The LHC often boasts energies in the TeV, but hadrons are much heavier than electrons, which I assume accounts for why this machine is so much smaller and why the energies are something like 8 orders of magnitude different.
Can someone tell me if these are reasonable assumptions?
https://en.wikipedia.org/wiki/Large_Electron%E2%80%93Positro...
But that was a ring five miles across, not a sphere roughly the size of a sumo wrestler.
I know almost nothing about particle physics.
> An electron-volt is the amount of energy gained or lost by a single electron when it moves through an electric potential difference of one volt.
I feel like if it was actually an electron-volt-second, that would appear in the definition. So I'm thinking that once the electron has traveled from a place of higher voltage to a place of lower voltage, it has gained energy according to the potential difference, and it doesn't actually matter whether it took a second or a year to do so.
It's easy to think of voltage as something like field strength, to be held more or less constant as the particle accelerates, but really it's a difference between two points, start and end, so the trip length has already been accounted for in the voltage measurement, and doesn't need to be further accounted for by measuring the time it took? Unsure, but that's my feeling anyhow.
What I was getting at is that there's only so much energy a field can put into a particle. Either the voltage will drop because the machine can't "keep up," or the particle will reach a terminal velocity where the force applied by the field is insufficient to accelerate the particle any more. But, barring those two things, the particle will continue to accelerate.
The voltage is the gradient across which the electron moves and gains (or loses) momentum, similar to a ball rolling up- or downhill. Once the electron has moved to the positive side of the electric field ("to the bottom of the hill" so to say), it can only gain more energy from that same field by first losing the equivalent in energy by moving back to the negatively charged segment ("the top of the hill").
Again, I'm no expert.
There are two solutions to gain the energy again from "the same" field.
1) Lost the energy it won in the last pass.
2) With the field to avoid losing the energy in the return trip.
You have a ring around which charged particles can travel. The voltage at the start is V, and the voltage just behind the start (the end) is defined to be zero. A charged particle "falls" down the potential until it gets to right before where it started. But then it will momentarily feel a large force in the opposite direction as it "climbs" back up to V from zero.
I don't know how particle accelerators avoid this, but the wiki on cyclotrons refers to a "rapidly varying electric field."
The trick to undestand the Cyclotron is in the graphic: chttps://en.wikipedia.org/wiki/Cyclotron#/media/File:Cyclotro...
It has two semicircular parts A an B, that act like a big capacitor. The electrons travel in circles, sometime it's inside A and sometimes is in B.
If the voltage between A and B is constant, it accelerates and the decelerates and then accelerates and the decelerates and then accelerates and the decelerates and then ... that is quite boring.
The trick is that when it's inside A you make A negative and B positive, so it accelerates going from A to B. While the electron is inside B you switch the voltages and now B negative and A positive, so it accelerates going from B to A. Now, While the electron is inside A you switch the voltages and now A negative and B positive, so it accelerates going from A to B. ...
So the trick to gain energy in both directions is to change the voltages of the semicircular parts. This change cost energy, so there is no magic creation of energy.
Probably the article in Wikipedia explains it better.