accelerating charged particle radiation reaction


by roboticmehdi
Tags: charged particle, electromagnetism, energy, radiation, self-interaction
roboticmehdi
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#1
Mar20-12, 05:04 AM
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It is known that if a charged particle accelerates then it emits electromagnetic wave (energy). If so then this means that the work we do on particle, W=F*s, doesn't all go to particles kinetic energy, E=0.5*m*v^2. Then this means that Newton's F=m*a doesn't hold for charged objects, particles, masses, etc.. Is that true? If yes, then what resists particle to accelerate to the speed it deserves ( F*s=0.5*m*v^2, solve for v ). I hope i could explain my point. I am sorry i ask a lot about electromagnetism but it is so damn confusing to me, i cant find peace if i dont understand it properly.
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tiny-tim
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Mar20-12, 07:25 AM
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hi roboticmehdi!

(try using the X2 button just above the Reply box )
Quote Quote by roboticmehdi View Post
if a charged particle accelerates then it emits electromagnetic wave (energy). If so then this means that the work we do on particle, W=F*s, doesn't all go to particles kinetic energy, E=0.5*m*v^2. Then this means that Newton's F=m*a doesn't hold for charged objects, particles, masses, etc.. Is that true?
that's correct

some of the work goes into the electromagnetic field, whose energy density increases

F = ma still holds, but you have to include the force from the field!
however, the effect is negligible in practice a lot less than the air resistance which we also usually ignore!
roboticmehdi
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Mar20-12, 10:38 AM
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Quote Quote by tiny-tim View Post
hi roboticmehdi!

(try using the X2 button just above the Reply box )


that's correct

some of the work goes into the electromagnetic field, whose energy density increases

F = ma still holds, but you have to include the force from the field!
however, the effect is negligible in practice a lot less than the air resistance which we also usually ignore!
Why negligible? i think it is not negligible, for example in antennas, which try to deliver as much work as possible to electromagnetic waves, and for example a negatively charged sphere orbiting positively charged sphere of much bigger mass, the orbiting sphere would eventually lose all of its orbit energy to electromagnetic waves. then there are particle accelerators and etc. anyway, thanks for reply. can you tell me more about this ? or give good sources? what is that force acting on a particle? how it works ? thank you.

tiny-tim
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Mar20-12, 11:59 AM
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accelerating charged particle radiation reaction


Quote Quote by roboticmehdi View Post
i think it is not negligible, for example in antennas, which try to deliver as much work as possible to electromagnetic waves,
but we wouldn't use F = ma for an antenna
what would be the body with mass m in that equation?
and for example a negatively charged sphere orbiting positively charged sphere of much bigger mass
"orbiting"? how would that happen?
then there are particle accelerators
again, we don't use F = ma, we use more complicated field equations
roboticmehdi
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Mar20-12, 12:28 PM
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Quote Quote by tiny-tim View Post
but we wouldn't use F = ma for an antenna
what would be the body with mass m in that equation?
electrons are accelerated in antenna by electric field in the conductor. that electric field applies force on electron and it accelerates. electron has mass. so theoreticly it is possible to apply F=ma. to make things simpler you can assume the conductor to be a superconductor so that there is no resistance of wire to electrons. its like ignoring air resistance and making the environment airless, i.e. vacuum.


"orbiting"? how would that happen?

the same way as earth orbits sun. the negatively charged sphere would be attracted to positively charged sphere. having the right initial velocity, it would circle around positively charged sphere. but of cource the orbit would decay and lose all energy to electromagnetic waves. the circling body would eventually fall on positively charged sphere.


again, we don't use F = ma, we use more complicated field equations
ok maybe not F=ma here but definitely F=d(mv)/dt
roboticmehdi
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Mar20-12, 12:29 PM
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i dont how to do this sorry. in my upper post some of my comments are inside the quoted text. dont miss it.
roboticmehdi
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#7
Mar20-12, 02:21 PM
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anybody has other answers to my question ?
chrisbaird
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#8
Mar20-12, 02:24 PM
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We don't typically use F = m a directly with electromagnetic waves, but it is still there in principle. We talk more in terms of energies than forces. The power radiated by an antenna, for instance, is not calculated as the force times velocity, but rather as the energy flow rate. The conservation of energy equation in electromagnetics accounts for both a force giving kinetic energy to a charged particle and also causing energy to radiate away.
vanhees71
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#9
Mar21-12, 04:10 AM
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This is a problem that is unsolved in classical electrodynamics. Classical point particles interacting with their own radiation field leads to equations with weird properties, predicting among other things self-acceleration.

The best source to learn about this difficult problem is the book

F. Rohrlich, Classical Charged Particles, World Scientific 2007


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