3 ms·

One thing that comes to mind is that the voltage difference determines the force on the charged particle, but in principle if the voltage can be maintained, the

One thing that comes to mind is that the voltage difference determines the force on the charged particle, but in principle if the voltage can be maintained, then the same force could be used put an arbitrary amount of energy into the particle. So, at most 350 keV the first time around, but then at most 350 keV the second time around, and the third... I know almost nothing about particle physics.

__MatrixMan__ 2y ago
I'm no specialist either, and initially I had the same thought, but wikipedia says this about an electron volt: > 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.
- MathMonkeyMan 2y ago
  Strictly speaking, the energy gained by a charged particle along any path is the integral of the dot product between the electric field and the line element along the path (edit: times the charge). In the absence of changing magnetic fields, this is always the difference between the voltage at the starting and ending points. 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.
mattashii 2y ago
> So, at most 350 keV the first time around, but then at most 350 keV the second time around, and the third... 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").
- MathMonkeyMan 2y ago
  I don't think that this is true. The velocity of the particle entering the "high" side of the field does not affect the force applied on the particle by the field. Check out the wiki article on [cyclotrons][1]. I think the trick is turning the field on and off. Again, I'm no expert. [1]: https://en.wikipedia.org/wiki/Cyclotron https://en.wikipedia.org/wiki/Cyclotron
  - gus_massa 2y ago
    I tbink both of you agree! 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.
    - MathMonkeyMan 2y ago
      I interpreted mattashii's point as being this: 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."
      - gus_massa 2y ago

__MatrixMan__ 2y ago

I'm no specialist either, and initially I had the same thought, but wikipedia says this about an electron volt: > 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.

MathMonkeyMan 2y ago
Strictly speaking, the energy gained by a charged particle along any path is the integral of the dot product between the electric field and the line element along the path (edit: times the charge). In the absence of changing magnetic fields, this is always the difference between the voltage at the starting and ending points. 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.

mattashii 2y ago

> So, at most 350 keV the first time around, but then at most 350 keV the second time around, and the third... 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").

MathMonkeyMan 2y ago
I don't think that this is true. The velocity of the particle entering the "high" side of the field does not affect the force applied on the particle by the field. Check out the wiki article on [cyclotrons][1]. I think the trick is turning the field on and off. Again, I'm no expert. [1]: https://en.wikipedia.org/wiki/Cyclotron https://en.wikipedia.org/wiki/Cyclotron
- gus_massa 2y ago
  I tbink both of you agree! 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.
  - MathMonkeyMan 2y ago
    I interpreted mattashii's point as being this: 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."
    - gus_massa 2y ago