Suposse we have a charged particle placed in some region with an uniform electric field. It's clear that a force $$\vec F_e =q\vec E $$ will appear on the particle due to Lorentz force. However, the particle will radiate with a power of $$P=q^2a^2/6\pi\epsilon_0c^3,$$ where $a$ is the magnitude of the acceleration, according to Larmor formula for power radiated. For energy conservation, a recoil force should appear on the particle. So my question is, the acceleration of the particle will be $$\vec a=q\vec E/m$$ or will it be lower?
Asked
Active
Viewed 592 times
1
-
Now enough clear, break it up into round sentences and give more details. – peterh May 09 '17 at 21:13
-
It won't work on this way. If you can't edit your question, if you can't formulate in round sentences, if you can't write a clear text about your problem, it won't work. It is not a chatroom. Delete your comment, and edit its content into your question. – peterh May 09 '17 at 21:41
-
A rigorous treatment of this problem is given in this article – Count Iblis May 10 '17 at 05:11
-
1Possible duplicate of Does a constantly accelerating charged particle emit EM radiation or not? – tparker May 10 '17 at 05:35
-
1This question has been asked many, many times on this site. See https://physics.stackexchange.com/questions/70915/does-a-constantly-accelerating-charged-particle-emit-em-radiation-or-not, https://physics.stackexchange.com/questions/21830/does-a-charged-particle-accelerating-in-a-gravitational-field-radiate, https://physics.stackexchange.com/questions/3247/should-a-neutron-fall-faster-than-a-proton, and the questions that those questions link to. – tparker May 10 '17 at 05:36
-
Does your acceleration formula contain the momentum or the energy of the particle? The particle will be loosing some momentum/energy but this does not affect the acceleration until it reaches relativistic energies where E=mc^2 has to be considered and the inertial mass changes. – anna v May 10 '17 at 05:42
1 Answers
0
If the charged particle is composed of smaller charged elements, the Lorentz-Abraham force is a validly derived approximate expression of internal force the charged particle will experience. However, magnitude of this force is so small that its effect on the motion of the charged particle has so far been unmeasurable.
If the particle is a point, so it is not made of smaller elements, the Poynting theorem cannot be interpreted as work-energy theorem and the usual derivations of the Lorentz-Abraham force fail. Such particle should not experience any internal force.
Ján Lalinský
- 37,229
-
Why does the Lorentz-Abraham force fail if the particle is a point? I thought that it was derived of Larmor formula, which is valid for puntual particles. Could you explain this with more detail? – Eduardo González May 09 '17 at 22:11
-
Larmor formula is not valid for point particles, because Larmor formula is derived based on the work-energy interpretation of the Poynting theorem. For point particles, this interpretation is not valid, as the Poynting theorem, although valid in vacuum between the particles, breaks down at the points where the particles are. – Ján Lalinský May 10 '17 at 19:16