Markus Hanke Posted February 24, 2021 Posted February 24, 2021 23 hours ago, studiot said: We cannot do this without an external agent. Indeed, and that’s the crux - there simply is no meaningful notion of the “shape” of the atom until such time when it is interacted with in some way. 19 hours ago, John Cuthber said: But the sum of the trio of P orbitals (one on each axis) is symmetrical. Yes, this makes sense now, and it is essentially what I was thinking about in my last post. Also, it’s important to remember that the wave function is a probability density distribution, so it needs to be volume-integrated first in order to become a probability distribution; and for an isolated atom in free space, one is free to choose the orientation of the volume form in whatever way one wants. 18 hours ago, swansont said: You can't say which p orbital the electron is in, so it's in all of them until you measure. Indeed.
studiot Posted February 24, 2021 Posted February 24, 2021 11 minutes ago, Markus Hanke said: Indeed, and that’s the crux - there simply is no meaningful notion of the “shape” of the atom until such time when it is interacted with in some way. Yes, this makes sense now, and it is essentially what I was thinking about in my last post. Also, it’s important to remember that the wave function is a probability density distribution, so it needs to be volume-integrated first in order to become a probability distribution; and for an isolated atom in free space, one is free to choose the orientation of the volume form in whatever way one wants. Indeed. It's also worth remembering that the Schroedinger equation is about the motion of the electron in the field of the nucleus. And the field of the nucleus is spherically symmetrical. This can be derived from the solution to the equation being varaibles separable ie the product of independent functions of R theta and phi. You can have the math if you like.
swansont Posted February 24, 2021 Posted February 24, 2021 5 hours ago, studiot said: It's also worth remembering that the Schroedinger equation is about the motion of the electron in the field of the nucleus. Not motion, as such. There’s no trajectory information there. Location and momentum (probabilities) and energy and angular momentum (eigenvalues) can be found.
studiot Posted February 24, 2021 Posted February 24, 2021 (edited) 11 minutes ago, swansont said: Not motion, as such. There’s no trajectory information there. Location and momentum (probabilities) and energy and angular momentum (eigenvalues) can be found. [math]\frac{{{d^2}\psi }}{{d{x^2}}} + \frac{{8{\pi ^2}m}}{{{h^2}}}\left( {E - U} \right)\psi = 0[/math] ?? And yes, I know this is only the one dimensional version, but it's easier to write. Edited February 24, 2021 by studiot
swansont Posted February 24, 2021 Posted February 24, 2021 9 minutes ago, studiot said: d2ψdx2+8π2mh2(E−U)ψ=0 ?? And yes, I know this is only the one dimensional version, but it's easier to write. Yes. Energy terms. Position. Not trajectories - not motion.
studiot Posted February 24, 2021 Posted February 24, 2021 3 hours ago, swansont said: Yes. Energy terms. Position. Not trajectories - not motion. So you wouldn't describe the equation for the motion of a falling weight Kinetic energy gained = loss of potential energy as an equation of motion? Fair enough.
John Cuthber Posted February 24, 2021 Posted February 24, 2021 4 minutes ago, studiot said: So you wouldn't describe the equation for the motion of a falling weight Kinetic energy gained = loss of potential energy as an equation of motion? Fair enough. Nice bait and switch there. The thing about "falling" is that it implies that the direction is "down" so there is information about a trajectory.
swansont Posted February 24, 2021 Posted February 24, 2021 1 minute ago, studiot said: So you wouldn't describe the equation for the motion of a falling weight Kinetic energy gained = loss of potential energy as an equation of motion? Fair enough. In classical physics, one can often ascribe a trajectory to the solution using energy. Not so much with QM. The kinematics equation s = v0t + 1/2 at^2 explicitly has a velocity in it, but there is no corresponding QM equation.
studiot Posted February 24, 2021 Posted February 24, 2021 15 minutes ago, John Cuthber said: Nice bait and switch there. Nasty little snide remark there. I did not say that equations of motion do not contain information about trajectory. Swansont and I were discussing affably the possibility of such an equation containing information about energy. Not every conceivable equation of motion contains trajectory information or for that matter information about every conceivable variable. So the equation distance = speed times time contains no trajectory information and has other 'missing' variables as does the equation acceleration = time rate of change of speed.
Alex Mercer Posted February 24, 2021 Author Posted February 24, 2021 Wow it got really technical, now I don't understand!! Oh well... 1
beecee Posted February 24, 2021 Posted February 24, 2021 16 minutes ago, Alex Mercer said: Wow it got really technical, now I don't understand!! Oh well... I havn't followed this thread, but I reckon the first three answers are pretty good. https://www.google.com/search?q=what+would+an+atom+actually+look+like&oq=what+would+an+atom+actually+look+like&aqs=chrome..69i57j0i22i30l2j0i390l2.23543j0j7&sourceid=chrome&ie=UTF-8 Physicist: Actual pictures of atoms aren't actually pictures at all. There are a few good rules of thumb in physics. Among the best is: light acts like you'd expect on scales well above its wavelength and acts weird on scales below.
Markus Hanke Posted February 25, 2021 Posted February 25, 2021 23 hours ago, studiot said: It's also worth remembering that the Schroedinger equation is about the motion of the electron in the field of the nucleus. And the field of the nucleus is spherically symmetrical. This can be derived from the solution to the equation being varaibles separable ie the product of independent functions of R theta and phi. You can have the math if you like. Thanks! I do have a copy of Griffiths “Introduction to Quantum Mechanics” here, he goes through the maths in quite some detail. I’d look at the SE as an eigenvalue equation for the system’s Hamiltonian though, not so much as an equation of motion.
Curious layman Posted February 25, 2021 Posted February 25, 2021 17 hours ago, Alex Mercer said: Wow it got really technical, now I don't understand!! Oh well... Lol, I feel ya bro. I agree with beecee, first three answers are good for an amateur. If you do try to understand the technical stuff, Swansont is the person to stick with. No offence to the others, but Swansont is top dog around here.
dimreepr Posted February 25, 2021 Posted February 25, 2021 What does a particle actually look like? Imagine a stone that's much smaller than a mountain?
Markus Hanke Posted February 26, 2021 Posted February 26, 2021 18 hours ago, dimreepr said: Imagine a stone that's much smaller than a mountain? The problem is that both the stone and the mountain are classical objects and as such share the same fundamental properties. The same is not true, however, for the chair I am sitting on and the elementary particles of which it is ultimately composed - you can’t describe the properties and interactions of these particles with Newtonian mechanics, and conversely the chair as a whole will not exhibit any quantum effects. So these are distinct categories of objects, even though there is a definite relationship between them. 1
studiot Posted February 28, 2021 Posted February 28, 2021 On 2/25/2021 at 8:11 AM, Markus Hanke said: I’d look at the SE as an eigenvalue equation for the system’s Hamiltonian though, not so much as an equation of motion. Hamilton's equations of Motion https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Book%3A_Classical_Mechanics_(Tatum)/14%3A_Hamiltonian_Mechanics/14.03%3A_Hamilton's_Equations_of_Motion Lagrange's equations of Motion https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-61-aerospace-dynamics-spring-2003/lecture-notes/lecture7.pdf They are all equations of motion, realising that even a stationary state is a state of motion. I will leave the last word to Hamilton himself. Quote Hamilton The theoretical of the laws of motion of bodies is of such interest and importance, that it has engaged the of all the most eminent mathematicians, since the invention of Dynamics as a mathematical science by Galileo. Amongst the successors, Lagrange has done more than perhaps than any other analystto give extent and harmony to such researches, by showing that the most varied consequences respecting the motions of systems of bodies, may be derived from one radical formula; the beauty of the method so suiting the dignity of the results, as to make his great work a kind of scientific poem.
Markus Hanke Posted March 1, 2021 Posted March 1, 2021 19 hours ago, studiot said: Lagrange's equations of Motion Well ok, point taken I guess it’s ultimately just a convention of terminology anyway.
joigus Posted March 1, 2021 Posted March 1, 2021 On 2/23/2021 at 3:00 PM, swansont said: What it "looks" like really only makes sense for objects where diffraction is not important. Exactly. And even for macroscopic objects, what they look like strongly depends on the kind of radiation we use to look at them: https://www.wired.com/2014/04/the-world-looks-different-when-you-see-in-infrared/ When you look at this person covered by a plastic bag, couldn't we say that "normal" reflected light is deceiving you, by offering a picture of the object that is actually less faithful than emitted infrared light? In order to get an infrared picture of the object, we also need to map these colours somewhere in our visual cortex. So there is a neurological aspect about the whole question too. Most of the light we see is reflected light, and the way our brains process this information gives rise to the usual "palette" of visual concepts, such as shiny, matte, red or blue, fuzzy, etc.
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