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GoodNickTaken
2007-Jan-27, 02:31 PM
I've spent the last few days searching endlessly after a simple(?) answer to a simple(?) question. What makes a tungsten filament in an incandescent bulb heat up and shine.

I know it's the EM-field (energy) produced by the E-field (voltage) and B-field (current), but what physically takes place inside the filament?

-It's not frictional heat from vibrating electrons, supposedly this is too small to contribute significantly.
-Somehow the EM-field is converted into heat. HOW? I just can't see how this can be accomplished *without* friction, one way or another, so which way is it?

The reason I signed up and posted here is because I found this 8 page long thread from 2005:

http://www.bautforum.com/showthread.php?t=20365

I've read it all. Some posts are close to asking the right question, but they are not picked up. For example, see this post:

http://www.bautforum.com/showpost.php?p=558330&postcount=205



Why isn't the KE enough? If that's not what actually causes the bulb to heat up and create light, what does it? How does the EM transform into heat and light if not through movement of particles inside the bulb?


This is for me the crux og the problem. I've read countless articles explaining how

1. the EM-field "dives" into the resistor
2. ...
3. LIGHT!

but what's happening at #2?

my_wan
2007-Jan-28, 03:10 AM
This really belongs in questions and answers.

The element does not shine because it got hot. Heat is mostly an energy wastefull byproduct. The heat is produce by the tungsten resistance to the flow of electrons through it much like friction. To produce light one of the electrons circling a tungsten atom starts orbiting at a higher level, it absorbs energy to do this, then when the electron returns to its usual orbit it releases this energy in the form of light.

korjik
2007-Jan-28, 06:26 AM
Read up on Ohm's law. That is what causes the heating. Then read up on blackbody radiation, that causes the light

GoodNickTaken
2007-Jan-28, 10:58 AM
The element does not shine because it got hot. Heat is mostly an energy wastefull byproduct. The heat is produce by the tungsten resistance to the flow of electrons through it much like friction. To produce light one of the electrons circling a tungsten atom starts orbiting at a higher level, it absorbs energy to do this, then when the electron returns to its usual orbit it releases this energy in the form of light.

Thanks for answering. You're probably correct in that this thread should be in another section. Apologies.

What you are saying is that the EM energy, described by the Poynting vector field, excites the atoms in the filament and upon de-excitation, another photon is created. Yes, I know about this stuff, but it doesn't ring well when you think about the photoelectric effect.

First of all, the EM-field has a frequency of ~60 hertz. The energy of a photon is E=hf and 60 hertz is very very low. The photoelectric effect shows us that nomatter how many photons you bombard an object, no matter the intensity, you need a certain energy to excite the atoms. I interpret it this way, that a photon with energy E1, can never excite an atom to produce a photon with energy E2>E1. Obvisously, visible light has a much higher frequency than 60 hertz, so something else must be the mechanism.



Read up on Ohm's law. That is what causes the heating. Then read up on blackbody radiation, that causes the light

Black body radiation is the frequency distribution of photons due to a specific temperature of the body.

According to Ohm's law, a filament heats up because of impurities in the material. The charged electrons bump into these impurities, and heats the thing up through friction.

I could understand that. Perfectly valid. But where does the poynting field vector enter all this? What's the purpose of the EM-field at *all*, if the thing shines due to friction between electrons and atoms?

It can't be that the EM-field excites atoms, it has too little energy. I've also read that electrons themselves has way too little kinetic energy to cause much friction. Somehow the EM-field must add to the friction. How. Does it tug on the electrons themselves, raising the velocity (thus causing a "more" friction)?

EDIT: I suspect it's something I'm missing here. Consider when you hold a fluorescent tube near a powerline. The tube will light up. Since the EM photons from the powerline has lower energy than the EM photons streaming from the tube, this process is not allowed due to the photoelectric effect. The conclusion must be that the EM photons from the power line give the atoms in the gass thermal motion, and that it's the thermal excitation that makes the tube shine.

Similarly, in a tungsten filament, the EM photons push and pull at the atoms, not the electrons around the atoms. When the thermal energy of the atoms is large enough, they shine through black body radiation (because they are hot). Please, is this correct?

Nereid
2007-Jan-28, 11:01 AM
Moved from ATM section.

Jeff Root
2007-Jan-28, 01:12 PM
I can understand that you may want a more detailed and in-depth
description than mine. But here is mine:

The electric field pushes and pulls on free electrons in the metal
conductor. Those electrons bump into the electron clouds of the
atoms in the conductor. In a good conductor there is relatively
little bumping going on, so the atoms don't heat up much. In the
lightbulb filament, though, the conductor is less good, so a lot of
bumping takes place, heating the atoms enough to emit something
closely approximating blackbody radiation.

-- Jeff, in Minneapolis

Delvo
2007-Jan-28, 03:16 PM
60 Hz doesn't represent the amount of power/energy in the electric field or electric current. For that, you'd have to measure amps, volts, and/or watts. "Frequency" in electrical systems refers to how quickly an alternating current changes direction. A direct current doesn't change direction so it has no frequency at all, but it still carries energy and power (and thus can still have real quantities for amps, volts, and watts).

So when you take a current's "frequency" and try to connect or equate that to the resulting radiation's "frequency", you're jumping between two disconnected and unrelated phenomena based on a mere verbal coincidence. It's like looking at one of the "cars" that make up a train and wondering where its engine and rubber tires are because it's called a "car". You'll realize that you already know this yourself if you ponder the fact that one electrical power source can yield more than one frequency of light.

When an electron "falls" back to a lower energy state within its atom, the amount of energy it emits is the difference between the two electron energy states, and, as you've pointed out, the photon's frequency is directly related to its energy, but not to how the electron got into the higher state in the first place. The photon is a brand new thing that's created then, not a holdover from whatever energy source first moved the electron (whether it's heat or electricity); the electron's changes of state separate what happens before them from what happens after them.

trinitree88
2007-Jan-28, 11:54 PM
When a single atom is considered, you can draw an energy level diagram, and correlate specific energy level jumps with emission of specific bright spectral lines. Considering the tungsten filament, you'd think that only certain colors would appear, and if your filament was only made of 1 atom that's quantum mechanics at it's best.
But, metallic bonding changes things. When one metal atom bonds with another, the two can share their valence electrons, while the kernel electrons remain with the individual atoms. These two electron "sea" electrons cause something funny in the energy level diagram of the two atoms taken together....each level splits slightly, enabling a doubling of light frequencies from a doubling of energy transitions. Since a true filament contains ~ 1020metallic atoms....every one of which causes more slight energy shifts, an entire conduction band, rather than level, exists where the valence electrons were.
The friction decribed about impurities in the metal causes the electrons to boil off from the metallic sea electrons...so called thermionic emission. (you see it as an incandescence of the filament).They have almost a continuous range of levels to drop to because of the bands, so the spectrum is near black-body because it's not limited like gas discharge tubes are. Should work for almost any metal, but physical properties of tungsten, particularly it's high melting point make it ideal for longevity.
It's interesting that T.A. Edison noted the current flow from thermionic emission to a second wire (diode) when positively charged, but noted in his notebook that he " doubted the effect could ever be of commercial importance"...the foundation of our electronics industry. We all do it.:shifty:
pete.

GoodNickTaken
2007-Jan-29, 01:43 AM
Thanks a lot for all the answers!!

Let me see if I got this right. The E-field, due to the voltage, pulls all the electrons at once. They move at a snail's pace, but nonetheless induces a B-field. When you cross these fields, ExB, you get electromagnetism, energy, in the form of photons.

The kinetic energy of the electrons is too small to heat the filament that much. Most of the heating comes from the photons in the EM field wreacking havoc between atoms and electron clouds in the metal, resulting in emitted photons in a wide range of frequencies.

I was asked to step in and give a lecture on electricity on short notice. I'm a MSc in physics, but I never could get the hang of this subject, you could even say I "hated" it. I turned into an astrophysicist instead. Anyway, thanks for the help. I doubt I'll go into this much detail, it's physics on a very low level, but I couldn't go on stage teaching things I didn't have a half-good understanding of myself - I won't be a teacher like *that*.

swansont
2007-Jan-29, 12:04 PM
The friction decribed about impurities in the metal causes the electrons to boil off from the metallic sea electrons...so called thermionic emission. (you see it as an incandescence of the filament).They have almost a continuous range of levels to drop to because of the bands, so the spectrum is near black-body because it's not limited like gas discharge tubes are. Should work for almost any metal, but physical properties of tungsten, particularly it's high melting point make it ideal for longevity.


Wait, are you claiming that the blackbody spectrum comes from thermionic emission and recombination? I don't think that's right.

I don't see how you would get the blackbody spectrum, since it's still a band structure, and why you would still have a blackbody spectrum at all temperatures. AFAIK you get the blackbody emission from the same mechanism as everything else that emits blackbody radiation, the acceleration of charged particles that occurs during a collision.

trinitree88
2007-Jan-29, 08:03 PM
Wait, are you claiming that the blackbody spectrum comes from thermionic emission and recombination? I don't think that's right.

I don't see how you would get the blackbody spectrum, since it's still a band structure, and why you would still have a blackbody spectrum at all temperatures. AFAIK you get the blackbody emission from the same mechanism as everything else that emits blackbody radiation, the acceleration of charged particles that occurs during a collision.

Sawnsont. Well, I don't think there's going to be line spectra here, not from a metal surface. Solid state's not my best area, but I think the incandescence of heated metals involves wavelength shifting lambda-max. Metals emit pretty much blackbody at all temperatures, you just notice it more when it begins to be vigorous and lambda -max moves first to (IR) and then visible. Their shinyness is thought to be due to the fact that they make good absorbers and re-emitters at all temperatures. No? Pete.

swansont
2007-Jan-29, 11:52 PM
Sawnsont. Well, I don't think there's going to be line spectra here, not from a metal surface. Solid state's not my best area, but I think the incandescence of heated metals involves wavelength shifting lambda-max. Metals emit pretty much blackbody at all temperatures, you just notice it more when it begins to be vigorous and lambda -max moves first to (IR) and then visible. Their shinyness is thought to be due to the fact that they make good absorbers and re-emitters at all temperatures. No? Pete.

Yes, metals are good blackbodies, but I don't think that has anything to do with promoting electrons out of the conduction band and having them fall back in. It's collisions of electrons that are relatively free to move about because of the conduction band, and the resulting atomic lattice vibrations, which then radiate.

trinitree88
2007-Jan-30, 02:46 AM
Yes, metals are good blackbodies, but I don't think that has anything to do with promoting electrons out of the conduction band and having them fall back in. It's collisions of electrons that are relatively free to move about because of the conduction band, and the resulting atomic lattice vibrations, which then radiate.

Swansont. I'm thinking the electrons have to be bumped up and out of their collective grounds states, in effect leaving holes for so many electron/hole pairs, but at a near continuum of energy levels because of the bandwidth, and then some of them statistically recombining yielding photons of all energies following a Planck distribution. The metal nuclei are not nearly as localized as in an ionic solid, as evidenced by the relatively low bond strength, and the malleability and ductility of the metals, so I'd think the phonons have a great range of energies too. No? Pete

swansont
2007-Jan-30, 11:48 AM
Swansont. I'm thinking the electrons have to be bumped up and out of their collective grounds states, in effect leaving holes for so many electron/hole pairs, but at a near continuum of energy levels because of the bandwidth, and then some of them statistically recombining yielding photons of all energies following a Planck distribution. The metal nuclei are not nearly as localized as in an ionic solid, as evidenced by the relatively low bond strength, and the malleability and ductility of the metals, so I'd think the phonons have a great range of energies too. No? Pete

I'm not a solid-state guy either, but conduction electrons are not the ones in the ground state. They are the highest-energy electrons, essentially free to move (hence conduction) which is why conductors are very close to being blackbodies. electron-hole recombination is "un-blackbody-like"

GoodNickTaken
2007-Jan-30, 03:11 PM
I like this discussion, it's enlightning.

I just had a thought I wanted to share.

In a AC current, the E-field changes direction X times a second (60?). This again, makes the electrons jitter back and forth, also reversing the magnetic fields 60 times a second. However, the crossproduct still has the same direction! This will lead to a sort of modulation. Where the E and B are "fully extended" in one direction, then they start to change direction, while the resulting EM-field shrinks, until they are "zero" and starts growing to a new maximum. From what I've read in this post, this is not the "frequency" of the light, it's more along the intensity of light.

The question: The switching og direction won't affect the intensity that much, since the fields spread out so fast. However, there must be a tiny weeny bit of energy "lost". Anyone up to the challenge of calculating how much energy is lost due to this flickering of direction? I guess it is, for all purposes, unmeasurable.

swansont
2007-Jan-30, 04:20 PM
I like this discussion, it's enlightning.

I just had a thought I wanted to share.

In a AC current, the E-field changes direction X times a second (60?). This again, makes the electrons jitter back and forth, also reversing the magnetic fields 60 times a second. However, the crossproduct still has the same direction! This will lead to a sort of modulation. Where the E and B are "fully extended" in one direction, then they start to change direction, while the resulting EM-field shrinks, until they are "zero" and starts growing to a new maximum. From what I've read in this post, this is not the "frequency" of the light, it's more along the intensity of light.

The question: The switching og direction won't affect the intensity that much, since the fields spread out so fast. However, there must be a tiny weeny bit of energy "lost". Anyone up to the challenge of calculating how much energy is lost due to this flickering of direction? I guess it is, for all purposes, unmeasurable.

You will get a 60Hz signal from the wire; it's basically a radio antenna. You run the return along the same path so you do not radiate this in the far-field. But make a loop antenna on your oscilloscope input, or put some sensitive electronic equipment nearby, and you'll pick it up.

But that's different from the blackbody radiation. You'll have a Maxwell-Boltzmann distribution of speeds of your electrons and atoms, and they will be undergoing collisions, and that's the source of the continuous, temperature-dependent spectrum.

trinitree88
2007-Feb-02, 12:57 AM
I'm not a solid-state guy either, but conduction electrons are not the ones in the ground state. They are the highest-energy electrons, essentially free to move (hence conduction) which is why conductors are very close to being blackbodies. electron-hole recombination is "un-blackbody-like"

Swansont. OK, I learned something. Thanks. Pete

mugaliens
2007-Feb-02, 07:24 PM
Thanks a lot for all the answers!!

Let me see if I got this right. The E-field, due to the voltage, pulls all the electrons at once.

Actually, it pulls only a tiny fraction. If it pulled all the electrons at once, that tiny filament would explode with the force of a rather large bomb. Try putting 120 volts across a 3-volte lightbulb and you'll see what I mean.


The kinetic energy of the electrons is too small to heat the filament that much.

Actually, due to their sizeable numbers (trillions upon trillions up trillions...), it's exactly what heats the filament, due to friction. By the way, ordinary friction occurs at the subatomic level, as electrons are captured, transferred, bumped, etc., when you rub your hands together. Using a voltage potential to cause electrons to do it without mechanical contact with another device produces the same effect as putting a grinder to a piece of steel. The sparks are hot because of friction, and emit light because they're hot (the electrons are in an excited state, and as the electrons collapse back into their normal orbits, they emit photons).

trinitree88
2007-Feb-02, 11:30 PM
I'll ask for pointers on this one. I checked several sites for clarity. Most of them refer to electrons having a statistical mechanical velocity distribution involving e-{W/kT}, where W is the Work function (the energy necessary to free up an electron from the metal surface...also involved in Einstein's interpretation of the photoelectric effect)...and k is the Boltzmann constant, and T is the temperature in Kelvins.
Because of the high velocity "tail' in the distribution, some electrons are endowed statistically with sufficient velocity to exceed W, but since they must also be near the surface of the filament, and traveling in a direction favorable to leave it, only some of those can actually do so. This is more likely as the temperature rises and phonons of kinetic energy kick more electrons to higher average velocities, but it is not nil at low temperatures. So thermionic emission can cause a light cloud of electrons to form near the metallic surface at all T's.
Return of these thermionic electrons to their vacated states releases some of the photons "seen" from the filament. Others can form from electrons near the metallic surface. The energy density of the thermionic electrons is seen to be proportional to ~ T2 up until red heat where it becomes proportional to ~T4. The total bolometric emission of photons is also proportional to T4...from there on out. I had thought that that was true at all temperatures..the Stefan-Boltzmann Law...No? pete.

papageno
2007-Mar-09, 08:36 PM
Yes, I know that thread necromancy is a bad thing, but the ATM forum is a bit slow in this period.
So, here are my two (euro)cents.


The question:


What makes a tungsten filament in an incandescent bulb heat up and shine?


The short answer:


Read up on Ohm's law. That is what causes the heating. Then read up on blackbody radiation, that causes the light


The a-bit-less-short answer:

1) What makes the filament heat up?

Joule heating.

The power source produces an electric field which acts on the electrons in the filament. In metals and other conductors, the conduction electrons - which are responsible for the electrical properties - are relatively free to move within the volume of the conductor. So the electric field can accelerate them, doing work on them and increasing their kinetic energy.
However, since a typical conductor is far from being a perfect and ideal crystal, the average distance an electron can travel before bumping into an impurity, stumbling on a lattice defect or scatter on the lattice vibrations is rather small (for copper at room temperature it is of the order of 30 nm). With each scattering event, an electron exchanges energy with the atoms forming the conductor, and the energy it gained from the external electric field tends to be transferred to the atoms and increase the energy of their vibrations in the lattice.
This increase of the energy of the lattice vibrations in the tungsten filament is the increase of temperature we observe.

This is the step #2 you were asking about: the electric field does work on the electrons which pass it on to the atoms because of the scattering. There is a net flow of energy from the electric field established by the power supply, and this is why the Poynting vector is pointing towards the filament.


2) What makes the filament shine?

Thermal radiation.
You would get the same if you used an oven to heat the filament.

The filament is made of charged particles in non-uniform motion (atoms vibrating, conduction electron "zipping" around being scattered), which means that the particles emit and absorb EM radiation (photons). On typical timescales, a steady state is reached quite quickly and the system (charged particles+photons) is fairly close to thermal equilibrium. Some of the photons manage to leave the filament and that's the light we see.
This, of course, does not exclude other processes (for example, fluorescence, thermionic emission, etc.).




I'm not a solid-state guy either, but conduction electrons are not the ones in the ground state. They are the highest-energy electrons...

Well, the conduction electron are the ones with the highest energy, but they are also in, or at least very close to, the ground state.
Since electrons are fermions, the Pauli exclusion principle applies: only two electrons can occupy an energy level. So, even in the ground state, the electrons are stacked in the available energy levels form the lowest level up to a maximum (the Fermi level), which depends on how many electrons there are.
For the typical electron densities in metals, the temperature necessary to take the electrons far from the ground state is of the order of thousands of Kelvin.




The switching of direction won't affect the intensity that much, since the fields spread out so fast. However, there must be a tiny weeny bit of energy "lost". Anyone up to the challenge of calculating how much energy is lost due to this flickering of direction? I guess it is, for all purposes, unmeasurable.

In typical metals the average time between scattering events is of the order of tens of femtoseconds. So within a a cycle of the 50-60 Hz mains power, the electrons undergo a large number of scattering events.




If it pulled all the electrons at once, that tiny filament would explode with the force of a rather large bomb.

Actually the electric field does act on all electrons at once, it's just that the electric fields in the atoms are far stronger than the fields we produce normally.

trinitree88
2007-Mar-10, 06:59 AM
Yes, I know that thread necromancy is a bad thing, but the ATM forum is a bit slow in this period.
So, here are my two (euro)cents.


The question:


The short answer:


The a-bit-less-short answer:

1) What makes the filament heat up?

Joule heating.

The power source produces an electric field which acts on the electrons in the filament. In metals and other conductors, the conduction electrons - which are responsible for the electrical properties - are relatively free to move within the volume of the conductor. So the electric field can accelerate them, doing work on them and increasing their kinetic energy.
However, since a typical conductor is far from being a perfect and ideal crystal, the average distance an electron can travel before bumping into an impurity, stumbling on a lattice defect or scatter on the lattice vibrations is rather small (for copper at room temperature it is of the order of 30 nm). With each scattering event, an electron exchanges energy with the atoms forming the conductor, and the energy it gained from the external electric field tends to be transferred to the atoms and increase the energy of their vibrations in the lattice.
This increase of the energy of the lattice vibrations in the tungsten filament is the increase of temperature we observe.

This is the step #2 you were asking about: the electric field does work on the electrons which pass it on to the atoms because of the scattering. There is a net flow of energy from the electric field established by the power supply, and this is why the Poynting vector is pointing towards the filament.


2) What makes the filament shine?

Thermal radiation.
You would get the same if you used an oven to heat the filament.

The filament is made of charged particles in non-uniform motion (atoms vibrating, conduction electron "zipping" around being scattered), which means that the particles emit and absorb EM radiation (photons). On typical timescales, a steady state is reached quite quickly and the system (charged particles+photons) is fairly close to thermal equilibrium. Some of the photons manage to leave the filament and that's the light we see.
This, of course, does not exclude other processes (for example, fluorescence, thermionic emission, etc.).



Well, the conduction electron are the ones with the highest energy, but they are also in, or at least very close to, the ground state.
Since electrons are fermions, the Pauli exclusion principle applies: only two electrons can occupy an energy level. So, even in the ground state, the electrons are stacked in the available energy levels form the lowest level up to a maximum (the Fermi level), which depends on how many electrons there are.
For the typical electron densities in metals, the temperature necessary to take the electrons far from the ground state is of the order of thousands of Kelvin.



In typical metals the average time between scattering events is of the order of tens of femtoseconds. So within a a cycle of the 50-60 Hz mains power, the electrons undergo a large number of scattering events.



Actually the electric field does act on all electrons at once, it's just that the electric fields in the atoms are far stronger than the fields we produce normally.


Papageno. Thanks. So, I'm thinking, if I ultrazone-refine the copper to reduce impurities, or the tunsten in a filament, I get a minute delay in the temperature rising ? ...That's mynute, not minit.order of nanoseconds?

papageno
2007-Mar-10, 10:18 AM
So, I'm thinking, if I ultrazone-refine the copper to reduce impurities, or the tunsten in a filament, I get a minute delay in the temperature rising ? ...That's mynute, not minit.order of nanoseconds?


Even if you have higly-pure and defect-free materials, the electrons would still scatter on the lattice vibrations (electron-phonon scattering).

If you want to reduce this scattering, the easiest way is to reduce the number of phonons (which are the excitations of the lattice) by decreasing the temperature.
It is possible to increase the average scattering rate to nanoseconds, though the electron density in copper and other metals might be too high for this to happen.

But another effect can occur at low temperatures. By reducing the number of phonons, the rate by which the electrons can transfer energy to the lattice is reduced. In extreme cases, this energy relaxation rate is too low , and the electrons cannot transfer to the lattice all the energy they gain from the electric field that is driving the current.

Since the electrons exchange with each other energy much faster than with the lattice, they reach a thermal equilibrium with themselves but not with the lattice. We end up with the lattice being at some temperature and the electron gas being at a higher temperature (search for hot electrons).