Let's say that you have two electrodes in vacuum and apply a potential difference to them. This will cause electrons to sometimes be "pulled out" of one of them. How will that impact the temperature of both electrodes?
I would expect the negative electrode to be heated by incoming electrons: after all they've had at least ~0 kinetic energy after leaving the positive electrode, so they bring in extra kinetic energy due to the potential difference.
I'm not sure what should happen to the positive electrode. On one hand, I would intuitively expect ones that have higher kinetic energy of thermal motion to be more likely pulled out, which would decrease the temperature (because we're skimming the top of the kinetic energy distribution, which should decrease the expected value). On the other hand, that would mean that this mechanism is a heat pump that should work regardless of the current temperatures of the electrodes, which is obvious violation of 2nd law of thermodynamics.
Where am I being foolish?
I started thinking about this after reading that one should not apply DC to fluorescent tubes, "because otherwise one filament cools off while the other overheats, evaporates and darkens one end of the tube" (http://ludens.cl/Electron/Fluolamp/fluolamp.html).
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@robryk Both electrodes would heat up, and to a lesser extent than if they were connected directly electrically.
The reason for this is that the kinetic energy you describe as it flies through a vacuum is no different than the connect energy it acquired as it flies through the wire itself.
The reason the heating effect is less due to a gap than it would be whiel electrically connected is because the gap effectively adds resistance, higher resistance for the same electrical potential means less current flow, less current flow means fewer electrons, fewer electrons means less heat.
Also dont forget there is an additional effect that both a wire and a spark gap would exhibit whereby the accelerating electrons would shed photons which would reduce the acceleration it would undergo due to the force applied to it alone. This would effectively decrease its kinetic energy and cause less heating, but in this case the loss of that energy would be directed as EM waves. Depending on the conditions most of this EM energy is likely to be at a lower frequency than visible light.
@freemo Why should the positive electrode heat up at all?
@robryk because as electrons escape from its surface new electrons move to take their place. Ergo there is a current flow through the positive electrode. Anytime current is flowing through a wire, presuming it isnt a super conductor, it will heat up.
@freemo Ok, I forgot to explicitly state that I assume everything is perfectly conductive~~. But what's wrong with the reasoning that the electrons that escape are the ones that have the highest kinetic energy, and thus their escape will lower the average kinetic energy of electrons in the electrode and thus the temperature of the electrons in the electrode, so also the temperature of the electrode?
@robryk if they are perfectly conducting then that does change things partially... in that case you wont have heating as I had suggested, but you also wont have cooling.. you have to remember that when electricity is flowing in this configuration you dont just have electrons flowing through a pipe like water.. you have two counter forces moving in opposite directions.. electrons flow in one direction, electron holes (positive charge) flows in the other.. each electron is paired with an electron hole and is in balance..
You can think of it in convention terms rather than QM terms to over simplify it (though bear in mind it isnt entierly accurate).. if we talk semi-classically you can imagine that any movement of electrons is paired with a recoil force applied to a proton. Protons are far more massive so they move much less for the same force exerted, but the overall momentum is the same.
So you wind up with a balanced system where the cooling and heating you imagine when thinking of only the electrons is matched and countered by the same reaction of the protons, in the end you will see the same effects on both electrodes more or less.
Are you predicting that both electrodes will heat up equally?
@robryk if they had non-zero resistance they should heat up equally.
if they are superconductors 100% of the voltage will be seen as dropped across the gap itself. Yes you are correct the energy would go somewhere, it would be emitted as EM radiation as per my first comment due to the acceleration of the electrons through the gap.
@robryk bear in mind in a ultra high vacuum the effective resistance of a spark gap would be very high, meaning even at very high voltages there would be very little current flow.. so even if you modeled the spark gap as a resistor and thus you would know energy would be radiated, the amount of energy radiated would be quite small at any reasonable voltage.
But yes, effectively what energy is loss and radiated across the gap would be as EM radiation and not as heat in the traditional sense (other than EM based IR of course).
@freemo The gap, assuming it's of macroscopic size, should have a V/I characteristic similar to diode's: as long as the voltage is lower than the depth of the conduction band in the negative electrode, no current will flow. If the voltage is higher, I'm not sure what would actually limit the current.
Do you expect the "threshold voltage" to be lower if the gap is filled with unionized gas? The threshold voltage for unionized gas is easily reached by starters of fluorescent tubes.
@robryk The V/I curve is actually rather complicated when we talk about this subject because it depends a great deal on just how hard a vacuum we are talking. So its not quite as simple as a one description fits all...
what happens is you get a runaway effect when there is any gas present, like a fluorescent tube. In other words it takes a high voltage to initiate the flow of electrons but once they begin flowing the ionized gas that creates a path between electrodes is of a much lower resistance and thus requires a much lower voltage to actually maintain. The voltage needed to trigger the spark gap (which is much higher than the voltage needed to maintain it) is called the breakdown voltage, it is the ionizing voltage of the material.
What you will see is the breakdown voltage, the max voltage needed to initiate the runaway effect decreases as pressure decreases for any gas, but only to a point. Once you react a certain pressure where the minimum breakdown voltage is seen if you lower the pressure any more the breakdown voltage quickly increases and escapes to infinity. By the time you reach a hard vacuum then there is no break down voltage of any kind and you are now following a different set of physical laws to model the flow of electrons.
When you are at a hard vacuum as described the resistance to the flow of electrons in a hard vacuum is purely the "radiation resistance" (technical term) which is the resistance caused due to the emission of EM by an accelerating charged particle, which causes recoil and resists acceleration as a consequence of energy being lost as EM as it attempts to accelerate.
For some technical terms, the term of the curve that relates pressure to breakdown voltage is called the Paschen curve, it is material specific.
The recoil force that causes the radiation resistance that I mentioned is called the Abraham–Lorentz force, which applies to non-relativistic speeds, which I assume is what we are discussing
@robryk to be clear if i wasnt in my message... at hard vacuum there is no idea of a threshold voltage, that only occurs when some gas is present. ie paschen curve in that case.
@freemo This is surprising for me. I would expect the potential energy of an electron inside a conductor to be lower than that of a electron in vacuum, and thus would expect that this difference has to be overcome by the applied voltage (and then would expect the anode to be heated by the incoming electrons, when their potential energy drops). (I'm ignoring tunneling, which is why I added the macroscopic distance caveat.) Is there an obvious reason why this is not so, or am I thinking in completely incorrect terms?
@robryk well before we can answer that you have to understand some fundamentals that it seems you may have mixed up...
So you are talking about the potential energy of an electron and this needing to be overcome by voltage.. this kind of confuses fundamentally what a voltage is, or for that matter the role it plays in the potential energy for an electron.
The potential energy of an electron wouldnt suddenly shift when it escapes or enters the conductor.. ill give a git of an analogy to show why...
You have a bowling ball sitting on the ground near the edge of a cliff. It has some potential energy determined by the force of gravity, its own mass, and the distance from the top of the cliff to the bottom.
Here is the thing though, the potential energy of the bowling ball is exactly the same when its sitting ont he ground motionless at the top of the cliff as it is in the first instant it rolls off (and is at the same height as it has just now begun falling).. it doesnt have a immediate change in potential energy when it goes from being at rest to begining to fall.
Similarly the bowling balls potential energy when it is in free fall a few moments before it hits the ground below is the same exact potential energy it has when it hits the ground, there is no transition.
Same with the electron, the potential energy is no different when it shifts into or out of the metalconductor.
Also, just for clarity voltage does not represent energy and tells you nothing about energy on its own, the voltage across electrodes should not be thought of as an energy, in electronics energy would be measures in watts not volts.
@robryk Great question by the way, hope I helped.
@freemo I'm thinking how to set up an experiment where I can tell if both electrodes would heat equally, because I still doubt that :) (my best approach so far is to buy and modify a fluorescent lamp by inserting a bridge after the inductor).
@freemo Ah, does your prediction of equal heating extend to the situation when there is ionized gas in the gap?
@robryk well ionized gas would definately behave differently and produce signficantly more heat than a hard vacuum.. however off hand I cant think of any reason why one electrode would heat up any more than the other in that environment. But you would have a very hard time measuring it considering youd have convention currents moving all sorts of heat around in that situation.
@freemo Voltage is the energy per unit of charge needed to move a charge between two points, no? If there's a potential difference of 50eV to overcome for electrons, I need to apply 50V across it to cause a nontrivial current to flow
I'm not sure why that analogy should fit the situation better than an analogy where the ball is constrained to the surface (e.g. it's on a wire that follows the shape of the potential).
@robryk Perhaps i misunderstood your question but it sounded like you were saying the potential energy of the electron would exhibit a sudden change when going from the gap to the conductor. The analogy is to show that the potential energy would not change when it goes from one medium to another anymore than a bowling ball moving from air to being stopped completely by the ground.
Yes you are correct voltage is energy per unit charge. But as I understand it you were describing the electron as having a potential energy that is different when it is in the gap than when it transitions into the conductor. Unless im misunderstanding there seems to be a disconnect between what this means.
The voltage is a gradient across the gap and ac onstant in the conductor. The potential energy of the electrode is at its maximum when its inside the positive electrode, is at the same value int he first instance when it transfers into the gap, and then decreases as is moves across the gap to 0 when it reaches the other electrode in the instant before entering the conductor, where it remains at 0 inside the conductor. There is no sudden change in potential energy when transitioning into or out of the conductor in this model or in any model I know of.
As for the example of your 50V being needed to move 50eV "nontrivially".. i guess it depends on what you mean by nontrivially... 50Vwould not be special in any way at moving 50eV, twice the voltage would get you twice the current, half the voltage would get you half the current,l 50V would not in anyway look special in terms of a voltage in that regard.
@freemo If they actually have nonzero resistance and are mirror images of each other, should they heat up equally?
If they're superconductors, what will be the voltage drop between them? If it's not zero, we're sinking that energy (current*voltage drop per unit of time) somewhere; where?