What do we mean by "exist?" That something is in some way present? Assuming the conservation of energy, it would seem that whatever exists requires some quantity of this property called energy. Even if it is the passing speculation of a mind.
Could a platonic realm be said to exist? By definition, no. That would cover what we refer to as laws, as well. Which would seem to be forms abstracted from deeper processes(energy again) in the first place.
Joe Fisher replied on Sep. 25, 2016 @ 15:21 GMT
Dear Thomas, Georginina, John, and Steves Dufurny and Agnew.
The most concise definition you will ever read concerning the physical condition for "to exist, is my irrefutable declaration that the physical construction of the observable Universe must be of the simplest nature permissible. All I am asking you to do is to notice that no matter in which direction you look, you will only ever see a plethora of seamlessly enmeshed varied colored surfaces. You cannot see invisible empty space. There must only be one unified infinite surface that am always illuminated by infinite non-surface light. When one uses the word “is,” one implies a finite state arising from a finite state of was. The word “am” is truly descriptive of a state of infinity.
If you disagree with me, kindly provide me with an explanation for your disagreement.
Thank you for reading my spellbinding material,
Joe Fisher, Realist
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Steve Dufourny replied on Sep. 25, 2016 @ 17:18 GMT
Hi Joe,
I don't know what tell you.I beleive that you have several interesting things to say.I beleive also that your interpretation is philosophical.But You do not give us détails of your thoughts.Speak about the dark matter, the consciousness, the standard model,this and that, the cosmology ....develop your ideas in function of matters.We do not know well your ideas infact.An infinite surface, me I want well but we need more explainations Joe.You can do it,I am persuaded that you know what are the sciences so explain us differently.What is the gravitation for you Joe for example ?
Regards
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John Brodix Merryman replied on Sep. 25, 2016 @ 23:25 GMT
Steve Agnew,
We seem to be in general agreement here(I think). The platonic realm exists as software running on the hardware of neural operations.Which is very much energy.
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Joe Fisher replied on Sep. 26, 2016 @ 15:25 GMT
Dear Steve and John.
Infinite visible surface that is always illuminated by infinite non-surface light does not have any finite details. Dark matter has visible surface. Any philosopher’s complex abstract thoughts about finite invisible consciousness are entertaining, but they have nothing to do with reality. The physical construction of the observable Universe must be of the simplest nature permissible. All I am asking you to do is to notice that no matter in which direction you look, you will only ever see a plethora of seamlessly enmeshed varied colored surfaces. You cannot see invisible empty space. There must only be one unified infinite surface that am always illuminated by infinite non-surface light. When one uses the word “is,” one implies a finite state arising from a finite state of was. The word “am” is truly descriptive of a state of infinity.
If you disagree with me, kindly provide me with an explanation for your disagreement.
Thank you for reading my spellbinding material,
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Akinbo Ojo replied on Oct. 1, 2016 @ 10:22 GMT
John M,
Belief in the assumption of conservation of energy must first tackle without ambiguity what energy is.
We appreciate its effect and what it does but what it is we have not fully understood or grasped.
The second point I will like to make is what happens to this topic if infinity is banished from physics? That is, if space and time are not infinite. Then space would not be infinite in extent nor will it be infinitely divisible. And for time there will be no eternity of duration or an infinitely small duration for things to occur, but a finite smallest time.
This state of affairs would have far reaching consequences for physics, cosmology, etc as I have already discussed on this forum topic. A space that cannot be infinitely divided will have substantial properties.
Regards,
Akinbo
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Joe Fisher replied on Oct. 1, 2016 @ 14:10 GMT
Dear Akimbo and Steve,
The physical construction of the observable Universe must be of the simplest nature permissible. The real Universe must only consist of unified visible infinite surface that am always illuminated by infinite non-surface light. There is no need to define finite invisible energy for nothing invisible has ever existed. Abstract finite invisible matter has never existed, nor has any amount of finite invisible space. Please think utter simplicity.
Joe Fisher, Realist
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Gary D. Simpson replied on Oct. 1, 2016 @ 21:19 GMT
Akinbo,
Conservation of energy is not an assumption or a belief. It is a direct result of time symmetry as viewed by the Noether Theorem.
Best Regards,
Gary Simpson
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Akinbo Ojo replied on Oct. 2, 2016 @ 11:49 GMT
Gary,
I have heard of Noether Theorem even if I have not studied it as a mathematician. Since you have, perhaps I should give you an assignment based on your statement.
We know how to calculate the potential energy (P.E. = -GMm/r) and kinetic energy (K.E. = GMm/2r) of satellites and find the total energy at any position in an orbit (T = -GMm/r + GMm/2r = -GMm/2r). From this we know that satellites lose energy when their orbital radius reduces. We are told such lost energy is radiated away as heat, and for General relativitists as gravitational waves (e.g. pulsar PSR 1913 +16).
When the orbital radius of Earth reduces till it reaches perihelion, the orbital system has lost energy according to energy conservation laws as currently understood. From whence is this orbital energy replenished such that the orbital radius is regained and the Earth rises in altitude again till it reaches aphelion? And if this energy is stored somewhere, where?
Secondly, I have calculated and can defend that the mass of the universe has been increasing based on the Big bang model temperatures in the early era. About 10
-8kg at the Planck time, 10
37kg at about 3 minutes ABT and 10
52kg today, how does this fit with Noether's theorem? Note that I am not against the conservation of energy principle but its better understanding. For instance my answer to the question I am asking you here is to consider radius as 'negative' energy, so that its increase can make total energy in a Big bang model sum to and be conserved at zero.
Joe Fisher,
I agree that the physical construction of the observable Universe must be of the simplest nature permissible.
Regards,
Akinbo
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Eckard Blumschein replied on Oct. 2, 2016 @ 14:17 GMT
Emilia is more known than her brother Hans because she provided a welcome mathematical background for the tenet of time symmetry including the denial of time's arrow and zero of elapsed time by Hilbert. Mirror symmetry of time is obviously nonsensical. Let me therefore put a naive questions: Wirkung (effect) is an integral over time. Did anybody seriously bother about a possible inversion of the direction of this integration? I don't doubt that fictitious negative Wirkung was easily fabricated as also was negative conserved energy. To me, conservation of negative energy sounds a bit too artificial.
Incidentally, I also wonder why "point-like particles" cannot rotate. Well points in the sense of Euclidean definition cannot rotate, however do "point-like particles" actually exist or are they just ideal fictions?
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Joe Fisher replied on Oct. 2, 2016 @ 15:44 GMT
Dear Gary, Akimbo, and Eckard,
Infinite visible surface is not reliant on finite invisible energy. Visible infinite surface does not consist of invisible expanding mass. Infinite visible surface is not made up of invisible particles or invisible atoms.
One more time, The physical construction of the observable Universe must be of the simplest nature permissible. All I am asking you to do is to notice that no matter in which direction you look, you will only ever see a plethora of seamlessly enmeshed varied colored surfaces. You cannot see invisible empty space. There must only be one unified infinite surface that am always illuminated by infinite non-surface light. When one uses the word “is,” one implies a finite state arising from a finite state of was. The word “am” is truly descriptive of a state of infinity.
If you disagree with me, kindly provide me with an explanation for your disagreement.
Joe Fisher, Realist
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Akinbo Ojo replied on Oct. 3, 2016 @ 10:01 GMT
Eckard,
I don't have answer to your question about Wirkung (effect), but I have comment on your wonder about "point-like particles" and whether they can rotate.
To answer, you must first clarify whether point-particle is extended or of zero dimension.
In my opinion, to rotate means for the center of an object to have a different alignment to a position on the object with reference to an external reference frame. From this, something of zero dimension cannot rotate. Also the center of an object that has no boundary on which there can be a position to align with cannot also rotate. That is, a center without a circumference cannot rotate.
Joe Fisher,
Can you provide an equation that cab be used to understand your "one unified infinite surface illuminated by infinite non-surface light"
Regards,
Akinbo
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Anonymous replied on Oct. 3, 2016 @ 15:33 GMT
Dear Akimbo,
All symbolic equations are finite and complex, and only allude to the possible constructions of invisible materials operating in invisible spaces. The real Universe am simple. The real Universe consists only of one unified visible infinite surface that am always illuminated by infinite non-surface light.
Joe Fisher, Realist
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Eckard Blumschein replied on Oct. 4, 2016 @ 04:09 GMT
Yes Akinbo,
Something that has no parts cannot rotate.
I also feel uncomfortable wth Joe's "infinite surface". A surface is something like a limit. Infinite means endless, without a surface. Transfinite is G. Cantor's fabrication as to naively illustrate Leibniz' pragmatism.
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Akinbo Ojo replied on Oct. 6, 2016 @ 09:27 GMT
Gary,
Have you done the calculation and do you have a response?
RE: "We know how to calculate the potential energy (P.E. = -GMm/r) and kinetic energy (K.E. = GMm/2r) of satellites and find the total energy at any position in an orbit (T = -GMm/r + GMm/2r = -GMm/2r). From this we know that satellites lose energy when their orbital radius reduces. We are told such lost energy is radiated away as heat, and for General relativitists as gravitational waves (e.g. pulsar PSR 1913 +16).
When the orbital radius of Earth reduces till it reaches perihelion, the orbital system has lost energy according to energy conservation laws as currently understood. From whence is this orbital energy replenished such that the orbital radius is regained and the Earth rises in altitude again till it reaches aphelion? And if this energy is stored somewhere, where?"
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Gary D. Simpson replied on Oct. 6, 2016 @ 16:00 GMT
Akinbo,
My apologies for the tardiness of my reply. I was away from the internet and did not see your post.
Regarding things in orbit, that is simply a classical mechanics problem. Energy moves between potential energy and kinetic energy. Total energy is conserved. For me, that was taught in Physics 101 and ME 213.
Regarding the Noether Theorem ... I cannot claim to have studies it in detail. My understanding of it is very modest. Ms. Noether died fairly young and did not publish any textbooks although she did publish several papers. As a result, most of the stuff available on the Noether Theorem was not written by Emmy Noether. To me, that makes it more difficult to understand.
When I think of conservation of energy, I am thinking about a specific event or a moment of time. You are thinking more along the lines of things that are the age of the universe. The Second Law of Thermodynamics tells me that time symmetry is not always true. Since conservation of energy requires time symmetry, you might be correct regarding your variable mass-energy statements. The problem is that conservation of energy has never been observed not to be true. Of course, all of the experiments associated with it have a very short time scale.
Best Regards,
Gary Simpson
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Akinbo Ojo replied on Oct. 7, 2016 @ 09:16 GMT
Gary,
Thanks for your reply. It is true that the interchange of energy between potential and kinetic energy was thought in Physics 101. At that time you simply accept what the teacher says so that you can pass your exams. If you ask questions that will make teachers scratch their heads you may be termed naughty and sent out of the class.
Regarding things in orbit, when the satellite crashes into the hub, it is said that all potential energy is converted to kinetic energy just before hitting the ground, at which point it energy gets converted into heat.
If Total energy is conserved in orbits in the claimed way, what we will have is a circle, not an ellipse with the hub at one focus. When orbital radius reduces, potential energy is converted to kinetic energy, but the total energy of the orbit is lost. This is simply classical mechanics as well as General relativity and can be seen from the equation for the Total energy above, i.e. T = -GMm/2r.
If you have the time, you may use real life example of the earth and sun and do the calculation of the interchange between potential and kinetic, but it is much simpler to use the equation for their sum (T).
So the question is, at perihelion (smaller radius) the TOTAL energy is lower than that at aphelion (larger radius), where did this energy replenishing the orbit after perihelion come from? This loss (after aphelion) and replenishment (after perihelion) keeps repeating itself cycle after cycle.
To preserve energy conservation laws (and Noether's theorem that you invoke), an energy store keeper must of necessity be present in the vicinity.
Regards,
Akinbo
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Gary D. Simpson replied on Oct. 7, 2016 @ 13:20 GMT
Akinbo,
You are not applying conservation of energy correctly. When the radius changes, the velocity also changes. Kepler made this observation and expressed it as follows: a line segment sweeps out equal areas during equal time intervals.
When the radius increases, the velocity decreases. When the radius decreases the velocity increases. That is conservation of total energy. The energy does not go anywhere. I think of potential energy as being stored energy. So when the radius is large, the particle has a lot of stored energy and little kinetic energy ... and vis-versa.
Regards,
Gary Simpson
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Akinbo Ojo replied on Oct. 7, 2016 @ 14:47 GMT
Gary,
I get you. I also know of Kepler's laws and the variation of orbital speed with radius.
When a satellite's orbital radius reduces some of its Potential energy (NOT ALL) is converted to Kinetic energy and the rest is radiated away as heat. It appears you are not too familiar with the potential and kinetic energy equations I posted? You can see that with change in radius, only half of the P.E. is converted to K.E.
If the orbital energy is not replenished, the satellite continuously spirals inwards at higher and higher speeds till it eventually crashes. This is what Newtonian theory as well Einstein's theory say as earlier referred to.
There are other related issues, but probably I have not made myself clear enough to discuss them. For instance, when you say, "The energy does not go anywhere. I think of potential energy as being stored energy..., the particle has a lot of stored energy...", one may ask where in the particle is the energy stored. Is the Earth hotter at aphelion than perihelion (disregarding radiation from the sun) as a result of storing energy?
In my opinion, the store of the potential energy is in the space between the orbiting bodies and not in the bodies themselves, just like with the simple pendulum.
Regards,
Akinbo
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Gary D. Simpson replied on Oct. 7, 2016 @ 16:33 GMT
Akinbo,
You are not applying conservation of energy correctly. The conversion between potential energy and kinetic energy is 100% ... not 50%. If your statement were true then there would be virtually no planets or satellites. Only perfectly circular orbits would survive.
I have not studied the equations that you are using, but if they predict only 50% conversion then they are not correct. The apogee/perigee problem is strictly classical mechanics and is a Physics 101 level problem. Total energy is conserved.
Regards,
Gary Simpson
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Akinbo Ojo replied on Oct. 8, 2016 @ 13:59 GMT
Gary,
Can't lay my hands on a good internet reference. I didn't invent the equation for energy in orbits. They are in my tattered Physics 101 book by Nelson and Parker.
You may however check some internet links like
this and
that to verify the equations.
The Total energy sum, both the potential and kinetic is given by -GMm/2r and you can see that this reduces as r reduces and increases as r increases. When the total reduces at perigee, what is lost at that orbital position is no longer in kinetic or potential energy form. And what is present in both kinetic and potential energy forms can also be separately calculated for that orbital position and summed up.
The fact that there are planets and satellites rather than discredit my statement is supposed to spur the search for the agency by which orbits maintain stability. So the question is what is the source of replenishment? Where is the missing energy stored and subsequently released? Note that orbits exist as well on the quantum scale and if we can apprehend a storer and replenisher of energy, such an agent may be doing same work in atomic orbits, in place of ad hoc mechanisms like Pauli's exclusion principle and stationary waves.
Akinbo
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Steve Agnew replied on Oct. 8, 2016 @ 14:51 GMT
I really like the hard intuitive nature of the question about an elliptical versus circular gravity orbit. The classical interpretation seems so simple since the total energy of the orbit is conserved and yet there is a
magical continuous exchange between orbital kinetic and potential energies. In fact, Newton was very leery of using any magic in his gravity especially since Hooke was...
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I really like the hard intuitive nature of the question about an elliptical versus circular gravity orbit. The classical interpretation seems so simple since the total energy of the orbit is conserved and yet there is a
magical continuous exchange between orbital kinetic and potential energies. In fact, Newton was very leery of using any magic in his gravity especially since Hooke was really the person who first supposed the inverse square law of gravity. Hooke was always on Newton about his mysticism and really Newton was a mystic at heart.
The intuitive argument asks by what magic is one form of energy converted into another, but GR gravity answers by simply distorting space and time so orbits follow "linear" geodesics in the distortion of GR spacetime.
However, elliptical orbits do radiate gravity waves and so there is a natural albeit very slow decay of an elliptical orbit into the minimum radiation of a circular orbit. First of all, though, there are tidal forces for elliptical orbits that heat up the two orbiting bodies and that heat radiates as light and tidal heating is usually much more significant than gravity radiation and also means that an elliptical orbit will decay into a circular one as well.
The quantum explanation is analogous. An elliptical orbit represents an excited biphoton state that undergoes biphoton exchange and emission that eventually decays an elliptical orbit into a circular orbit. Entangled biphotons are the exchange particle of gravity and biphoton exchange is how neutral matter moves with gravity. However, even a circular orbit decays because of the much larger effect of tidal friction and heat emission by each of the two bodies. Thus neutral matter can move with single photon exchange as well.
The quantum biphoton exchange of gravity force is what binds all matter to the universe and it is from these bonds that space and time emerge. The gravity orbit of two hydrogen atoms represents the simplest example of the biphoton exchange of gravity. Two orbiting hydrogen atoms transition from quantum single photon exchange to gravity biphoton exchange at some characteristic radius, but the transition is continuous.
Gravity is simply the entangled exchange of a biphoton between the two hydrogen atoms as a quadrupole while charge is the exchange of a single photon between the two neutral atoms as a dipole.
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Gary D. Simpson replied on Oct. 8, 2016 @ 17:06 GMT
Akinbo,
I'm not going to find the error for you. You need to do that to convince yourself that it is an error. Total energy is conserved. It moves between potential and kinetic. I will be away from the internet for several days and will be unable to reply to posts. I suggest that you attempt to derive the equations in question from basic principles. When you use the rule for centripetal force, make sure that you realize that the force of gravity and the centripetal force might not be balanced. If they are not, then the net force will produce an acceleration.
Steve,
Yes ... what you state is generally true ... BUT, the situation described by Akinbo is FAR removed from a situation where GR or gravity waves need to be invoked.
Best Regards,
Gary Simpson
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Robert H McEachern replied on Oct. 8, 2016 @ 17:40 GMT
Akinbo,
Your equation: "We know how to calculate the potential energy (P.E. = -GMm/r) and kinetic energy (K.E. = GMm/2r) of satellites..." has a value for the K.E., that is only valid for (assumes) a circular orbit.
Rob McEachern
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Eckard Blumschein replied on Oct. 9, 2016 @ 06:22 GMT
Steve A,
Your "Existence exists..." sounds funny. I also like your lesson on elliptic and circular movement. Let me add that the circular one corresponds to an oscillation while the elliptical one additionally includes a transient component i.e. the arrow of time.
I am bold enough as to see mirror symmetry of an arrow as Hilbert/Einsteinian nonsense. In a stable system, the transient component decays. Divergence belongs to a closed system.
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Steve Agnew replied on Oct. 9, 2016 @ 15:49 GMT
There is a lot of magic in the action of matter in fields that we tend to forget about, but intuition reminds us of the magic of fields. In the old days, Newton supposed there to be an aether that was the medium that effected all gravity action. Einstein came along and showed that no, there is no medium like aether and it is simply the distortion of space and time that makes objects follow...
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There is a lot of magic in the action of matter in fields that we tend to forget about, but intuition reminds us of the magic of fields. In the old days, Newton supposed there to be an aether that was the medium that effected all gravity action. Einstein came along and showed that no, there is no medium like aether and it is simply the distortion of space and time that makes objects follow geodesic paths, which are straight lines in spacetime.
Thus gravity source and observer do not exchange matter to bond, they each simply follow the determinate geodesic paths of a distorted space and time and therefore commute. However quantum field action occurs by exchange of photons between source and observer. The medium of exchange are aether-like vacuum oscillators called virtual photons and particles, the infinite energy of which is the renormalization that Feynman called
dippy. Source and observer do not commute with single photon exchange, but do with biphoton exchange.
The intuition that elliptical orbits are a little
dippy is actually correct. The gravity field action is limited by the speed of light and GR is one way to show that limit. Another way to show that limit is for a quantum gravity to also be due to exchange and a biphoton exchange provides the right symmetry. Others invent graviton particles but those particles hit the brick wall of GR geodesics. With biphoton exchange just like with single photon exchange, the universe primitives are matter and action and time and space simply emerge from the duality of matter and action.
The orbit of two neutral hydrogen atoms is a useful exercise. Attraction can be due to single photon exchange, which is called dipole-induced-dipole or dispersion and goes as 1/r6 power. Gravity goes as 1/r2 and so even though gravity is 1e39 weaker than charge, eventually gravity biphotons determine the orbit, not dispersive single photons.
Tidal forces are dispersive and so no longer result in heat emission for two orbiting hydrogens and it is only gravity waves that emit and result in decay of elliptical orbit transients. Once the binary hydrogens decay to circular orbits, they no longer emit or decay by themselves and it takes interaction with other matter to decay further into a dispersive orbit and into dust clouds and stars and such. Eventually two hydrogens bond into H2 molecules, which of course is the penultimate quantum entanglement since chemical bonds are the penultimate manifestation of entanglement.
Black holes are the ultimate manifestation of quantum entanglement and represent the destiny of all matter in the decaying aether universe.
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Akinbo Ojo replied on Oct. 10, 2016 @ 14:52 GMT
Gary,
Actually my motivation for bringing up this example was your dogmatic position concerning what constitutes energy conservation or how it is to be interpreted. There is no error to find if you really do the calculation. It is a bitter pill that has remained unresolved but variously explained away as due to perturbation from other planets (despite its predictable and non-random...
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Gary,
Actually my motivation for bringing up this example was your dogmatic position concerning what constitutes energy conservation or how it is to be interpreted. There is no error to find if you really do the calculation. It is a bitter pill that has remained unresolved but variously explained away as due to perturbation from other planets (despite its predictable and non-random occurrence), or a kind of conservation of angular momentum thereby slower speed with longer radius and faster speed with shorter radius (the angular momentum equations don't add up and if forced to do so invalidate the inverse square law), etc. In a gravitational orbit, the force of gravity IS the centripetal force and the acceleration is towards the centre. Where there is deceleration from the centre Newtonian Physics 101 tells us that this cannot be due to gravity but some other force or tendency. It cannot also be a tangential force from inertia (as some claim) as this will have no component in the radial direction where it must act. In short, Deceleration CANNOT be caused by the continuing action or stoppage of action of an accelerating force.
Rob,
I note your comment. In my opinion, the equation applies to all gravitational orbits, both circular and elliptical. If you have any references to the contrary I will love to check them up. In any case in a circular orbit, there is no interchange between potential and kinetic energy and such orbits are rare if at all they exist. From the example of the simple pendulum, where we see such interchange between K.E. and P.E., we know that force MUST be involved to convert P.E. to K.E. and also K.E. to P.E. and the force or agency CANNOT be the same for each process. To convert P.E. to K.E. in the simple pendulum we have the force of gravity. To convert K.E. back to P.E., we have the tension in the pendulum string. Gravity is a force that converts P.E. to K.E.. In gravitational orbits, K.E. is converted to P.E. in an oscillating fashion just as with the simple pendulum. That force converting K.E. to P.E. is waiting to be investigated. It cannot be gravity. I see you have great interest in quantum physics, the possible further importance of such force is that it may have a role in preventing the negatively charged electron from crashing into positively charged atomic nucleus, preventing their overdue embrace under electromagnetic force of attraction. That is while electromagnetic attraction acts to convert P.E. to K.E., some agent acts to convert K.E. to P.E. resulting in a harmonic/ oscillatory phenomenon. I give my suggested agent in my e-book,
Hypotheses Fingo and it is not exotic. The P.E. between galactic clusters is increasing seen as the Hubble flow. Is it conceivable that something that can heave galactic clusters apart will have no effect, no matter how tiny on the quantum level? Can "Pauli's exclusion principle" and "stationary electron wave" convert K.E. to P.E.?
Steve A,
Between a positively charged proton and a negatively charged electron both constituting a neutral hydrogen atom, which kind of photon exchange exists between them, attractive or repulsive type? If attractive, why has the atom not collapsed? If repulsive, why has the electron not escaped? Thanks.
Regards,
Akinbo
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Robert H McEachern replied on Oct. 10, 2016 @ 16:55 GMT
Gary D. Simpson replied on Oct. 10, 2016 @ 22:15 GMT
Akinbo,
You just ain't gettin' it .... Total Energy is conserved. The reason I am dogmatic is because the topic requires dogma. There has never been an observed violation of conservation of energy. That makes it a physical law.
Rather than attempt to rationalize the equation that you are using, I will show you how correctly to apply conservation of energy to the apogee/perigee problem ...
Let subscript A denote apogee. Let subscript P denote perigee.
(PE)subA + (KE)subA = (PE)subP + (KE)subP
The two kinetic energy terms are simply (1/2)mv^2 terms.
The potential energy terms are more complex. This is not because of any deep philosophical problem but rather because the force associated with the potential energy is a function of position. Essentially, it is necessary to integrate between the radius at the perigee to the radius at the apogee. Doing so gives -GMm(1/RsubA - 1/RsubP). Which is similar to what you presented. I would take the datum as being the radius at perigee and define that as zero potential energy.
Total energy is conserved. For this problem, you can specify 3 of the values and the fourth value is then determined by conservation of energy.
So, at perigee, the particle is moving too fast for the circular orbit and it rises. At apogee, the particle is moving too slowly for the circular orbit and it falls. And it can do this FOREVER because total energy is conserved.
Best Regards,
Gary Simpson
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Akinbo Ojo replied on Oct. 11, 2016 @ 10:32 GMT
Rob,
The Orbital Energy link you posted has not been functional for a long time. I tried it before you posted it when looking for a reference to show Gary that the equations are not mine and are well known to those familiar with the topic. Not the best, but I got another link, which after much beating about the bush gives the equation at the "Bottom line". I hope Gary will have time to view...
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Rob,
The Orbital Energy link you posted has not been functional for a long time. I tried it before you posted it when looking for a reference to show Gary that the equations are not mine and are well known to those familiar with the topic. Not the best, but I got another link, which after much beating about the bush gives the equation at the
"Bottom line". I hope Gary will have time to view and appraise with his math skills.
Gary,
Thanks for taking the effort. I provided the conventional equation for TOTAL energy (i.e. sum of both the kinetic and potential energy terms) as -GMm/2r, but you did not provide your own version. If you did this you may better appreciate my point of view. But take note of the following in the interim.
- I am not quarreling with energy conservation law rather I am telling you that for it to be fulfilled there is an agency providing replenishment/ energy storage to sustain the orbit.
- In gravitational orbits (and even atomic), when a satellite moves to a lower orbital radius, total energy is reduced. For an electron in classical orbit, total energy = e
2/8πεr. Since we cannot see atomic orbits, it is easier there to invoke adhoc energy storers and replenishers on that scale in form of Pauli's exclusion principle and stationary waves or even claim the electron has no orbital radius but is rather an 'electron cloud' without position. The classical equation however shows that like the gravitational case, if the atomic orbit is elliptical the total energy of the electron is less at periapsis than at apsis.
- When you say,
"at perigee, the particle is moving too fast for the circular orbit and it rises. At apogee, the particle is moving too slowly for the circular orbit and it falls..."The speed does not determine the rise and fall. A satellite crashing to earth spirals inwards increasing in speed continuously as it does so, it does not rise unless boosted by some force. I think for artificial satellites they use thrusters or so. Using the simple pendulum again for illustration, the pendulum bob 'moves fast' till it reaches "perigee", then instead of continuing to fall it rises
because a force, i.e. the tension in the string makes it to rise. I am sure if that string was invisible, it would have presented a similar dilemma to physics as that which I am trying to illustrate. A particle cannot act on itself according to Newton's first law of motion. Anything that increases or decreases in speed must have been the subject of a force. And again from a combination of Newton's first and second laws (it is not for nothing I use his image as cover in my ebook), the force that causes increase in speed CANNOT be the same force causing a slowing down because forces have direction in which they act. In both the simple pendulum and in gravitational orbits, the force that causes increase in speed and reduction in P.E. is gravity, I am urging you to examine the physics and let us contemplate the force causing the increase in gravitational potential and the reduction in speed either on this forum topic or elsewhere. Deceleration to gravity can certainly not be due to gravity.
- You continue referring to circular orbits, the reality is that orbits are elliptical and not circular. Indeed Kepler wished so much that they would be circular, which seemed to symbolize 'perfection'. Legend has it that it was in a moment of inspiration when he transported himself to the sun that he had to let go and came up with his first law that gravitational orbits are elliptical in shape AS A LAW, not a coincidence.
- Reduction in orbital radius as a sign of total energy loss cuts across Newtonian and General relativity. It is on that basis that it is claimed that the reducing radius (signified by shorter periods) of pulsar PSR 1913 +16) represented the emission of gravitational waves. Two people got a Nobel prize for this. I do not necessarily accept gravitational waves as currently conceived.
Whatever it is, if you still insist that a reduction in orbital radius is not tantamount to a loss in total energy I have tried enough to make my case and we can let the matter rest.
Best regards,
Akinbo
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Robert H McEachern replied on Oct. 11, 2016 @ 12:39 GMT
Akinbo,
At the bottom of the "bottom line" web-site, it clear states:
"Potential and Kinetic Energy in a Circular Orbit"
The equations given for the K.E., only apply to circular orbits.
Rob McEachern
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Gary D. Simpson replied on Oct. 11, 2016 @ 19:10 GMT
Akinbo,
I am not willing to give up on you yet.
If you will provide the specific numbers you are using, the exact form of the equations that you are using, and a step by step description of how the numbers and equations interact, then I will review what you present in detail. I will advice you what - if anything - that I perceive as incorrect.
BTW, the way that an orbit is changed is to fire thrusters in the TANGENTIAL direction. The spacecraft then accelerates and moves to a higher orbit because gravity and centripetal force are not in balance. The resulting orbit will then be elliptical unless retro-thrusters are applied to reduce speed as needed to match the desired orbit.
Best Regards,
Gary Simpson
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Gary D. Simpson replied on Oct. 12, 2016 @ 01:58 GMT
Akinbo,
I think I know what you are doing. You are comparing the energies of two separate orbits rather than making a comparison of the energy of one orbit at two different positions.
Regards,
Gary Simpson
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Akinbo Ojo replied on Oct. 12, 2016 @ 10:19 GMT
Rob,
"The equations given for the K.E., only apply to circular orbits."No sir. The major assumption that made Newton finally find an explanation to Kepler's first and second laws that orbits are elliptical (varying radius) and not circular (constant radius) was the singular postulation of an inverse square relationship. It is this that also explained Kepler's second law...
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Rob,
"The equations given for the K.E., only apply to circular orbits."No sir. The major assumption that made Newton finally find an explanation to Kepler's first and second laws that orbits are elliptical (varying radius) and not circular (constant radius) was the singular postulation of an inverse square relationship. It is this that also explained Kepler's second law whereby the satellite moves faster at smaller radius and slower at larger radius. In a circular orbit, satellite speeds will be constant. From Newton's law, the satellite
irrespective of its radius from the centre obeys the inverse square law throughout the orbital cycle. My further comments to Gary may elaborate further.
Gary,
"If you will provide the specific numbers you are using, the exact form of the equations that you are using,..."Thanks to technology, truth can no longer be hidden or swept under the carpet. I simply took a photo of the relevant page in my old tattered Nelkon and Parker, and pronto converted it to pdf :) (see attached). Hope it is clear.
If not, in shortened form from Newtonian equations:
F = GMm/r
2F = mv
2/r
Therefore GMm/r
2 = mv
2/r,
K.E. = ½mv
2 = GMm/2r
Potential energy can be similarly derived.
P.E. = mgh, and acceleration due to gravity, g is GM/r
2So, P.E. = GMm/r.
To fit the convention that potential energy increases with height and reduces with height we have the minus sign. Thus,
P.E. = -GMm/r
Rob, should note that mv
2/r operates throughout the cycle irrespective of increasing or reducing radius as we see in elliptical orbits. If mv
2/r operates thus, it will be illogical that the further mathematical derivations of it would not operate. In any case these equations are well used in our space exploits both for circular, elliptical or parabolic orbital paths.
"I think I know what you are doing... comparing the energies of two separate orbits rather than making a comparison of the energy of one orbit at two different positions."No sir. One orbit, with the satellite at different positions. Cyclical inter-conversions between P.E. and K.E. occurs only in elliptical orbits.
The question I posed above that you both have not examined or volunteered an opinion to is that given Newton's laws of motion and gravity, whether a mass can increase in speed without any force acting on it, and secondly whether if it increases in speed due to the action of a force whether that same force can be held responsible if it is observed to subsequently slow down. All these are related to the energy question as it indicates the presence of a force operating in orbits, what some may call, "a ghost in Newton's clockwork theory", preventing gravitational orbits from collapsing. Or more figuratively preventing the union of the earth and moon despite billions of years of affection for each other. And for Rob, who is interested in quantum physics that same ghost may be responsible for the stability of atomic orbits. Hear Penrose on this:
"A fundamental and seemingly insurmountable problem is that as an orbiting electron circles the nucleus it should, in accordance with Maxwell's equations, emit electromagnetic waves of an intensity increasing rapidly to infinity, in a tiny fraction of a second, as it spirals inwards and plunges into the nucleus! However, nothing like this is observed. Indeed, what is observed is quite inexplicable on the basis of classical theory...", p.295,
The Emperor's New Mind.
Regards,
Akinbo
*It has been a nice discussion. I don't know if we should continue...
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attachments:
orbit_equations.pdf
this post has been edited by the author since its original submission
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Gary D. Simpson replied on Oct. 12, 2016 @ 13:18 GMT
Akinbo,
I have downloaded the .pdf file and will study it shortly.
Based upon the discussion contained within your post, you have made the error that I suspected you made. You do not understand centripetal force. This is a fairly significant error that you must correct prior to having any realistic chance of understanding more advanced subject matter.
The equations that you are using are in fact for circular orbits irrespective of how much you claim that they are not. Specifically, you have set F equal to F ... or at least that is what you think you have done. In truth, what you have done is to set the force of gravity equal to the centripetal force due to motion. This is only true for circular orbits. You need to understand in your own mind why this is so.
For a circular orbit, centripetal force and gravity are exactly in balance. For such a case, velocity and tangential velocity are equal and the radial velocity is zero. For non-circular orbits, the radial velocity is zero only at apogee and perigee, but the associated velocity is not the same as it would be for a circular orbit. Because of all this, the kinetic energy terms that you are using are not correct.
You absolutely must come to understand this.
Best Regards,
Gary Simpson
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John R. Cox replied on Oct. 12, 2016 @ 15:51 GMT
Akinbo,
I've looked in on this just a couple of times and haven't tried to follow the arguments, so pardon my intrusion. But just superficially it seems to me that what you are fascinated with is that 'old perpetual motion problem'. I mean, how can it be that the music of the spheres can apparently continue if we accept a closed universe where all the mass/energy that now exists, all came into being all at once like some book of genesis. Did you ever consider that it was consciously because of that puzzle, that Einstein chose to remove Force from the equation and sought a formulation using just the terms of mass and acceleration to describe inertial motion? Note that removing Force as a term, does not remove the Field from any region of space. So in GR, an elliptical orbital does not physically change in velocity, it simply follows a line of equilibrium between the two masses along which the speed of time is constant. The elephant in the room is; why G? :-) jrc
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Robert H McEachern replied on Oct. 12, 2016 @ 18:37 GMT
Akinbo,
The equations you are using, are not Newton's. That is the problem. The equation you are using, for the K.E., ONLY applies to circular orbits - read your own references, they clearly state that to be the case.
Rob McEachern
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Gary D. Simpson replied on Oct. 12, 2016 @ 19:22 GMT
Akinbo,
The reference that you provide clearly states "circular orbit".
This might make it more understandable ... when the gravity and centrifugal forces are equal in magnitude with opposite directions, they sum to zero. This physically means that the system is at "static equilibrium" in the radial direction. That means that there is no acceleration in the radial direction and that the radius is constant.
It is very important that you understand these concepts.
Best Regards,
Gary Simpson
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Akinbo Ojo replied on Oct. 12, 2016 @ 19:43 GMT
Gary,
It is my opinion that the centripetal force is analogous to gravity in gravitational orbits but I will not belabor the issue.
RE:
"Because of all this, the kinetic energy terms that you are using are not correct."I had previously asked you to supply me your own terms for P.E., K.E. and their sum total, T for elliptical orbits so that we can lay this matter to rest. In doing so, only the terms G, the gravitational constant, M, the mass of the central body, m the mass of the orbiting body and r, their distance apart preferrably needs to feature. But if you are uncomfortable with that restriction just provide your formula for the total energy sum of an elliptical orbit.
JRC,
It is quite possible that Einstein perceived that something was amiss in gravitational orbits. An anti-gravity force seemed to be required to counter gravity and prevent orbital decay. His cosmological constant idea if applied to the problem might have shed some light. It is not in contention even in GR that in elliptical orbits there is physical change in velocity. I don't really understand what you mean by a line of equilibrium between the two masses. But if I may interpret it my way, that line keeps shortening and extending in a cyclical fashion. We can allude the shortening of the line to an attraction tendency. It is the repulsion or extension that needs to be explained for completion of our understanding.
Rob,
If you will be kind enough, like I asked Gary, can you supply your own equations for an elliptical orbit, not just for K.E. but also for the total energy (P.E. + K.E.). This should hopefully lay this matter to rest.
Regards,
Akinbo
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Akinbo Ojo replied on Oct. 12, 2016 @ 20:00 GMT
Gary,
Just saw your post on Oct. 12, 2016 @ 19:22 GMT. It is night time here, but I discerned something from,
"... when the gravity and centrifugal forces are equal in magnitude with opposite directions, they sum to zero. This physically means that the system is at "static equilibrium" in the radial direction. That means that there is no acceleration in the radial direction and that the radius is constant."What of when the radius is NOT CONSTANT but oscillates about an equilibrium value resulting in an elliptical shape? From what we were taught about harmonic motion, would this not suggest that there are two return forces maintaining the orbital stability as your post hints. If one prevails, the radius shortens till the bodies collapse, if it is the other the radius lengthens till the orbiting body escapes. But the two tendencies are capable of setting up a vibratory phenomenon which is what we observe in reality and not a static phenomenon. It is that centrifugal force that I want you to educate me about!
It is the force that can increase P.E. in the orbit. I think we are getting somewhere. Even if you can't tell me everything about the centrifugal force, from your post it looks like you agree that it acts radially OUTWARDS and not tangentially. A tangential force would have no component in a radial direction.
Akinbo
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Robert H McEachern replied on Oct. 12, 2016 @ 23:20 GMT
Akinbo,
Classically (non-relativistic) the Total Energy, E, is given by:
E=(mv^2)/2 - GMm/r.
The first term is the K.E. In general the K.E. has no simple expression as a function of r. However, in the special case of a circular orbit, (mv^2)/2 = GMm/2r.
Rob McEachern
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Gary D. Simpson replied on Oct. 13, 2016 @ 00:33 GMT
Akinbo,
You are being willfully obstinate. I am leaving town shortly and will be away from the internet.
Let us review a few relevant posts.
1. You have claimed that total energy for an elliptical orbit is not conserved. You have stated that the total energy at apogee is not equal to the total energy at perigee. I posted to this thread because that statement is incorrect. This...
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Akinbo,
You are being willfully obstinate. I am leaving town shortly and will be away from the internet.
Let us review a few relevant posts.
1. You have claimed that total energy for an elliptical orbit is not conserved. You have stated that the total energy at apogee is not equal to the total energy at perigee. I posted to this thread because that statement is incorrect. This was my only interest in the thread. I do not want for you to waste your time/effort on an erroneous line of reasoning. Your claim is false and you absolutely must understand why it is false.
2. The basis for the claim listed in 1) is the set of equations presented in the file named "orbit equations.pdf". That file clearly states that it is applicable to circular orbits. It makes this statement in lines 2 and 3.
3. The reason that the equations presented in the reference in 2) are limited to circular orbits is because the velocity used in the kinetic energy calculation is the velocity associated with a circular orbit. The equation for potential energy is correct for any orbit. It can even be used to determine escape velocity.
4. I use (1/2)mv^2 for KE same as everyone else. The equation that you present for PE is acceptable to me. The reason that I have not presented a special equation as you have requested is because it is not useful. Instead, the problem is solved as follows:
Total Energy = PE + KE = CONSTANT
Get the velocity and radius at a point in the orbit ... usually either apogee or perigee. Determine the Total Energy. This energy is constant at all points in the orbit. Now pick another point in the orbit. You may specify either the velocity or the radius. This allows you to calculate either the kinetic energy or the potential energy. Now solve for the other energy term using the equation for total energy. You may now solve for either radius or velocity.
5. Of course centripetal force is in the radial direction ... do not try to obfuscate the discussion. The velocity used to calculate the centripetal force is the tangential velocity. The radial velocity contributes to kinetic energy but not to centripetal force. When gravity and centripetal force are not in balance then there is acceleration in the radial direction and Newton's Second Law is used. This is the essence of your mistake.
You absolutely must understand these concepts. Otherwise, your thinking will be complete gibberish.
Best Regards,
Gary Simpson
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Gary D. Simpson replied on Oct. 13, 2016 @ 00:41 GMT
Akinbo,
Forgot to mention ... in your reference file, the author discusses two different orbits. The second orbit is the result of friction taking energy out of the first orbit. So, the energy difference is equal to the work performed on the satellite by friction with the atmosphere. That is not a single elliptical orbit. Instead it is two circular orbits.
Best Regards,
Gary Simpson
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Akinbo Ojo replied on Oct. 13, 2016 @ 10:25 GMT
Rob,
Thanks for volunteering an equation for Total energy that we can use for elliptical orbits, i.e. E=(mv^2)/2 - GMm/r
There is actually a simple expression for v as a function of r applicable to all orbits, both circular and elliptical and it is:
v
2 = GM/r
This equation is in use to find the earth's velocity to be higher at perigee and at apogee, knowing the radius, r and the sun's mass, M. So that resolves the question of whether the equation is applicable only in circular orbits and not elliptical. If v
2 = GM/r is correct then the total energy, K.E. + P.E. from your (mv^2)/2 - GMm/r is equal to -GMm/2r
If v then has a simple expression as a function of r and the mass of the satellite m is unchanged then I have no ready response to the claim that K.E. has no simple expression as a function of r.
Gary,
Some of my previous responses already address the points you raise. With regard to
"3). The reason that the equations presented in the reference in 2) are limited to circular orbits is because the velocity used in the kinetic energy calculation is the velocity associated with a circular orbit. The equation for potential energy is correct for any orbit. It can even be used to determine escape velocity", my response to Rob addresses this. The expression v
2 = GM/r
applies to elliptical orbits and you can substitute the sun's mass, perihelion and aphelion distance to respectively find that v is 30,124m/s at perihelion and 29,625m/s at aphelion (take sun's mass, M = 2 x10
30kg, perihelion radius =1.47 x10
11m, aphelion radius = 1.52 x10
11m).
Then in 4), you say,
"Get the velocity and radius at a point in the orbit ... usually either apogee or perigee. Determine the Total Energy HOW?". Thanks for the exchanges, even if we are yet to reach a consensus. Enjoy the break.
Regards,
Akinbo
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Robert H McEachern replied on Oct. 13, 2016 @ 14:02 GMT
Akinbo,
"This equation is in use to find the earth's velocity to be higher at perigee and at apogee, knowing the radius, r and the sun's mass."
It does not work. Think about it. According to your false statement, ALL orbiting objects, regardless of the differing shapes of their orbits, would have to have EXACTLY the same velocity, whenever they are the same distance from the object they are orbiting. They do not.
Rob McEachern
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Akinbo Ojo replied on Oct. 13, 2016 @ 18:25 GMT
Very brilliant inference Rob! Indeed ALL objects in a gravitational orbit would have the same velocity whenever they are the same distance from the object they are orbiting. It is counter-intuitive at first glance but it is the reality borne out of the math formula applicable. Legend has it that Galileo was similarly astonished that a feather and a ball of lead thrown down from the Leaning tower of Pisa both reached the ground at the same time (if we discountenance air resistance).
This state of affairs is also found in Kepler's second and third laws, whereby the area of the ellipse swept and the period of an orbit (time taken to complete a cycle) is only dependent on the mass, M of the central body and not the nature or mass, m of the orbiting body.
This is the mathematical reason:
F = GMm/r
2F = mv
2/r
Thus,
GMm/r
2 = mv
2/r
m cancels out on both sides to give
GM/r
2 = v
2/r
As we are the only two left in the boxing ring (Gary is off the topic for now), perhaps we live to "fight" another day.
Regards,
Akinbo
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Robert H McEachern replied on Oct. 13, 2016 @ 19:23 GMT
Akinbo,
If you drop two objects straight down they are in the same orbit, so move at the same speed. If you drop one and throw the other horizontally, they are not in the same orbit and do not travel at the same speed, even though they are at the same radius from the center of the earth, as they fall; they have the same vertical velocity component, but differing horizontal components. Hence, they have different kinetic energies.
You need to actually READ a physics book, rather than just pulling out a equation that you do not understand when it is applicable, and when it is not.
Rob McEachern
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Steve Agnew replied on Oct. 14, 2016 @ 04:05 GMT
This is a very basic question about quantum mechanics and has to do with Bohr's argument that avoided Planck's UV catastrophe. By assuming that the electron accelerated to the speed of light and therefore became a wave that was in a resonance with itself between that wave and the electron is what is quantum mechanics is all about.
Akinbo Ojo replied on Oct. 10, 2016 @ 14:52 GMT as "Steve A, Between a positively charged proton and a negatively charged electron both constituting a neutral hydrogen atom, which kind of photon exchange exists between them, attractive or repulsive type? If attractive, why has the atom not collapsed? If repulsive, why has the electron not escaped? Thanks.Regards,Akinbo.Electrons do not collapse into nuclei because of quantum resonance...at least until electrons collapse into protons to form neutrons, but that is another story.
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Akinbo Ojo replied on Oct. 14, 2016 @ 11:06 GMT
Rob,
Whether you drop two objects straight down or you drop one and throw the other, they will all fall and reach the ground at the SAME time. This is because they are subject to the same vertical acceleration due to gravity. A bullet fired horizontally will reach the ground same time as a dropped object.
The kinetic energies will be different because their masses are different. The fact that they move at the same speed (v = √GM/r) does not imply same kinetic energy. In all the expressions of kinetic energy I have given, be it ½mv
2 or GMm/2r which all come from textbooks, the mass, m of the falling or orbiting object counts. Gary posted some textbook stuff you may crosscheck.
Steve,
I believe by resonance you are referring to Bohr's stationary wave model? If so, there is an absurdity with the model which has probably motivated its replacement by a so-called electron cloud rather than a classical orbit. Stationary or standing waves occur when two equal waves travel opposite to each other and interfere. In an hydrogen atom how its single electron will travel in two opposite directions simultaneously is mysterious, if not absurd.
Akinbo
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Steve Agnew replied on Oct. 15, 2016 @ 14:28 GMT
The resonance between the electron and proton is set by the Schrodinger equation, the foundation of quantum. The electron is not a classical particle when it is bound in an atom, but an electron does radiate photons in all EM field gradients just like a gravity body radiates gravity waves in all gravity field gradients.
In hydrogen, the electron and proton are bound to each other by a game of catch with the Rydberg energy photon at 13 eV that represents a mass defect or loss. A Rydberg photon was emitted at the CMB creation when that hydrogen first formed and that photon paired with the binding photon is called biphoton gravity. The entangled exchange of these two photons defines quantum gravity and what we call the time size of the universe and sets the scale between charge and gravity at 1e39th power.
It is better to think of the action of bodies as first of all an exchange of photons and simply let the notions of motion, space, and time all emerge from the action of that exchange. Two bodies in orbit exchange the biphotons of gravity with each other and that is what provides acceleration. Velocity and kinetic energy are simply the result of acceleration and the exchange of potential and kinetic energy and constant total energy is simply a result of the virial theorem for a closed system.
Constant total energy is a very good approximation for orbit geodesics as Newton's law and the virial theorem, but does not include gravity waves. Gravity waves from GR along with tidal friction and heat emission affect gravity orbits too and work well from the Planck scale to the BH event horizon. Quantum gravity is then what describes the rest of the universe from the scale of discrete aether to the final destiny of all matter as the quantum spin of BH's.
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