# E/m ratio?

1. Nov 17, 2005

### 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

2. Nov 25, 2005

### 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.

3. Nov 26, 2005

### Staff: Mentor

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.

4. Nov 26, 2005

### 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".

5. Nov 26, 2005

### samalkhaiat

6. Nov 26, 2005

### samalkhaiat

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!! :rofl:
Where did I make a mistake?

sam

Last edited: Nov 26, 2005
7. Nov 26, 2005

### 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 spacial extent itself. The most common example of an extended field collapsing (almost) completely is in photon absorption.

8. Nov 26, 2005

### 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 ...

9. Nov 28, 2005

### 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.

10. Nov 29, 2005

### samalkhaiat

11. Nov 29, 2005

### samalkhaiat

12. Dec 4, 2005

### 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.

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).

Last edited: Dec 4, 2005
13. Dec 6, 2005

### dextercioby

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.

14. Dec 6, 2005

### samalkhaiat

Last edited: Dec 6, 2005