I am thinking an explicit list of all energy forms in a system of charged particle (take electron or proton, for instance). It has at least potential energy and kinetic energy by Newton.

Characteristics of the setting for Timaeus's answer

There is a drift velocity $v$ of the setting:

\begin{equation} v = \mu \, \frac{\mathscr{E}}{p} = \frac{ \mu V }{ p \ln(b/a) } \cdot \frac{1}{r} \end{equation} where $v$ is drift velocity, $\mathscr{E}$ electric field strength, $\mu$ mobility constant, $p$ gas pressure, $b$ cathode wire radius, and $r$ is practically about the thickness of ionization. The mobility can be assumed to be constant with the modest electric field strength. Is velocity and magnetic potential energy still relevant?

Other things

  • How you can express the force here, since it cannot apparently be expressed purely as a function of position? The detector gives the position of ionizing radiation (charged particles, photons, X-rays and neutrons) which probably is not enough to deduce the force.
  • Which energy forms are relevant in the regions: proportional region and Geiger-Mueller, see the following figure? There is diffusion processes in all regions but dual paths in regions below Geiger-Mueller.

enter image description here

where Geiger-Mueller is at the right. Y-axis is amplitude A, and x-axis is the applied voltage between the anode and cathode.

Attempts to define the total energy

The total energy of the system includes mostly kinetic energy, electric potential energy, magnetic potential energy, and rest energy. When the particle moves closer the anode, the total energy increases mostly because of the increased kinetic energy.

where the order of relevant energies is not clear to me in the context. However, I would like to be more explicit what is included in the potential energy here. Electron has a spin so probably I should include here some spin energy. One weak claim about the context:

$r-\mathscr{E}$ presentation is some form of the total energy presentation about electron avalanche in the GEM system, where $r$ is the distance from the anode and $\mathscr{E}$ is the electric field strength.

where the total energy presentation is wanted to be visualized somehow. However, I must understand the total energy better in the context, to understand the statement better.

Processes in GEM system

Ionization of the gas. Can you estimate how the energy is used here? How can you estimate which type of energy is used here? Or just energy?

How can you define the energy forms of a charged particle in an electric field?


2 Answers 2


Potential energy is wrong. Even in Newtonian Mechanics it only works if the force is 1) purely a function of position and also 2) is conservative.

Magnetic forces depend on velocity so they fail. And electric forces are not conservative if you aren't in statics.

What you really have is kinetic energy and rest energy for the charged particle, some energy associated with the interaction of the spin (magnetic moment) and the magnetic field and finally the electric and magnetic fields themselves have some energy density spread throughout space.

So the kinetic energy is the dominating force here when you go closer to the anode. Right?

Step one, realize that fields are real things, not just a conveniently pretty method to draw pictures about forces. Step two, realize that you have an interacting system of fields and charges and that each of them have energy. Step three, study the mutual dynamics of the fields and the charges. Much like particles bouncing around can pass energy from this particle to that particle until an equilibrium distribution is achieved (about how many particles are on each place and what energy and momentum they have), so too can fields and particles exchange energy until you get an equilibrium. Ignoring the energy of the fields and pretending that the particles have it (and that there is potential energy) is entirely 100% missing the point.

  • 1
    $\begingroup$ @Masi Your new question repeats all the same mistakes, basically you assumed potential energy was a thing in situations where it wasn't. Which is fine. My answer addresses the energy involved. $\endgroup$
    – Timaeus
    Nov 11, 2015 at 16:23
  • $\begingroup$ Thank you for your edit! So I propose that the kinetic energy, rest energy and the energy of fields are dominating in the total energy here in reaching the equilibrium. $\endgroup$ Nov 11, 2015 at 19:56
  • $\begingroup$ @Masi If there are magnetic fields then there is also energy associated with the magnetic dipole (magnetic moment) of the charges. And the rest mass might be pointless if you aren't creating or destroying particles. $\endgroup$
    – Timaeus
    Nov 11, 2015 at 20:00
  • $\begingroup$ In the electron avalanche, there are created many new electrons from the electromagnetic field energy. This fact proposes me that the rest mass in relevant in the multiplication region but not outside of it. What do you think? $\endgroup$ Nov 11, 2015 at 20:42
  • 1
    $\begingroup$ @Masi Absolutely not. If you truly created electrons then you would have to create positrons too. What happens in most devices is you rip an electron out of a sea of free electrons in a metal or out of an atom and make it move into a new region (often a region where another eectron just vacated). And unless the number of charged particles in a region is changing over time, then just as much rest mass is flowing into a region as is flowing out. $\endgroup$
    – Timaeus
    Nov 11, 2015 at 20:45
  • Don't forget about the rest energy, $E=mc^2$.

  • Since a particle's intrinsic spin cannot be changed, it doesn't make sense to distinguish "intrinsic spin energy" $\frac12 I\omega^2$ (as you would compute for, say, a spinning flywheel) and the rest energy.

  • A particle which is moving in an electric field will see a motional magnetic field. Charged particles with spin have nonzero magnetic moment $\vec \mu$; the magnetic potential energy is $\vec \mu \cdot \vec B$.

There are others; please edit them in.

  • $\begingroup$ I would like get more explanation why you cannot separate the intrinsic spin energy and the rest energy. What are the standard deviations and means of these two quantities? They are probably very close to each other. Can you please point out which other energies here are mostly relevant? I will update your answer when I receive better view of the case. There is one Red-big book of some famous particle physicist whose name I have forgotten. I explains this topic well. I will find it out. $\endgroup$ Nov 10, 2015 at 19:10
  • 1
    $\begingroup$ @Masi Imagine you had a body that was always spinning at the same rate so you can't slow it down to extract that energy because it always spins at that rate. So you could destroy the whole thing to extract its energy and then it wouldn't be spinning, but you'd get all the energy back, not just the energy from spinning. That energy you get from destroying something and that you need to provide to make it is called the rest energy. But in this case it has to spin, but since that is a fixed unchanging thing, it becomes part of the cost of making it and part of what you get back from destroying it $\endgroup$
    – Timaeus
    Nov 11, 2015 at 16:17
  • $\begingroup$ Excellent explanation! So the electrons are changing their energy form from spinning energy to rest energy, while the kinetic energy is dominating. Right? - - Timaeus's answer points out the magnetic forces fail because they depend on velocity. I added the drift velocity equation to the body of the question. How do you think the magnetic force behaves? $\endgroup$ Nov 11, 2015 at 19:49
  • 1
    $\begingroup$ @Masi The energy of spinning does not change into rest energy. Since the particle must spin, the only way to get rid of whae energy of spinning is to get rid of the whole particle so you get rid of the rest energy too. And to make a particle you have to provide the rest energy and the energy for the spinning. So the two kinds of energy always appear together and always leave together. So we don't actually know how much one is versus the other one. Most people call the total of both of them the "rest energy" and don't worry about it $\endgroup$
    – Timaeus
    Nov 11, 2015 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.