. Which is why people think that those "forces" are "active" when in reality they a only "consequences". This is exactly my point.
Mordred Posted July 18, 2015 Posted July 18, 2015 (edited) Why would that make any difference??. All forces are a result of some form of interaction. Regardless, how they relate to changes in motion remains the same. I fail to see your objection. All forces is a descriptive of how a interacts with b. It is not something that exists without a and b. Regardless of what a and b represents. In other words its a defined unit of measure. One with a magnitude and direction, Edited July 18, 2015 by Mordred
Andre Lefebvre Posted July 18, 2015 Author Posted July 18, 2015 All forces is a descriptive of how a interacts with b. It is not something that exists without a and b. Regardless of what a and b represents. So what is the strong nuclear force then?
Mordred Posted July 18, 2015 Posted July 18, 2015 It is a measure of the interaction between two or more quarks . Yes the gluon mediates the interaction. However that's due to energy not existing on its own. All forms of energy is a property of either a particle or object.
Andre Lefebvre Posted July 18, 2015 Author Posted July 18, 2015 (edited) It is a measure of the interaction between two or more quarks How come, then, that 75% of the time it decays in a quark and antiquark top if it's a mesure of interaction between quarks? Yes the gluon mediates the interaction. What does "mediate" means really? However that's due to energy not existing on its own. Is the total energy of the universe doesn't exist on its own? All forms of energy is a property of either a particle or object. And that's why you have to imagine dark matter that accounts for 26.8% of the energy of the universe; plus 4.9% for normal matter. So there's still 68.3% of dark energy that doesn't have a particle or object "proprietor". On the other hand, the expansion of the universe dilutes its energy because it is invariant; curiously this invariance applies also to photons. They don't diminish either with expansion; they are also diluted. But what appeared at the big bang if there was no matter, only "radiance"? The Big bang did not liberated energy that don't exist on its own? What made the universe if not energy? Edited July 18, 2015 by Andre Lefebvre
Strange Posted July 19, 2015 Posted July 19, 2015 I guess you're right; but if a galaxy is not a "hollow" ball with "object inside,", what use is that theorem that says "no net gravitational force is exertedby the shell on any object inside, regardless of the object's location within the shell.” What use is any mathematical theorem? It is an interesting and counterintuitive result. That's not the point. The point is that the "gravitational force" doesn't exist because gravitation is a consequence of space-time deformation. Yes, that is the way it is modelled in GR. If you keep on thinking of forces, you end up saying that at mid-point inside the Eartrh, the gravitation "pull" is equal in front and in back of you so you feel no gravitation. Which is true. When, in fact, there's no "pull" whatsoever anywhere in or around a massive object whatever the equation that says otherwise. It is not unreasonable to think of the gravity pulling on you to stop you drifting off into the air. So what if that isn't how GR describes it? So I agree that if I keep on basing my opinion in this fact, I will not get the same results that if I based it on the "attraction" of mass. Whether you use the words "force" or "pull" or not, makes no difference at all to the results you get. The first of all basic physics is that gravity is a consequence of the deformation of the geometry of space-time. That is not basic at all. It is a very complex idea. I certainly wouldn't claim to understand it at anything other than a vague conceptual level. But bringing equations that cancels gravity anywhere in the Earth because those equations are based on "attraction" of masses, won't do, that's for sure. I'm sorry. You can, of course, derive the shell theorem using GR. It is just a lot more complex to do. Don't you agree that there's something wrong on the notion of gravity "force" diminishing at the square or the distance when you have stars distributed on great distances from the center of a galaxy that orbit at all the same speed? So now you are saying that GR is wrong? And it is. Which is why people think that those "forces" are "active" when in reality they a only "consequences". This is exactly my point. But it doesn't matter. You are worrying about terminology, instead of understanding the physics.
Mordred Posted July 19, 2015 Posted July 19, 2015 (edited) How come, then, that 75% of the time it decays in a quark and antiquark top if it's a mesure of interaction between quarks? You asked what the strong force is. The answer to that question is above. The gluon doesn't define the strong force. The gluon is the exchange particle between two or more quarks. In point of detail its a field of gluons not necessarily an individual gluon. What does "mediate" means really? mediate in this application means exchange of energy, charge or color etc between two quarks. One quark emits say it's color property the gluon carries that color charge to the next quark. The same thing happens to to the energy for the strong force. Is the total energy of the universe doesn't exist on its own? And that's why you have to imagine dark matter that accounts for 26.8% of the energy of the universe; plus 4.9% for normal matter. So there's still 68.3% of dark energy that doesn't have a particle or object "proprietor". On the other hand, the expansion of the universe dilutes its energy because it is invariant; curiously this invariance applies also to photons. They don't diminish either with expansion; they are also diluted. But what appeared at the big bang if there was no matter, only "radiance"? The Big bang did not liberated energy that don't exist on its own? What made the universe if not energy? This isn't the reason for dark matter or for that matter dark energy. Those weren't added to our models on a whim but based on observation evidence. Based on matter distribution of visible matter (baryonic) galaxies should rotate slower the farther you get from the center. They didn't. It took over 60 years of alternate model fighting to try to explain this without dark matter. I recall some of the elaberate arguments as I was a member of another forum before dark matter was finally accepted. The fact is only the existsnce of mass/ matter in a halo distribution enveloping galaxies could explain the rotation curve. Then on top of this observations spotted gravitational lensing where there shouldn't be any. There was no nearby baryonic matter of sufficient densities. We can measure and detect baryonic matter quite easily. We can even give estimated values of how much in a given region and type. Dark energy aka the Cosmological constant is needed to explain expansion in particular the accelerating rate of expansion. Based on baryonic matter and dark matter distributions our universe shouldn't be expanding as fast as it is. The late time integrated Sachs Wolfe effect is used to measure the density of the cosmological constant as the universe evolves after the CMB. Now as to how to go from the beginning of the universe to the particles we know? Well you have to understand how virtual particles work. Particles can pop in and out of existance all the time and at any time in particular virtual particles. You have quasi particles. These are used to describe particle like interactions. One example is your inflaton. Quasi particles usually only describe a specific particle property. In the case of the inflaton that is energy. However this isn't a real particle. It's more of a placeholder till the real cause or particle responsible is determined. However virtual particles are different than quasi particles in that they act like specific particles. However they are what's called "off shell" basically means not quite a real particle. One reason is they lack the energy to form a real particle this leads them to decay quicker than the real particle. Particles form in matter/antimatter pairs due to the conservation of charge, color and energy laws. To fully understand that would take more than I can post. As far as gluons forming top quarks 75% of the time. Well quite frankly you can't trust that pop media article you posted that in. We can't accelerate qluons in any LHC for one. Secondly there are no unbound quarks in the universe today. They are bound in protons, neutrons etc. The LHC accelerate magnetically charged particles. The gluon doesn't interact via the electromagnetic field. What we do is accelerate protons. The protons gain inertial mass, this results in highly energetic gluons within the proton along with highly energetic quarks that make up the proton. This reaction is then possible 75% of the time. This does not mean this occurs 75% of the time in ordinary conditions. One should never trust pop media style articles. They always tend to mislead or misinform. Edited July 19, 2015 by Mordred
Andre Lefebvre Posted July 19, 2015 Author Posted July 19, 2015 This does not mean this occurs 75% of the time in ordinary conditions.One should never trust pop media style articles. They always tend to mislead or misinform. I don't get my infos from pop media I go get them in the reports of experiences; yhe papers that give the results. Pop media dont tell you that during experuiences 75% of decay of gluon was Top and antiTop quarks.
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