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alan.b

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alan.b

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- #2

f95toli

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

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

- #3

SpectraCat

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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

alan.b

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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...

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

alan.b

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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.

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

SpectraCat

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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.

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

jfy4

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[tex]P=Kq^{2}a^{2}[/tex]

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

i hope this helps a little...

- #8

varga

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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.

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?

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.

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

varga

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jfy4 said: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.

[tex]P=Kq^{2}a^{2}[/tex]

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.

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?

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 usenergy eigenstateswhich 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...

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

jfy4

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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

f95toli

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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.

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

Frame Dragger

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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/

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]

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- #13

varga

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jfy4 said: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 ;)

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.

... energy has to be something. For charges, the energy comes from the kinetic energy of the charge, so the charge slows down.

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".

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.

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

Frame Dragger

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

- #15

PhaseShifter

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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.

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

SpectraCat

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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.Show me the same thing in vacuum and you will convince me.

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?

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

jfy4

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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?

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 ;)

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

with

[tex]I^{ij}= \mathcal{I}^{ij}-\frac{1}{3}\delta^{ij}I^{k}_{k}[/tex]

with

[tex]\mathcal{I}^{ij}=\int d^{3}x \mu x^{i}x^{j}[/tex]

where [tex]\mu[/tex] is the mass density

I hope this helps clear some things up.

- #18

varga

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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.

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

- "...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

varga

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PhaseShifter said: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.

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.

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.

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

varga

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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.

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.

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

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.

- #21

PhilDSP

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

What's wrong with virtual photons? They're an ingenious, almost legitimate way to begin to model and understand sub-quantum levels of near-field radiation and energy (from within a framework that does not allow you to do that)

- #22

varga

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jfy4 said: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.

Ok, we're getting close. I agree with that, but before I answer your question let me just boil it down a little bit more. If electron's mass is to stay constant, then the "change in kinetic energy" can only mean "change in velocity".

Conclusion:

- The radiation (if any) from a charge can only come at the expense of its VELOCITY.

Your question:

- Why electron should not lose kinetic energy to radiation?

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

* Potential energy is energy stored within a physical system as a

* According to the principle of conservation of energy, energy cannot be created or destroyed; hence this energy cannot disappear.

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

* Electric potential energy aka "electrostatic potential energy" is a potential energy associated with the

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

* By definition, it is equal to the amount of kinetic energy gained by a single unbound electron when it

Conclusion:

- The definition of the most basic unit of energy, 'electron volt', would not work.

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.

It is clear, but not possible according to classical physics and electrodynamics.

My questions:

- ELECTRON VOLT EXPERIMENT: Let's put an electron two volts of potential difference "above" the nucleus, let's 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".

As it gets closer it accelerates and exchanges its potential energy for kinetic energy, its position for velocity. After some time it completely passes through electric potential difference of one volt. Now, tell us what is its velocity (kinetic energy) at this point? When do electrons radiate (or absorb), when they slow down or speed up?

- #23

f95toli

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It is clear, but not possible according to classical physics and electrodynamics.

Of course it is possible. Synchrotron radiation is fairly easy to explain once you take into account relativistic effects; this has been know for about a hundred years (which is why people knew that the "orbit model" could not be correct even back then) and has nothing to do with with quantum mechanics. Most of chapter 14 of Jackson (which is THE book on classical electrodynamics) deals with radiation by moving charges (and sections 14.5-14.7 deals with synchrotron radiation).

Also, remember that we are talking about charges travelling at

- #24

PhilDSP

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As it gets closer it accelerates and exchanges its potential energy for kinetic energy, its position for velocity. After some time it completely passes through electric potential difference of one volt. Now, tell us what is its velocity (kinetic energy) at this point? When do electrons radiate (or absorb), when they slow down or speed up?

An electron (attached to an atom) loses energy to the photon when it radiates. So it drops down to a lower energy state which is closer to the nucleus. In doing so it must travel more quickly around the nucleus to balance the attraction of the nucleus so its velocity increases.

When absorbing radiation, the electron gains energy and so jumps further from the nucleus and reduces its velocity.

But those are the passive effects (not the cause of radiation according to the Bohr model).

I think the apparent contradiction in terms of kinetic energy gain or loss is that linear momentum is exchanged for angular momentum and vice versa.

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- #25

jfy4

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Ok, we're getting close. I agree with that, but before I answer your question let me just boil it down a little bit more. If electron's mass is to stay constant, then the "change in kinetic energy" can only mean "change in velocity".

Conclusion:

- The radiation (if any) from a charge can only come at the expense of its VELOCITY.

Your question:

- Why electron should not lose kinetic energy to radiation?

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

* Potential energy is energy stored within a physical system as aresult of the positionor configuration of the different parts of that system. It has the potential to be converted into other forms of energy, such as kinetic energy, and to do work in the process.

* According to the principle of conservation of energy, energy cannot be created or destroyed; hence this energy cannot disappear.Instead, it is stored as potential energy.

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

* Electric potential energy aka "electrostatic potential energy" is a potential energy associated with theconservative Coulomb forceswithin a defined system of point charges.

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

* By definition, it is equal to the amount of kinetic energy gained by a single unbound electron when itaccelerates through an electric potential differenceof one volt.

Conclusion:

- The definition of the most basic unit of energy, 'electron volt', would not work.

It is clear, but not possible according to classical physics and electrodynamics.

My questions:

- ELECTRON VOLT EXPERIMENT: Let's put an electron two volts of potential difference "above" the nucleus, let's 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".

As it gets closer it accelerates and exchanges its potential energy for kinetic energy, its position for velocity. After some time it completely passes through electric potential difference of one volt. Now, tell us what is its velocity (kinetic energy) at this point? When do electrons radiate (or absorb), when they slow down or speed up?

Sorry about the order of this,

Your progress through, and conclusion to your middle argument about the functionality of an electron volt is unorganized, and I am unable to follow it at all, would you mind making your argument more explicit. The conclusion seems to have nothing to do with the content cited before hand, save the last, which defined an electron volt accurately.

The first part of your post talked about the radiation reaction and it seems you accept now that a charges radiation is at its kinetic energy's expense. Im not sure what else you are talking about after this part. if we both agree that a particle's speed reduces because of the radiation reaction, then what seems to be the problem? You do understand that a particle can be subject to multiple forces correct? It can both be accelerated, arbitrarily forward and backwards, if you will, and still have a net acceleration in one direction.

The energy radiated has nothing to do with the particle's potential energy which, as you stated, is a function of position alone, but its kinetic energy which is radiated away in the form of photons, thus slowing the particle down.

yes potential energy can be converted to kinetic, and kinetic energy is radiated away by the charge, causing the charge to accelerate less.

Is that clear?

Im really at wits end here, especially because you seem to be almost saying the same thing i am, but somehow interpreting as a violation of the conservation of energy or some bad definition of an electron volt...

all the energy of the system is still there, not destroyed or created, but in a different form.

Please be more clear if you can :)

- #26

varga

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jfy4 said:Im really at wits end here, especially because you seem to be almost saying the same thing i am, but somehow interpreting as a violation of the conservation of energy or some bad definition of an electron volt...

all the energy of the system is still there, not destroyed or created, but in a different form.

Please be more clear if you can :)

I think it's very clear and that you should respond more directly to citations from Wikipedia. For example, what part of

Can you please at least answer my question directly so I can make my point and respond to the rest of your message without making anyone repeat themselves. ELECTRON VOLT EXPERIMENT: Let's put an electron two volts of potential difference "above" the proton and measure the energy of the system at this initial position where both are stationary, then let the electron "free fall". After some time it passes through electric potential difference of one volt, at this point, what is its velocity and kinetic energy?

- #27

varga

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f95toli said:Of course it is possible. Synchrotron radiation is fairly easy to explain once you take into account relativistic effects; this has been know for about a hundred years (which is why people knew that the "orbit model" could not be correct even back then) and has nothing to do with with quantum mechanics. Most of chapter 14 of Jackson (which is THE book on classical electrodynamics) deals with radiation by moving charges (and sections 14.5-14.7 deals with synchrotron radiation).

Also, remember that we are talking about charges travelling atrelativisticspeeds, the normal expressions for kinetic energy etc are not applicable.

I did not count relativistic effects as a part of classical physics and electrodynamics.

I'm pretty sure bubble chamber trajectories need no any relativistic corrections to be accurately predicted, classical Coulomb and Lorentz forces in combination with kinematics equation (Newton's laws of motion) will do the job. But I do not really know what effects are you talking about, I don't have that book, can you throw some keywords so I can look into it at Wikipedia? I know about relativistic equations for kinetic energy, I do not think any of that is part of the definition for 'electron volt'.

- #28

SpectraCat

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I think it's very clear and that you should respond more directly to citations from Wikipedia. For example, what part of"conservative Coulomb forces"is not clear to you?

Can you please at least answer my question directly so I can make my point and respond to the rest of your message without making anyone repeat themselves. ELECTRON VOLT EXPERIMENT: Let's put an electron two volts of potential difference "above" the proton and measure the energy of the system at this initial position where both are stationary, then let the electron "free fall". After some time it passes through electric potential difference of one volt, at this point, what is its velocity and kinetic energy?

If we ignore the radiation, and treat the system completely classically, then at the point to which you refer, it's kinetic energy is 1 eV .. it's velocity is sqrt(2eV/m

We are pretty far afield from your original question/point .. are we going to get back there any time soon?

By the way, don't worry about the definition of an electron volt .. it is not phenomenologically defined, as you seem to be assuming. It is simply a collection of fundamental constants corresponding to the *ideal* case from your quote. An actual free-electron cannot be accelerating, because then it isn't free, it is interacting with a potential.

- #29

PhaseShifter

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What part of electroI think it's very clear and that you should respond more directly to citations from Wikipedia. For example, what part of"conservative Coulomb forces"is not clear to you?

- #30

jfy4

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I think it's very clear and that you should respond more directly to citations from Wikipedia. For example, what part of"conservative Coulomb forces"is not clear to you?

Can you please at least answer my question directly so I can make my point and respond to the rest of your message without making anyone repeat themselves. ELECTRON VOLT EXPERIMENT: Let's put an electron two volts of potential difference "above" the proton and measure the energy of the system at this initial position where both are stationary, then let the electron "free fall". After some time it passes through electric potential difference of one volt, at this point, what is its velocity and kinetic energy?

The part that is unclear to me is what the conservative coulomb force has to do with conservation of energy violations, and the problem with the definition of an electron volt...

Spectra cat has just answered your electron volt experiment, which we already knew, trivially, because we knew the definition of an electron volt... and you must have been able to get the velocity from that.

It is unclear to me what is even the problem you are addressing... please write it out so i can help, im really trying to.

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varga

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PhilDSP said:An electron (attached to an atom) loses energy to the photon when it radiates. So it drops down to a lower energy state which is closer to the nucleus. In doing so it must travel more quickly around the nucleus to balance the attraction of the nucleus so its velocity increases.

When absorbing radiation, the electron gains energy and so jumps further from the nucleus and reduces its velocity.

But those are the passive effects (not the cause of radiation according to the Bohr model).

I think the apparent contradiction in terms of kinetic energy gain or loss is that linear momentum is exchanged for angular momentum and vice versa.

Yes, I think the contradiction is apparent too. I see more than one:

1.) Electron slows down when it radiates photons...

- it's velocity (trajectory) would jerk at specific interval, because the energy packets have a certain minim and are discrete - quantified. we do not observe this sudden jerks in electron trajectories, I believe.

- does it slow down BECAUSE it radiates a photon, or it radiates because it slows down?

- if electron in orbit has the constant velocity, is its "acceleration" due to 'change in direction' actually 'acceleration' or 'deceleration' and is it suppose to radiate, absorb, or both in this case?

2.) Electron slows down when it radiates photons, so it drops down to a lower energy state which is closer to the nucleus.

- closer to nucleus may be "lower energy" state in terms of the whole system and principle of the least resistance, but the *actual energy* of the electron (and the whole system) should stay the same and is compensated by increased velocity, there is no room for radiation here.

- put proton and electron 1mm apart where electron has energy Ep=A, move electron to 2mm distance where Ep=B, now move the electron with acceleration or with constant velocity from point A to point B, note that energy of the system and electron energy must not change due to any derivative of the position separately, but strictly according to the relation with the distance, i.e. position itself.

3.) Electron

- which one, slows down or speeds up? causality, i do not see logical sequence there, is it all happening simultaneously?

- if its velocity actually increases at the end, does that not mean it absorbed a photon?

- again, is acceleration causing radiation or is absorption of radiation causing acceleration?

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PhaseShifter

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3.) Electronslows downwhen it radiates photons, drops down to a lower energy state, in doing soits velocity increases.

- which one, slows down or speeds up? causality, i do not see logical sequence there, is it all happening simultaneously?

- if its velocity actually increases at the end, does that not mean it absorbed a photon?

- again, is acceleration causing radiation or is absorption of radiation causing acceleration?

Emitting the photon and dropping to a lower energy state are simultaneous. There is no "slowing down".

Although, to be more technical a wave packet that resembles an orbiting electron is actually a superposition of many states rather than a single state, and would actually merely be changing the relative amplitudes of component wavefunctions.

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- #33

PhilDSP

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Emitting the photon and dropping to a lower energy state are simultaneous. There is no "slowing down".

Since the ejection of the photon is going to be outward from the center of the atom, the photon's momentum will be away from the center. The reactive change of the electron's momentum will correspondingly be inward forcing the drop to the lower energy state.

Although, to be more technical a wave packet that resembles an orbiting electron is actually a superposition of many states rather than a single state, and would actually merely be changing the relative amplitudes of component wavefunctions.

Yes, we should point out that we're working with the Bohr model which we know to be a caricature of the real situation.

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jfy4 said:Your progress through, and conclusion to your middle argument about the functionality of an electron volt is unorganized, and I am unable to follow it at all, would you mind making your argument more explicit.

I expected you would know about potential energy. I'll address that in relation to my example of electron volt experiment. Meanwhile, I'd like you to say something about these three explicit and specific questions:

a) when electron speeds up towards proton, does it radiate or absorb radiation?

b) when electron flies away from proton and it slows down, does it radiate or absorb?

c) we say electron in orbit with constant velocity still "accelerates" due to change in direction, but is this really 'acceleration' or 'deceleration'?

if we both agree that a particle's speed reduces because of the radiation reaction, then what seems to be the problem?

The energy radiated has nothing to do with the particle's potential energy which, as you stated, is a function of position alone, but its kinetic energy which is radiated away in the form of photons, thus slowing the particle down.

yes potential energy can be converted to kinetic, and kinetic energy is radiated away by the charge, causing the charge to accelerate less.

Is that clear?

No, it's slightly funny.

F= k* m1m2/r2 or F= k* q1q2/r2 -> acceleration= F/m

Force due to interaction of field potentials is what is causing acceleration and deceleration. If there was any loss or gain of energy due to radiation, then some of the most basics and tested principles in physics would turn out to be invalid, like 'conservation of energy'. So, do you mean to say these equations have error proportional to "synchrotron radiation" or what?

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PhaseShifter

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Yes, we should point out that we're working with the Bohr model which we know to be a caricature of the real situation.

Yes, this is where his question is a little confusing--orbiting packets aren't eigenfunctions of the Hamiltonian operator, and thus don't have a well-defined energy. But a classical system with orbiting particles doesn't have quantized energy states, and the electrons would spiral into the nucleus as they continuously radiated EM waves into the surroundings.

Historically the Bohr model is important for first explaining the various series of lines in atomic spectra. The mathematics indicated quantized energy states for the bound electron, but the model itself couldn't explain why angular momentum had to be quantized, and that's why it was replaced after just a decade.

I sometimes suspect that the only reason it's still taught is that people have problems visualizing an electron as a complex wavefunction rather than a particle with well-defined position and momentum.

Even now most textbooks don't attempt to explain that the odd-shaped orbitals with multiple lobes are actually superpositions of wavefunctions with opposite angular momentum, so the imaginary part of the complex numbers conveniently cancels out.

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