Electrons and Inertial Frames

J_B_

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My eventual purpose is to ask: Is an electron orbiting a neutron in an inertial frame. What I can gather from what I've read so far, the answer is, "It's complicated."

So, I'll start with something simpler - just a simple mass on a circular trajectory traveling at speed c. Is this mass in an inertial frame? I would think it is not, but what do I know?
 

Hans Blaster

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My eventual purpose is to ask: Is an electron orbiting a neutron in an inertial frame. What I can gather from what I've read so far, the answer is, "It's complicated."

Electrons don't orbit neutrons, they do "orbit" protons. (It's called hydrogen.)

But the "orbit" model is very suspect, and a better model is quantum mechanics. The relativistic version of QM (Dirac equation rather than Schrödinger equation) includes the relativistic effects. Fun fact: the yellow color of gold comes from a relativistic effect on the "inner" electrons.
 
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Monksailor

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I studied a little bit about which you refer as just yesterday I was contemplating what I had leaned 45 years ago in Chemistry about the nucleus and the surrounding "spheres" of orbiting electrons; the farther down the chart, heavier, the more electrons and consequential spheres and each sphere had a certain number of electrons that it "liked" to have which resulted in pos. or neg. valances of atoms, I believe, and affected the stability and or volatility or reactionary properties thereof; ionic bonding, I believe. I think that they had letter designations such as "s", "d", "f" and such.

Anyway, in the past I did not need to get past Chemistry or into Quantum Physics so I just delved into a tiny tip for an attempt to help you here. It appears that there are different scientific schools of thought regarding electrons in orbit being characterized by being in an inertial reference frame or a non-inertial reference frame.(SEE: https://www.quora.com/Is-the-movement-of-electrons-in-an-orbit-inertial) Additionally, It is not possible or plausible to simplify what you want to find out, as I see it. There are just too many variable factors and forces which are being ignored in your simplification, as I see it. An inertial reference frame has to do with measurement of relative objects, forces, and time, if i am not mistaken. One cannot even give you an answer to your simplified quest as you provide insufficient data, as I see it.
 
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FrumiousBandersnatch

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The classical 'Rutherford model' of the atom as a sort of tiny solar system with electron 'planets' orbiting a nucleus 'sun' died in the early 20th century with the advent of quantum mechanics (it was known to be problematic almost from the start). So I don't think the question really makes sense with electrons; but if they were like little balls orbiting, they would be accelerating, so not themselves in an inertial frame.

Sean Carroll's informal talk on quantum mechanics explains why the Rutherford atom doesn't work (at 16 mins 30 seconds):
 
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J_B_

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A few typos in that post, so not a good start. Probably makes me look like an idiot. But I'll press on. First of all, I'll clear up the typos.

If by 'c' you mean the speed of light, then it doesn't happen -- a mass can't move at c. If you mean a lower velocity, then the rest frame of the electron is not an inertial frame.

Yeah, sure. All I really meant to say was that it's moving fast enough to have significant relativistic effects.

Electrons don't orbit neutrons, they do "orbit" protons. (It's called hydrogen.)

But the "orbit" model is very suspect, and a better model is quantum mechanics. The relativistic version of QM (Dirac equation rather than Schrödinger equation) includes the relativistic effects. Fun fact: the yellow color of gold comes from a relativistic effect on the "inner" electrons.

Right. I meant to say "nucleus" to keep it general rather than a specific "neutron". We'll get to the orbit thing in a moment.

The classical 'Rutherford model' of the atom as a sort of tiny solar system with electron 'planets' orbiting a nucleus 'sun' died in the early 20th century with the advent of quantum mechanics (it was known to be problematic almost from the start). So I don't think the question really makes sense with electrons; but if they were like little balls orbiting, they would be accelerating, so not themselves in an inertial frame.

Sean Carroll's informal talk on quantum mechanics explains why the Rutherford atom doesn't work (at 16 mins 30 seconds):

Actually, I'm reading through Carroll's book on general relativity. That's the birth of my question because I take all kinds of side trips as I read - not to help me with the math, but to help me interpret what the math means. I'm aware the "orbit" idea is problematic, but it seemed the best approach for the moment. I expect to shed that term in due time.

Circular motion requires acceleration, so no, it's not an inertial frame.

Good. It seems to be a consensus that's the answer, so at least my head's in the right place so far.

Believe it or not, the math is not the problem. I'm confident I can handle the math. Rather, it's the interpretation that trips me up. We start with an observation: the speed of light is constant. OK. How do we explain that? Lorentz Transformations. OK. Not that tough. A brilliant insight for those who developed it, but for those of us with some math skills who are trailing behind the band wagon it's simple enough to understand.

For my example of a simple mass, let's make it's rest mass equivalent to an electron, since that's where I'm headed. Even then, it's still going to experience spacetime dilation, yes?

This is where things went off the rails for me. I simply wanted to know, when an electron is in orbit about a nucleus, does it experience spacetime dilation? It seems it would, but every reference I read started talking about charge and spin and comoving frames, etc. etc. etc. and I could never tell if they were trying to explain a "yes" answer to my question or a "no" answer. Or maybe they're telling me it's a stupid question.
 
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essentialsaltes

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This is where things went off the rails for me. I simply wanted to know, when an electron is in orbit about a nucleus, does it experience spacetime dilation?

I think it must. I don't know that you could tell directly, since electrons don't age. But relativistic effects provide little tweaks to chemistry. These are generally due to the increased mass of the electrons. And if you ask me, if your electron is getting fat from relativity, then it's clock is also slowing.
 
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Of the Kingdom

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You might be interested in this question on stack exchange:

How fast do electrons travel in an atomic orbital?

Although not a classical orbit, the motion can be approximated by one. An answer gives the computed velocity as 1/137 of c, or roughly proportional to that. Relativistic effects exist, but would not be dramatic.

It would not be an inertial frame, but neither is the Earth. Both could be considered approximately inertial for some purposes.
 
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J_B_

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I think it must. I don't know that you could tell directly, since electrons don't age. But relativistic effects provide little tweaks to chemistry. These are generally due to the increased mass of the electrons. And if you ask me, if your electron is getting fat from relativity, then it's clock is also slowing.

Thanks for keeping it short and sweet. That's all I really needed. But, since we're here, might I ask about the electron's orbit? some refer instead to its energy level. OK. But even at the quantum level I thought it was possible to know either position or momentum. If we know position, and that position always lies on a particular closed surface about the nucleus, why is is wrong to call it an orbit? Or is it not true that the position lies on a closed surface?
 
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sfs

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Or is it not true that the position lies on a closed surface?
It is not true that the position lies on a closed surface. The position (if measured) will lie in a fuzzy cloud, the precise shape of which depends on the atom and the orbital that the electron is occupying. (Google for images of 'orbitals'.)
 
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essentialsaltes

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some refer instead to its energy level.

I think better yet, we speak of the electron being in a particular quantum state (which is associated with a particular quantum wave-function), and yes each quantum state is associated with a particular energy. The quantum state is described by the usual 4 quantum numbers:

n = 'electron shell'
l = angular momentum
lz = angular momentum with respect to z-axis
s = spin (up or down)

But even at the quantum level I thought it was possible to know either position or momentum. If we know position, and that position always lies on a particular closed surface about the nucleus, why is is wrong to call it an orbit?

I'm not quite sure what you mean by a 'closed surface'. If you map out the wavefunctions (yes equivalent to taking position measurements on an electron in that state over and over again) they are three-dimensional 'clouds' of probability. They do not look like circular orbits.

Worse yet, many of the allowed quantum states allow the angular momentum to be zero. So most of the spherically symmetric solutions (that might look kinda like orbits) have no angular momentum, so the electron is not 'going around' the nucleus if we try to make a classical picture of the quantum state.
 
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sjastro

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Thanks for keeping it short and sweet. That's all I really needed. But, since we're here, might I ask about the electron's orbit? some refer instead to its energy level. OK. But even at the quantum level I thought it was possible to know either position or momentum. If we know position, and that position always lies on a particular closed surface about the nucleus, why is is wrong to call it an orbit? Or is it not true that the position lies on a closed surface?
If electrons did follow an orbit then as a charged particle undergoing acceleration it will lose energy through electromagnetic radiation and spiral into the nucleus.
It's one of the reasons along with blackbody radiation that led to the development of Quantum Mechanics.
 
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Radagast

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But even at the quantum level I thought it was possible to know either position or momentum.

No. The product of the two uncertainties is always at least h/(4π). Neither uncertainty can be zero (although one uncertainty can be very low if the other is very high).

If we know position, and that position always lies on a particular closed surface about the nucleus, why is is wrong to call it an orbit? Or is it not true that the position lies on a closed surface?

Typical regions in which electrons might be found are like this. They are 3D regions, not surfaces, and they are fuzzy. They are not "orbits."

yCfP7dNnS5yhsw0hi8bc_orbitals.jpg
 
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J_B_

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It is not true that the position lies on a closed surface. The position (if measured) will lie in a fuzzy cloud, the precise shape of which depends on the atom and the orbital that the electron is occupying. (Google for images of 'orbitals'.)

OK. I can get the idea of a cloud.

Typical regions in which electrons might be found are like this. They are 3D regions, not surfaces, and they are fuzzy. They are not "orbits."

yCfP7dNnS5yhsw0hi8bc_orbitals.jpg

Being a visual learner, the pictures help. That's part of why I always want to understand how to interpret the math. In school things often moved so fast that I just memorized the equations and spit out the answers. Then, in later years, when I had time to reflect it was, "Oh, so that's what they meant. I get it now."

No. The product of the two uncertainties is always at least h/(4π). Neither uncertainty can be zero (although one uncertainty can be very low if the other is very high).

I tend to be a "close enough" kind of guy, which doesn't always serve me well, but sure, I understand we never know anything perfectly.

I think better yet, we speak of the electron being in a particular quantum state (which is associated with a particular quantum wave-function), and yes each quantum state is associated with a particular energy. The quantum state is described by the usual 4 quantum numbers:

n = 'electron shell'
l = angular momentum
lz = angular momentum with respect to z-axis
s = spin (up or down)

I'm not quite sure what you mean by a 'closed surface'. If you map out the wavefunctions (yes equivalent to taking position measurements on an electron in that state over and over again) they are three-dimensional 'clouds' of probability. They do not look like circular orbits.

Worse yet, many of the allowed quantum states allow the angular momentum to be zero. So most of the spherically symmetric solutions (that might look kinda like orbits) have no angular momentum, so the electron is not 'going around' the nucleus if we try to make a classical picture of the quantum state.

Well, if it's a variable, it can vary. So of course it can be zero. But if angular momentum is the only motion and it's zero, are you saying the electron isn't moving?

I don't get "spin" yet. I mean, it's easy to imagine an object spinning about an axis, but I get the feeling the primary purpose of quantum mechanics is to take words that have obvious meanings and use them to mean something completely different. For example, given the cloud is only a probability cloud, I'm not even sure angular momentum means angular momentum.

Physics should have taken a clue from biology and concocted Latin words that could be imbued with the necessary meaning rather than stealing perfectly good English words from classical mechanics and insisting they mean something completely different.

OK, I'm done with my rant now. You can proceed to explain.
 
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Radagast

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Being a visual learner, the pictures help.

Same here. That's why I posted them.

I tend to be a "close enough" kind of guy, which doesn't always serve me well, but sure, I understand we never know anything perfectly.

Well, if it's a variable, it can vary. So of course it can be zero.

Well, no, actually. As I said, the product of the two uncertainties is always at least h/(4π). Neither uncertainty can be zero (although one uncertainty can be very low if the other is very high).

This is a rock-solid truth. It also applies to other pairs of complementary variables, like energy and time.

Case in point: electrons in a laser spend a lot of time in a high-energy state. Therefore the time uncertainty is high. Therefore the energy uncertainty is low (but not zero). Therefore photons coming out of a laser have almost exactly the same frequency.

But if angular momentum is the only motion and it's zero, are you saying the electron isn't moving?

It would be closer to the truth to say that it's jiggling around.

I don't get "spin" yet. I mean, it's easy to imagine an object spinning about an axis, but I get the feeling the primary purpose of quantum mechanics is to take words that have obvious meanings and use them to mean something completely different.

Yeah, pretty much. It would be more correct to say that an electron behaves, in its interaction with magnetic fields, kind of like a large charged object spinning on its axis. So we use the term "spin," although an electron isn't really "spinning on its axis."

For example, given the cloud is only a probability cloud, I'm not even sure angular momentum means angular momentum.

It doesn't. It means behaviour which, mathematically, resembles angular momentum in large objects.

Physics should have taken a clue from biology and concocted Latin words that could be imbued with the necessary meaning rather than stealing perfectly good English words from classical mechanics and insisting they mean something completely different.

I take your point. But the words are a clue as to which traditional physics equations are (to some extent) going to be useful.
 
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Radagast

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Physics should have taken a clue from biology and concocted Latin words that could be imbued with the necessary meaning rather than stealing perfectly good English words from classical mechanics and insisting they mean something completely different.

And, of course, there's the computer science approach of taking ordinary English words, giving them a special technical meaning, and spelling them differently to flag the difference. Hence "byte."
 
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essentialsaltes

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Well, if it's a variable, it can vary. So of course it can be zero.

It's not exactly variable. In the hydrogen atom, the angular momentum of the electron can only take on certain quantized values: 0, 1, 2, 3... [in the appropriate units]

But if angular momentum is the only motion and it's zero, are you saying the electron isn't moving?

It certainly can't be revolving 'around' the nucleus in a stable orbit. But more importantly, what we learn from quantum mechanics is that reality is not what we expected. We think we know what matter is. A rock is a rock. And an electron might be a tiny piece of matter: a tiny little rock with an electric charge. But it isn't.

Thinking of an electron as orbiting a nucleus is not merely 'unhelpful' since it doesn't generate the correct energy levels, etc. It is not a good model, because it doesn't reflect the facts of quantum reality.

I don't get "spin" yet. I mean, it's easy to imagine an object spinning about an axis

It's like that, in that it is a different kind of angular momentum that a particle can have, but it is 'intrinsic'. It's not due to any actual little rock that is spinning, but just an inherent property like charge. I mean, what is charge? Again, it's a property assigned to the particle. We can say a capacitor plate is negatively charged because it has an excess of electrons on it. But we can hardly say an electron is charged because there is one more electron on it. Or I guess we do, since it's an inherent property of the electron. Just as spin is.

Physics should have taken a clue from biology and concocted Latin words that could be imbued

Just put quantum in front of everything!
 
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Physics would be so much easier if the universe was Newtonian. Of course then life itself would be impossible. We depend upon the Sun for energy and fusion is beyond the scope of Newtonian physics. My point is that if one tries to interpret the motion of electrons using Newtonian physics one will be wrong. Concepts such as spin and momentum are not the same as in the macro world. An orbital is a region where one is likely to find an electron. One cannot say very much about the movement of an electron in an orbital.
 
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