Special Relativity and momentum

[SDG]

Originally Posted by Ziggurat

...
A decrease in mass produces a decrease in momentum when the velocity is constant. So the sideways linear momentum HAS decreased in this scenario, and it hasn't decreased uniformly across the flashlight.
...

Are you sure?





I did not use 'prime' for P in the diagrams but the second figure P is bigger than the first figure P.
This is Doppler effect so when transformation happens from the moving frame to the rest frame the value is correct.
There is P_recoil x component in the moving frame and y components are equal.

[SDG]

Originally Posted by Reformed Offlian

Wouldn't aberration cause an apparent rotation in the orientation of the flashlight, too?

What apparent rotation?
There is an inertial observer who is part of a bigger grid of inertial observers, no relative motion between them.
It appears to me that if talk about an apparent rotation we are making one preferred inertial observer from his grid of inertial observers.
The angular velocity of apparent rotation is not the same for all observers in the grid.
The grid is broken.
The preferred observer is not even inertial anymore, the observer rotates, to follow the apparent rotation.

[SDG]

Originally Posted by Reformed Offlian

I hope that the OP takes into account that two relatively-moving observers will not agree on the value of P (the momentum of the photon). Because Doppler shift.

Right, so are we going to ignore the P_recoil x component when the Doppler effect takes care of the y component?

[Dave Rogers]

Originally Posted by SDG

Are you sure?

https://i.imgur.com/HpboITi.png

https://i.imgur.com/5p3Twdt.png

I did not use 'prime' for P in the diagrams but the second figure P is bigger than the first figure P.
This is Doppler effect so when transformation happens from the moving frame to the rest frame the value is correct.
There is P_recoil x component in the moving frame and y components are equal.

As someone pointed out up thread, this is equally true of classical mechanics; your description of the experiment would be equally applicable to a bullet fired from a rifle at non-relativistic speed.

Dave

[Ziggurat]

Originally Posted by SDG

A hydrogen atom emits a photon going from n=2 to n=1.
Does the rest mass of the hydrogen atom change?

Yes.

Quote:

Even better, just simple electron, Bremsstrahlung, what is the electron mass change when a photon is emitted?

Just an electron cannot emit Bremsstrahlung radiation. It needs to interact with other matter. And in this case, you are trading off kinetic energy to create the photon, not rest mass energy.

Quote:

As you wrote E² = p²c² + m²c⁴.
Is this equation true?

Yes. Even more importantly, though, it is part of the theory of special relativity. So if you want to evaluate SR, you need to account for it.

Quote:

Let us assume it is.
There is a spontaneous hydrogen atom emission, m is constant,

m is not constant. That's the whole point. You have started with a false assumption.

[Ziggurat]

Originally Posted by SDG

Are you sure?

Yes, I am sure.

You made a mistake. You didn't realize that the mass changed. And you didn't account for the effects of this change. That is all that's going on here. You did not find a flaw in special relativity.

[Ziggurat]

SDG: do this problem with a gun and a bullet, using Newtonian mechanics. Figure out why the gun doesn't twist when you move to a translating frame. The basics are the same, you need to account for the loss of mass.

[Reformed Offlian]

Originally Posted by SDG

Objects (electrons) emitting photons do not lose mass, they lose momentum.

Yes, objects *do* lose mass when they emit photons. Electrons do not; as Ziggurat points out, they trade off kinetic energy for momentum. However, the atoms to which they are bound *do* change mass, even if their constituent particles do not. Avoid the fallacy of composition here.

[Ziggurat]

Originally Posted by SDG

A hydrogen atom emits a photon going from n=2 to n=1.
Does the rest mass of the hydrogen atom change?

Let me follow up with another example where the numbers are more dramatic and have been directly measured.

Deuterium (one proton, one neutron, one electron - an isotope of hydrogen) has a mass of 2.0141 atomic units. Helium (two protons, two neutrons, two electrons) has a mass of 4.0026 au.

If you squish two deuterium atoms together, they will fuse to form helium. But two deuterium atoms have a total mass of 4.0282 au. That's 0.0256 au more than helium. Squishing them together makes them lose mass. Where does that mass go?

It gets converted into energy. E=mc². 0.0256 atomic units converted to energy might not look like much at first glance, but it's actually a huge amount. And that's basically what powers a hydrogen bomb.

It's very hard to directly measure the mass loss of hydrogen going from an n=2 excited state to an n=1 ground state because the mass loss is so tiny, but it's still happening.

[The Man]

Originally Posted by SDG

A hydrogen atom emits a photon going from n=2 to n=1.
Does the rest mass of the hydrogen atom change?

Yep, as a bound state the hydrogen atom has less rest mass than the sum of the free rest masses of all it components. By emitting a photon and dropping to a lower energy state the electron and subsequently the hydrogen atom as a whole loses mass. As this involves the atom as a whole it reduces the atoms rest mass.


Please see Mass defect in regard to binding energy.

[ponderingturtle]

Originally Posted by smartcooky

Photons have mass? I thought they were zero-mass particles?

Now we get into various definitions of mass. They do not have a rest mass but some people use relativistic mass which they do have.

[The Man]

Originally Posted by ponderingturtle

Originally Posted by smartcooky

Photons have mass? I thought they were zero-mass particles?

Now we get into various definitions of mass. They do not have a rest mass but some people use relativistic mass which they do have.

As already noted in this thread photons have both energy and momentum. In emitting a photon a particle losses both energy and momentum. As momentum is mass time velocity, if velocity of the emitting particle remains constant then mass must change. For complex assemblages like an atom where an electron emits a photon by dropping into a lower energy state both the energy and momentum of the electron change (hence the lower state) but the overall binding energy goes up. It now takes even more energy to free that electron, as such the rest mass of the atom drops by the mass equivalent of that photon energy. So while rest mass reduces the photon only carries energy and momentum no rest mass itself as it has no rest or comoving state.

[Reformed Offlian]

Originally Posted by SDG

Right, so are we going to ignore the P_recoil x component when the Doppler effect takes care of the y component?

I don't understand this question. My point was simply that you have to define in which reference frame P is being measured. Even a transversely moving observer will measure it differently (transverse doppler shift is a thing in SR).

[SDG]

Originally Posted by Ziggurat

And... here's the mistake.

In order to emit a photon, the mass of the flashlight must decrease. A decrease in mass produces a decrease in momentum when the velocity is constant. So the sideways linear momentum HAS decreased in this scenario, and it hasn't decreased uniformly across the flashlight. The previous center of mass is no longer the current center of mass. And relative to the stationary point located where the center of mass was when the photon was emitted, the flashlight now has nonzero angular momentum.

And that's the tricky bit, and you don't need special relativity for it. A non-rotating uniformly moving body can have angular momentum, depending on where you measure from.

As I mentioned, there is a lot said in this post.
2. A decrease in mass produces a decrease in momentum when the velocity is constant.

Momentum has to be conserved.
If we consider the photon energy leaving the flashlight system as system mass reduction then the velocity v will be bigger.
Conservation of momentum is the input factor, the velocity is the output effect.


3. So the sideways linear momentum HAS decreased in this scenario, and it hasn't decreased uniformly across the flashlight.
Well, and the force does not propagate instantaneously in SR.
The flashlight has to start rotation at the point of emission regardless where the theoretical center of mass is located.


The force will propagate through the center of mass in the rest frame eventually, not causing any rotation.
A rotation starts right after emission in any other moving frame.

[Reformed Offlian]

Originally Posted by SDG

What apparent rotation?
There is an inertial observer who is part of a bigger grid of inertial observers, no relative motion between them.
It appears to me that if talk about an apparent rotation we are making one preferred inertial observer from his grid of inertial observers.
The angular velocity of apparent rotation is not the same for all observers in the grid.
The grid is broken.
The preferred observer is not even inertial anymore, the observer rotates, to follow the apparent rotation.

My point (and I believe Myriad's) is that the axis of the "recoil" trajectory will still appear to point back through G, even for a laterally-moving observer.

[Ziggurat]

Originally Posted by SDG

As I mentioned, there is a lot said in this post.
2. A decrease in mass produces a decrease in momentum when the velocity is constant.

Momentum has to be conserved.

Total momentum is conserved. But it can be apportioned differently between different parts of the system. The momentum of just the flashlight changes. It is not conserved.

In the moving frame, some of the momentum to the side is carried by the photon after it is emitted. But the photon carried no momentum before it was emitted. Therefore, the flashlight must have less momentum to the side after emitting the photon in order to keep total momentum conserved. But it can have less momentum to the side without changing velocity because its mass decreased.

Quote:

3. So the sideways linear momentum HAS decreased in this scenario, and it hasn't decreased uniformly across the flashlight.
Well, and the force does not propagate instantaneously in SR.
The flashlight has to start rotation at the point of emission regardless where the theoretical center of mass is located.

It doesn't rotate at all. That's the whole point. The contradiction you imagined doesn't exist. The change in momentum is due to a change in mass, and because that change in mass is not uniform across the body of the flashlight (at least not instantly), the change in momentum isn't either. The "torque" accounts for how ONE SIDE of the flashlight lost mass and therefore momentum, and acquired angular momentum about its FORMER center of mass while not rotating. The former center of mass is not the new center of mass.

You can't apply rules for rigid mass-conserved bodies to a body whose mass is changing and expect everything to be the same, even in Newtonian mechanics. Nor do you seem to understand how angular momentum works when you're looking at something other than the center of mass. That's excusable, since it's tricky business even in Newtonian mechanics, but you're still getting it wrong.

Quote:

A rotation starts right after emission in any other moving frame.

No. A rotation doesn't start at all, in any frame. You don't understand what's happening yet.

[Ziggurat]

Originally Posted by Reformed Offlian

My point (and I believe Myriad's) is that the axis of the "recoil" trajectory will still appear to point back through G, even for a laterally-moving observer.

No, there is in fact a torque on the system in this scenario. He's right about that, but wrong about basically everything else.

The flashlight experiences a change in angular momentum without rotating, because its mass changes and so does its center of gravity. When viewed from the initial center of gravity, the flashlight had no angular momentum before emission, but it has angular momentum after emission, because there's more mass on one side than the other after (well, mass times length, but I'm simplifying). When viewed from the final center of gravity, the flashlight has angular momentum before emission, because (again) there's more mass on one side than the other, but no angular momentum after. Either way, the angular momentum changes, due to the torque, without any rotation occurring. What you aren't allowed to do is change where you're measuring your angular momentum from without accounting for that change, which is what SDG is doing without even realizing it. And again, this isn't peculiar to special relativity, the exact same problem comes up in Newtonian physics.

[Reformed Offlian]

Originally Posted by Ziggurat

No, there is in fact a torque on the system in this scenario. He's right about that, but wrong about basically everything else.

The flashlight experiences a change in angular momentum without rotating, because its mass changes and so does its center of gravity. When viewed from the initial center of gravity, the flashlight had no angular momentum before emission, but it has angular momentum after emission, because there's more mass on one side than the other after (well, mass times length, but I'm simplifying). When viewed from the final center of gravity, the flashlight has angular momentum before emission, because (again) there's more mass on one side than the other, but no angular momentum after. Either way, the angular momentum changes, due to the torque, without any rotation occurring. What you aren't allowed to do is change where you're measuring your angular momentum from without accounting for that change, which is what SDG is doing without even realizing it. And again, this isn't peculiar to special relativity, the exact same problem comes up in Newtonian physics.

Ah, OK, thanks.

[Ziggurat]

This is one of those "a little knowledge is a dangerous thing" situations. SDG knows just enough to set up a problem with some interesting subtleties, but not enough to actually figure out those subtleties.

It's actually not a bad problem to illustrate some of these subtle issues (sort of like the barn door/ladder problem for showing relativity of simultaneity). I could probably give this problem to upper division physics majors in college on a test and they wouldn't all get it right. Probably only half of them would figure it out without an explanation. And basically nobody who hasn't studied physics is going to figure this out. So getting it wrong is not a sign of stupidity on his part, not by any stretch of the imagination. It really isn't a mark against him.

But when you run into a problem like this where the answer doesn't appear to make sense to you, or you think you've found some sort of contradiction, it's rather arrogant to assume that you not only got it right, but that you discovered something that generations of the brightest minds failed to notice or understand. The far more likely explanation, in every case, is that you're missing something, not that everyone else did. And indeed, SDG did miss something. Which, again, is completely understandable. But that's all it is, nothing more. This isn't the flaw in Special Relativity that SDG thought, it's just an example of his own imperfect understanding of Special Relativity.

[slyjoe]

Originally Posted by Ziggurat

This is one of those "a little knowledge is a dangerous thing" situations. SDG knows just enough to set up a problem with some interesting subtleties, but not enough to actually figure out those subtleties.

It's actually not a bad problem to illustrate some of these subtle issues (sort of like the barn door/ladder problem for showing relativity of simultaneity). I could probably give this problem to upper division physics majors in college on a test and they wouldn't all get it right. Probably only half of them would figure it out without an explanation. And basically nobody who hasn't studied physics is going to figure this out. So getting it wrong is not a sign of stupidity on his part, not by any stretch of the imagination. It really isn't a mark against him.

But when you run into a problem like this where the answer doesn't appear to make sense to you, or you think you've found some sort of contradiction, it's rather arrogant to assume that you not only got it right, but that you discovered something that generations of the brightest minds failed to notice or understand. The far more likely explanation, in every case, is that you're missing something, not that everyone else did. And indeed, SDG did miss something. Which, again, is completely understandable. But that's all it is, nothing more. This isn't the flaw in Special Relativity that SDG thought, it's just an example of his own imperfect understanding of Special Relativity.

Nice polite explanation

[SDG]

Originally Posted by Ziggurat

Yes.



Just an electron cannot emit Bremsstrahlung radiation. It needs to interact with other matter. And in this case, you are trading off kinetic energy to create the photon, not rest mass energy.

'My bad', I was not specific enough.
I wanted to point the rest mass of electron and proton do not change.
Then there is biding energy of hydrogen atom.
When electron emits a photon from hydrogen atom it is undergoing a 'Bremsstrahlung' trading off kinetic energy for photon emission in the hydrogen atom proton, electron system.

Quote:

Yes. Even more importantly, though, it is part of the theory of special relativity. So if you want to evaluate SR, you need to account for it.



m is not constant. That's the whole point. You have started with a false assumption.

You are making an assumption that 'mass change', binding energy change is happening at the tip of the LED flashlight.
That is not the case, it is happening somewhere in batteries and these can be placed right in the center for the simplicity of argument.
If we do this then your argument about the mass change in relation to center of mass change is mute.

Edit: This is a thought experiment and we can have three LEDs arranged in a triangle at the tip to avoid any torquing when electrons do their job at the emission (if we have to go to such a detail )

Edit 2: If we have to go there ...
If we are worried about EM field being transferred along wire from the batteries to the LEDs then we can have the same setup, creating symmetry, towards the opposite side.
Having said that, the LEDs at the back would not emit photons out but towards each other into a 'triangle' receiver to eliminate torquing at the back.
The back photons stay in the flashlight.

[tusenfem]

Originally Posted by SDG

When electron emits a photon from hydrogen atom it is undergoing a 'Bremsstrahlung' trading off kinetic energy for photon emission in the hydrogen atom proton, electron system.

No, that is not "bremsstrahlung", maybe you should freshen up your physics a bit. (https://en.wikipedia.org/wiki/Bremsstrahlung)
Bremsstrahlung is a "free-free" process where electrons are braked (from the German bremsen), whereas an electron going from one shell to another is a bound-bound process and the electron goes to a lesser energy shell around the nucleus (for simplicity using the Bohr atomic model).
Please don't start making things even more confusing.

[Pixel42]

Originally Posted by tusenfem

Please don't start making things even more confusing.

The only way to make it look like there might be a problem with relativity is to make your thought experiment as confusing as possible.

[SDG]

Originally Posted by tusenfem

No, that is not "bremsstrahlung", maybe you should freshen up your physics a bit. (https://en.wikipedia.org/wiki/Bremsstrahlung)
Bremsstrahlung is a "free-free" process where electrons are braked (from the German bremsen), whereas an electron going from one shell to another is a bound-bound process and the electron goes to a lesser energy shell around the nucleus (for simplicity using the Bohr atomic model).
Please don't start making things even more confusing.

When electron goes to lesser energy shell it means it is closer to the proton.
Electron has lower potential energy.
When this would happen to an orbiting body in gravity then lower potential energy means higher kinetic energy in orbit, conservation of energy.
The electron cannot increase its kinetic energy in lower potential.
This is 'Bremsstrahlung', kinetic energy that should have been there, taken away from the electron through the photon emission.

Edit: I am using 'Bremsstrahlung', using quotes, in a meaning Ziggurat mentioned it: 'trading off kinetic energy to create the photon'.

[tusenfem]

Originally Posted by SDG

When electron goes to lesser energy shell it means it is closer to the proton.
Electron has lower potential energy.
When this would happen to an orbiting body in gravity then lower potential energy means higher kinetic energy in orbit, conservation of energy.
The electron cannot increase its kinetic energy in lower potential.
This is 'Bremsstrahlung', kinetic energy that should have been there, taken away from the electron through the photon emission.

I am sorry but this makes no sense at all, because we know that the Bohr atomic model is incorrect, the electrons are not running rings around a nucleus, This is only to envisualise.
The energy difference between the levels is radiated away by a photon.
The last sentence in your explanation is nonsense.

[Ziggurat]

Originally Posted by SDG

'My bad', I was not specific enough.
I wanted to point the rest mass of electron and proton do not change.

The rest mass of an individual electron and an individual proton do not add up to the rest mass of a hydrogen atom. And the rest mass of the hydrogen atom changes depending upon its state. If you are talking about what happens to a hydrogen atom that undergoes a transition, then it's the hydrogen atom's rest mass, not the rest mass of an isolated electron or proton, which matters. So it's irrelevant that an electron and a proton have fixed rest masses when considered in isolation, because they are not in isolation.

Quote:

Then there is biding energy of hydrogen atom.

Which is negative, and thus decreases the mass of hydrogen compared to the sum of an electron and a proton in isolation.

Quote:

You are making an assumption that 'mass change', binding energy change is happening at the tip of the LED flashlight.

It must happen wherever the photon is created. If you create it at the tip of the flashlight, that's where it happens. If you are using an LED, that's where the electron which created it loses its energy and thus mass.

Quote:

That is not the case, it is happening somewhere in batteries and these can be placed right in the center for the simplicity of argument.

That is incorrect. Mass/energy does flow from the battery, but it flows to the front BEFORE the photon is emitted. Energy cannot teleport.

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If we do this then your argument about the mass change in relation to center of mass change is mute.

If you do that, you will be making an error.

Quote:

Edit: This is a thought experiment and we can have three LEDs arranged in a triangle at the tip to avoid any torquing when electrons do their job at the emission (if we have to go to such a detail )

I don't really understand what you mean, but I'm not going to bother figuring it out since it's bound to be wrong. Your understanding of the initial setup is wrong, altering it won't fix your understanding.

Quote:

Edit 2: If we have to go there ...
If we are worried about EM field being transferred along wire from the batteries to the LEDs then we can have the same setup, creating symmetry, towards the opposite side.

Won't make any difference. Mass is lost by the electron which changes state inside the LED to emit the photon. That happens at the location of the emission. Any other mass flow within the flashlight is a separate issue, and we don't need to consider it in order to solve your problem. But it too will obey conservation of momentum.

Quote:

The back photons stay in the flashlight.

If they stay with the flashlight, then they contribute to its mass.

[Ziggurat]

Originally Posted by SDG

The electron cannot increase its kinetic energy in lower potential.

Once again, you are wrong (notice a pattern yet?). An electron in the n=1 state has a HIGHER kinetic energy than an electron in the n=2 state. It has a lower overall energy because the loss of potential energy is bigger than the gain in kinetic energy, but the the lower energy state actually has more kinetic energy. This is standard intro level quantum mechanics. When an electron drops from n=2 to n=1, it doesn't lose kinetic energy, it gains kinetic energy. The reason it still emits energy is that it loses even more potential energy than it gains.

In fact, it's not even just quantum mechanics. The exact same thing holds true classically too. Mercury orbits the sun at a velocity of about 47 km/s, but Pluto orbits at a velocity of only about 4.7 km/s. So Mercury has about 100 times the kinetic energy per kg as Pluto. But it's in a lower energy state, because the potential energy of Mercury is also much, much lower than Pluto's.

tl;dr: even under your expanded definition of 'Bremsstrahlung', hydrogen atom transitions don't count.

[Ziggurat]

Originally Posted by tusenfem

I am sorry but this makes no sense at all, because we know that the Bohr atomic model is incorrect, the electrons are not running rings around a nucleus, This is only to envisualise.

Yes and no. The Bohr model is wrong, but you can still calculate a kinetic energy even for a bound particle, and the kinetic energy does change when you transition between states. The problem is that he's got the signs reversed: the more tightly bound the electron, the higher its kinetic energy.

[SDG]

Originally Posted by Ziggurat

...
It must happen wherever the photon is created. If you create it at the tip of the flashlight, that's where it happens. If you are using an LED, that's where the electron which created it loses its energy and thus mass.
...

How LED works: https://lamphq.com/functional-principle-of-leds/



When a free electron interacts with cavity, going back to anode it emits photon.
The free electron was part of the cathode before.
Where is the energy coming from to release the electron?

A free electron has 'more mass' (more energy) then an electron bound to a system, correct?
It is right at the LED where the electron becomes 'heavier' and 'lighter' in a very short time when a photon is released.
The free electron trades its kinetic energy for photon.

It is illogical to claim the electron in LED became free out of nothing and therefore there is a decrease of mass at the LED.
The original energy cause is in batteries.

[wollery]

Originally Posted by SDG

How LED works: https://lamphq.com/functional-principle-of-leds/

https://i.imgur.com/3WPt3YJ.png

When a free electron interacts with cavity, going back to anode it emits photon.
The free electron was part of the cathode before.
Where is the energy coming from to release the electron?

A free electron has 'more mass' (more energy) then an electron bound to a system, correct?
It is right at the LED where the electron becomes 'heavier' and 'lighter' in a very short time when a photon is released.
The free electron trades its kinetic energy for photon.

It is illogical to claim the electron in LED became free out of nothing and therefore there is a decrease of mass at the LED.
The original energy cause is in batteries.

Until the photon escapes the flashlight the sum of the mass-energy is constant for the flashlight, so any mass / energy exchanges before this are irrelevant. As soon as the photon is emitted there is a drop in mass, which occurs at the position where the photon is emitted.

[Ziggurat]

Originally Posted by SDG

How LED works: https://lamphq.com/functional-principle-of-leds/

https://i.imgur.com/3WPt3YJ.png

When a free electron interacts with cavity, going back to anode it emits photon.
The free electron was part of the cathode before.
Where is the energy coming from to release the electron?

A free electron has 'more mass' (more energy) then an electron bound to a system, correct?
It is right at the LED where the electron becomes 'heavier' and 'lighter' in a very short time when a photon is released.
The free electron trades its kinetic energy for photon.

It is illogical to claim the electron in LED became free out of nothing and therefore there is a decrease of mass at the LED.
The original energy cause is in batteries.

You really don’t understand what you are reading. The anode and cathode in this circuit do not act like an anode and cathode in a vacuum tube. The actual transition from high energy state to low energy state happens WITHIN the LED chip. The battery is what pushes the electron into the high energy state, but the transition from high energy state to low energy state doesn’t happen at the battery. And that is the transition that creates the photon, NOT the transition from low energy to high energy which happens at the battery. The battery doesn’t make the electron LOSE energy. That is still happening locally, in the LED chip component. I could tell you about semiconductor band theory, n vs p doping, and n-p junctions, but that would all go over your head. Suffice to say, as usual, you have all of this backwards.

It is becoming increasingly clear that you are not interested in learning anything here. Which is a shame, because there’s really a lot you could learn. But I can’t make you learn when you don’t want to, no matter how hard I might try.

[slyjoe]

I would like to know what EM field is being generated in a flashlight.

ETA: From SDG's post:

Quote:

...
Edit 2: If we have to go there ...
If we are worried about EM field being transferred along wire from the batteries to the LEDs then we can have the same setup, creating symmetry, towards the opposite side.
Having said that, the LEDs at the back would not emit photons out but towards each other into a 'triangle' receiver to eliminate torquing at the back.
The back photons stay in the flashlight.

[Ziggurat]

Originally Posted by slyjoe

I would like to know what EM field is being generated in a flashlight.

ETA: From SDG's post:

One of the simplest scenarios to do the calculations on is the case of a coaxial cable for the battery to whatever it’s running. You get a radial electric field between the high voltage side and the low voltage side, and you get a circumferential magnetic field from the currents through the wires. And you even get momentum being carried down the wires by the field. It’s interesting stuff, but well beyond SDG’s level of understanding.

And none of that changes the fact that the electron is losing mass at the point of emission.

[SDG]

Originally Posted by Ziggurat

You really don’t understand what you are reading. The anode and cathode in this circuit do not act like an anode and cathode in a vacuum tube. The actual transition from high energy state to low energy state happens WITHIN the LED chip. The battery is what pushes the electron into the high energy state, but the transition from high energy state to low energy state doesn’t happen at the battery. And that is the transition that creates the photon, NOT the transition from low energy to high energy which happens at the battery. The battery doesn’t make the electron LOSE energy. That is still happening locally, in the LED chip component. I could tell you about semiconductor band theory, n vs p doping, and n-p junctions, but that would all go over your head. Suffice to say, as usual, you have all of this backwards.

It is becoming increasingly clear that you are not interested in learning anything here. Which is a shame, because there’s really a lot you could learn. But I can’t make you learn when you don’t want to, no matter how hard I might try.

Does the flow electron have more energy compared to bound electron part of the cathode?
Does energy have to be added to free electron from the cathode?

Edit: Why would I insist on LED flashlight in my original post? Hmmm....

[Ziggurat]

Originally Posted by SDG

Does the flow electron have more energy compared to bound electron part of the cathode?
Does energy have to be added to free electron from the cathode?

I have no idea what you mean by "flow electron". But again, you obviously don't understand how the LED works. Hell, I'm skeptical at this point that you know how ordinary conduction in a wire works.

Conduction electrons in the cathode are already in a high energy state. Is that what you mean by "flow electron", just a conduction electron? If so, that's weird terminology. At any rate, they do not need additional energy to enter into the LED chip. Inside the LED chip, they lose energy after crossing the junction, but this is still inside the LED chip. They then leave the LED chip via the bond wire, and travel along that wire to the anode.

Quote:

Edit: Why would I insist on LED flashlight in my original post? Hmmm....

Likely because you have some misconception of how an LED works. Do you think that the electron is jumping the gap between the cathode and the anode? It isn't.

[SDG]

Originally Posted by Ziggurat

I have no idea what you mean by "flow electron". But again, you obviously don't understand how the LED works. Hell, I'm skeptical at this point that you know how ordinary conduction in a wire works.

Conduction electrons in the cathode are already in a high energy state. Is that what you mean by "flow electron", just a conduction electron? If so, that's weird terminology. At any rate, they do not need additional energy to enter into the LED chip. Inside the LED chip, they lose energy after crossing the junction, but this is still inside the LED chip. They then leave the LED chip via the bond wire, and travel along that wire to the anode.



Likely because you have some misconception of how an LED works. Do you think that the electron is jumping the gap between the cathode and the anode? It isn't.

From the linked text:

Quote:

In one semiconductor layer there is an excess of positive charge carriers. In the other layer the negative charge carriers are in the majority. If the anode and cathode are supplied with voltage, an electron flow occurs between the semiconductor layers. As a result, energy is released, resulting in small flashes of light. The LED emits photons, which we then perceive as visible light.

How is it possible the flow occurs?
What is the cause?
It is right in the text.
'If the anode and cathode are supplied with voltage'
Where does the energy come from to generate the voltage delta?
Does battery lose 'mass' when it depletes?

[W.D.Clinger]

Originally Posted by SDG

It is right in the text.
'If the anode and cathode are supplied with voltage'
Where does the energy come from to generate the voltage delta?

Voltage is a measure of electromotive force, not energy.

From your question, I'm guessing you do not understand the difference between force and energy.

Is that confusion the essence of your question?

[SDG]

Originally Posted by W.D.Clinger

Voltage is a measure of electromotive force, not energy.

From your question, I'm guessing you do not understand the difference between force and energy.

Is that confusion the essence of your question?

What is the cause of the electromotive force?

Quote:

If the anode and cathode are supplied with voltage, an electron flow occurs between the semiconductor layers. As a result, energy is released, resulting in small flashes of light.

What is the relation between voltage and energy released?

[Ziggurat]

Originally Posted by SDG

Does battery lose 'mass' when it depletes?

Yes. E=mc². That applies to the batteries too.

Where does that mass go? It doesn't teleport from the battery to the photon. It travels with the electrons which the battery put into a higher energy (and thus higher mass) state. These electrons then lose mass at the LED diode when emitting the photon. So the creation of the photon still involves an electron losing mass at the LED.

[W.D.Clinger]

Originally Posted by SDG

Originally Posted by W.D.Clinger

Voltage is a measure of electromotive force, not energy.

From your question, I'm guessing you do not understand the difference between force and energy.

Is that confusion the essence of your question?

What is the cause of the electromotive force?

The LED doesn't care, and neither should you. Although we could mention dozens of possible causes, such digressions would only confuse you further. The LED isn't confused by such irrelevancies, because the physics is local. What matters at the LED is the voltage, not the causes of that voltage.

By asking such an irrelevant question instead of clarifying your apparent confusion between force and energy, I assume you are implicitly admitting your confusion on that rather basic matter of physics, just as the question you ask below reveals your ignorance of rather basic electronics and physics.

Originally Posted by SDG

Quote:

If the anode and cathode are supplied with voltage, an electron flow occurs between the semiconductor layers. As a result, energy is released, resulting in small flashes of light.

What is the relation between voltage and energy released?

Please make some effort to learn how the joule is defined in terms of SI base units such as second and ampere, or in terms of derived SI units such as volt, coulomb, watt.