Mike C 1,139 Posted November 6, 2007 Author Report Share Posted November 6, 2007 The math doesn't lack any credibility. It can be performed, tested and verified by many which is what makes it credible. Visionaries on the other hand cannot so they are strictly limited to speculation and even that it questionable when they try to use excuses like denouncing the rigors of math. Examples of unproven math is the 'string' theory, inflation theory and IMO, Einsteins mass energy formula.The last one is obviously false since we should all know that FORCES create the energies. Yet, in spite of this lack of proof, all the above are accepted as factual. Photographic observations are 'technical' visualizations. However, here the interpretations can be erroneous as were the Doppler observations that were refuted because they implied a repeat of the Geocentric concept for the universe.However, the false interpretation of these observations were still accepted to promote the BBU. So, photo data and visualizations can be in error too.That is why I explain my opinions clearly with 'basic' physics. Mike C Quote Link to post Share on other sites

Mike C 1,139 Posted November 6, 2007 Author Report Share Posted November 6, 2007 Well thats silly, it doesnt make much difference so lets just ignore it.. the fact is it would make a difference and so it should be considered otherwise its not a model of the universe. Back to the one angstrom sine wave, if it produces a standing wave we should still be able to detect it. Standing waves appear to not be moving but are actually the superposition of multiple waves moving in opposite directions. I do not know where you got that information but the only wave in opposition to the electrons wave is the one the protons make around the 'center of gravity' and this wave would have a very small magnitude in conparison to the electrons .The electrons make a much wider sweep around the protons to have a greater influence between the atoms than the protons do. Mike C Quote Link to post Share on other sites

astrocuriousstudent 10 Posted November 9, 2007 Report Share Posted November 9, 2007 The reason that electrons would lose energy under classical physics is that it is a charge traveling in a circular motion, and a circular motion is an accelerated motion. So I have on question for this thread, how can the electro move in a cicular motion, when it is moving at nearly, if not actually at, the speed of light. the orbit would have to be elliptical, and constantly changing bit by bit because of the incredible momentum generated. I dont know if this really has any bearing on this subject though... I was just wondering if maybe people who replied overlooked a detail... feel free to critisize as much as you want, I would like to learn as much as I can, and you can only really learn from your mistakes, right? Quote Link to post Share on other sites

CraigD 398,736 Posted November 10, 2007 Report Share Posted November 10, 2007 So I have on question for this thread, how can the electro move in a cicular motion, when it is moving at nearly, if not actually at, the speed of light.The simple, classical physics answer is: electrons around atoms aren’t moving at anywhere near the speed of light. Try calculating the speed that the electron in a hydrogen atom would orbit the nucleus. You’ll need just a couple of formulae and data:for the force between 2 point charges, known as Coulomb’s lawfor Centripetal forcethe classical distance of an electron from a proton, about [math]2.5 \times 10^{-11} \,\mbox{m}[/math]the mass of an electron, about [math]9 \times 10^{-31} \,\mbox{kg}[/math]the charge of an electron, about [math]1.6 \times 10^{-19} \,\mbox{C}[/math]Given all that, you can calculate the speed of a circular orbit of an electron around a proton. Give it a try - I believe you’ll be shocked at how small it is, until you consider how small it’s orbit is, and hence how many orbits it completes in a second. Post your results, or let us know if you get stuck in any step of the calculation. Although some might disagree, IMHO one of the best way to visualize physics is by working through its formulae with simple examples. It’s common for electrons to be accelerated to speeds very close to the speed of light, but only in particle accelerators and some high-energy celestial object, such as supernovae and black holes, and such electrons are a form of ionizing radiation, not anything that could be part of an atom, unless greatly slowed. Keep in mind that none of these classical mechanical calculations are, according to current best theory, a very good description of what’s really happening with electrons in atoms. On this tiny scale, notions such as position and velocity are no longer intuitively sensible, becoming a sort of “smear” or probabilities rather than definite things that can be calculated and measured. Applied to atoms, classical physics predicts a nonsensical occurrence known as the ultraviolet catastrophe. This failure is, in large part, the reason why quantum physics was developed. This thread is an attempt by Mike C to provide an alternate explanation for why the ultraviolet catastrophe doesn’t occurr, without using quantum physics. Though I and most of the other members who have posted in the thread don’t think his explanation is correct, everyone is in agreement that some explanation for why it doesn’t occur needs to exist. Quote Link to post Share on other sites

astrocuriousstudent 10 Posted November 10, 2007 Report Share Posted November 10, 2007 Given all that, you can calculate the speed of a circular orbit of an electron around a proton. Give it a try - I believe you’ll be shocked at how small it is, until you consider how small it’s orbit is, and hence how many orbits it completes in a second. Post your results, or let us know if you get stuck in any step of the calculation. Although some might disagree, IMHO one of the best way to visualize physics is by working through its formulas with simple examples. It’s common for electrons to be accelerated to speeds very close to the speed of light, but only in particle accelerators and some high-energy celestial object, such as supernovae and black holes, and such electrons are a form of ionizing radiation, not anything that could be part of an atom, unless greatly slowed. Keep in mind that none of these classical mechanical calculations are, according to current best theory, a very good description of what’s really happening with electrons in atoms. On this tiny scale, notions such as position and velocity are no longer intuitively sensible, becoming a sort of “smear” or probabilities rather than definite things that can be calculated and measured. Applied to atoms, classical physics predicts a nonsensical occurrence known as the ultraviolet catastrophe. This failure is, in large part, the reason why quantum physics was developed. This thread is an attempt by Mike C to provide an alternate explanation for why the ultraviolet catastrophe doesn’t occurr, without using quantum physics. Though I and most of the other members who have posted in the thread don’t think his explanation is correct, everyone is in agreement that some explanation for why it doesn’t occur needs to exist. I have not taken any upper level physics classes, nor have I taken any upper level math classes, I am merely a high school student trying to learn outside of what his instructors have taught him, therefore I do not know how to use the information you have given me to find the average speed of an electron. To make up for that, I have been researching on the internet an found in other places that the speed is roughly about 4000 km/s. Actually, if you would be good enough to show me how to go through the process, that would be fantastic, but if you don't feel like being an instructor, that fine as well. The reason I asked was because; That is what I was taught, that the average speed of an electron was around "C," Nobody told me that such circumstances only exist within particle accelerators and highly excited atoms. Now that I think more about it though, it makes more sense. In the process of looking for the average speed of the electron however, I was relieved to find that I am not the only person with that misconception. Other people on multiple forums have also posted about the speed of an electron being close to or equivalent to that of "C". Thank you for correcting me. Also, I only deleted the information you gave me to do the calculations with because I am not yet allowed to post links on my replies. Quote Link to post Share on other sites

CraigD 398,736 Posted November 10, 2007 Report Share Posted November 10, 2007 Actually, if you would be good enough to show me how to go through the process, that would be fantastic, but if you don't feel like being an instructor, that fine as well. I was afraid you might say that! The calculation’s not hard, and I’d already done it before suggesting you try it, but it takes a bit of effort to write it into a decently pretty post. Here goes… Coulomb’s law, give the force between 2 “point particles” as:[math]F = \frac1{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}[/math]where: [math]\varepsilon_0[/math] is the permittivity of free space, about [math]8.85 \times 10^-12[/math] s^3 C / m^2 kg[math]q_1[/math] and [math]q_2[/math] are the charges of the 2 particles – in the case of a single electron and proton, the elementary charge, about [math]1.6 \times 10^{-19} \,\mbox{C}[/math], positive for the proton, negative for the electron. and [math]r[/math] is the distance between them. Picking a value for r is tricky, as there are several definitions of it. For our purposes of just getting a rough value, we don’t need to worry about this, just pick one. From the wikipedia page, we take the smallest value listed for the radius of a hydrogen atom, [math]25 \,\mbox{pm} = 2.5 \times 10^{-11} \,\mbox{m}[/math]. (It’s often necessary to convert between different SI units, so you’ll need to keep a reference of them handy, either in your head, or a table like this) The force required to cause a body to move at constant speed in a circle is known as centripetal force. It can be derived using geometry, and is:[math]F = \frac{m v^2}{r}[/math]where: m is the mass of the body, for an electron, about [math]9 \times 10^{-31} \,\mbox{kg}[/math] v is its speed (what we intend to calculate) and r is the radius of the circle, the same r as used in the previous equation. Most people familiar with physics and math would set the two equations equal and simplify them algebraically, but we can get the same result by substituting our known values into the charged particle force equation to get a numeric value for it, then solving the other for v. [math]F = \frac1{4\pi\varepsilon_0}\frac{q_1q_2}{r^2} \dot= \frac1{4 \cdot 3.1415 \cdot 8.85 \times 10^{-12}}\frac{-1.6 \times 10^{-19} \cdot 1.6 \times 10^{-19}}{(2.5 \times 10^{-11})^2} \dot= 3.683 \times 10^{-7} \,\mbox{N}[/math] Because the unit of force kilogram meters per second per second is used a lot, it has a special name, the Newton, abbreviated N. Note that I was lazy and didn’t include units (kg, m, s, C, etc.) in the above until the end, and ignored the sign of the charges. You can get away with this if you sure all the units being substituted into the equation combine and cancel to the units of the result, but should make a habit of doing calculations with all the units and signs until you confident with such time-saving tricks. Now we can take our equation for centripetal force, and solve it for v:[math]F = \frac{m v^2}{r}[/math] [math]v^2 = \frac{F r}{m}[/math] [math]v = \sqrt{\frac{F r}{m}}[/math] Substituting the value of F we calculated previously, we get[math]v = \sqrt{\frac{3.683 \times 10^{-7} \cdot 2.5 \times 10^{-11}}{9 \times 10^{-31}}} \dot= 3200000 \,\mbox{m/s}[/math] Divide it by the circumference of the circle its traveling to calculate its rate of revolution.[math]d = 2 \pi r \dot= 1.571 \times 10^{-10} \,\mbox{m}[/math] We get [math]\frac{3200000}{1.571 \times 10^{-10}}\dot= e \times 10^16[/math] revolutions/s. In more familiar units, this is about a million trillion RPM, a pretty blurring value. It’s important to keep in mind that, while fun to do these sorts of classical calculations – and a valid way to show that electrons in an atom are not traveling at nearly c - they’re not the best way to visualize electrons in atoms. For that, we need a basic understanding of quantum physics. You’ll notice I link to wikipedia a lot, and that hypography even has a tag with a button in its post editor for doing this. Along with several other online sources, it’s a great place to browse science and other information. I recommend any student who has not already done so make wikipedia his friend :) Quote Link to post Share on other sites

Mike C 1,139 Posted November 11, 2007 Author Report Share Posted November 11, 2007 So I have on question for this thread, how can the electro move in a cicular motion, when it is moving at nearly, if not actually at, the speed of light. the orbit would have to be elliptical, and constantly changing bit by bit because of the incredible momentum generated. I dont know if this really has any bearing on this subject though... I was just wondering if maybe people who replied overlooked a detail... feel free to critisize as much as you want, I would like to learn as much as I can, and you can only really learn from your mistakes, right? I decided to answer here because of your assumption that the electron is moving at the 'speed of light'. In the 'ground state orbit, the electron velocity is moving at 2 million meters per second.This is just a fraction of 'c'. Most all physics books provide you with the formulas regarding the Bohr model of the hydrogen atom. In the central region of stars, the plasma open orbital velocities of the electrons may approach c because of the density and closeness of the particles in that environment.That is my opinion. Mike C Quote Link to post Share on other sites

CraigD 398,736 Posted November 11, 2007 Report Share Posted November 11, 2007 In the 'ground state orbit, the electron velocity is moving at 2 million meters per second.Mike C, can you back up this claim with a link or reference? Quote Link to post Share on other sites

Mike C 1,139 Posted November 16, 2007 Author Report Share Posted November 16, 2007 Mike C, can you back up this claim with a link or reference? Craig That repremand was a bit premature. A couple of formulas regarding the hydrogen atom characteristics in my files are listed below: The radius of the ground state orbit of the HA is: a, sub o = h^2 / 4 pi^2, m sub e, k sub e, e^2 = 5.3 x 10^-11 meters. The velocity of the electron in this orbit is: v sub e = h / 2 pi, a sub o (Rad.#), m sub e = 2.19 x 10^ 6 meters/second. m sub e = 9.109 x 10^ 31 kgrsk sub e = 8.987 x 10^9 N-m^2/c^2e = 1.602 x 10 ^-19 Coulombsh = 6.626 x 10-34 J/s The 1st formuls can be confirmed in most all physics books.I had not been able to find the source for the second formula but I am certain that it is correct. Mike C Quote Link to post Share on other sites

CraigD 398,736 Posted November 16, 2007 Report Share Posted November 16, 2007 Craig That repremand was a bit premature.I agree, and have had the infraction removed (for those who missed it, I gave Mike a “failure to back up/defend claims” infraction). In post #23, I made a mistake, confusing the constant for the Vacuum permittivity [math]\varepsilon_0[/math] with it’s form as the electrostatic constant ([math]\frac1{4 \pi \varepsilon_0}[/math]), so instead of using [math]9 \times 10^9[/math] kg m/s m^2 / C^2, I was using [math]\frac1{4 \pi \cdot 9 \times 10^9}[/math]. :doh: The result should have been about [math]v \dot= 3200000[/math] m/s, close to Mike’s “2 million meters per second”. I’ve edited post #23 to corrected the error. v sub e = h / 2 pi, a sub o (Rad.#), m sub e = 2.19 x 10^ 6 meters/second.I’m unable to follow this, as I don’t know what operation the commas represent. As I read it, using the constants provided, it evaluates to 3 separate statements, “v sub e equals about ten to the negative thirty-four watts”, “a sub o times Rad dot poundsign equals nothing”, and “m sub equals two point one nine times ten to the six meters per second” – none of which sense. Though it takes a bit of effort to learn, I recommend mastering the basics of LaTeX necessary to use the [math] tags available at hypography. They’re described in 6576 and 12086, and I highly recommend this tutorial website. If you don’t want to use LaTeX, I recommend using the conventional text math symbols: + - * / ^ ( ) and =. Quote Link to post Share on other sites

Mike C 1,139 Posted November 17, 2007 Author Report Share Posted November 17, 2007 I’m unable to follow this, as I don’t know what operation the commas represent. As I read it, using the constants provided, it evaluates to 3 separate statements, “v sub e equals about ten to the negative thirty-four watts”, “a sub o times Rad dot poundsign equals nothing”, and “m sub equals two point one nine times ten to the six meters per second” – none of which sense. I only went trough high school in the # 1 high school in Detroit. Other than that, I have studied astronomy and cosmology for more than 20 years. We also had discussion groups in our astronomy club on cosmology and physics. The second formula regarding the electron velocity was copied from some credible source but I cannot find that source since I went through two moves that reduced my library by at least 50%. The commas separate the components of the formulas.The slash bars represent division.Sub represents 'subscript' and provide further descriptions of the component symbols. v sub e refers to the electron velocity.h, of course represents the Planck constant.a sdub o (Rad, #) is the ground state orbit = to .5 x 10^- 11 metersm sub e equals the mass of the electron. So, v = 6.626 x 10^-34 / 2 pi x 5.3 x 10^-11 x 9.109 x 10^-31 = 2.18 10^6. I used 5.3 x 10^-11 as the orbital dimension. The above mistake for a sub o definition should be corrected to 5.3 x 10^-11. Mike C Quote Link to post Share on other sites

Mike C 1,139 Posted November 20, 2007 Author Report Share Posted November 20, 2007 Craig That pound sign I used also represents ‘numbers’. I guess the proper way was to use the letter n as the subscript for the R (radius numbers).So R sub n, that is.All the other components have the subscripts that define the components . h^2 h α sub o = --------------- ve = ----------- 4 pi^2mekeme 2piRnme The above formulas did NOT reproduce (paste) properly. On my home computer, I managed to reproduce the subsripts and superscripts accurately along with the Greek letters that also did not reproduce here. Mike C Quote Link to post Share on other sites

Erasmus00 86,950 Posted November 20, 2007 Report Share Posted November 20, 2007 I’m unable to follow this, as I don’t know what operation the commas represent. As I read it, using the constants provided, it evaluates to 3 separate statements, “v sub e equals about ten to the negative thirty-four watts”, “a sub o times Rad dot poundsign equals nothing”, and “m sub equals two point one nine times ten to the six meters per second” – none of which sense. What he is doing is using the Bohr atom. Angular momentum is quantized in units of hbar, so [imath] L = mvr = \hbar [/imath]. Here L is angular momentum. Combine this with the fact that the radius is about a bohr radius, we can solve for v. -Will Quote Link to post Share on other sites

Bernard 1 Posted October 5 Report Share Posted October 5 (edited) For all students and physics laymen (and women) who happen to read the discussion above, I would like to give the following hint: Since 1920 - 1927 it is known to all physicists and university physics students that the electron in the hydrogen atom groundstate does NOT move in an orbit! Its angular momentum L = 0!!! There ist definitely no angular velocity! Rather the solution of the Schrödinger equation shows, that its kinetic energy is based on a pure RADIAL movement! So the whole discussion above is totally unreal. The electron is kept in a most probable radial distance of 1 Bohr radius (52.9 pm) from the proton simply by its repulsive kinetic energy near the nucleus, counteracting the Coulomb attraction potential. Thus the modern more realistic picture would be an electron bouncing a bit radially. As this is known since nearly 100 years now, I still can't understand that highschool physics still speaks of the old Bohr model of a rotating electron. Perhaps the difference between orbit and orbital is not well understood! Orbital means the space where electrons exist on average, also called shells. This does not necessarily mean that they orbit. in the short excited states they do that additionally, but definitly not in the groundstate of the H atom. Dr. Bernhard Kup Edited October 5 by Berhard typo OceanBreeze 1 Quote Link to post Share on other sites

OceanBreeze 392 Posted October 9 Report Share Posted October 9 On 10/5/2020 at 10:54 PM, Bernard said: For all students and physics laymen (and women) who happen to read the discussion above, I would like to give the following hint: Since 1920 - 1927 it is known to all physicists and university physics students that the electron in the hydrogen atom groundstate does NOT move in an orbit! Its angular momentum L = 0!!! There ist definitely no angular velocity! Rather the solution of the Schrödinger equation shows, that its kinetic energy is based on a pure RADIAL movement! So the whole discussion above is totally unreal. The electron is kept in a most probable radial distance of 1 Bohr radius (52.9 pm) from the proton simply by its repulsive kinetic energy near the nucleus, counteracting the Coulomb attraction potential. Thus the modern more realistic picture would be an electron bouncing a bit radially. As this is known since nearly 100 years now, I still can't understand that highschool physics still speaks of the old Bohr model of a rotating electron. Perhaps the difference between orbit and orbital is not well understood! Orbital means the space where electrons exist on average, also called shells. This does not necessarily mean that they orbit. in the short excited states they do that additionally, but definitly not in the groundstate of the H atom. Dr. Bernhard Kup I think these days it is generally accepted that the electron is a quantum object. That is, an electron is partially particle-like and partially wave-like, but it may well be something more complex that is neither a wave nor a particle. So, while Bohr’s model was/is simplistic and wrong, he did manage to exactly calculate the ground state energy of hydrogen as well as the correct size of the hydrogen ataom. This may have been made possible by the strange fact that in the hydrogen atom, the radial excitation and an angular excitation have the same energy. Furthermore, the Bohr mathematical formulas are still used because they do give correct results. For example, the Bohr radius formula remains central in atomic physics calculations, due in part to its simple relationship with other fundamental constants. As I understand it, scientists are more interested in making the equations match the result of experiments than trying to describe what is the most realistic physical model. In fact, we may never know what is actually physically real at the quantum level. Welcome to the forum Dr. Bernhard Kup Mutex 1 Quote Link to post Share on other sites

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