Re: Accelerating a particle and photons


"srp" <srp2@xxxxxxxxxxxxxxxx> wrote in message
news:43B96DC3.1090606@xxxxxxxxxxxxxxxxxxx
> Ron Baker, Pluralitas! a écrit :
>> "srp" <srp2@xxxxxxxxxxxxxxxx> wrote in message
>> news:43B96044.5020402@xxxxxxxxxxxxxxxxxxx
>>
>>>vze2vt56@xxxxxxxxxxx a écrit :
>>>
>>>>If I oscillate a charged particle with some frequency f, I'll be
>>>>generating EM radiation of that frequency.
>>>
>>>Maybe not that simple. Oscillating a charged particle at a
>>>"mechanical" frequency f in no way implies that the EM radiation
>>>emitted will have that frequency.
>>
>> Really? Suppose you have a charged object (say a pith ball
>> on the end of a wooden stick) and you cause it to have motion
>> described by x = cos(2*pi*t), y = 0, z = 0, -oo < t < oo.
>> (You could also suppose the charge is 9e-31 Coulombs.)
>> The photons don't have a frequency of 1Hz?
>> What frequency do they have?
>> (And how many of them are there?)
>
> I have no idea if photons would be released in such a case.
> I was talking about charged particles, like electrons.

Then did you really have an idea about the OP's question.
The OP was not asking specifically about electrons.

How is the pith ball not a charged particle?
The charge on the pith ball is an electron.
Disregard the pith ball if you like.
You cause an electron to have the motion described
above. What frequency do the photons have?
How many of them are there?

>
>> How about the electromagnetic field? Does it have
>> a frequency? Is it different than the photons?
>
> Well, yes and no. It is not that simple.
>
> If the EM field is the result of coherent EM emission like
> lasers, I guess a frequency could theoretically be assigned.

What about the electromagnetic field for an electron
with the cos() motion described above?
Can you offer a guess on that?

>
> For photons, a frequency can be definitely be associated.
>
>> <snip>
>>
>>>>Does this mean no photons are emmitted until I change
>>>>the particle's direction?
>>>
>>>Yes, a photon will be emitted each time the particle stops prior
>>>to reaccelerating in the reverse direction (this is called
>>>"bremsstrahlung"), but the frequency of the released photon will
>>>be that of the energy that sustained the velocity the electron
>>>had peaked at before stopping.
>>
>>
>> Suppose an electron is going .9 c and we slow it down
>> to 1 m/s. It doesn't stop. No photon? No radiation?
>
> Yes a photon would be emitted with the energy in excess of that
> required to maintain a 1 m/s velocity.

Previously you said the (one?) photon is emitted when it 'stops'.
Are you now saying it is emitted when the ac/deceleration
stops? i.e. at the moment when d^2 x / dt^2 becomes 0 and
not when dx/dt becomes 0?

So if you brake an electron from .999999c to 1m/s linearly over
a period of 10 years there is no radiation/photon until
exactly the end of that 10 years? And that one photon
will have millions of eV energy?

>
> André Michaud

--
rb


.

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