Special Relativity and momentum

[SDG]

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.

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]

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]

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]

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]

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]

'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.

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.

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.

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.

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.

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.

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.

The back photons stay in the flashlight.

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

 

[Ziggurat]

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]

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]

...
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]

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

[qimg]https://i.imgur.com/3WPt3YJ.png[/qimg]

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]

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

[qimg]https://i.imgur.com/3WPt3YJ.png[/qimg]

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:

...
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]

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]

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]

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.

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]

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:

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]

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]

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?

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]

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]

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.

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.