| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
01 Jan 2006 07:07:02 PM |
| Object: |
Accelerating a particle and photons |
If I oscillate a charged particle with some frequency f, I'll be
generating EM radiation of that frequency. But this EM radiation is
just a stream of photons, each with an energy corresponding to
frequency f. But for a photon of that energy to be emmitted, it must
know the rate I'm oscillating the charged particle at (since that
determines f). Does this mean no photons are emmitted until I change
the particle's direction?
Another of phrasing my question - the photon must be emmitted with some
specific frequency. This frequency is only determined once I change the
direction of the particle I'm waving. So how is the photon emmitted
before I do that, with that frequency?
Thanks,
Brian
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| User: "FrediFizzx" |
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| Title: Re: Accelerating a particle and photons |
01 Jan 2006 10:40:41 PM |
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<vze2vt56@verizon.net> wrote in message
news:1136164022.412164.53260@o13g2000cwo.googlegroups.com...
| If I oscillate a charged particle with some frequency f, I'll be
| generating EM radiation of that frequency. But this EM radiation is
| just a stream of photons, each with an energy corresponding to
| frequency f. But for a photon of that energy to be emmitted, it must
| know the rate I'm oscillating the charged particle at (since that
| determines f). Does this mean no photons are emmitted until I change
| the particle's direction?
|
| Another of phrasing my question - the photon must be emmitted with
some
| specific frequency. This frequency is only determined once I change
the
| direction of the particle I'm waving. So how is the photon emmitted
| before I do that, with that frequency?
How exactly are you oscillating the charged particle? Is it not by
photons? What do you think oscillating means?
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.vacuum-physics.com
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| User: "" |
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| Title: Re: Accelerating a particle and photons |
01 Jan 2006 11:00:43 PM |
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Why does it matter how I'm oscillating the charged particle? The
charged particle can be a metal ball, and I can be holding it with
tweezers for all I care.
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| User: "Ron Baker, Pluralitas!" |
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| Title: Re: Accelerating a particle and photons |
02 Jan 2006 10:39:00 AM |
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|
<vze2vt56@verizon.net> wrote in message
news:1136164022.412164.53260@o13g2000cwo.googlegroups.com...
If I oscillate a charged particle with some frequency f, I'll be
generating EM radiation of that frequency. But this EM radiation is
just a stream of photons, each with an energy corresponding to
frequency f. But for a photon of that energy to be emmitted, it must
know the rate I'm oscillating the charged particle at (since that
determines f). Does this mean no photons are emmitted until I change
the particle's direction?
I used to think that was an easy question.
When a charge accelerates it radiates.
But then I ran across:
http://www.mathpages.com/home/kmath528/kmath528.htm
Another of phrasing my question - the photon must be emmitted with some
specific frequency. This frequency is only determined once I change the
direction of the particle I'm waving.
How do you figure that?
So how is the photon emmitted
What do you mean "the" photon?
before I do that, with that frequency?
Thanks,
Brian
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| User: "The Ghost In The Machine" |
|
| Title: Re: Accelerating a particle and photons |
02 Jan 2006 02:00:07 PM |
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In sci.physics, Ron Baker, Pluralitas!
<stoshu@bellsouth.net.pa>
wrote
on Mon, 02 Jan 2006 16:39:00 GMT
<EGcuf.10058$hI1.7447@tornado.socal.rr.com>:
<vze2vt56@verizon.net> wrote in message
news:1136164022.412164.53260@o13g2000cwo.googlegroups.com...
If I oscillate a charged particle with some frequency f, I'll be
[snip for brevity]
So how is the photon emmitted
What do you mean "the" photon?
Well, there's only one. What, you still have it? Give it back!
Other people want to use it, you know.
:-)
On a slightly more realistic level, a perfect 100 W 500
nm light source (pea green) would generate about
2.517 * 10^20 of them per second, streaming in all
directions -- or maybe in a tight, collimated beam, if
it's a laser source.
"The" photon indeed. :-)
[rest snipped]
--
#191,
It's still legal to go .sigless.
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| User: "srp" |
|
| Title: Re: Accelerating a particle and photons |
02 Jan 2006 11:18:13 AM |
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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.
But this EM radiation is just a stream of photons,
Yes.
each with an energy corresponding to frequency f.
No.
But for a photon of that energy to be emmitted, it must
know the rate I'm oscillating the charged particle at (since that
determines f).
It doesn't need to know that since the EM frequency of the
generated photons has nothing to do with the mechanical frequency
you are oscillating the particle at.
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.
Another of phrasing my question - the photon must be emmitted with some
specific frequency. This frequency is only determined once I change the
direction of the particle I'm waving. So how is the photon emmitted
before I do that, with that frequency?
The photon is emitted with all the energy that was required to
maintain the peak velocity that the particle reached before
stopping. It will have the frequency of that energy.
This is a very mechanical process.
André Michaud
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| User: "Ron Baker, Pluralitas!" |
|
| Title: Re: Accelerating a particle and photons |
02 Jan 2006 12:08:17 PM |
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"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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?)
How about the electromagnetic field? Does it have
a frequency? Is it different than the photons?
<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?
<snip>
--
rb
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| User: "Jan Panteltje" |
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| Title: Re: Accelerating a particle and photons |
02 Jan 2006 01:19:03 PM |
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On a sunny day (Mon, 02 Jan 2006 18:08:17 GMT) it happened "Ron Baker,
Pluralitas!" <stoshu@bellsouth.net.pa> wrote in
<l_duf.10065$hI1.8853@tornado.socal.rr.com>:
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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?)
How about the electromagnetic field? Does it have
a frequency? Is it different than the photons?
I am going to reply to this, but I know I do not know the exact answer.
Thought about it some time ago, and came to some picture that satisfies me
enough as to explain' it.
In the case of the electron moving back and forwards at 1Hz between 2 plates,
maybe the analogy of you pushing a stick up and down in a pool (water) is
close.
From the point of view of 'photon' little bits of energy is transferred
between the water molecules, so in a way energy transfer only happens if a
'bump' takes place.
The [energy] levels, in case of the one 1Hz sine are all the ones from
the part of the sine that moves fast to the top and bottom where there is
almost no motion.
So the wave is made up of these little bumps....
Well we know there is no water, but clearly wave is not localized.. just
'made' of photons.
Because of inertia in the fluid any time when changing direction we have
a high energy exchange.
So.. we need an ether, may something is there that we have not detected yet.
Other things do not make sense, but of course I am used to think 'billiard
balls' and in 3D.. even so : remember 200 years ago 'radio' waves would have
gotten you burned by the Church(tm).
So glad spring theory explains it all :-)
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| User: "srp" |
|
| Title: Re: Accelerating a particle and photons |
02 Jan 2006 12:15:01 PM |
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Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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.
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.
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.
André Michaud
.
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|
|
| User: "Ron Baker, Pluralitas!" |
|
| Title: Re: Accelerating a particle and photons |
02 Jan 2006 01:08:56 PM |
|
|
"srp" <srp2@globetrotter.net> wrote in message
news:43B96DC3.1090606@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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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| User: "srp" |
|
| Title: Re: Accelerating a particle and photons |
02 Jan 2006 02:52:57 PM |
|
|
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96DC3.1090606@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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.
Well, read back yourself. He was specifically talking about
"a charged particle", and to my knowledge, electrons are
charged particles.
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
The charge on the pith ball is an electron.
I doubt that the charge on a charged pith ball would
be a single electron.
Disregard the pith ball if you like.
You cause an electron to have the motion described
above. What frequency do the photons have?
They would have the frequency of the energy released
each time the electron stops in one direction before
it starts accelerating in the reverse direction.
In other words, it will depend on the velocity you
will impart to the electron to maintain the
oscillating motion. If the oscillation is stable, the
photons will all have the same frequency.
How many of them are there?
As many as there will be changes in direction of the
electron during the experiment.
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?
Note the the cos() function represents a to and fro motion in
reality.
Can you offer a guess on that?
Possibly.
I can see a charged particle in motion, so the fields will
be composite, meaning that the composite fields will be made
up of the fields of the electron itself plus the fields of
the energy carrying it (the energy that gives it its velocity).
I see that the latter will increase as the velocity increases
and moving on as a free photon as the electron stops, leaving
the latter behind. The electron will maintain its own
intrinsic fields.
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?
Since the electron moves at a velocity determined by its energy
in excess of its rest mass, it cannot move at any velocity lower
than the energy that it possesses. For it to slow down, it has
lose the energy in excess of that required to maintain the new
lower velocity. If its stops completely, it has to lose all
of its energy in excess of rest mass.
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?
I don't know. Hypothetical. Impossible to do.
We do not have the technology to follow linearly an electron
moving at near light speed to start with, let alone slowly
braking it from .999999c to 1m/s over a period of 10 years
in the process.
And that one photon
will have millions of eV energy?
I have no idea. Hypothetical. It couldn't be more anyway than
the energy that was required to maintain it at .999999c to
start with if no outside factor cause it to lose any for other
causes.
André Michaud
.
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| User: "Ron Baker, Pluralitas!" |
|
| Title: Re: Accelerating a particle and photons |
03 Jan 2006 11:09:11 AM |
|
|
"srp" <srp2@globetrotter.net> wrote in message
news:43B992C5.9020404@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96DC3.1090606@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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.
Well, read back yourself. He was specifically talking about
"a charged particle", and to my knowledge, electrons are
charged particles.
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
Equivocation.
Do you have a dictionary? What does it have
for 'particle'?
The charge on the pith ball is an electron.
I doubt that the charge on a charged pith ball would
be a single electron.
Why couldn't it be? And what difference does
it make?
Are you saying you can only speak about
individual free electrons?
Disregard the pith ball if you like.
You cause an electron to have the motion described
above. What frequency do the photons have?
They would have the frequency of the energy released
each time the electron stops in one direction before
it starts accelerating in the reverse direction.
And what frequency is that? The conditions have
be specified and are pretty simple. Are you not
able to calculate it?
In other words, it will depend on the velocity you
will impart to the electron to maintain the
oscillating motion.
It is specified. A single electron in 1 Hz sinewave motion
along one axis. 1 m amplitude. That gives a
peak velocity of 1 m/s. Pretty simple.
What is the frequency of the photon(s)?
If the oscillation is stable, the
photons will all have the same frequency.
How many of them are there?
As many as there will be changes in direction of the
electron during the experiment.
So, two photons per second?
But didn't you say in the last post that it is
when the ac/deceleration stops that the photon
is emitted? Which is it? When the acceleration
stops or when the velocity goes to zero?
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?
Note the the cos() function represents a to and fro motion in
reality.
Can you offer a guess on that?
Possibly.
I can see a charged particle in motion, so the fields will
be composite, meaning that the composite fields will be made
up of the fields of the electron itself plus the fields of
the energy carrying it (the energy that gives it its velocity).
I see that the latter will increase as the velocity increases
and moving on as a free photon as the electron stops, leaving
the latter behind. The electron will maintain its own
intrinsic fields.
So what is the frequency of the EM waves?
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?
Since the electron moves at a velocity determined by its energy
in excess of its rest mass, it cannot move at any velocity lower
than the energy that it possesses. For it to slow down, it has
lose the energy in excess of that required to maintain the new
lower velocity. If its stops completely, it has to lose all
of its energy in excess of rest mass.
You didn't answer the question.
When is 'the' photon emitted? When the acceleration
becomes 0 or when the velocity 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?
I don't know. Hypothetical. Impossible to do.
You've given many answers to hypotheticals.
Did you not really know then?
The physics is known, isn't it?
A physicist could calculated it, couldn't they?
Physicists deal with hypotheticals all the time.
We do not have the technology to follow linearly an electron
moving at near light speed to start with, let alone slowly
braking it from .999999c to 1m/s over a period of 10 years
in the process.
But we have the technology to brake it from .99 c to 10 m/s
over say 0.1 s. What would happen in that case?
And that one photon
will have millions of eV energy?
I have no idea. Hypothetical. It couldn't be more anyway than
the energy that was required to maintain it at .999999c to
start with if no outside factor cause it to lose any for other
causes.
André Michaud
--
rb
.
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| User: "srp" |
|
| Title: Re: Accelerating a particle and photons |
03 Jan 2006 11:36:58 AM |
|
|
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B992C5.9020404@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96DC3.1090606@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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.
Well, read back yourself. He was specifically talking about
"a charged particle", and to my knowledge, electrons are
charged particles.
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
Equivocation.
Do you have a dictionary? What does it have
for 'particle'?
Sure. But I was answering in context of the OP.
No pith ball in sight.
Charged particle boiled down to electron.
It you disagree, feel free to sue me.
The charge on the pith ball is an electron.
I doubt that the charge on a charged pith ball would
be a single electron.
Why couldn't it be? And what difference does
it make?
Are you saying you can only speak about
individual free electrons?
In context, in the frame of what the OP wanted, yes.
Disregard the pith ball if you like.
You cause an electron to have the motion described
above. What frequency do the photons have?
They would have the frequency of the energy released
each time the electron stops in one direction before
it starts accelerating in the reverse direction.
And what frequency is that? The conditions have
be specified and are pretty simple. Are you not
able to calculate it?
Answer:
In other words, it will depend on the velocity you
will impart to the electron to maintain the
oscillating motion.
It is specified. A single electron in 1 Hz sinewave motion
along one axis. 1 m amplitude. That gives a
peak velocity of 1 m/s. Pretty simple.
What is the frequency of the photon(s)?
The OP made no such request. Besides, if you succeed in
having an electron move at that mechanical frequency,
feel free to measure the outcome. For now, this seems
highly hypothetical and irrelevant.
If the oscillation is stable, the
photons will all have the same frequency.
How many of them are there?
As many as there will be changes in direction of the
electron during the experiment.
So, two photons per second?
But didn't you say in the last post that it is
when the ac/deceleration stops that the photon
is emitted?
I said it is when the electron is forced to stop.
Which is it? When the acceleration
stops or when the velocity goes to zero?
Why should it be when the acceleration stops ?
Do you really understand acceleration ?
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?
Note the the cos() function represents a to and fro motion in
reality.
Can you offer a guess on that?
Possibly.
I can see a charged particle in motion, so the fields will
be composite, meaning that the composite fields will be made
up of the fields of the electron itself plus the fields of
the energy carrying it (the energy that gives it its velocity).
I see that the latter will increase as the velocity increases
and moving on as a free photon as the electron stops, leaving
the latter behind. The electron will maintain its own
intrinsic fields.
So what is the frequency of the EM waves?
Which one ? That of the electron or that of the carrying energy ?
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?
Since the electron moves at a velocity determined by its energy
in excess of its rest mass, it cannot move at any velocity lower
than the energy that it possesses. For it to slow down, it has
lose the energy in excess of that required to maintain the new
lower velocity. If its stops completely, it has to lose all
of its energy in excess of rest mass.
You didn't answer the question.
When is 'the' photon emitted? When the acceleration
becomes 0 or when the velocity becomes 0?
I'd say when the velocity tends towards zero. Certainly not
when the acceleration become zero, why should it ?
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?
I don't know. Hypothetical. Impossible to do.
You've given many answers to hypotheticals.
Did you not really know then?
I answered only to the OP's interrogation, which was not
hypothetical.
The physics is known, isn't it?
Is it really ? You seem to have problems with acceleration.
Are you a physicist ?
A physicist could calculated it, couldn't they?
For individual localized particles?
I'd like to see the maths.
Any refs?
Physicists deal with hypotheticals all the time.
Maybe they should deal more with reality for a change.
We do not have the technology to follow linearly an electron
moving at near light speed to start with, let alone slowly
braking it from .999999c to 1m/s over a period of 10 years
in the process.
But we have the technology to brake it from .99 c to 10 m/s
over say 0.1 s. What would happen in that case?
I think you can't without extensive equipment. Much too slow
for meaningful real life experimentation.
André Michaud
.
|
|
|
| User: "Ron Baker, Pluralitas!" |
|
| Title: Re: Accelerating a particle and photons |
03 Jan 2006 04:27:57 PM |
|
|
"srp" <srp2@globetrotter.net> wrote in message
news:43BAB633.6020607@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B992C5.9020404@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96DC3.1090606@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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.
Well, read back yourself. He was specifically talking about
"a charged particle", and to my knowledge, electrons are
charged particles.
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
Equivocation.
Do you have a dictionary? What does it have
for 'particle'?
Sure. But I was answering in context of the OP.
No pith ball in sight.
Charged particle boiled down to electron.
It you disagree, feel free to sue me.
No. Anybody reading this can see
that you don't know what you are talking about.
--
rb
The charge on the pith ball is an electron.
I doubt that the charge on a charged pith ball would
be a single electron.
Why couldn't it be? And what difference does
it make?
Are you saying you can only speak about
individual free electrons?
In context, in the frame of what the OP wanted, yes.
Disregard the pith ball if you like.
You cause an electron to have the motion described
above. What frequency do the photons have?
They would have the frequency of the energy released
each time the electron stops in one direction before
it starts accelerating in the reverse direction.
And what frequency is that? The conditions have
be specified and are pretty simple. Are you not
able to calculate it?
Answer:
In other words, it will depend on the velocity you
will impart to the electron to maintain the
oscillating motion.
It is specified. A single electron in 1 Hz sinewave motion
along one axis. 1 m amplitude. That gives a
peak velocity of 1 m/s. Pretty simple.
What is the frequency of the photon(s)?
The OP made no such request. Besides, if you succeed in
having an electron move at that mechanical frequency,
feel free to measure the outcome. For now, this seems
highly hypothetical and irrelevant.
If the oscillation is stable, the
photons will all have the same frequency.
How many of them are there?
As many as there will be changes in direction of the
electron during the experiment.
So, two photons per second?
But didn't you say in the last post that it is
when the ac/deceleration stops that the photon
is emitted?
I said it is when the electron is forced to stop.
Which is it? When the acceleration
stops or when the velocity goes to zero?
Why should it be when the acceleration stops ?
Do you really understand acceleration ?
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?
Note the the cos() function represents a to and fro motion in
reality.
Can you offer a guess on that?
Possibly.
I can see a charged particle in motion, so the fields will
be composite, meaning that the composite fields will be made
up of the fields of the electron itself plus the fields of
the energy carrying it (the energy that gives it its velocity).
I see that the latter will increase as the velocity increases
and moving on as a free photon as the electron stops, leaving
the latter behind. The electron will maintain its own
intrinsic fields.
So what is the frequency of the EM waves?
Which one ? That of the electron or that of the carrying energy ?
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?
Since the electron moves at a velocity determined by its energy
in excess of its rest mass, it cannot move at any velocity lower
than the energy that it possesses. For it to slow down, it has
lose the energy in excess of that required to maintain the new
lower velocity. If its stops completely, it has to lose all
of its energy in excess of rest mass.
You didn't answer the question.
When is 'the' photon emitted? When the acceleration
becomes 0 or when the velocity becomes 0?
I'd say when the velocity tends towards zero. Certainly not
when the acceleration become zero, why should it ?
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?
I don't know. Hypothetical. Impossible to do.
You've given many answers to hypotheticals.
Did you not really know then?
I answered only to the OP's interrogation, which was not
hypothetical.
The physics is known, isn't it?
Is it really ? You seem to have problems with acceleration.
Are you a physicist ?
A physicist could calculated it, couldn't they?
For individual localized particles?
I'd like to see the maths.
Any refs?
Physicists deal with hypotheticals all the time.
Maybe they should deal more with reality for a change.
We do not have the technology to follow linearly an electron
moving at near light speed to start with, let alone slowly
braking it from .999999c to 1m/s over a period of 10 years
in the process.
But we have the technology to brake it from .99 c to 10 m/s
over say 0.1 s. What would happen in that case?
I think you can't without extensive equipment. Much too slow
for meaningful real life experimentation.
André Michaud
.
|
|
|
| User: "srp" |
|
| Title: Re: Accelerating a particle and photons |
03 Jan 2006 04:55:57 PM |
|
|
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43BAB633.6020607@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B992C5.9020404@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96DC3.1090606@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net 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.
Well, read back yourself. He was specifically talking about
"a charged particle", and to my knowledge, electrons are
charged particles.
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
Equivocation.
Do you have a dictionary? What does it have
for 'particle'?
Sure. But I was answering in context of the OP.
No pith ball in sight.
Charged particle boiled down to electron.
It you disagree, feel free to sue me.
No. Anybody reading this can see
that you don't know what you are talking about.
--
rb
Wow! What a rebuttal!
What about acceleration ?
What about physicists being able to calculate ?
Where are the maths ?
Typical behavior of quackers with no backbone.
Fine with me.
André
.
|
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| User: "Ron Baker, Pluralitas!" |
|
| Title: Re: Accelerating a particle and photons |
03 Jan 2006 10:54:32 PM |
|
|
"srp" <srp2@globetrotter.net> wrote in message
news:43BB00E2.2020605@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43BAB633.6020607@globetrotter.net...
<snip>
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
Equivocation.
Do you have a dictionary? What does it have
for 'particle'?
Sure. But I was answering in context of the OP.
No pith ball in sight.
Charged particle boiled down to electron.
It you disagree, feel free to sue me.
No. Anybody reading this can see
that you don't know what you are talking about.
--
rb
Wow! What a rebuttal!
What about acceleration ?
What about physicists being able to calculate ?
Where are the maths ?
Typical behavior of quackers with no backbone.
Fine with me.
André
I've started a new top level thread with a
collection of your statements and my claim
that they are nonsense. Let's see who
gets shot down.
--
rb
.
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|
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| User: "srp" |
|
| Title: Re: Accelerating a particle and photons |
04 Jan 2006 07:00:33 AM |
|
|
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43BB00E2.2020605@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43BAB633.6020607@globetrotter.net...
<snip>
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
Equivocation.
Do you have a dictionary? What does it have
for 'particle'?
Sure. But I was answering in context of the OP.
No pith ball in sight.
Charged particle boiled down to electron.
It you disagree, feel free to sue me.
No. Anybody reading this can see
that you don't know what you are talking about.
--
rb
Wow! What a rebuttal!
What about acceleration ?
What about physicists being able to calculate ?
Where are the maths ?
Typical behavior of quackers with no backbone.
Fine with me.
André
I've started a new top level thread with a
collection of your statements and my claim
that they are nonsense.
Oops! I must have killed the thread. I kill most threads
not to overload my newsreader. I keep only questions of
interest (typically by newbees). Better luck next time.
Let's see who gets shot down.
So you were trying to shoot me down, not really wanting
to clarify the issue. That's what I had gathered.
But why not do it here, in context, so people can
observe your inadequacy in context.
I still think you are not even a physicist, just an ignorant
quacker that was out to impress some chick in your school.
I suggest you go play with kids your age.
André Michaud
.
|
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| User: "Mark Fergerson" |
|
| Title: Re: Accelerating a particle and photons |
02 Jan 2006 08:50:04 PM |
|
|
srp wrote:
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96DC3.1090606@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
vze2vt56@verizon.net a écrit :
If I oscillate a charged particle with some frequency f, I'll be
generating EM radiation of that frequency.
Conditional yes.
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.
Why not?
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?
They bloody well better (assuming there are any photons to begin
with; just saying you're waving a charge around is not adequate. You
must specify if conditions allow free EM waves to form and escape.) Er,
either of you ever built an antenna?
What frequency do they have?
(And how many of them are there?)
It depends which direction you look at it from, and how large a view
you take. Assuming, etc.
I have no idea if photons would be released in such a case.
I was talking about charged particles, like electrons.
Sigh.
Then did you really have an idea about the OP's question.
The OP was not asking specifically about electrons.
Well, read back yourself. He was specifically talking about
"a charged particle", and to my knowledge, electrons are
charged particles.
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
The charge on the pith ball is an electron.
I doubt that the charge on a charged pith ball would
be a single electron.
Why not? In the Millikan oil drop experiment, is the oil drop not a
"charged particle"? It's a composite particle, but that just localizes
it better than say a single electron.
What do you think the pith ball being "charged" means in the first
place? It too is a composite particle, but so what?
Disregard the pith ball if you like.
You cause an electron to have the motion described
above. What frequency do the photons have?
They would have the frequency of the energy released
each time the electron stops in one direction before
it starts accelerating in the reverse direction.
In other words, it will depend on the velocity you
will impart to the electron to maintain the
oscillating motion. If the oscillation is stable, the
photons will all have the same frequency.
They can't; kT. Close though...
How many of them are there?
It depends which direction you look at it from, and how large a view
you take.
As many as there will be changes in direction of the
electron during the experiment.
Do you not see that the direction of motion of the charge affects how
many photons an observer will see, depending on the observer's
orientation WRT that motion? Assuming, etc.
Also, is the observer in motion WRT that cewnter-of-oscillation of
the charge? That affects the observed frequency. Assuming, etc.
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.
You can only get coherent radiation from many sources (IOW many
oscillating charges), not from one.
You cannot assign "a frequency" to the field from a single oscillating
charge, but you can get as close as kT allows.
What about the electromagnetic field for an electron
with the cos() motion described above?
Note the the cos() function represents a to and fro motion in
reality.
Can you offer a guess on that?
Possibly.
I can see a charged particle in motion, so the fields will
be composite, meaning that the composite fields will be made
up of the fields of the electron itself plus the fields of
the energy carrying it (the energy that gives it its velocity).
I see that the latter will increase as the velocity increases
and moving on as a free photon as the electron stops, leaving
the latter behind. The electron will maintain its own
intrinsic fields.
Can you name a few of the factors determining whether or not photons
will be emitted?
(hint; "near field" and "far field")
For photons, a frequency can be definitely be associated.
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.
Sigh. It means you may detect a photon if you stand in the right
place WRT the charge's direction of motion, assuming etc.
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.
Sigh. There are equations for that, but we need to know the rate of
deceleration, among other things.
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?
Since the electron moves at a velocity determined by its energy
in excess of its rest mass, it cannot move at any velocity lower
than the energy that it possesses. For it to slow down, it has
lose the energy in excess of that required to maintain the new
lower velocity. If its stops completely, it has to lose all
of its energy in excess of rest mass.
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?
I don't know. Hypothetical. Impossible to do.
Horseshit; the galactic magnetic field does it all the time. Oops,
did I give something away?
We do not have the technology to follow linearly an electron
moving at near light speed to start with, let alone slowly
braking it from .999999c to 1m/s over a period of 10 years
in the process.
And that one photon
will have millions of eV energy?
I have no idea. Hypothetical. It couldn't be more anyway than
the energy that was required to maintain it at .999999c to
start with if no outside factor cause it to lose any for other
causes.
Sigh.
Mark L. Fergerson
.
|
|
|
| User: "Ron Baker, Pluralitas!" |
|
| Title: Re: Accelerating a particle and photons |
03 Jan 2006 10:15:06 AM |
|
|
"Mark Fergerson" <nunya@biz.ness> wrote in message
news:UMluf.5284$jR.3191@fed1read01...
srp wrote:
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96DC3.1090606@globetrotter.net...
Ron Baker, Pluralitas! a écrit :
"srp" <srp2@globetrotter.net> wrote in message
news:43B96044.5020402@globetrotter.net...
<snip>
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?
They bloody well better (assuming there are any photons to begin
It was a socratic question to Mr. Srp.
Maybe you noticed.
with; just saying you're waving a charge around is not adequate. You must
specify if conditions allow free EM waves to form and escape.) Er, either
of you ever built an antenna?
What frequency do they have?
(And how many of them are there?)
It depends which direction you look at it from, and how large a view you
take. Assuming, etc.
I have no idea if photons would be released in such a case.
I was talking about charged particles, like electrons.
Sigh.
Then did you really have an idea about the OP's question.
The OP was not asking specifically about electrons.
Well, read back yourself. He was specifically talking about
"a charged particle", and to my knowledge, electrons are
charged particles.
How is the pith ball not a charged particle?
Because it is a pith ball, made up of I don't know how
many atoms that can be ionized. A pith ball is not a
charged particle, it can only be a charged pith ball.
The charge on the pith ball is an electron.
I doubt that the charge on a charged pith ball would
be a single electron.
Why not? In the Millikan oil drop experiment, is the oil drop not a
"charged particle"? It's a composite particle, but that just localizes it
better than say a single electron.
What do you think the pith ball being "charged" means in the first
place? It too is a composite particle, but so what?
With two people pointing that out to srp, he
seems to have gotten quiet.
Disregard the pith ball if you like.
You cause an electron to have the motion described
above. What frequency do the photons have?
They would have the frequency of the energy released
each time the electron stops in one direction before
it starts accelerating in the reverse direction.
In other words, it will depend on the velocity you
will impart to the electron to maintain the
oscillating motion. If the oscillation is stable, the
photons will all have the same frequency.
They can't; kT. Close though...
How many of them are there?
It depends which direction you look at it from, and how large a view you
take.
If you consider all directions you can do total
energy calculations and show (to the OP and srp)
by conservation that it doesn't make sense to talk
about 'the' photon.
<snip comments to srp>
--
rb
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| User: "srp" |
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| Title: Re: Accelerating a particle and photons |
03 Jan 2006 11:10:56 AM |
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Ron Baker, Pluralitas! a écrit :
With two people pointing that out to srp, he
seems to have gotten quiet.
I am quiet only because you brought nothing more up
worth commenting about.
If the OP wants more info, I will interact.
André Michaud
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| User: "Gregory L. Hansen" |
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| Title: Re: Accelerating a particle and photons |
02 Jan 2006 07:39:56 PM |
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In article <1136164022.412164.53260@o13g2000cwo.googlegroups.com>,
<vze2vt56@verizon.net> wrote:
If I oscillate a charged particle with some frequency f, I'll be
generating EM radiation of that frequency. But this EM radiation is
just a stream of photons, each with an energy corresponding to
frequency f. But for a photon of that energy to be emmitted, it must
know the rate I'm oscillating the charged particle at (since that
determines f). Does this mean no photons are emmitted until I change
the particle's direction?
Another of phrasing my question - the photon must be emmitted with some
specific frequency. This frequency is only determined once I change the
direction of the particle I'm waving. So how is the photon emmitted
before I do that, with that frequency?
Thanks,
Brian
You shouldn't think of it as a little billiard ball flying out at
some point with no knowledge of the particle's past or future, although
that would be an understandable mental image. Quantum field theory is a
theory of fields. If you oscillate a charged particle at some frequency,
you will generate a field that oscillates with that frequency. It is a
quantum field that will exist in a superposition of states. At any given
interval it may or may not transfer momentum to your detector. When a
ping sounds out, "a photon was emitted".
To turn the question around, in the *classical* theory, how does the
emitted field know to be sinusoidal and have the same frequency as the
oscillating source? When the problem is solved in the textbooks, a
sin(wt) form is assumed and plugged in. But, to anthropomorphize a bit,
at any given moment in time the field local to the particle only knows
the instantaneous acceleration of the particle-- it doesn't know that the
acceleration was and will continue to be sinusoidal. It certainly makes
sense that the field will have some kind of structure that has the same
period as the particle, but can you prove it is sin(wt)?
--
"Never argue with a fool. They will drag you down to their level and win
by experience."
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