Sayonara Posted November 11, 2004 Posted November 11, 2004 What you mean is we can observe their effects and make predictions as to the outcome of certain experiments.But we dont even know what they are' date='or how they work,what energy source they draw upon.As for calling the author Mark McCutcheon a moron who knows nothing of science,get a reality check!! Your a 'nobody' on an insignificant forum,you have obviously shown to people here that your knowledge on any science topic is fundamentally average,and thats being complimentary.Yet you think by merely spouting Utter crap that you carry some kind of aura.This person who has shown he understands physics far in advance than you,can get his work published and make significant amounts of money.And you think your words to the contrary are important....a word of advice GET A LIFE.And if you think thats trolling,or flaming...please dont reply because i dont want you too,Im just pointing out facts.Compared to him your nappy rash...[/quote'] That's it Philbo - you're out of here. This time, don't come back.
Severian Posted November 11, 2004 Posted November 11, 2004 What you mean is we can observe their effects and make predictions as to the outcome of certain experiments.But we dont even know what they are' date='or how they work,what energy source they draw upon.[/quote'] Even though that post is beneath a reply, I still feel obliged to reply for the sake of the other posters. In physics we test the theories that we have very rigourously. The theories make predictions for what we should see in experiments (observables) and if we do not see what the theory predicts the theory is discarded or modified. This is repeated with lots and lots of different experiments and observables. The Standard Models of particle physics contains as a subset of it a complete description of electromagnetism (and thus magnetism). It not only correctly models the effects of electromagnetism in everyday physics around us, but it correctly describes the quantum effects too. In fact, the Standard Model is now the best tested theory that the human race has ever produced. It will be interesting to see if it survives the LHC experiment which is designed to test the Standard Model in even more extreme conditions. Gravity is not well undertood at the quantum level but on macroscopic distance scales it is exceptionally well described by Einstein's GR. The fundamental forces of nature don't need 'energy' sources. What they are doing is transfering momentum between particles by exchanging a 'force mediator particle'. The photon is the force mediator for electromagnetism. The exchange of a phton does not add energy into the sysytem - it merely redistributes it. As for calling the author Mark McCutcheon a moron who knows nothing of science,get a reality check!! Your a 'nobody' on an insignificant forum,you have obviously shown to people here that your knowledge on any science topic is fundamentally average,and thats being complimentary. Yet you think by merely spouting Utter crap that you carry some kind of aura.This person who has shown he understands physics far in advance than you,can get his work published and make significant amounts of money. Publishing a book is not a test of scientific merit. Shall we look and see how many science papers Mr McCutcheon has published in scientific journals: http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=FIND+A+MCCUTCHEON%2C+m Doesn't inspire much confidence, does it?
swansont Posted November 11, 2004 Posted November 11, 2004 coming back to the subject... does gravity possess some sort of energy source? it must' date=' according to the fundamental laws of work and energy.... to create the centripetal force, even if the work done by the force tangentially on objects is zero. The idea of gravitons is just as hypothetical as tachyons... neither of them have experimental evidence, although gravitons are predicted by the string theory, which, itself, is in dire need for experimental evidence. But even if we assume gravitons to exist, I don't see how their existence compensates the need for an energy source. Gravitons are simply thought to [i']transmit[/i] the attractive force of gravity, not act as a substitute for an energy source that creates the force. just my thoughts... -mak10 You appear to be thinking of gravity as some sort of engine. Energy doesn't get "used up" so there doesn't need to be a source. Gravity (and E&M and nuclear) exterts a force, and what we measure as energy is a result of that - a force is equivalent to a potential gradient, and vice versa. You also may be running into the problem that physics explains how nature behaves, but at a fundamental level, it does not explain why.
mak10 Posted November 11, 2004 Author Posted November 11, 2004 You appear to be thinking of gravity as some sort of engine. Energy doesn't get "used up" so there doesn't need to be a source. Gravity (and E&M and nuclear) exterts a force' date=' and what we measure as energy is a result of that - a force is equivalent to a potential gradient, and vice versa. You also may be running into the problem that physics explains how nature behaves, but at a fundamental level, it does not explain why.[/quote'] I take gravity to be a force not an engine... so theres no problem there. What I am trying to say here, and I don't think I was clear enough, is that the production of this force, as with any other force, requires an energy source. I know energy doesn't get "used up", in the general sense, but it is required (however small) to produce the force necessary for the attractive properties of gravity. And if we consider gravity to emanate from the center of the earth... then the energy gained or work done on the objects as a result of the exertion is zero, since the force is tangential to motion of objects/people etc. So I am not interested in that. The fundamental forces of nature don't need 'energy' sources. What they are doing is transfering momentum between particles by exchanging a 'force mediator particle'. The photon is the force mediator for electromagnetism. The exchange of a phton does not add energy into the sysytem - it merely redistributes it. could you give me some good links where I can look this up in more detail?? and as far as I know, momentum transfer by photons (or gravitons) involves kinetic energy... which are, then, transferred to other photons. But where did that kinetic energy come from?? from other photons?? where did they get the energy from?? is energy, itself, composed of these photons (as i hear from womewhere, they were energy packets?) or the photons derive their energy from some source?? and it is this source, if it exists, that i am sooooo much interested in.... thanks! -mak10
Severian Posted November 11, 2004 Posted November 11, 2004 could you give me some good links where I can look this up in more detail?? and as far as I know' date=' momentum transfer by photons (or gravitons) involves kinetic energy... which are, then, transferred to other photons. But where did that kinetic energy come from?? from other photons?? where did [i']they[/i] get the energy from?? is energy, itself, composed of these photons (as i hear from womewhere, they were energy packets?) or the photons derive their energy from some source?? and it is this source, if it exists, that i am sooooo much interested in.... thanks! Yes and no. You are right in saying that the photons or gravitons transfer energy between the two bodies which are interacting, but it is a two way process. There is no energy being 'used up' and it can be transfered back and forwards as often as you like. So the Earth and the Sun are presumably exchanging gravitons (and thus energy) backwards and forward between each other (I say 'presumably' since gravitons have not yet been observed). You can find a lot of information sites about particle physics on the web. For example: http://particleadventure.org/index.html http://public.web.cern.ch/Public/Content/Chapters/AboutCERN/WhyStudyPrtcles/WhyStudyPrtcles-en.html http://www.cpepweb.org/ http://www.aps.org/units/dpf/quarks_unbound/
mak10 Posted November 11, 2004 Author Posted November 11, 2004 There is no energy being 'used up' and it can be transfered back and forwards as often as you like. there is a series of particle interactions and energy is transferred via this process with no net loss of energy as such... i understand that perfectly well. but since the interaction mechanisms like collision require energy... the photons, thus, must derive their energy source from somwhere other than their neighbouring photons themselves. to be more explicit, consider this quote from the cern website thats speaking about the elementary particles: Since then, only the enormous concentration of energy that can be reached in an accelerator at CERN can bring them back to life. so energy is required to start the interaction mechanisms which, in turn, transmit energy to other photons/gravitons by elastic collisions, with no net loss of energy. that is the energy.... the first, shall i say, stimulating energy is what I am looking for. so far i havent found any good explanations for even its existence although its existence is absolutely necessary. i will, in the meantime, have a look at the websites some more and see if i can find any info on this. thanx severian! -mak10
Severian Posted November 11, 2004 Posted November 11, 2004 Yes, you are right - in order to transfer energy from one to the other, there has to be energy there in the first place. But each photon is transmitting such a small amount of energy that it can just take it from the matter particles which is feeling the force (which will have plenty energy of its own). For example, the electromagnetic interation between two electrons is the exchange of a photon. One of the electrons emits a photon, whose energy is thus coming from the electron itself, and the photon travels over to the other electron who absorbs it and its energy. This will be happening all the time, with energy passed backward and forwards between the two electrons. PS: This also answers the question in the other thread about whether or not one could still have gravity at zero temperature. If zero temperature were really possible, no particle in the object could emit a graviton since it has no energy to give it and there would be no gravitational force. Since zero temp is impossible, this is a bit academic though...
swansont Posted November 11, 2004 Posted November 11, 2004 PS: This also answers the question in the other thread about whether or not one could still have gravity at zero temperature. If zero temperature were really possible, no particle in the object could emit a graviton since it has no energy to give it and there would be no gravitational force. Since zero temp is impossible, this is a bit academic though... Zero temp doesn't mean zero energy, though. Electrons are still in their orbitals and have energy, so there is still an EM force. It's the center-of-mass KE that goes to zero.
Severian Posted November 12, 2004 Posted November 12, 2004 Hmmmm.... but that isn't free energy. They are not allowed to drop into lower energy states if they are already filled.
swansont Posted November 12, 2004 Posted November 12, 2004 Right. No excess energy, but they still have energy. But virtual particle "borrow" energy from [math] \Delta E \Delta T > \hbar[/math] The EM force isn't going to stop. Why would the gravitational?
Bald Wonder Posted November 12, 2004 Posted November 12, 2004 Question. Using the bowling ball on a trampoline analogy, what force keeps the earth from spiraling into the sun? If you actually place a bowling ball on a trampoline and roll a tennis ball across the surface, the tennis ball's orbit becomes progressively smaller until it collides with the bowling ball. Is it centrifugal force?
Severian Posted November 12, 2004 Posted November 12, 2004 Yes, you are right I think. You would have two photon exchange processes: the electron emits a photon, becoming virtual (ie. having borrowed energy) and then would have to emit/absorb another photon before the [math]\Delta t[/math] was up. It is interesting that that is a higher order process though, so will have a smaller probability of happening. That means that this should be testable: reduce the temperature to 'near' zero and see if the effective charge is reduced. I doubt it is actually visable though since you also have the change in effective charge coming from the variation in [math]\alpha_{em}[/math] from the change of the energy scale.
Severian Posted November 12, 2004 Posted November 12, 2004 Question. Using the bowling ball on a trampoline analogy, what force keeps the earth from spiraling into the sun? If you actually place a bowling ball on a trampoline and roll a tennis ball across the surface, the tennis ball's orbit becomes progressively smaller until it collides with the bowling ball. Is it centrifugal force? Nothing stops it. In principle the orbit will decay eventually. It is just that space is so empty and the Earth is so big (compared to objects which would steal its momentum in collisions) that it will take an incredibly long time to decay - longer than the lifetime of the sun. That is why satelittes have finite lifetimes: they are no so big, so collisions with small dust particles steal a sizable chuck of their momentum and energy, and eventually down they come. Edit: incidentally, in a perfect vacuum, with no other forces involved, the orbit would last 'forever'. It is interesting that this is only true for a law of gravity which is 1/r2. No other distance law gives stable orbits, so it is fortunate indeed.
Bald Wonder Posted November 12, 2004 Posted November 12, 2004 So the principle involved is simply Newton's first law? The earth's inertial centrifugal force equals the sun's centripetal force of gravity, and therefore the net force is zero. Since the external forces acting on the earth are neglible, it continues along it's current path. Is that correct? The bowling ball/trampoline analogy is misleading...
Severian Posted November 12, 2004 Posted November 12, 2004 Yes and no. If you take a Newtonian viewpoint, it is not 'continuing along its currect path' - it is being pulled towards the Earth by gravity. The force is not 'balanced' so the Earth is in 'free-fall' about the Sun. If the Earth did not have any angular momentum it would simply fall in (or alternatively imagine the Sun was a 2d-plane rather than a sphere, again the Earth would just fall down to the surface). It is only because the Earth has so much momentum that by the time it 'falls' toward the sun, it it is already passed the sun, so wings round into empty space again (so to speak). In the Einstein view, you are correct that the Earth has no net force on it. It is travelling in a straight line with no deviation, and only appears to be attracted to a point because of the warped topology of space time. The bowling ball/trampoline analogy is misleading only because of the disappative effects: the tennis ball loses energy too fast and falls into the centre too quickly. Otherwise it is fine.
swansont Posted November 12, 2004 Posted November 12, 2004 Yes' date=' you are right I think. You would have two photon exchange processes: the electron emits a photon, becoming virtual (ie. having borrowed energy) and then would have to emit/absorb another photon before the [math']\Delta t[/math] was up. It is interesting that that is a higher order process though, so will have a smaller probability of happening. That means that this should be testable: reduce the temperature to 'near' zero and see if the effective charge is reduced. I doubt it is actually visable though since you also have the change in effective charge coming from the variation in [math]\alpha_{em}[/math] from the change of the energy scale. I think that's shown to be untrue, though, since the Bose-Einstein condensate theory (AFAIK) doesn't modify the EM interaction, and it behaves as advertised. But I'm not sure of the magnitude of the hypothetical effect.
swansont Posted November 12, 2004 Posted November 12, 2004 So the principle involved is simply Newton's first law? The earth's inertial centrifugal force equals the sun's centripetal force of gravity, and therefore the net force is zero. Since the external forces acting on the earth are neglible, it continues along it's current path. Is that correct? The bowling ball/trampoline analogy is misleading... No. The force is not zero - the "centrifugal force" does not act on the earth. There must be a force acting on the earth, else it would travel in a straight line, per Newton's first law. Centrifugal force is a pseudo force that appears to exist when one is in an accelerating frame of reference, and one still expects Newton's laws to hold. It's also the reaction force to a centripetal force, so it's exerted by the earth, not on the earth.
mak10 Posted November 12, 2004 Author Posted November 12, 2004 There must be a force acting on the earth, what is this force called? what is its source?? are gravitons from the sun responsible for this and how do they manage to hold the earth in its orbit?? another question I had was regarding the attractive force of gravity. from the informative sites Severian gave (for which I am immensely grateful to him!!), this particular site writes: Gravity affects matter and antimatter the same way because gravity is not a charged property and a matter particle has the same mass as its antiparticle. if gravity is not a charged property, then why does it attract?? -mak10
Severian Posted November 12, 2004 Posted November 12, 2004 what is this force called? what is its source?? are gravitons from the sun responsible for this and how do they manage to hold the earth in its orbit?? I am not sure if this is really what you are asking' date=' but the force is gravity and it is coming from the sun. Gravitons haven't been experimentally confirmed but the assuption is that they pass momentum between the sun and the Earth, causing the force on the Earth. the analogy which is often made is two ice skaters throwing a ball between each other. It is not a terribly good analogy since the ice skaters will move away from one another where-as the exchange of gravitons from Earth to sun is supposed to make them attract one another. another question I had was regarding the attractive force of gravity. from the informative sites Severian gave (for which I am immensely grateful to him!!), this particular site writes: if gravity is not a charged property, then why does it attract?? They don't mean it isn't charged in that sense. They mean that there is not a positive and negative charge like there is for electromagnetism. Under gravity objects with mass always attract and by an amount proportional to their mass.
swansont Posted November 12, 2004 Posted November 12, 2004 if gravity is not a charged property' date=' then why does it attract??[/quote'] It's a property of mass itself that causes attraction by gravity/curvature of space It is a property of charges that they attract or repel via the electromagnetic force. It is a property of quarks and hadrons that they attract with the strong nuclear force. It is a property of quarks and leptons that they interact via the weak nuclear force.
bloodhound Posted November 12, 2004 Posted November 12, 2004 Edit: incidentally' date=' in a perfect vacuum, with no other forces involved, the orbit would last 'forever'. It is interesting that this is only true for a law of gravity which is 1/r[sup']2[/sup]. No other distance law gives stable orbits, so it is fortunate indeed. im sorry to butt in, in the middle of nowwhere, but i had to do a question of closed orbits, and it had a note at the end saying [For F proportional to r^n, only Hooke's law, n=1, and the law of gravitational attraction, n=-2, give closed orbits for all bounded motions. J Bertrand, 1873]
Severian Posted November 12, 2004 Posted November 12, 2004 im sorry to butt in' date=' in the middle of nowwhere, but i had to do a question of closed orbits, and it had a note at the end saying [For F proportional to r^n, only Hooke's law, n=1, and the law of gravitational attraction, n=-2, give closed orbits for all bounded motions. J Bertrand, 1873'] Lol. Fair enough, I stand corrected. But Hooke's law is the law which governs a spring - if gravity obeyed Hooke's law it would mean that a planet at the other end of the universe would feel a stronger gravitational attraction to the sun than the Earth would. So I stand by my statement that it is lucky that we have 1/r2... Incidentally, QCD, the force which binds quarks into protons behaves very like Hooke's law. Fortunately the force is not long range because when you pull the quarks apart you put so much energy into the system that you create a quark-antiquark pair straight out of the vacuum. In essence you break the spring.
bloodhound Posted November 12, 2004 Posted November 12, 2004 Yes, my vector calculus lecturer also pointed that we are very lucky in that respect. it does make you wonder...
gib65 Posted November 16, 2004 Posted November 16, 2004 The fundamental forces of nature don't need 'energy' sources. What they are doing is transfering momentum between particles by exchanging a 'force mediator particle'. The photon is the force mediator for electromagnetism. The exchange of a phton does not add energy into the sysytem - it merely redistributes it. According to this, might it be possible (hypothetically) for two bodies who are in the process of exchanging gravitons, to be so far apart that they run out of gravitons to exchange before they collide? Suppose their masses were incredibly different from each other so that the larger one emits gravitons at such a high rate that, even though the smaller one is trying to compensate by exchanging gravitons at as high a rate as it possibly can, it could not possibly keep supplying the larger one with the gravitons it needs to replenish its stock. Is it possible that the larger one could actually run out?
TWJian Posted November 16, 2004 Posted November 16, 2004 F= M1 x M2 x G /D Where M1 and M2 are the respective masses of both objects,G is a constant which equals 7 power of -11 once rounded up(i forgot the exact value[6.98******* or something) and D is the distance between the two masses. Therefore,the weight of a object on earth is around 10N per kg of mass
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