# Atom orbitals

1. Mar 8, 2010

### alan.b

Atom orbitals, I mean the ones where electrons orbit proton with continuous trajectory like planets orbit around the sun. I do not mean to disagree with modern teachings nor to speculate, I simply do not see how QM actually excludes this possibility. I just want someone walk me through several most convincing experiments and show me how did humanity itself changed their opinion from "electron orbitals" to "electron clouds". I know there are couple of explanations like "electrons would radiate and spiral down to nucleus", but I do not know of any experimental evidence to support that. Also, I was wondering, are there some fields in physics that still use this or similar model of atom based on classical mechanics?

2. Mar 8, 2010

### f95toli

This is how synchrotrons work. They are basically rings where electrons are accelerated in circular orbits which in turn makes the electrons radiate light.

So there is actually a field of technology which is based upon the fact that orbiting electrons loose energy...

3. Mar 8, 2010

### SpectraCat

An electron in a classical orbit would have a non-zero angular momentum. However, the angular momentum of the H-atom ground state has been measured to have an angular momentum of exactly zero, in accordance with the "cloud-like" version of atomic orbitals predicted by quantum mechanics.

4. Mar 8, 2010

### alan.b

Ok, thank you.

- why electrons in QM not radiate, or do they?

- how does synchrotron read electron radiation?

- what kind of radiation electrons emit, how often?

- is it not electron the most elementary amount of electric charge, so how can the smallest amount of something, that has no size, radiate anything and what would be left of a single electron if it radiates even a single photon?

5. Mar 8, 2010

### alan.b

I suppose all the solar system planets, moons, binary stars or whatever classical mechanics systems have this non-zero angular momentum? Can you say where does this non-zero angular momentum come from? How does QM predict zero angular momentum? Thank you.

6. Mar 8, 2010

### SpectraCat

The fact that you are asking these questions indicates to me that I cannot answer them in a way that you can appreciate. I would suggest that you read up on some basic physics and reconsider your questions.

I will just say now that angular momentum *is* the cross product of a particle's position and momentum vectors. In qualitative terms, since there is always curvature of the particle's (or planet's) trajectory in the orbiting systems you mention, then the cross product above must always be non-zero. Where angular momentum "comes from" is a much deeper question that I cannot really answer.

The QM prediction of zero angular momentum for the H-atom ground state arises from the perfect spherical symmetry of the probability distribution describing the ground state atomic orbital.

7. Mar 8, 2010

### jfy4

The radiation of photons from an electron is the direct result of an equation, which says that accelerating charges radiate. The formula looks something like this for low velocities.

$$P=Kq^{2}a^{2}$$

where P is the power radiated and a is the acceleration, q is the charge, and K is just a lot of constants together.

The type of radiation is electromagnetic radiation, or light.

The radiation of light from an accelerating charge, must come from the particles kinetic energy... so a dimensionless particle, that we both can agree can have momentum, loses it kinetic energy to radiation.

These next words, im not sure how much they'll resonate but here i go...
Solutions for the hydrogen atom using the time independent Schrodinger equation give us energy eigenstates which are stationary. Only specific situations where evolution in time is a concern matters when it comes to a particle radiating.

i hope this helps a little...

8. Mar 8, 2010

### varga

It's me, Alan. Major meltdown here, moving to a new address, I broke my laptop, changing internet provider... in short, I have no idea what was my password and I can't retrieve it via my old e-mail, so I got my flatmate to open this account.

Ok, I had no idea what momentum you were talking about, was it individual, combined and what relation did it have to what point of reference. I see now that by "angular momentum" you simply mean 'non-linear' or 'curved trajectory'. Out of curiosity, if something was moving left-right in a straight line, making 180 degree turns, what would be its angular momentum?

What symmetry has to do with angular momentum?

How was this angular momentum measured for the H-atom?

Do planetary orbits not have symmetrical probability distribution?

9. Mar 8, 2010

### varga

Thank you, but is this radiation actually directly measured in any experiment or is it only indirectly derived? I understand electrons slow down in particle accelerators, but this is not actual measurement of radiation.

Would it be unreasonable to suppose electrons in orbit around the nucleus do not accelerate as much as assumed, but rather follow some "curvature of space-time" and really traversing rather straight line from their point of view and their reference frame?

I see what you are saying, exchange of potential and kinetic energy - lift a ball up from the ground and you will give it potential energy, let it go and as it falls down it losses potential energy and gains kinetic energy in a form of velocity, thus the energy is conserved. So, where does radiation come into this, again?

Do planets radiate, and if so why do planets not spiral in to the Sun?

There is only ONE explanation why would electrons in QM atom model not radiate:

- they appear at one location and reappear at another location(s), but the sequence of these points do not connect with a continuous line, it's not a trajectory you could draw without lifting a pen from a paper (the time when electron is nowhere), rather its trajectory is a collection of unconnected dots. Is this how it is supposed to be according to QM?

10. Mar 8, 2010

### jfy4

For your first part yes, you can measure that radiation for sure. eject electrons into a uniform magnetic field at right angles to each other and you will have electrons going in a circle radiating a lot, just make sure they have a large centripetal acceleration ;)

You seem to have misunderstood my last part about radiation, which is my fault for not being clear, i will try harder here.

As an electron is accelerated, it radiates energy. That energy has to be something. For charges, the energy comes from the kinetic energy of the charge, so the charge slows down.

Planets do radiate, although its a little silly to talk about it because it is undetectable. only large sources produce gravitational radiation profound enough for us to detect. The radiation from orbiting planets is almost nothing, and so does extremely little to affect orbit.

i hope this is clearer... and helps

11. Mar 9, 2010

### f95toli

There are a number of huge synchrotron facilities around the world where they generate X-Rays using this technique, many modern measurement techniques in materials science, biotech etc are based on radiation emitted from accelerating electrons. The "hard" X-rays generated in synchrotron are ideal for e.g. determining the structure of proteins etc.

Here is a link to one such facility
http://www.maxlab.lu.se/

12. Mar 9, 2010

### Frame Dragger

I would just add that you can SEE synchroton light. Here, it's a bit Čerenkov-ish http://upload.wikimedia.org/wikipedia/commons/4/4a/Synchrotron_radiation.jpg [Broken]

Last edited by a moderator: May 4, 2017
13. Mar 9, 2010

### varga

http://en.wikipedia.org/wiki/Lorentz_force

Like that? However, the radiation there is due to electrons colliding with gas molecules in the bulb. Show me the same thing in vacuum and you will convince me.

Yes, energy of charges (electrons) comes from their mass and its velocity too, but primarily from their electromagnetic potentials. The energy stored in electromagnetic fields does not degrade over time or due to any interaction, just like energy stored in gravitational potential is constant in relation to its mass, and herein lies the problem.

Let's take a proton to be our model of Earth and lets take an electron to represent the Moon, and let's imagine these are the only two particle in the universe. Lets put this electron "above" the proton, lets measure the energy of our system at this initial position where both are stationary, and now let us turn our simulation on, let the electron "free fall" towards the proton.

As it gets closer it will accelerate and exchange field potential energy to gain this velocity, just as free falling objects in gravity field. The closer it gets the more it accelerates, but then you say it also radiates more, or does it absorb photons in this case? So, what happens when this electron gets really close and is supposed to have the greatest velocity, does it actually slow down? If it did slow down, then Coulomb's law would be seriously flawed, however it seem to work just fine, just as well as Newton's law of gravitation. And, if we started with only two particles, how many photons did they manage to emit, how many particles are there now, and how is the energy conserved?

Now, let us shift this electron to one side a bit just enough so they miss each other. As the electron flies by and moves away it is supposed to decelerate according to Coulomb's law, but you say it is also supposed to radiate, or again, perhaps absorb photons in this case, but what is there to absorb if nothing else is anywhere near?

I hope we are not talking about "virtual photons".

Let's calculate some real numbers, lets take the number of electrons in Earth and plug it in that formula, take acceleration from Earth spin, rotation around Sun and rotation of solar system around galaxy, what is the P equal to?

If that's nothing, than perhaps it's nothing for electrons in atom orbit too. Has anyone tried to plug in some real numbers and see what would be the speed, radius, acceleration and radiation of an electron in classical orbit around a proton?

14. Mar 9, 2010

### Frame Dragger

I love that photograph of the electron beam circling in the gas... so lovely.

15. Mar 9, 2010

### PhaseShifter

Let's start here and make one small change--give the electron just enough sideways momentum that the attraction of the proton should pull it into a circular orbit.

This model is essentially a rotating dipole, which will produce an oscillating electric field. These oscillations will radiate through space, carrying energy away from the location of the electron-proton system. In a classical system, this means the electron must be losing an equal amount of energy, and the orbit must therefore degrade towards one with a lower potential.

Now the electron is circling the proton even faster, and therefore emitting higher frequency radiation. This means it loses energy faster and faster until the radius of the orbit is zero.

This would be an infinite change in potential, which obviously doesn't happen with real atoms.

16. Mar 9, 2010

### SpectraCat

Try googling "free-electron laser" ... those generate stimulated emission of radiation due to the controlled acceleration of electrons in (for example) a periodically-poled magnetic field.

Yes, in the early 1900's. Try googling "Bohr atomic model" for the details.

Or check here: http://en.wikipedia.org/wiki/Atomic_theory#First_steps_towards_a_quantum_physical_model_of_the_atom

and here: http://en.wikipedia.org/wiki/Larmor_formula

17. Mar 9, 2010

### jfy4

Im still not understanding your issue with the radiation reaction force... Why can an electron not lose kinetic energy to radiation? In fact, they way i see it, you said the correct thing which compliments what i said before. The energy of the charge cannot come from its potential, that leaves really one other source for losing energy, from kinetic. The radiation from a charge come at the expense of its kinetic energy.

Conservation of energy comes about because the energy lost by the electron is the energy radiated. Is that clear? Im not sure how else to explain this... I really hope this is making sense.

For gravitational radiation, im not going to do the calculation for you, because i believe it. I believe it because many experimentalist are, as we speak, working to detect gravitational radiations from sources, and hence, have worked out the math them selves, hoping to match prediction with observation. however, i will give you the formula and you can do it becuase you are the skeptical one ;)

$$\frac{dE}{dt}=\frac{1}{5}\frac{G}{c^{5}}\left<\dddot{I}_{ij}\dddot{I}^{ij}}\right>$$

with

$$I^{ij}= \mathcal{I}^{ij}-\frac{1}{3}\delta^{ij}I^{k}_{k}$$

with

$$\mathcal{I}^{ij}=\int d^{3}x \mu x^{i}x^{j}$$

where $$\mu$$ is the mass density

I hope this helps clear some things up.

18. Mar 9, 2010

### varga

@Frame Dragger

Thanks, if it weren't for you I would never realize this is actually called "synchrotron radiation", even if that was kind of the 1st thing f95toli said, and even your phrasing "synchrotron light" did not directly lead me to this realization.

- "...immediately signaled to turn off the synchrotron as "he saw an arc in the tube." The vacuum was still excellent...."

Ok, I admit these people are smart since their first suspicion was the same as mine, hehe - that something is wrong with the vacuum. If that photo of yours shows radiation that is in vacuum and not due to any collisions, then I'm convinced there is some weird radiation there, I'll call it "synchrotron radiation" and I will accept all the equations of this theory.

19. Mar 10, 2010

### varga

We have two smallest amounts of electric charge there can be, these two electric fields can not create any more electric fields, because the field energy is constant, and discrete. I agree these two fields, including all the magnetic fields that go along, will oscillate (orbit) around, the question is for how long.

Ok, lets say they do radiate and lose energy, so how long it would take for it to spiral down - million years, ten minutes, a second? In any case, how do you describe electron motion of QM atom model, how is it possible for them to not radiate, how can they move in a certain area without acceleration, without making any corners? How can they move without making curved trajectories like in bubble chambers and that photo of pink electron beam?

20. Mar 10, 2010

### varga

Ok, I'm now very familiar with with these accelerators but I'm still suspicious whether there is any collision going on there, because I could not find a picture where it is obvious electrons are not colliding like the one I posted where electrons in gas go circles.

Ok, but I'm not sure if they actually modeled the interaction by integrating Coulomb's law and practically solving a two-body problem by numerical integration and modeling motion with Newton's laws of motion and kinematics equations, or did they use some more statistical equations with sin/cos, harmonics, spring-like oscillations or something like that?

But wait the second, that theory ignores this radiation for the most part, except when electron jumps orbits, it is actually still working and applicable model as I understand.