Explaining E/M Ratios of Charged Particles without Math

[samalkhaiat]

According to GR as well as Maxwell's theory, when charged particles are accelerated they radiate energy by electromagnetic and gravitational radiation. This energy, according to special relativity, is equivalent to mass.
I wonder, how can elementary particles preserve their fixed e/m ratios when subject to external forces?
I can show this using the Einstein-Maxwell field equations , but can you explain it without using mathematics?


sam

 

[lightgrav]

Easy:
Charge and mass are intrinsic properties of an object.
Radiated Electro-Magnetic waves (or photons) have zero total charge;
the Energy they contain comes from the Kinetic Energy of the object,
or from the Field Energy whose decrease accompanies the acceleration,
or from both.

I expect that gravitational radiation (if it exists) has its energy source
similarly as the interaction of the system components.

 

[jtbell]

According to GR as well as Maxwell's theory, when charged particles are accelerated they radiate energy by electromagnetic and gravitational radiation. This energy, according to special relativity, is equivalent to mass.
I wonder, how can elementary particles preserve their fixed e/m ratios when subject to external forces?

The energy that the particles radiate is supplied to them via the work done by external forces. Their intrinsic energy (corresponding to the rest mass) does not change.

 

[lightgrav]

jtbell:

In many instances of charged particle acceleration, the external Force does negative Work; in many other instances, the external Force does zero Work.

In such cases (most of them) the radiated energy obviously is NOT supplied to the radiation via the Work done by the external Forces.

An "explanation" that is incorrect in most cases, I suspect, is never really "right".

 

[samalkhaiat]

Easy

It is. Isn't it

Charge and mass are intrinsic properties of an object.

Yes, and they are the sources of the radiated fields

Radiated Electro-Magnetic waves (or photons) have zero total charge

Now, this is easy

the Energy they contain comes from the Kinetic Energy of the object

Theorm: the change in the kinetic energy is equal to the work done by the external forces. I thought jtbell suggested this, but you thought it "is incorrect"

or from the Field Energy whose decrease accompanies the acceleration,or from both.

So if the acceleration of a charge lasts long enough, will the field loses all its energy? Is it possible to have a system of charged particles with no field, or with zero field's energy?

sam

 

[samalkhaiat]

The energy that the particles radiate is supplied to them via the work done by external forces. Their intrinsic energy (corresponding to the rest mass) does not change.

The work done by the fields on the particles (E.d) is equal to the rate of change of the kinetic energy. In relativity, kinetic energy includes the rest energy

In the approximation: the vilocities of the charges are small compared with the speed of light(v<<c), Maxwell theory says; the radiation is determined by the second time derivative of the dipole moment d;

d = Sum[e.r] = Sum[m.r.e/m]
If e/m is the same for all particles, then

d= e/m.Sum[m.r] = e/m.R.Sum[m] =M.R.e/m,
where M is the total mass,and R is the radius vector of the centre of mass of the system.
Therefore the 2nd time derivative of d is proportional to the acceleration of the centre of mass which is zero. Thus the system can not radiate!
Where did I make a mistake?

sam

 

[lightgrav]

The change in Kinetic Energy is equal to the Sum of all Work done by all External Forces applied to the object.
I intended "external Forces" to mean external to the charged object
(the "source of radiation") and external to the radiation that is emitted from the charged object. jtbell's wording indicated a similar intent.

The Work-Energy Theorem applied only to the charged particle must include its interaction with the radiated photon in the Sum of external Forces, so that approach cannot be used to describe the photon separately from the rest of the fields.

The Work-Energy Theorem applied to the charged particle and photon as a system does not yield the Energy of the photon separately.

In order for the field to be responsible for an extended acceleration, the field must have considerable spatial extent itself. The most common example of an extended field collapsing (almost) completely is in photon absorption.

 

[lightgrav]

Sam, your BIG mistake was presuming that the center-of-mass cannot accelerate!

Minor issues are (1) the peculiar model with only one kind of charged object
(2) looking only for electric dipole radiation
(3) treating all particles as fixed in position
(4) expecting radiation from a static situation ...

 

[Ich]

No, the BIG mistake was (1) - only one kind of charged object. Such a system could never emit dipole radiation, the lowest one would be quadrupole radiation just like gravitational waves.

 

[samalkhaiat]

I intended "external Forces" to mean external to the charged object
(the "source of radiation") and external to the radiation that is emitted from the charged object. jtbell's wording indicated a similar intent.

What is that? God!
Look, for a system of n charged particles interacting by their EM-field, the equation of motion for the ith particle is, (v<<c);
m(i)dv/dt = e(i)[E + v X H] --(1)
the right-hand side is the external force,(E,H) come from the remaining (n-1) particles.
To get the "work-energy" theorem, just take the scalar product with v;
d[KE(i)]/dt = e(i) v.E
"the change in KE = work done by F(ext)"

The Work-Energy Theorem applied only to the charged particle must include its interaction with the radiated photon in the Sum of external Forces

BE VERY CAREFUL, you are stepping into a very deep water.
Let us do what you want us to do and add to EQ(1), the force of the radiated field (the reactive effect of radiation), which (one can show) is proportional to the 2nd time derivative of v[f(rad)=const da(t)/dt] ( it is also called the "self-force"):
ma(t) = F(ext) + const. da(t)/dt.
This is the so-called Abraham-Lorentz equation.
SORRY, no "work-energy" theorem from this equation. Not just this, the equation leads you to all sort of troubles.
A completely satisfactory treatment of the reactive effects of radiation does not exist even in Quantum electrodynamics.The difficulties with this equation touch one of the most fundamental aspects of physics, the nature of e and m of an elementary particle and their renormalizations.
In 1904, Lorentz investigated the above equation by making a purely electromagnetic (classical) model of a charged particle.(Lorentz, Theory of Electrons, P.252)
In case you do not know, the Abraham-Lorentz equation leads to the absurd result that a free charge increases its energy without limit.To see this, set F(ex)=0, the resulting equation describes the action of the charge "on itself". This equation has, in addition to the trivial solution:
v = constant, another solution (runaway solution) in which the acceleration is;
a(t) = a(0) exp(t/T),
where T=const=6.3x10^(-24) sec. for electron.
As you see, a(t) increases indefinitly with the time.This means that a charge passing through EM-field, upon emergence from the field, would have to be infinitely "self-accelerated"!
The "runaway" solution can only be avoided if one violates causality! YOU CHOOSE!

The Work-Energy Theorem applied to the charged particle and photon as a system does not yield the Energy of the photon separately.

WRONG, We determine the form of the radiative reaction force (self-force) by demanding that "the work done by this force on the particle in some time interval be equal to the negative of the energy radiated in that interval", So energy will be conserved, at least over that time interval.


sam

 

[samalkhaiat]

Sam, your BIG mistake was presuming that the center-of-mass cannot accelerate!

My "intended" BIG mistake is a "good mistake", because it points to the fact that a "closed" system of particles with equal e/m, can not emits dipole radiation.

Minor issues are (1) the peculiar model with only one kind of charged object

What is peculiar about system of 239 electrons?

(2) looking only for electric dipole radiation

Dipole radiation is simpler than Coulomb and Quadrupole radiation.

(3) treating all particles as fixed in position

FIXED?

(4) expecting radiation from a static situation .

I don't understand this.

sam

 

[lightgrav]

"An explanation can be incorrect" even if it uses some of the right words.
I never implied that the Work-Energy Theorem was incorrect; in fact, I implicitly used it in my own first post in this thread.

What is that? God!
Look, for a system of n charged particles interacting by their EM-field, the equation of motion for the ith particle is, (v<<c);
m(i)dv/dt = e(i)[E + v X H] --(1)
the right-hand side is the external force,(E,H) come from the remaining (n-1) particles.
To get the "work-energy" theorem, just take the scalar product with v;
d[KE(i)]/dt = e(i) v.E
"the change in KE = work done by F(ext)"

If you don't understand a statement that I make, just ask.

I don't know why you wrote that quote ... what was it about the Work-Energy Theorem that you thought I needed to see again?

Treating the emitting object and the emitted photon as a single system (which is isolated enough to conserve Energy all by itself) while it interacts with external Forces strongly enough to cause the radiation is at best a very rough approximation.

Treating the emitting object as the entire system requires that the Force by the emitted photon be included as en external Force. I thought this would be obvious ... .
Otherwise momentum won't be conserved during emission (photons carry momentum).

 

[dextercioby]

According to GR as well as Maxwell's theory, when charged particles are accelerated they radiate energy by electromagnetic and gravitational radiation. This energy, according to special relativity, is equivalent to mass.
I wonder, how can elementary particles preserve their fixed e/m ratios when subject to external forces?
I can show this using the Einstein-Maxwell field equations , but can you explain it without using mathematics?
sam

Well, first, the electric charge of a particle (electron for simplicity) is energy dependent and "blows up" at the Landau pole. So for an interacting particle "phsyical electric charge" divided by "physical" (QED renormalized) mass is energy dependent...

Daniel.

 

[samalkhaiat]

If you don't understand a statement that I make, just ask.
I don't know why you wrote that quote ... what was it about the Work-Energy Theorem that you thought I needed to see again?

If you don't understand the physics involved in the radiation processes, then you should not teach them to somebody with 10 years experience in the field of theoretical physics!

Treating the emitting object and the emitted photon as a single system (which is isolated enough to conserve Energy all by itself) while it interacts with external Forces strongly enough to cause the radiation is at best a very rough approximation.

Even though, this statement is confused, Yes it is an approximation, but not rough approximation.
This is how it is done in classical electrodynamics(CED).
Do you have a better treatment?
Good explanation of this approximation is given by:
J.D.Jackson, Classical Electrodynamics,P.780.
Have a look at this book (if you good in math), you will learn alot!

Treating the emitting object as the entire system requires that the Force by the emitted photon be included as en external Force. I thought this would be obvious ... .
Otherwise momentum won't be conserved during emission (photons carry momentum).

I thought, I went through this garbage in some details and saw the trubles with Abraham-Lorentz equation .
The reactive effects of radiation on the motion of charged particle can be expected to be important in times of order of T~10^-24sec.At this order of magnitude CED breaks down, we must use QED.
For times t>>T (i.e in the domain of CED) the reactive effects are sufficiently small and have a negligible effect on the motion.
The treatment in QED leads to the so-called infra-red divergence(look up Bremsstrahung)
For all processes in QED, the infrared divergences can be made to cancel, to all orders of perturbation theory, if one introduces the bare charge and mass (Renormalizations):
e' = e[1 + O(e^2)],
m' = m + O(e'^2).
So, only when we neglect the Quantum processes, we could say:
The e/m-problem is EASY, you thought it is, didn't you

cheers

sam