D
Need help with maths for new theory of physics
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I am sure that this interaction has been quite a learning experience for me. However, since you are all such savants in this field, I would like you to demonstrate just this once, using your theories and your skills in mathematics. A single problem that I will set forth here:
There is a point source of light powered by a 100 W lamp and yielding about 1600 lumens radiating non-collimated light. For ease of obtaining a solution, assume that the light source is isotropic. Using your mathematics and explaining every step of the way how you reach your conclusions. Could you state the density of light at (a) 50 metres from the source (b) 100 m from the source, (c) 150 m from the source and (d) 200 m from the source. Explain (a) how the light travels to each of these points using the quantum wave function (b) explain how the wave function collapses at each of these points when the light is detected and the effect of this collapse (c) explain the intensity of the light at each of these points,(d) explain how far the light spreads out at each of these points and (e) explain the energy of the photons that is detected at each of these points. Jeff you are on.
Topsquark, claims to favour QED, in which case let him explain, radio-waves by stating how an electron begins to vibrate in a wire giving rise to the emission of a photon of a certain energy, that photon spontaneously undergoes annihilation, in the process giving rise to an electron and a positron. The electron and the positron, since they represent matter and antimatter, very soon undergo mutual annihilation giving rise in the process to a photon with guess what? Yes, exactly the same energy as the photon that was originally emitted. This is how radio-waves propagate according to quantum Electrodynamics. If you don’t believe me look it up.
Finally, if you perform the above task and perform it sincerely to the best of your abilities. If you like, I can then solve all of the questions, using the formula I had written. It might not be the prettiest mathematical formula but it works!
There is a point source of light powered by a 100 W lamp and yielding about 1600 lumens radiating non-collimated light. For ease of obtaining a solution, assume that the light source is isotropic. Using your mathematics and explaining every step of the way how you reach your conclusions. Could you state the density of light at (a) 50 metres from the source (b) 100 m from the source, (c) 150 m from the source and (d) 200 m from the source. Explain (a) how the light travels to each of these points using the quantum wave function (b) explain how the wave function collapses at each of these points when the light is detected and the effect of this collapse (c) explain the intensity of the light at each of these points,(d) explain how far the light spreads out at each of these points and (e) explain the energy of the photons that is detected at each of these points. Jeff you are on.
Topsquark, claims to favour QED, in which case let him explain, radio-waves by stating how an electron begins to vibrate in a wire giving rise to the emission of a photon of a certain energy, that photon spontaneously undergoes annihilation, in the process giving rise to an electron and a positron. The electron and the positron, since they represent matter and antimatter, very soon undergo mutual annihilation giving rise in the process to a photon with guess what? Yes, exactly the same energy as the photon that was originally emitted. This is how radio-waves propagate according to quantum Electrodynamics. If you don’t believe me look it up.
Finally, if you perform the above task and perform it sincerely to the best of your abilities. If you like, I can then solve all of the questions, using the formula I had written. It might not be the prettiest mathematical formula but it works!
The Double Slit Experiment is probably the most famous experiment that proves that quantum mechanics is right or maybe it is the experiment that proves that QM is wrong. In the double slit experiment two narrow slits are made in a screen. When only one of the slits is open what was expected to be seen was the single slit, sharply reproduced on the projection screen. Instead what was seen was a diffraction patter that spread out in a kind of blob of light. When both slits were open, the expectation was that there would be a much bigger blob of light of the same shape that was seen when only one slit was open. To everyone’s surprise what was seen when both the slits were open was an interference pattern. This proved that light was not a particle as Isaac Newton had surmised, but a wave. Only waves can undergo interference. Quantum mechanics , modified the double slit experiment by making the slits much smaller and arranging for a suitable light source that could send out photons at single file intervals.
When the experiment was tried with only a single slit open the expected diffraction pattern was seen but when the experiment was repeated with only one slit open an amazing result was seen. Even though the photons were directed at a single slit, what was seen after a suitable passage of time was an interference pattern. This raised a number of very interesting questions. Did the photons know that both slits were open? Was it possible that some of the photons were able to pass through both slits simultaneously? How did the photons know to go to certain places and avoid others in order to build up an interference pattern? It was an absolute conundrum and was proof beyond doubt of the quantum strangeness of things.
An even stranger situation was created when photon detectors were placed next to each of the slits. A strange thing now happened, as soon as the detectors were switched on, the interference pattern disappeared and the diffraction pattern re-appeared. This raised the even more hair raising possibility that the photons were actually aware that they were being observed. What could be happening?
Taking the first example, what appears to be the absolute proof of quantum quaintness and strangeness, could in fact turn out to be the exact opposite, namely the proof that quantum mechanics was wrong and that the aether does in fact exist. Consider, the aether, think of its properties. One of its most prominent traits is its very low interaction with matter. This means that if the aether exists, it will naturally flow through all the objects in its vicinity, more often than not taking the easiest route. Imagine then what would happen if the aether encountered the two open slits; it would naturally pass through both slits as being the path of easiest access. The aether being what it is would also as a result form an invisible interference pattern on the screen. When the photons were released, what could be more natural than that they would follow the path of the aether as being the path of least resistance, with the result that an interference pattern is built up. This is similar to Bohm’s pilot wave theory but is in actual fact very different since it has nothing to do with pilot waves and everything to do with the presence of an aether.
Interestingly this result also explains why the interference pattern disappears when the photon detectors are switched on. When the detectors are switched on they create their own pattern and the pattern created by the aether is disrupted resulting in the disappearance of the interference pattern and the reappearance of the diffraction pattern.
Which explanation sounds more likely, that: that photons are aware of where to go, that photons can be in two places at once, that photons somehow know when they are being observed, that they are precogniscant OR that they are merely following the pattern made by the all pervasive aether.
When the experiment was tried with only a single slit open the expected diffraction pattern was seen but when the experiment was repeated with only one slit open an amazing result was seen. Even though the photons were directed at a single slit, what was seen after a suitable passage of time was an interference pattern. This raised a number of very interesting questions. Did the photons know that both slits were open? Was it possible that some of the photons were able to pass through both slits simultaneously? How did the photons know to go to certain places and avoid others in order to build up an interference pattern? It was an absolute conundrum and was proof beyond doubt of the quantum strangeness of things.
An even stranger situation was created when photon detectors were placed next to each of the slits. A strange thing now happened, as soon as the detectors were switched on, the interference pattern disappeared and the diffraction pattern re-appeared. This raised the even more hair raising possibility that the photons were actually aware that they were being observed. What could be happening?
Taking the first example, what appears to be the absolute proof of quantum quaintness and strangeness, could in fact turn out to be the exact opposite, namely the proof that quantum mechanics was wrong and that the aether does in fact exist. Consider, the aether, think of its properties. One of its most prominent traits is its very low interaction with matter. This means that if the aether exists, it will naturally flow through all the objects in its vicinity, more often than not taking the easiest route. Imagine then what would happen if the aether encountered the two open slits; it would naturally pass through both slits as being the path of easiest access. The aether being what it is would also as a result form an invisible interference pattern on the screen. When the photons were released, what could be more natural than that they would follow the path of the aether as being the path of least resistance, with the result that an interference pattern is built up. This is similar to Bohm’s pilot wave theory but is in actual fact very different since it has nothing to do with pilot waves and everything to do with the presence of an aether.
Interestingly this result also explains why the interference pattern disappears when the photon detectors are switched on. When the detectors are switched on they create their own pattern and the pattern created by the aether is disrupted resulting in the disappearance of the interference pattern and the reappearance of the diffraction pattern.
Which explanation sounds more likely, that: that photons are aware of where to go, that photons can be in two places at once, that photons somehow know when they are being observed, that they are precogniscant OR that they are merely following the pattern made by the all pervasive aether.
topsquark
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Just to be clear I do not "favor" QED. I consider it to be correct because it has been tested and verified.
The problem you wanted me to resolve using QED really can't be. The Math for doing QED over large distances is too opaque even for the experts in the field. Fortunately we don't need QM for this one. All you need is the intensity and the fact that the intensity falls off by the inverse square law. As to the photon field collapsing that's what we call it. We don't really have any idea what happens when the field collapses... all we know is that the wavefunction falls into one of its eigenstates. If this sounds like I am saying we don't really know what's going on, you are right. We don't. But we know the rules and can calculate it and get the correct answers.
As to the double slit experiment I had to read your analysis over a few times. The words "interference" and "diffraction pattern" mean the same thing! When both slits are open and we don't detect the particles then there is a diffraction pattern. When we detect which slit the photon goes through we lose the diffraction pattern. That being said the experiment acts as if light were acting like a wave as you are suggesting. Young's original double silt experiment toppled Newton's idea of a "corpuscle," what we now call a photon. This experiment does indeed suggest some kind of an aether.
BUT, in 1900 Max Planck was able to explain the blackbody radiation curve using, not waves, but little packets of light energy. Light was behaving like a particle, not a wave. (Actually he thought the quantized aspect of the photon energies had something to do with the cavity walls but Einstein was able to use the energy packets in his photoelectric effect paper, showing that it actually is light energy that is quantized.) A particle does not need an aether to propagate.
It turns out if we try to measure light as a wave we get a wave. If we try to measure it as a particle we get a particle. It's one of the weirder aspects of QM and it's just the first one to crop up. So far as I know no one has been able to describe just what an elementary particle is.
There is a lot about the History of QM that you don't know. There's a whole bunch of books out there... They are worth a read.
-Dan
The problem you wanted me to resolve using QED really can't be. The Math for doing QED over large distances is too opaque even for the experts in the field. Fortunately we don't need QM for this one. All you need is the intensity and the fact that the intensity falls off by the inverse square law. As to the photon field collapsing that's what we call it. We don't really have any idea what happens when the field collapses... all we know is that the wavefunction falls into one of its eigenstates. If this sounds like I am saying we don't really know what's going on, you are right. We don't. But we know the rules and can calculate it and get the correct answers.
As to the double slit experiment I had to read your analysis over a few times. The words "interference" and "diffraction pattern" mean the same thing! When both slits are open and we don't detect the particles then there is a diffraction pattern. When we detect which slit the photon goes through we lose the diffraction pattern. That being said the experiment acts as if light were acting like a wave as you are suggesting. Young's original double silt experiment toppled Newton's idea of a "corpuscle," what we now call a photon. This experiment does indeed suggest some kind of an aether.
BUT, in 1900 Max Planck was able to explain the blackbody radiation curve using, not waves, but little packets of light energy. Light was behaving like a particle, not a wave. (Actually he thought the quantized aspect of the photon energies had something to do with the cavity walls but Einstein was able to use the energy packets in his photoelectric effect paper, showing that it actually is light energy that is quantized.) A particle does not need an aether to propagate.
It turns out if we try to measure light as a wave we get a wave. If we try to measure it as a particle we get a particle. It's one of the weirder aspects of QM and it's just the first one to crop up. So far as I know no one has been able to describe just what an elementary particle is.
There is a lot about the History of QM that you don't know. There's a whole bunch of books out there... They are worth a read.
-Dan
Think about what you are saying Dan. If what you claim is true then we could all go home perfectly satisfied with nothing to disagree about or discuss. Have you forgotten that light (EM Radiation) is a wave? And that when it is required it can be a particle just like that? Like pinnochio changing from a wooden puppet into a real boy. Nothing in between, no half wave, half particle but a complete solid particle when the need arises and a completley anomalous wave when it is needed. If all that were needed were to calculate the intensity and the area over which the light spread. Huygen's would be as good as anyone else. Will get back, have a little work to do.
topsquark
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EM radiation is not made of either waves or particles of light. It isn't half wave and half particle... Many call photons "wavicles" to make the distinction. Sorry, but that's the best theory we have. You aren't going to be able to change it simply by saying that light must be a wave. It just isn't true. It's not a matter of belief it's a matter of experimental results going back over a hundred years. Do you think that no one has ever thought as you do? Do you think that no one has ever tried to prove QM wrong? Einstein himself came up with a number of experiments in order to invalidate QM. All he ended up doing was to make the evidence for QM even stronger.
As I said before: There is no macroscopic analogue for what a quantum particle is. You simply aren't going to be able to understand it. No one does. But you can measure their properties and use QM (or more specifically QFT) to figure out what they are going to do in given situations. And the experimental measurements turn out to support QM.
-Dan
A characteristic of waves is that they spread out or disperse, as they do so, their shared energy is reduced inversely with the square of the distance they travel. This being so it is axiomatic that a wave which starts off with a certain energy at its source will not maintain that same energy when it reaches its destination. In the scenario proposed by yourself, to merely calculate the intensity of the wave from the inverse square of its distance, would leave too many questions unanswered, because each individual photon in that wave-front maintains it original energy. Imagine that the light in the question I had posed was red light with a wavelength of 650 nm. How is it that after travelling 50m, or 100 m or 150 m or 200 m the light still maintains its red identity? In order to emphasise the problem, the metres in the question will be replaced by kilometres. Surely, since with each increase in distance, the original light is spreading out over a larger and larger area, which means that the original energy of the wave is also spreading out over a larger and larger area and changing in character, no red light would be seen at the destination. The characteristic red light that was emitted at the source SHOULD no longer be present and some other coloured wave or energy form should have been detected? BUT amazingly, this is not what happens, true the overall intensity of the light does vary with the inverse of the square of the distance but the light that remains is still red. This is one of the of the great mysteries that quantum mechanics tried, and failed, to solve.
Using the formula I had devised, everything becomes very simple:
[MATH]\sum_{n=0}^{\infty}\left(\frac{\overset{\rightarrow}\gamma_e^\mathfrak{e}= n^2}{2n-1} \right)= \begin{Bmatrix} {\frac{\vec\gamma_e^\mathfrak{e}1}{1},\frac{\vec\gamma_e^\mathfrak{e}4}{2},\frac{\vec\gamma_e^\mathfrak{e}9}{3} ,\frac{\vec\gamma_e^\mathfrak{e}16}{4},\frac{\vec\gamma_e^\mathfrak{e}25}5},\frac{\vec\gamma_e^\mathfrak{e}36}{6},\frac{\vec\gamma_e^\mathfrak{e}49}{7},\frac{\vec\gamma_e^\mathfrak{e}64}{8}\vec\infty\end{Bmatrix}[/MATH]
It is possible to see that this relationship is like a table, if the number from the bottom row is taken, then the area over which the wave has spread out is given by the number in the top row. In this case the answers would be :
The inverse or reciprocal of these answers gives the figure for how much the intensity of the light has reduced. Best of all the symbol [MATH]\vec\gamma_e^\mathfrak{e} [/MATH] signifies that the energy with which the photon started out, is the energy that the photon still possesses at the end of its journey.
If the wavelength of the photon is 650 nm, then its frequency will be [MATH] 461 \times 10^12[/Math] Hz
And its energy will be [MATH]4.61 \times 10^{14} \times 6.62 \times 10^{-34} = 3.051 \times 10^{-19}[/MATH] J equals 1.9 eV approx.
And that is what no other theory can do.
n.b Dan you forgot to comment on my explanation of the Double Slit Experiment. And incidentally your wavicles or whatever they are, also follow the inverse square law of intensity AND spread out according to the square of the distance. Think about it, light waves are not some strange wavicles that are conjured out of space or a bag of tricks. They are somethig that exist.
Using the formula I had devised, everything becomes very simple:
[MATH]\sum_{n=0}^{\infty}\left(\frac{\overset{\rightarrow}\gamma_e^\mathfrak{e}= n^2}{2n-1} \right)= \begin{Bmatrix} {\frac{\vec\gamma_e^\mathfrak{e}1}{1},\frac{\vec\gamma_e^\mathfrak{e}4}{2},\frac{\vec\gamma_e^\mathfrak{e}9}{3} ,\frac{\vec\gamma_e^\mathfrak{e}16}{4},\frac{\vec\gamma_e^\mathfrak{e}25}5},\frac{\vec\gamma_e^\mathfrak{e}36}{6},\frac{\vec\gamma_e^\mathfrak{e}49}{7},\frac{\vec\gamma_e^\mathfrak{e}64}{8}\vec\infty\end{Bmatrix}[/MATH]
It is possible to see that this relationship is like a table, if the number from the bottom row is taken, then the area over which the wave has spread out is given by the number in the top row. In this case the answers would be :
- After travelling 50Km the wave would have spread out over an area of [MATH]2.5 \times 10^9[/MATH] sq m
- After travelling 100 Km the wave would have spread out over [MATH] 10 \times 10^9 [/MATH] sq m
- After travelling 150 Km the wave would have spread out over [MATH]225 \times 10^9[/MATH] sq m
- After travelling 200 Km the wave would have spread out over [Math]4 \times 10^{10}[/MATH] sq m
The inverse or reciprocal of these answers gives the figure for how much the intensity of the light has reduced. Best of all the symbol [MATH]\vec\gamma_e^\mathfrak{e} [/MATH] signifies that the energy with which the photon started out, is the energy that the photon still possesses at the end of its journey.
If the wavelength of the photon is 650 nm, then its frequency will be [MATH] 461 \times 10^12[/Math] Hz
And its energy will be [MATH]4.61 \times 10^{14} \times 6.62 \times 10^{-34} = 3.051 \times 10^{-19}[/MATH] J equals 1.9 eV approx.
And that is what no other theory can do.
n.b Dan you forgot to comment on my explanation of the Double Slit Experiment. And incidentally your wavicles or whatever they are, also follow the inverse square law of intensity AND spread out according to the square of the distance. Think about it, light waves are not some strange wavicles that are conjured out of space or a bag of tricks. They are somethig that exist.
Dear Members of Free Maths Help Forums,
I am glad to have participated in an invigorating discussion. Unfortunately, this thread seems to have gone on for too long, at least according to the minds of some users. Dan, has left the conversation, as he has other things to do and doesn’t have the time or the inclination to pursue this further. I too, feel that maybe the thread has become so long that the original purport has been lost. Before I end this thread however, I would like to make a final request.
Even a quick reading of the mathematical formula I had put up, shows that when taken in context, it works very well and serves its purpose. To save viewers scrolling through several pages, here it is:
[MATH]\dot\Psi = \sum_{n=1}^{\infty}\left(\frac{\overset{\rightarrow}\gamma_e^\mathfrak{e}=n^2}{2n-1} \right)= \begin{Bmatrix} {\frac{\vec\gamma_e^\mathfrak{e}1}{1},\frac{\vec\gamma_e^\mathfrak{e}4}{2},\frac{\vec\gamma_e^\mathfrak{e}9}{3} ,\frac{\vec\gamma_e^\mathfrak{e}16}{4},\frac{\vec\gamma_e^\mathfrak{e}25}5},\frac{\vec\gamma_e^\mathfrak{e}36}{6},\frac{\vec\gamma_e^\mathfrak{e}49}{7},\frac{\vec\gamma_e^\mathfrak{e}64}{8}\vec\infty\end{Bmatrix}[/MATH]
My request is that if anyone can suggest a more elegant mathematically 'correct’ expression for this formula, I would appreciate it. Thanks
I am glad to have participated in an invigorating discussion. Unfortunately, this thread seems to have gone on for too long, at least according to the minds of some users. Dan, has left the conversation, as he has other things to do and doesn’t have the time or the inclination to pursue this further. I too, feel that maybe the thread has become so long that the original purport has been lost. Before I end this thread however, I would like to make a final request.
Even a quick reading of the mathematical formula I had put up, shows that when taken in context, it works very well and serves its purpose. To save viewers scrolling through several pages, here it is:
[MATH]\dot\Psi = \sum_{n=1}^{\infty}\left(\frac{\overset{\rightarrow}\gamma_e^\mathfrak{e}=n^2}{2n-1} \right)= \begin{Bmatrix} {\frac{\vec\gamma_e^\mathfrak{e}1}{1},\frac{\vec\gamma_e^\mathfrak{e}4}{2},\frac{\vec\gamma_e^\mathfrak{e}9}{3} ,\frac{\vec\gamma_e^\mathfrak{e}16}{4},\frac{\vec\gamma_e^\mathfrak{e}25}5},\frac{\vec\gamma_e^\mathfrak{e}36}{6},\frac{\vec\gamma_e^\mathfrak{e}49}{7},\frac{\vec\gamma_e^\mathfrak{e}64}{8}\vec\infty\end{Bmatrix}[/MATH]
My request is that if anyone can suggest a more elegant mathematically 'correct’ expression for this formula, I would appreciate it. Thanks
topsquark
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AAARRRGGGHHH!!! I can't stand this!
Your summation notation makes NO SENSE. How many times do I need to say it? A summation is a number. You have listed a set [math]\{ \text{ ... } \}[/math] as your summation, which is NOT a number and simply cannot be done the way you want it to. And you have never stated just what the summation is supposed to represent. (And frankly your summation makes no sense with the = sign in it, either.)
If I knew what you were trying to say I'd've corrected it. What you wrote makes NO SENSE at all! Why are you not listening? Define your terms or you aren't going to get anywhere.
-Dan
topsquark
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If you are after the set for some reason you can define it as
[math]\dot{ \Psi} = \left \{ \dfrac{\gamma _e^{\mathfrak{e}} n^2}{2n - 1} | n \in \mathbb{Z}^+ \right \} = \left \{ \dfrac{\gamma _e^{\mathfrak{e}} (1)}{1} , ~ \dfrac{\gamma _e^{\mathfrak{e}} (4)}{3} , \dfrac{\gamma _e^{\mathfrak{e}} (9)}{5}, \text{ ...} \right \} [/math]
The [math]n \in \mathbb{Z}^+[/math] part says that this will make a list for all n that are positive integers.
But you have still given no Mathematical reason for why you want this expression.
-Dan
D
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And no PHYSIC(S)al reason either.....
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Dan,
Thank you for putting my maths formula in some sort of order, I am grateful for your help.
Subhotosh,
The problem is as follows:
Wave-particle duality demonstrates the limitations of using classical mechanics—Newtonian physics—to fully describe the properties of light and matter at the quantum scale.
My intention was to show that there does exist sufficient logic to demonstrate that a medium like the aether exists. After all 85% of the Universe is made up of dark matter. That is the existence of dark matter can at least be hypothesised by physical inference. Wave-particles have no such basis. It is a totally unsupported theory based solely on conjecture. I had stated that new, verifiable discoveries like the Lamb shift, which are reproducible through empirical experiments, lend credence to the fact that the electron in the atom self-stabilises its energy by continually emitting and absorbing ‘virtual’ photons. This negates any need for wave-particle duality. All of the interactions within the nucleus are mediated by ‘virtual’ particles like gluons or ‘virtual’ interactions through quarks. Here is what Wikipedia has to say about ‘virtual’ particles.
In physics, a virtual particle is a transient quantum fluctuation that exhibits some of the characteristics of an ordinary particle, while having its existence limited by the uncertainty principle.
Since quarks act over distances of about [MATH]10^{-18}[/MATH] m over times of About [MATH]10^{-20}[/MATH] s it seems reasonable to state that quarks are virtual particles.
My question to you is this; do you mean to keep ‘virtual’ interactions and virtual particles locked up and exclusive inside the nucleus? Surely, that is both short-sighted and narrow minded. The Gestalt Aether Theory, opens up the world of ‘virtual’ particles and ‘virtual’ interactions so that it includes the whole of the Universe.
Thank you for putting my maths formula in some sort of order, I am grateful for your help.
Subhotosh,
The problem is as follows:
Wave-particle duality demonstrates the limitations of using classical mechanics—Newtonian physics—to fully describe the properties of light and matter at the quantum scale.
My intention was to show that there does exist sufficient logic to demonstrate that a medium like the aether exists. After all 85% of the Universe is made up of dark matter. That is the existence of dark matter can at least be hypothesised by physical inference. Wave-particles have no such basis. It is a totally unsupported theory based solely on conjecture. I had stated that new, verifiable discoveries like the Lamb shift, which are reproducible through empirical experiments, lend credence to the fact that the electron in the atom self-stabilises its energy by continually emitting and absorbing ‘virtual’ photons. This negates any need for wave-particle duality. All of the interactions within the nucleus are mediated by ‘virtual’ particles like gluons or ‘virtual’ interactions through quarks. Here is what Wikipedia has to say about ‘virtual’ particles.
In physics, a virtual particle is a transient quantum fluctuation that exhibits some of the characteristics of an ordinary particle, while having its existence limited by the uncertainty principle.
Since quarks act over distances of about [MATH]10^{-18}[/MATH] m over times of About [MATH]10^{-20}[/MATH] s it seems reasonable to state that quarks are virtual particles.
My question to you is this; do you mean to keep ‘virtual’ interactions and virtual particles locked up and exclusive inside the nucleus? Surely, that is both short-sighted and narrow minded. The Gestalt Aether Theory, opens up the world of ‘virtual’ particles and ‘virtual’ interactions so that it includes the whole of the Universe.
topsquark
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Okay, this is a Physics post and you asked for Math help here. If you want to discuss Physics then start a new thread.
You really need to lay off the Particle Physics for now and learn some basics.
Yes, "wavicles" are real. Stop saying that they aren't and look it up.
No, an aether does not exist. Stop saying that it does and look it up.
A virtual particle is one that is traded between two particles in an interaction. It is virtual simply because it is never measured. Any particle can be virtual in an interaction but that doesn't mean that it always a virtual particle. Here are some examples.
Electron-electron scattering: The electrons exchange a photon between them. As it never gets beyond these two electrons the photon is a virtual particle.
Electron positron annihilation: An electron and positron exchange a virtual electron, resulting in two photons. The electron is never detected so it is a virtual electron.
Quarks and gluons in the nucleus. We can't measure the quarks and gluons due to color confinement, so they are virtual. The only way we can learn anything about the quark structure of the nucleus is to measure particles coming off them via the strong and weak nuclear forces. These are not single quark states so we have to accept that we have to "ignore" the structure of the nucleus in many cases. That's for another thread.
Beta decay: Typically the electron coming from the atom has a measured momentum and thus is not a virtual particle. The electron anti-neutrino coming from the nucleus is not measured so it is a virtual particle.
The result of having a virtual particle in an interaction means that we end up integrating over the momentum of the virtual particle as a way of getting rid of an extra degree of freedom in the equations. Yes, the integral is occasionally non-convergent. We deal with this using the procedure of renormalization... a feature of QFT that is unsettling in some ways. Again, that's for another thread.
By the way, the Wiki article is a bit off.... The way it is described is something I would call a "resonance." A resonance can be a virtual particle (it almost always is in the experiments) but it does not have to be an elementary particle, which is a point particle and has no internal structure.
-Dan
Addendum: I just looked up "Gestalt Aether Theory." For a student it is one of the most dangerous articles I've ever come across. No wonder your Physics is so lacking in anything resembling workable ideas. Study the Scientific Method... you'll learn Science better that way. Here's a reference that you might find to be at a good level for you.
You really need to lay off the Particle Physics for now and learn some basics.
Yes, "wavicles" are real. Stop saying that they aren't and look it up.
No, an aether does not exist. Stop saying that it does and look it up.
A virtual particle is one that is traded between two particles in an interaction. It is virtual simply because it is never measured. Any particle can be virtual in an interaction but that doesn't mean that it always a virtual particle. Here are some examples.
Electron-electron scattering: The electrons exchange a photon between them. As it never gets beyond these two electrons the photon is a virtual particle.
Electron positron annihilation: An electron and positron exchange a virtual electron, resulting in two photons. The electron is never detected so it is a virtual electron.
Quarks and gluons in the nucleus. We can't measure the quarks and gluons due to color confinement, so they are virtual. The only way we can learn anything about the quark structure of the nucleus is to measure particles coming off them via the strong and weak nuclear forces. These are not single quark states so we have to accept that we have to "ignore" the structure of the nucleus in many cases. That's for another thread.
Beta decay: Typically the electron coming from the atom has a measured momentum and thus is not a virtual particle. The electron anti-neutrino coming from the nucleus is not measured so it is a virtual particle.
The result of having a virtual particle in an interaction means that we end up integrating over the momentum of the virtual particle as a way of getting rid of an extra degree of freedom in the equations. Yes, the integral is occasionally non-convergent. We deal with this using the procedure of renormalization... a feature of QFT that is unsettling in some ways. Again, that's for another thread.
By the way, the Wiki article is a bit off.... The way it is described is something I would call a "resonance." A resonance can be a virtual particle (it almost always is in the experiments) but it does not have to be an elementary particle, which is a point particle and has no internal structure.
-Dan
Addendum: I just looked up "Gestalt Aether Theory." For a student it is one of the most dangerous articles I've ever come across. No wonder your Physics is so lacking in anything resembling workable ideas. Study the Scientific Method... you'll learn Science better that way. Here's a reference that you might find to be at a good level for you.